Termodinâmica - Uma Abordagem de Engenharia - 9ª Edição

Thermodynamics An Engineering Approach Ninth Edition Yunus A Çengel Michael A Boles Mehmet Kanoğlu McGraw Hill Education cen22672fmixxivindd i 111317 0645 PM T H E R M O D Y N A M I C S AN ENGINEERING APPROACH NINTH EDITION Final PDF to printer cen22672fmixxivindd ii 111317 0645 PM Final PDF to printer cen22672fmixxivindd iii 111317 0645 PM T H E R M O D Y N A M I C S AN ENGINEERING APPROACH NINTH EDITION YUNUS A ÇENGEL University of Nevada Reno MICHAEL A BOLES North Carolina State University MEHMET KANOĞLU University of Gaziantep Final PDF to printer cen22672fmixxivindd iv 111317 0645 PM THERMODYNAMICS AN ENGINEERING APPROACH NINTH EDITION Published by McGrawHill Education 2 Penn Plaza New York NY 10121 Copyright 2019 by McGrawHill Education All rights reserved Printed in the United States of America Previous editions 2015 2011 and 2008 No part of this publication may be reproduced or distributed in any form or by any means or stored in a database or retrieval system without the prior written consent of McGrawHill Education including but not limited to in any network or other electronic storage or transmission or broadcast for distance learning Some ancillaries including electronic and print components may not be available to customers outside the United States This book is printed on acidfree paper 1 2 3 4 5 6 7 8 9 LWI 21 20 19 18 ISBN 9781259822674 MHID 1259822672 Portfolio Manager Thomas M Scaife PhD Product Developer Jolynn Kilburg Marketing Manager Shannon ODonnell Director of Digital Content Chelsea Haupt PhD Content Project Managers Jane Mohr Tammy Juran and Sandy Schnee Buyer Susan K Culbertson Design Egzon Shaqiri Content Licensing Specialist Beth Thole Cover Image Source NASA Bill Ingalls Compositor SPi Global All credits appearing on page or at the end of the book are considered to be an extension of the copyright page Library of Congress CataloginginPublication Data Names Çengel Yunus A author Boles Michael A author Kanoğlu Mehmet author Title Thermodynamics an engineering approach Yunus A Çengel University of Nevada Reno Michael A Boles North Carolina State University Mehmet Kanoğlu University of Gaziantep Description Ninth edition New York NY McGrawHill Education 2019 Identifiers LCCN 2017048282 ISBN 9781259822674 acidfree paper ISBN 1259822672 acidfree paper Subjects LCSH Thermodynamics Classification LCC TJ265 C43 2019 DDC 6214021dc23 LC record available at httpslccnlocgov2017048282 The Internet addresses listed in the text were accurate at the time of publication The inclusion of a website does not indicate an endorsement by the authors or McGrawHill Education and McGrawHill Education does not guarantee the accuracy of the information presented at these sites mheducationcomhighered Final PDF to printer cen22672fmixxivindd v 111317 0645 PM Quotes on Ethics Without ethics everything happens as if we were all five billion passengers on a big machinery and nobody is driving the machinery And its going faster and faster but we dont know where Jacques Cousteau Because youre able to do it and because you have the right to do it doesnt mean its right to do it Laura Schlessinger A man without ethics is a wild beast loosed upon this world Manly Hall The concern for man and his destiny must always be the chief interest of all technical effort Never forget it among your diagrams and equations Albert Einstein To educate a man in mind and not in morals is to educate a menace to society Theodore Roosevelt Politics which revolves around benefit is savagery Said Nursi The true test of civilization is not the census nor the size of the cities nor the crops but the kind of man that the country turns out Ralph W Emerson The measure of a mans character is what he would do if he knew he never would be found out Thomas B Macaulay Final PDF to printer cen22672fmixxivindd vi 111317 0645 PM A B O U T T H E A U T H O R S Yunus A Çengel is Professor Emeritus of Mechanical Engineering at the University of Nevada Reno He received his BS in mechanical engineering from Istanbul Technical University and his MS and PhD in mechanical engineering from North Carolina State University His areas of interest are renewable energy energy efficiency energy policies heat transfer enhancement and engineering education He served as the director of the Industrial Assessment Center IAC at the University of Nevada Reno from 1996 to 2000 He has led teams of engineering students to numerous manufacturing facilities in Northern Nevada and California to perform industrial assessments and has prepared energy conservation waste minimization and productivity enhancement reports for them He has also served as an advisor for various government organizations and corporations Dr Çengel is also the author or coauthor of the widely adopted textbooks Heat and Mass Transfer Fundamentals and Applications 5th ed 2015 Fluid MechanicsFundamentals and Applications 4th ed 2018 Fundamentals of ThermalFluid Sciences 5th ed 2017 and Differential Equations for Engineers and Scientists 1st ed 2013 all published by McGrawHill Some of his textbooks have been translated into Chinese Long and Short Forms Japanese Korean Spanish French Portuguese Italian Turkish Greek Tai and Basq Dr Çengel is the recipient of several outstanding teacher awards and he has received the ASEE MeriamWiley Distinguished Author Award for excellence in authorship in 1992 and again in 2000 Dr Çengel is a registered Professional Engineer in the State of Nevada and is a member of the American Society of Mechanical Engineers ASME and the American Society for Engineering Education ASEE Michael A Boles is Associate Professor of Mechanical and Aerospace Engineering at North Carolina State University where he earned his PhD in mechan ical engineering and is an Alumni Distinguished Professor Dr Boles has received numerous awards and citations for excellence as an engineering educator He is a past recipient of the SAE Ralph R Teetor Education Award and has been twice elected to the NCSU Academy of Outstanding Teachers The NCSU ASME student section has consistently recognized him as the outstanding teacher of the year and the faculty member having the most impact on mechanical engineering students Dr Boles specializes in heat transfer and has been involved in the analytical and numerical solution of phase change and drying of porous media He is a member of the American Society of Mechanical Engineers ASME the American Society for Engineering Education ASEE and Sigma Xi Dr Boles received the ASEE MeriamWiley Distinguished Author Award in 1992 for excellence in authorship Mehmet Kanoğlu is Professor of Mechanical Engineering at University of Gaziantep He received his BS in mechanical engineering from Istanbul Technical University and his MS and PhD in mechanical engineering from University of Nevada Reno His research areas include energy efficiency refrigeration systems gas liquefaction hydrogen production and liquefaction renewable energy systems geothermal energy and cogeneration He is the author or coauthor of over 60 jour nal papers and numerous conference papers Final PDF to printer cen22672fmixxivindd vii 111317 0645 PM vii ABOUT THE AUTHORS Dr Kanoğlu has taught courses at University of Nevada Reno University of Ontario Institute of Technology American University of Sharjah and Uni versity of Gaziantep He is the coauthor of the books Refrigeration Systems and Applications 2nd ed Wiley 2010 and Efficiency Evaluation of Energy Systems Springer 2012 Dr Kanoğlu has served as an instructor in certified energy manager train ing programs and as an expert for United Nations Development Programme UNDP for energy efficiency and renewable energy projects He instructed numerous training courses and gave lectures and presentations on energy efficiency and renewable energy systems He has also served as advisor for state research funding organizations and industrial companies Final PDF to printer cen22672fmixxivindd viii 111317 0645 PM B R I E F C O N T E N T S C H A P T E R O N E INTRODUCTION AND BASIC CONCEPTS 1 C H A P T E R T W O ENERGY ENERGY TRANSFER AND GENERAL ENERGY ANALYSIS 51 C H A P T E R T H R E E PROPERTIES OF PURE SUBSTANCES 109 C H A P T E R F O U R ENERGY ANALYSIS OF CLOSED SYSTEMS 161 C H A P T E R F I V E MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES 211 C H A P T E R S I X THE SECOND LAW OF THERMODYNAMICS 271 C H A P T E R S E V E N ENTROPY 323 C H A P T E R E I G H T EXERGY 413 C H A P T E R N I N E GAS POWER CYCLES 475 C H A P T E R T E N VAPOR AND COMBINED POWER CYCLES 543 C H A P T E R E L E V E N REFRIGERATION CYCLES 597 C H A P T E R T W E L V E THERMODYNAMIC PROPERTY RELATIONS 643 C H A P T E R T H I R T E E N GAS MIXTURES 675 C H A P T E R F O U R T E E N GASVAPOR MIXTURES AND AIRCONDITIONING 711 C H A P T E R F I F T E E N CHEMICAL REACTIONS 747 C H A P T E R S I X T E E N CHEMICAL AND PHASE EQUILIBRIUM 791 C H A P T E R S E V E N T E E N COMPRESSIBLE FLOW 823 C H A P T E R E I G H T E E N W E B C H A P T E R RENEWABLE ENERGY Final PDF to printer cen22672fmixxivindd ix 111317 0645 PM ix BRIEF CONTENTS A P P E N D I X 1 PROPERTY TABLES AND CHARTS SI UNITS 881 A P P E N D I X 2 PROPERTY TABLES AND CHARTS ENGLISH UNITS 931 Final PDF to printer cen22672fmixxivindd x 111317 0645 PM C O N T E N T S Preface xvii C H A P T E R O N E INTRODUCTION AND BASIC CONCEPTS 1 11 Thermodynamics and Energy 2 Application Areas of Thermodynamics 3 12 Importance of Dimensions and Units 3 Some SI and English Units 6 Dimensional Homogeneity 8 Unity Conversion Ratios 9 13 Systems and Control Volumes 10 14 Properties of a System 12 Continuum 12 15 Density and Specific Gravity 13 16 State and Equilibrium 14 The State Postulate 14 17 Processes and Cycles 15 The SteadyFlow Process 16 18 Temperature and the Zeroth Law of Thermodynamics 17 Temperature Scales 17 The International Temperature Scale of 1990 ITS90 20 19 Pressure 21 Variation of Pressure with Depth 23 110 Pressure Measurement Devices 26 The Barometer 26 The Manometer 29 Other Pressure Measurement Devices 32 111 ProblemSolving Technique 33 Step 1 Problem Statement 33 Step 2 Schematic 33 Step 3 Assumptions and Approximations 34 Step 4 Physical Laws 34 Step 5 Properties 34 Step 6 Calculations 34 Step 7 Reasoning Verification and Discussion 34 Engineering Software Packages 35 Equation Solvers 36 A Remark on Significant Digits 37 Summary 38 References and Suggested Readings 39 Problems 39 C H A P T E R T W O ENERGY ENERGY TRANSFER AND GENERAL ENERGY ANALYSIS 51 21 Introduction 52 22 Forms of Energy 53 Some Physical Insight to Internal Energy 55 More on Nuclear Energy 56 Mechanical Energy 58 23 Energy Transfer by Heat 60 Historical Background on Heat 61 24 Energy Transfer by Work 62 Electrical Work 65 25 Mechanical Forms of Work 66 Shaft Work 66 Spring Work 67 Work Done on Elastic Solid Bars 67 Work Associated with the Stretching of a Liquid Film 68 Work Done to Raise or to Accelerate a Body 68 Nonmechanical Forms of Work 70 26 The First Law of Thermodynamics 70 Energy Balance 71 Energy Change of a System ΔEsystem 72 Mechanisms of Energy Transfer Ein and Eout 73 27 Energy Conversion Efficiencies 78 Efficiencies of Mechanical and Electrical Devices 82 28 Energy and Environment 85 Ozone and Smog 86 Acid Rain 87 The Greenhouse Effect Global Warming and Climate Change 88 Topic of Special Interest Mechanisms of Heat Transfer 91 Summary 96 References and Suggested Readings 97 Problems 97 C H A P T E R T H R E E PROPERTIES OF PURE SUBSTANCES 109 31 Pure Substance 110 32 Phases of a Pure Substance 110 Final PDF to printer xi CONTENTS cen22672fmixxivindd xi 111317 0645 PM 33 PhaseChange Processes of Pure Substances 111 Compressed Liquid and Saturated Liquid 112 Saturated Vapor and Superheated Vapor 112 Saturation Temperature and Saturation Pressure 113 Some Consequences of Tsat and Psat Dependence 114 34 Property Diagrams for PhaseChange Processes 116 1 The Tv Diagram 116 2 The Pv Diagram 118 Extending the Diagrams to Include the Solid Phase 118 3 The PT Diagram 120 The PvT Surface 121 35 Property Tables 122 EnthalpyA Combination Property 122 1a Saturated Liquid and Saturated Vapor States 123 1b Saturated LiquidVapor Mixture 125 2 Superheated Vapor 128 3 Compressed Liquid 129 Reference State and Reference Values 130 36 The IdealGas Equation of State 133 Is Water Vapor an Ideal Gas 135 37 Compressibility FactorA Measure of Deviation from IdealGas Behavior 136 38 Other Equations of State 139 van der Waals Equation of State 140 BeattieBridgeman Equation of State 140 BenedictWebbRubin Equation of State 141 Virial Equation of State 142 Topic of Special Interest Vapor Pressure and Phase Equilibrium 144 Summary 148 References and Suggested Readings 149 Problems 149 C H A P T E R F O U R ENERGY ANALYSIS OF CLOSED SYSTEMS 161 41 Moving Boundary Work 162 Polytropic Process 166 42 Energy Balance for Closed Systems 167 43 Specific Heats 172 44 Internal Energy Enthalpy and Specific Heats of Ideal Gases 174 Specific Heat Relations of Ideal Gases 176 45 Internal Energy Enthalpy and Specific Heats of Solids and Liquids 181 Internal Energy Changes 182 Enthalpy Changes 182 Topic of Special Interest Thermodynamic Aspects of Biological Systems 185 Summary 192 References and Suggested Readings 193 Problems 194 C H A P T E R F I V E MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES 211 51 Conservation of Mass 212 Mass and Volume Flow Rates 212 Conservation of Mass Principle 214 Mass Balance for SteadyFlow Processes 216 Special Case Incompressible Flow 216 52 Flow Work and the Energy of a Flowing Fluid 219 Total Energy of a Flowing Fluid 220 Energy Transport by Mass 221 53 Energy Analysis of SteadyFlow Systems 222 54 Some SteadyFlow Engineering Devices 225 1 Nozzles and Diffusers 226 2 Turbines and Compressors 229 3 Throttling Valves 232 4a Mixing Chambers 233 4b Heat Exchangers 235 5 Pipe and Duct Flow 237 55 Energy Analysis of UnsteadyFlow Processes 239 Topic of Special Interest General Energy Equation 244 Summary 247 References And Suggested Readings 248 Problems 248 C H A P T E R S I X THE SECOND LAW OF THERMODYNAMICS 271 61 Introduction to the Second Law 272 62 Thermal Energy Reservoirs 273 63 Heat Engines 274 Thermal Efficiency 275 Can We Save Qout 277 The Second Law of Thermodynamics KelvinPlanck Statement 279 Final PDF to printer xii THERMODYNAMICS cen22672fmixxivindd xii 111317 0645 PM 64 Refrigerators and Heat Pumps 279 Coefficient of Performance 280 Heat Pumps 281 Performance of Refrigerators Air Conditioners and Heat Pumps 282 The Second Law of Thermodynamics Clausius Statement 284 Equivalence of the Two Statements 285 65 PerpetualMotion Machines 286 66 Reversible and Irreversible Processes 288 Irreversibilities 289 Internally and Externally Reversible Processes 290 67 The Carnot Cycle 291 The Reversed Carnot Cycle 293 68 The Carnot Principles 293 69 The Thermodynamic Temperature Scale 295 610 The Carnot Heat Engine 297 The Quality of Energy 298 Quantity versus Quality in Daily Life 299 611 The Carnot Refrigerator and Heat Pump 300 Topic of Special Interest Household Refrigerators 303 Summary 307 References and Suggested Readings 308 Problems 308 C H A P T E R S E V E N ENTROPY 323 71 Entropy 324 A Special Case Internally Reversible Isothermal Heat Transfer Processes 327 72 The Increase of Entropy Principle 328 Some Remarks About Entropy 330 73 Entropy Change of Pure Substances 331 74 Isentropic Processes 334 75 Property Diagrams Involving Entropy 336 76 What is Entropy 337 Entropy and Entropy Generation in Daily Life 340 77 The T ds Relations 341 78 Entropy Change of Liquids and Solids 343 79 The Entropy Change of Ideal Gases 346 Constant Specific Heats Approximate Analysis 347 Variable Specific Heats Exact Analysis 347 Isentropic Processes of Ideal Gases 349 Constant Specific Heats Approximate Analysis 349 Variable Specific Heats Exact Analysis 350 Relative Pressure and Relative Specific Volume 350 710 Reversible SteadyFlow Work 354 Proof that SteadyFlow Devices Deliver the Most and Consume the Least Work When the Process Is Reversible 356 711 Minimizing the Compressor Work 357 Multistage Compression with Intercooling 358 712 Isentropic Efficiencies of SteadyFlow Devices 361 Isentropic Efficiency of Turbines 361 Isentropic Efficiencies of Compressors and Pumps 363 Isentropic Efficiency of Nozzles 365 713 Entropy Balance 367 Entropy Change of a System ΔSsystem 368 Mechanisms of Entropy Transfer Sin and Sout 368 1 Heat Transfer 368 2 Mass Flow 369 Entropy Generation Sgen 370 Closed Systems 371 Control Volumes 372 Entropy Generation Associated with a Heat Transfer Process 378 Topic of Special Interest Reducing the Cost of Compressed Air 380 Summary 389 References and Suggested Readings 390 Problems 390 C H A P T E R E I G H T EXERGY 413 81 Exergy Work Potential of Energy 414 Exergy Work Potential Associated with Kinetic and Potential Energy 415 82 Reversible Work and Irreversibility 417 83 SecondLaw Efficiency 422 84 Exergy Change of a System 425 Exergy of a Fixed Mass Nonflow or Closed System Exergy 425 Exergy of a Flow Stream Flow or Stream Exergy 428 85 Exergy Transfer by Heat Work and Mass 430 Exergy Transfer by Heat Q 431 Exergy Transfer by Work W 432 Exergy Transfer by Mass m 432 86 The Decrease of Exergy Principle and Exergy Destruction 433 Exergy Destruction 434 87 Exergy Balance Closed Systems 435 Final PDF to printer xiii CONTENTS cen22672fmixxivindd xiii 111317 0645 PM 88 Exergy Balance Control Volumes 446 Exergy Balance for SteadyFlow Systems 447 Reversible Work 447 SecondLaw Efficiency of SteadyFlow Devices 448 Topic of Special Interest SecondLaw Aspects of Daily Life 454 Summary 458 References and Suggested Readings 459 Problems 460 C H A P T E R N I N E GAS POWER CYCLES 475 91 Basic Considerations in the Analysis of Power Cycles 476 92 The Carnot Cycle and its Value in Engineering 478 93 AirStandard Assumptions 480 94 An Overview of Reciprocating Engines 481 95 Otto Cycle the Ideal Cycle for SparkIgnition Engines 482 96 Diesel Cycle the Ideal Cycle for CompressionIgnition Engines 489 97 Stirling and Ericsson Cycles 493 98 Brayton Cycle the Ideal Cycle for GasTurbine Engines 497 Development of Gas Turbines 499 Deviation of Actual GasTurbine Cycles from Idealized Ones 502 99 The Brayton Cycle with Regeneration 504 910 The Brayton Cycle with Intercooling Reheating and Regeneration 506 911 Ideal JetPropulsion Cycles 510 Modifications to Turbojet Engines 514 912 SecondLaw Analysis of Gas Power Cycles 516 Topic of Special Interest Saving Fuel and Money by Driving Sensibly 519 Summary 526 References and Suggested Readings 527 Problems 528 C H A P T E R T E N VAPOR AND COMBINED POWER CYCLES 543 101 The Carnot Vapor Cycle 544 102 Rankine Cycle the Ideal Cycle for Vapor Power Cycles 545 Energy Analysis of the Ideal Rankine Cycle 545 103 Deviation of Actual Vapor Power Cycles From Idealized Ones 548 104 How Can we Increase the Efficiency of the Rankine Cycle 551 Lowering the Condenser Pressure Lowers Tlowavg 551 Superheating the Steam to High Temperatures Increases Thighavg 552 Increasing the Boiler Pressure Increases Thighavg 552 105 The Ideal Reheat Rankine Cycle 555 106 The Ideal Regenerative Rankine Cycle 559 Open Feedwater Heaters 559 Closed Feedwater Heaters 561 107 SecondLaw Analysis of Vapor Power Cycles 567 108 Cogeneration 569 109 Combined GasVapor Power Cycles 574 Topic of Special Interest Binary Vapor Cycles 577 Summary 579 References and Suggested Readings 579 Problems 580 C H A P T E R E L E V E N REFRIGERATION CYCLES 597 111 Refrigerators and Heat Pumps 598 112 The Reversed Carnot Cycle 599 113 The Ideal VaporCompression Refrigeration Cycle 600 114 Actual VaporCompression Refrigeration Cycle 603 115 SecondLaw Analysis of VaporCompression Refrigeration Cycle 605 116 Selecting the Right Refrigerant 609 117 Heat Pump Systems 611 118 Innovative VaporCompression Refrigeration Systems 613 Cascade Refrigeration Systems 613 Multistage Compression Refrigeration Systems 615 Multipurpose Refrigeration Systems with a Single Compressor 617 Liquefaction of Gases 618 119 Gas Refrigeration Cycles 619 1110 Absorption Refrigeration Systems 622 Final PDF to printer xiv THERMODYNAMICS cen22672fmixxivindd xiv 111317 0645 PM Topic of Special Interest Thermoelectric Power Generation and Refrigeration Systems 626 Summary 628 References and Suggested Readings 628 Problems 629 C H A P T E R T W E L V E THERMODYNAMIC PROPERTY RELATIONS 643 121 A Little MathPartial Derivatives and Associated Relations 644 Partial Differentials 645 Partial Differential Relations 647 122 The Maxwell Relations 649 123 The Clapeyron Equation 650 124 General Relations for du dh ds cv and cp 653 Internal Energy Changes 654 Enthalpy Changes 654 Entropy Changes 655 Specific Heats cv and cp 656 125 The JouleThomson Coefficient 660 126 The Δh Δu and Δs of Real Gases 662 Enthalpy Changes of Real Gases 662 Internal Energy Changes of Real Gases 664 Entropy Changes of Real Gases 664 Summary 667 References and Suggested Readings 668 Problems 668 C H A P T E R T H I R T E E N GAS MIXTURES 675 131 Composition of a Gas Mixture Mass and Mole Fractions 676 132 PvT Behavior of Gas Mixtures Ideal and Real Gases 677 IdealGas Mixtures 678 RealGas Mixtures 679 133 Properties of Gas Mixtures Ideal and Real Gases 682 IdealGas Mixtures 683 RealGas Mixtures 687 Topic of Special Interest Chemical Potential and the Separation Work of Mixtures 690 Summary 700 References and Suggested Readings 701 Problems 702 C H A P T E R F O U R T E E N GASVAPOR MIXTURES AND AIRCONDITIONING 711 141 Dry and Atmospheric Air 712 142 Specific and Relative Humidity of air 713 143 DewPoint Temperature 715 144 Adiabatic Saturation and WetBulb Temperatures 717 145 The Psychrometric Chart 720 146 Human Comfort and AirConditioning 721 147 AirConditioning Processes 723 Simple Heating and Cooling ω constant 724 Heating with Humidification 725 Cooling with Dehumidification 727 Evaporative Cooling 728 Adiabatic Mixing of Airstreams 730 Wet Cooling Towers 732 Summary 734 References and Suggested Readings 736 Problems 736 C H A P T E R F I F T E E N CHEMICAL REACTIONS 747 151 Fuels and Combustion 748 152 Theoretical and Actual Combustion Processes 752 153 Enthalpy of Formation and Enthalpy of Combustion 758 154 FirstLaw Analysis of Reacting Systems 762 SteadyFlow Systems 762 Closed Systems 763 155 Adiabatic Flame Temperature 767 156 Entropy Change of Reacting Systems 769 157 SecondLaw Analysis of Reacting Systems 771 Topic of Special Interest Fuel Cells 776 Summary 778 References and Suggested Readings 779 Problems 779 C H A P T E R S I X T E E N CHEMICAL AND PHASE EQUILIBRIUM 791 161 Criterion for Chemical Equilibrium 792 Final PDF to printer xv CONTENTS cen22672fmixxivindd xv 111317 0645 PM 162 The Equilibrium Constant for IdealGas Mixtures 794 163 Some Remarks about the KP of IdealGas Mixtures 798 164 Chemical Equilibrium for Simultaneous Reactions 802 165 Variation of KP with Temperature 804 166 Phase Equilibrium 806 Phase Equilibrium for a SingleComponent System 806 The Phase Rule 807 Phase Equilibrium for a Multicomponent System 808 Summary 813 References and Suggested Readings 814 Problems 815 C H A P T E R S E V E N T E E N COMPRESSIBLE FLOW 823 171 Stagnation Properties 824 172 Speed of Sound and Mach Number 827 173 OneDimensional Isentropic Flow 829 Variation of Fluid Velocity with Flow Area 831 Property Relations for Isentropic Flow of Ideal Gases 833 174 Isentropic Flow Through Nozzles 836 Converging Nozzles 836 ConvergingDiverging Nozzles 840 175 Shock Waves and Expansion Waves 844 Normal Shocks 844 Oblique Shocks 850 PrandtlMeyer Expansion Waves 854 176 Duct Flow with Heat Transfer and Negligible Friction Rayleigh Flow 858 Property Relations for Rayleigh Flow 864 Choked Rayleigh Flow 865 177 Steam Nozzles 867 Summary 870 References and Suggested Readings 872 Problems 872 C H A P T E R E I G H T E E N W E B C H A P T E R RENEWABLE ENERGY 181 Introduction 182 Solar Energy Solar Radiation FlatPlate Solar Collector Concentrating Solar Collector Linear Concentrating Solar Power Collector SolarPowerTower Plant Solar Pond Photovoltaic Cell Passive Solar Applications Solar Heat Gain through Windows 183 Wind Energy Wind Turbine Types and Power Performance Curve Wind Power Potential Wind Power Density Wind Turbine Efficiency Betz Limit for Wind Turbine Efficiency 184 Hydropower Analysis of Hydroelectric Power Plant Turbine Types 185 Geothermal Energy Geothermal Power Production 186 Biomass Energy Biomass Resources Conversion of Biomass to Biofuel Biomass Products Electricity and Heat Production by Biomass Solid Municipality Waste Summary References and Suggested Readings Problems A P P E N D I X O N E PROPERTY TABLES AND CHARTS SI UNITS 881 Table A1 Molar mass gas constant and critical point properties 882 Table A2 Idealgas specific heats of various common gases 883 Table A3 Properties of common liquids solids and foods 886 Table A4 Saturated waterTemperature table 888 Table A5 Saturated waterPressure table 890 Table A6 Superheated water 892 Table A7 Compressed liquid water 896 Table A8 Saturated icewater vapor 897 Figure A9 Ts diagram for water 898 Figure A10 Mollier diagram for water 899 Table A11 Saturated refrigerant134a Temperature table 900 Table A12 Saturated refrigerant134aPressure table 902 Table A13 Superheated refrigerant134a 903 Final PDF to printer xvi THERMODYNAMICS cen22672fmixxivindd xvi 111317 0645 PM Figure A14 Ph diagram for refrigerant134a 905 Figure A15 NelsonObert generalized compressibility chart 906 Table A16 Properties of the atmosphere at high altitude 907 Table A17 Idealgas properties of air 908 Table A18 Idealgas properties of nitrogen N2 910 Table A19 Idealgas properties of oxygen O2 912 Table A20 Idealgas properties of carbon dioxide CO2 914 Table A21 Idealgas properties of carbon monoxide CO 916 Table A22 Idealgas properties of hydrogen H2 918 Table A23 Idealgas properties of water vapor H2O 919 Table A24 Idealgas properties of monatomic oxygen O 921 Table A25 Idealgas properties of hydroxyl OH 921 Table A26 Enthalpy of formation Gibbs function of formation and absolute entropy at 25C 1 atm 922 Table A27 Properties of some common fuels and hydrocarbons 923 Table A28 Natural logarithms of the equilibrium constant Kp 924 Figure A29 Generalized enthalpy departure chart 925 Figure A30 Generalized entropy departure chart 926 Figure A31 Psychrometric chart at 1 atm total pressure 927 Table A32 Onedimensional isentropic compress ibleflow functions for an ideal gas with k 14 928 Table A33 Onedimensional normalshock func tions for an ideal gas with k 14 929 Table A34 Rayleigh flow functions for an ideal gas with k 14 930 A P P E N D I X T W O PROPERTY TABLES AND CHARTS ENGLISH UNITS 931 Table A1E Molar mass gas constant and critical point properties 932 Table A2E Idealgas specific heats of various common gases 933 Table A3E Properties of common liquids solids and foods 936 Table A4E Saturated waterTemperature table 938 Table A5E Saturated waterPressure table 940 Table A6E Superheated water 942 Table A7E Compressed liquid water 946 Table A8E Saturated icewater vapor 947 Figure A9E Ts diagram for water 948 Figure A10E Mollier diagram for water 949 Table A11E Saturated refrigerant134a Temperature table 950 Table A12E Saturated refrigerant134aPressure table 951 Table A13E Superheated refrigerant134a 952 Figure A14E Ph diagram for refrigerant134a 954 Table A16E Properties of the atmosphere at high altitude 955 Table A17E Idealgas properties of air 956 Table A18E Idealgas properties of nitrogen N2 958 Table A19E Idealgas properties of oxygen O2 960 Table A20E Idealgas properties of carbon dioxide CO2 962 Table A21E Idealgas properties of carbon monoxide CO 964 Table A22E Idealgas properties of hydrogen H2 966 Table A23E Idealgas properties of water vapor H2O 967 Table A26E Enthalpy of formation Gibbs function of formation and absolute entropy at 77F 1 atm 969 Table A27E Properties of some common fuels and hydrocarbons 970 Figure A31E Psychrometric chart at 1 atm total pressure 971 INDEX 973 NOMENCLATURE 981 CONVERSION FACTORS 983 Final PDF to printer cen22672fmixxivindd xvii 111317 0645 PM P R E F A C E B A C K GR OUND Thermodynamics is an exciting and fascinating subject that deals with energy and thermodynamics has long been an essential part of engineering curricula all over the world It has a broad application area ranging from microscopic organisms to common household appliances transportation vehicles power generation systems and even philosophy This introductory book contains sufficient material for two sequential courses in thermodynamics Students are assumed to have an adequate background in calculus and physics O B J EC T IVES This book is intended for use as a textbook by undergraduate engineering students in their sophomore or junior year and as a reference book for practicing engineers The objectives of this text are To cover the basic principles of thermodynamics To present a wealth of realworld engineering examples to give students a feel for how thermodynamics is applied in engineering practice To develop an intuitive understanding of thermodynamics by empha sizing the physics and physical arguments that underpin the theory It is our hope that this book through its careful explanations of concepts and its use of numerous practical examples and figures helps students develop the necessary skills to bridge the gap between knowledge and the confidence to properly apply knowledge P H IL O S OPH Y AND G O AL The philosophy that contributed to the overwhelming popularity of the prior editions of this book has remained unchanged in this edition Namely our goal has been to offer an engineering textbook that Communicates directly to the minds of tomorrows engineers in a simple yet precise manner Leads students toward a clear understanding and firm grasp of the basic principles of thermodynamics Encourages creative thinking and development of a deeper understanding and intuitive feel for thermodynamics Is read by students with interest and enthusiasm rather than being used as an aid to solve problems Special effort has been made to appeal to students natural curiosity and to help them explore the various facets of the exciting subject area of thermodynamics The enthusiastic responses we have received from users of prior editionsfrom small colleges to large universities all over the world and the continued translations into new languages indicate that our objectives have largely been achieved It is our philosophy that the best way to learn is Final PDF to printer xviii THERMODYNAMICS cen22672fmixxivindd xviii 111317 0645 PM by practice Therefore special effort is made throughout the book to reinforce material that was presented earlier Yesterdays engineer spent a major portion of his or her time substituting values into the formulas and obtaining numerical results However formula manipulations and number crunching are now being left mainly to computers Tomorrows engineer will need a clear understanding and a firm grasp of the basic principles so that he or she can understand even the most complex problems formulate them and interpret the results A conscious effort is made to emphasize these basic principles while also providing students with a perspective of how computational tools are used in engineering practice The traditional classical or macroscopic approach is used throughout the text with microscopic arguments serving in a supporting role as appropriate This approach is more in line with students intuition and makes learning the subject matter much easier NEW IN THIS EDITIO N All the popular features of the previous editions have been retained A large number of the endofchapter problems in the text have been modified and many problems were replaced by new ones Also several of the solved example problems have been replaced Video ResourcesUsing the student response data from the eighth edition LearnSmartSmartBook 2D3D animation videos have been added to the ebook to help clarify challenging concepts In addition to these conceptual video resources worked example problem videos are included in the ebook to help students apply their conceptual understanding to problem solving LE ARNING TO O L S EARLY INTRODUCTION OF THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is introduced early in Chapter 2 Energy Energy Transfer and General Energy Analysis This introductory chapter sets the framework of establishing a general understanding of various forms of energy mechanisms of energy transfer the concept of energy balance thermo economics energy conversion and conversion efficiency using familiar settings that involve mostly electrical and mechanical forms of energy It also exposes students to some exciting realworld applications of thermodynamics early in the course and helps them establish a sense of the monetary value of energy There is special emphasis on the utilization of renewable energy such as wind power and hydraulic energy and the efficient use of existing resources EMPHASIS ON PHYSICS A distinctive feature of this book is its emphasis on the physical aspects of the subject matter in addition to mathematical representations and manipulations The authors believe that the emphasis in undergraduate education should remain on developing a sense of underlying physical mechanisms and a mastery of solving practical problems that an engineer is likely to face in the real world Developing an intuitive understanding should also make the course a more motivating and worthwhile experience for students Final PDF to printer xix PREFACE cen22672fmixxivindd xix 111317 0645 PM EFFECTIVE USE OF ASSOCIATION An observant mind should have no difficulty understanding engineering sciences After all the principles of engineering sciences are based on our everyday experiences and experimental observations Therefore a physi cal intuitive approach is used throughout this text Frequently parallels are drawn between the subject matter and students everyday experiences so that they can relate the subject matter to what they already know The process of cooking for example serves as an excellent vehicle to demonstrate the basic principles of thermodynamics SELFINSTRUCTING The material in the text is introduced at a level that an average student can follow comfortably It speaks to students not over students In fact it is selfinstructive The order of coverage is from simple to general That is it starts with the simplest case and adds complexities gradually In this way the basic principles are repeatedly applied to different systems and students master how to apply the principles instead of how to simplify a general formula Noting that the principles of sciences are based on experimental observations all the derivations in this text are based on physical arguments and thus they are easy to follow and understand EXTENSIVE USE OF ARTWORK Figures are important learning tools that help students get the picture and the text makes very effective use of graphics This edition features an enhanced art program done in four colors to provide more realism and pedagogical understanding Further a large number of figures have been upgraded to become threedimensional and thus more reallife Figures attract attention and stimulate curiosity and interest Most of the figures in this text are intended to serve as a means of emphasizing some key concepts that would otherwise go unnoticed some serve as page summaries LEARNING OBJECTIVES AND SUMMARIES Each chapter begins with an overview of the material to be covered and chapterspecific learning objectives A summary is included at the end of each chapter providing a quick review of basic concepts and important relations and pointing out the relevance of the material NUMEROUS WORKEDOUT EXAMPLES WITH A SYSTEMATIC SOLUTIONS PROCEDURE Each chapter contains several workedout examples that clarify the material and illustrate the use of the basic principles An intuitive and systematic approach is used in the solution of the example problems while maintaining an informal conversational style The problem is first stated and the objectives are identified The assumptions are then stated together with their justifications The proper ties needed to solve the problem are listed separately if appropriate Numerical values are used together with their units to emphasize that numbers without units are meaningless and that unit manipulations are as important as manipulating the numerical values with a calculator The significance of the findings is dis cussed following the solutions This approach is also used consistently in the solutions presented in the instructors solutions manual Final PDF to printer xx THERMODYNAMICS cen22672fmixxivindd xx 111317 0645 PM A WEALTH OF REALWORLD ENDOFCHAPTER PROBLEMS The endofchapter problems are grouped under specific topics to make prob lem selection easier for both instructors and students Within each group of problems are Concept Questions indicated by C to check the students level of understanding of basic concepts The problems under Review Problems are more comprehensive in nature and are not directly tied to any specific sec tion of a chapterin some cases they require review of material learned in previous chapters Problems designated as Design and Essay are intended to encourage students to make engineering judgments to conduct independent exploration of topics of interest and to communicate their findings in a pro fessional manner Problems designated by an E are in English units and SI users can ignore them Problems with the are comprehensive in nature and are intended to be solved with a computer using appropriate software Several economics and safetyrelated problems are incorporated throughout to promote cost and safety awareness among engineering students Answers to selected problems are listed immediately following the problem for conve nience to students In addition to prepare students for the Fundamentals of Engineering Exam that is becoming more important for the outcomebased ABET 2000 criteria and to facilitate multiplechoice tests over 200 multiple choice problems are included in the endofchapter problem sets They are placed under the title Fundamentals of Engineering FE Exam Problems for easy recognition These problems are intended to check the understanding of fundamentals and to help readers avoid common pitfalls RELAXED SIGN CONVENTION The use of a formal sign convention for heat and work is abandoned as it often becomes counterproductive A physically meaningful and engaging approach is adopted for interactions instead of a mechanical approach Subscripts in and out rather than the plus and minus signs are used to indicate the direc tions of interactions PHYSICALLY MEANINGFUL FORMULAS The physically meaningful forms of the balance equations rather than formu las are used to foster deeper understanding and to avoid a cookbook approach The mass energy entropy and exergy balances for any system undergoing any process are expressed as Mass balance m in m out Δ m system Energy balance E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Entropy balance S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy Exergy balance X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy Final PDF to printer xxi PREFACE cen22672fmixxivindd xxi 111317 0645 PM These relations reinforce the fundamental principles that during an actual process mass and energy are conserved entropy is generated and exergy is destroyed Students are encouraged to use these forms of balances in early chapters after they specify the system and to simplify them for the particular problem A more relaxed approach is used in later chapters as students gain mastery A CHOICE OF SI ALONE OR SIENGLISH UNITS In recognition of the fact that English units are still widely used in some industries both SI and English units are used in this text with an emphasis on SI The material in this text can be covered using combined SIEnglish units or SI units alone depending on the preference of the instructor The property tables and charts in the appendices are presented in both units except the ones that involve dimensionless quantities Problems tables and charts in English units are designated by E after the number for easy recognition and they can be ignored by SI users TOPICS OF SPECIAL INTEREST Most chapters contain a section called Topic of Special Interest where interesting aspects of thermodynamics are discussed Examples include Thermodynamic Aspects of Biological Systems in Chapter 4 Household Refrigerators in Chapter 6 SecondLaw Aspects of Daily Life in Chapter 8 and Saving Fuel and Money by Driving Sensibly in Chapter 9 The topics selected for these sections provide intriguing extensions to thermodynamics but they can be ignored if desired without a loss in continuity GLOSSARY OF THERMODYNAMIC TERMS Throughout the chapters when an important key term or concept is intro duced and defined it appears in boldface type Fundamental thermodynamic terms and concepts also appear in a glossary located on our accompanying website This unique glossary helps to reinforce key terminology and is an excellent learning and review tool for students as they move forward in their study of thermodynamics CONVERSION FACTORS Frequently used conversion factors and physical constants are listed at the end of the text S UPPL EM ENT S The following supplements are available to users of the book PROPERTIES TABLE BOOKLET ISBN 1260048993 This booklet provides students with an easy reference to the most important property tables and charts many of which are found at the back of the textbook in both the SI and English units Final PDF to printer xxii THERMODYNAMICS cen22672fmixxivindd xxii 111317 0645 PM COSMOS McGrawHills COSMOS Complete Online Solutions Manual Organization System allows instructors to streamline the creation of assignments quizzes and tests by using problems and solutions from the textbook as well as their own custom material COSMOS is now available online at httpcosmos mhhecom ACKNO W L EDG MENTS The authors would like to acknowledge with appreciation the numerous and valuable comments suggestions constructive criticisms and praise from the following evaluators and reviewers Edward Anderson Texas Tech University John Biddle Cal Poly Pomona University Gianfranco DiGiuseppe Kettering University Shoeleh Di Julio California State UniversityNorthridge Afshin Ghajar Oklahoma State University Harry Hardee New Mexico State University Kevin Lyons North Carolina State University Kevin Macfarlan John Brown University Saeed Manafzadeh University of IllinoisChicago Alex Moutsoglou South Dakota State University Rishi Raj The City College of New York Maria Sanchez California State UniversityFresno Kalyan Srinivasan Mississippi State University Robert Stiger Gonzaga University Their suggestions have greatly helped to improve the quality of this text We thank Mohsen Hassan Vand for his valuable suggestions and contributions We also would like to thank our students who provided plenty of feedback from students perspectives Finally we would like to express our appreciation to our wives and to our children for their continued patience understanding and support throughout the preparation of this text Yunus A Çengel Michael A Boles Mehmet Kanoğlu Final PDF to printer cen22672fmixxivindd xxiii 111317 0645 PM McGrawHill Connect Connect is a highly reliable easytouse homework and learning manage ment solution that utilizes learning science and awardwinning adaptive tools to improve student results Analytics Connect Insight Connect Insight is Connects oneofakind visual analytics dashboard Now available for both instructors and students that provides ataglance informa tion regarding student performance which is immediately actionable By presenting assignment assessment and topical performance results together with a time metric that is easily visible for aggregate or individual results Connect InSight generates easytoread reports on individual students the class as a whole and on specific assignments The Connect Insight dashboard delivers data on performance study behavior and effort Instructors can quickly identify students who struggle and focus on material that the class has yet to master Connect automatically grades assignments and quizzes providing easytoread reports on individual and class performance Find the following instructor resources available through Connect Student Study GuideThis resource outlines the fundamental concepts of the text and is a helpful guide that allows students to focus on the most important concepts The guide can also serve as a lecture outline for instructors Learning ObjectivesThe chapter learning objectives are outlined here Organized by chapter and tied to ABET objectives Correlation GuideNew users of this text will appreciate this resource The guide provides a smooth transition for instructors not currently using this text Image LibraryThe electronic version of the figures are supplied for easy integration into course presentations exams and assignments Instructors GuideProvides instructors with helpful tools such as sample syllabi and exams an ABET conversion guide a thermodynamics glossary and chapter objectives ErrataIf errors should be found in the solutions manual they will be reported here Solutions ManualThe detailed solutions to all text homework problems are provided in PDF form PowerPoint slidesPowerpoint presentation slides for all chapters in the text are available for use in lectures AppendicesThese are provided in PDF form for ease of use Online Resources for Students and Instructors Final PDF to printer xxiv THERMODYNAMICS cen22672fmixxivindd xxiv 111317 0645 PM COSMOS McGrawHills COSMOS Complete Online Solutions Manual Organization System allows instructors to streamline the creation of assignments quizzes and tests by using problems and solutions from the textbook as well as their own custom material COSMOS is now available online at httpcosmos mhhecom Adaptive SmartBook SmartBook helps students study more efficiently by delivering an interactive reading experience through adaptive highlighting and review Final PDF to printer cen22672ch01001050indd 1 110317 0710 AM 1 CHAPTER1 I NT R O D U C T I O N AN D B AS I C C O N C EPTS E very science has a unique vocabulary associated with it and thermo dynamics is no exception Precise definition of basic concepts forms a sound foundation for the development of a science and prevents possible misunderstandings We start this chapter with an overview of ther modynamics and the unit systems and continue with a discussion of some basic concepts such as system state state postulate equilibrium process and cycle We discuss intensive and extensive properties of a system and define density specific gravity and specific weight We also discuss temperature and temperature scales with particular emphasis on the International Temperature Scale of 1990 We then present pressure which is the normal force exerted by a fluid per unit area and we discuss absolute and gage pressures the variation of pressure with depth and pressure measurement devices such as manom eters and barometers Careful study of these concepts is essential for a good understanding of the topics in the following chapters Finally we present an intuitive systematic problemsolving technique that can be used as a model in solving engineering problems OBJECTIVES The objectives of Chapter 1 are to Identify the unique vocabulary associated with thermodynamics through the precise definition of basic concepts to form a sound foundation for the development of the principles of thermodynamics Review the metric SI and the English unit systems that will be used throughout the text Explain the basic concepts of thermodynamics such as system state state postulate equilibrium process and cycle Discuss properties of a system and define density specific gravity and specific weight Review concepts of temperature temperature scales pressure and absolute and gage pressure Introduce an intuitive systematic problemsolving technique Final PDF to printer 2 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 2 110317 0710 AM 11 THERMODYNAMICS AND ENERGY Thermodynamics can be defined as the science of energy Although every body has a feeling of what energy is it is difficult to give a precise definition for it Energy can be viewed as the ability to cause changes The name thermodynamics stems from the Greek words therme heat and dynamis power which is most descriptive of the early efforts to convert heat into power Today the same name is broadly interpreted to include all aspects of energy and energy transformations including power generation refrigera tion and relationships among the properties of matter One of the most fundamental laws of nature is the conservation of energy principle It simply states that during an interaction energy can change from one form to another but the total amount of energy remains con stant That is energy cannot be created or destroyed A rock falling off a cliff for example picks up speed as a result of its potential energy being converted to kinetic energy Fig 11 The conservation of energy principle also forms the backbone of the diet industry A person who has a greater energy input food than energy output exercise will gain weight store energy in the form of fat and a person who has a smaller energy input than output will lose weight Fig 12 The change in the energy content of a body or any other system is equal to the difference between the energy input and the energy out put and the energy balance is expressed as Ein Eout ΔE The first law of thermodynamics is simply an expression of the conser vation of energy principle and it asserts that energy is a thermodynamic property The second law of thermodynamics asserts that energy has quality as well as quantity and actual processes occur in the direction of decreasing quality of energy For example a cup of hot coffee left on a table eventu ally cools but a cup of cool coffee in the same room never gets hot by itself Fig 13 The hightemperature energy of the coffee is degraded trans formed into a less useful form at a lower temperature once it is transferred to the surrounding air Although the principles of thermodynamics have been in existence since the creation of the universe thermodynamics did not emerge as a science until the construction of the first successful atmospheric steam engines in England by Thomas Savery in 1697 and Thomas Newcomen in 1712 These engines were very slow and inefficient but they opened the way for the devel opment of a new science The first and second laws of thermodynamics emerged simultaneously in the 1850s primarily out of the works of William Rankine Rudolph Clausius and Lord Kelvin formerly William Thomson The term thermodynamics was first used in a publication by Lord Kelvin in 1849 The first thermody namics textbook was written in 1859 by William Rankine a professor at the University of Glasgow It is well known that a substance consists of a large number of particles called molecules The properties of the substance naturally depend on the behavior of these particles For example the pressure of a gas in a container is the result of momentum transfer between the molecules and the walls of the container However one does not need to know the behavior of the gas particles to determine the pressure in the container It would be sufficient to attach a pressure gage to the container This macroscopic approach to the FIGURE 11 Energy cannot be created or destroyed it can only change forms the first law Potential energy Kinetic energy PE 7 units KE 3 units PE 10 units KE 0 FIGURE 12 Conservation of energy principle for the human body Energy out 4 units Energy in 5 units Energy storage 1 unit FIGURE 13 Heat flows in the direction of decreasing temperature Heat Cool environment 20C Hot coffee 70C Final PDF to printer 3 CHAPTER 1 cen22672ch01001050indd 3 110317 0710 AM study of thermodynamics that does not require a knowledge of the behavior of individual particles is called classical thermodynamics It provides a direct and easy way to solve engineering problems A more elaborate approach based on the average behavior of large groups of individual particles is called statistical thermodynamics This microscopic approach is rather involved and is used in this text only in a supporting role Application Areas of Thermodynamics All activities in nature involve some interaction between energy and matter thus it is hard to imagine an area that does not relate to thermodynamics in some manner Therefore developing a good understanding of basic principles of thermodynamics has long been an essential part of engineering education Thermodynamics is commonly encountered in many engineering systems and other aspects of life and one does not need to go very far to see some application areas of it In fact one does not need to go anywhere The heart is constantly pumping blood to all parts of the human body various energy conversions occur in trillions of body cells and the body heat generated is constantly rejected to the environment Human comfort is closely tied to the rate of this metabolic heat rejection We try to control this heat transfer rate by adjusting our clothing to the environmental conditions Other applications of thermodynamics are right where one lives An ordi nary house is in some respects an exhibition hall filled with wonders of thermodynamics Fig 14 Many ordinary household utensils and appli ances are designed in whole or in part by using the principles of thermo dynamics Some examples include the electric or gas range the heating and air conditioning systems the refrigerator the humidifier the pressure cooker the water heater the shower the iron and even the computer and the TV On a larger scale thermodynamics plays a major part in the design and analy sis of automotive engines rockets jet engines and conventional or nuclear power plants solar collectors and the design of vehicles from ordinary cars to airplanes Fig 15 The energyefficient home that you may be living in for example is designed on the basis of minimizing heat loss in winter and heat gain in summer The size location and the power input of the fan of your computer is also selected after an analysis that involves thermodynamics 12 IMPORTANCE OF DIMENSIONS AND UNITS Any physical quantity can be characterized by dimensions The magnitudes assigned to the dimensions are called units Some basic dimensions such as mass m length L time t and temperature T are selected as primary or fundamental dimensions while others such as velocity V energy E and volume V are expressed in terms of the primary dimensions and are called secondary dimensions or derived dimensions A number of unit systems have been developed over the years Despite strong efforts in the scientific and engineering community to unify the world with a single unit system two sets of units are still in common use today the English system which is also known as the United States Customary System USCS and the metric SI from Le Système International d Unités which FIGURE 14 The design of many engineering systems such as this solar hot water system involves thermodynamics Solar collectors Hot water Heat exchanger Pump Shower Cold water Hot water tank Final PDF to printer 4 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 4 110317 0710 AM is also known as the International System The SI is a simple and logical system based on a decimal relationship between the various units and it is being used for scientific and engineering work in most of the industrialized nations including England The English system however has no apparent systematic numerical base and various units in this system are related to each other rather arbitrarily 12 in 1 ft 1 mile 5280 ft 4 qt 1 gal etc which makes it confusing and difficult to learn The United States is the only indus trialized country that has not yet fully converted to the metric system FIGURE 15 Some application areas of thermodynamics a McGrawHill EducationJill Braaten b Doug MenuezGetty Images RF c Ilene MacDonaldAlamy RF d Malcolm FifeGetty Images RF e Ryan McVayGetty Images RF f Mark EvansGetty Images RF g Getty ImagesiStockphoto RF h Glow Images RF i Courtesy of UMDE Engineering Contracting and Trading Used by permission g Wind turbines h Food processing i A piping network in an industrial facility a Refrigerator b Boats c Aircraft and spacecraft d Power plants e Human body f Cars Final PDF to printer 5 CHAPTER 1 cen22672ch01001050indd 5 110317 0710 AM The systematic efforts to develop a universally acceptable system of units dates back to 1790 when the French National Assembly charged the French Academy of Sciences to come up with such a unit system An early version of the metric system was soon developed in France but it did not find univer sal acceptance until 1875 when The Metric Convention Treaty was prepared and signed by 17 nations including the United States In this international treaty meter and gram were established as the metric units for length and mass respectively and a General Conference of Weights and Measures CGPM was established that was to meet every six years In 1960 the CGPM produced the SI which was based on six fundamental quantities and their units were adopted in 1954 at the Tenth General Conference of Weights and Measures meter m for length kilogram kg for mass second s for time ampere A for electric current degree Kelvin K for temperature and can dela cd for luminous intensity amount of light In 1971 the CGPM added a seventh fundamental quantity and unit mole mol for the amount of matter Based on the notational scheme introduced in 1967 the degree symbol was officially dropped from the absolute temperature unit and all unit names were to be written without capitalization even if they were derived from proper names Table 11 However the abbreviation of a unit was to be capitalized if the unit was derived from a proper name For example the SI unit of force which is named after Sir Isaac Newton 16471723 is newton not Newton and it is abbreviated as N Also the full name of a unit may be pluralized but its abbreviation cannot For example the length of an object can be 5 m or 5 meters not 5 ms or 5 meter Finally no period is to be used in unit abbre viations unless they appear at the end of a sentence For example the proper abbreviation of meter is m not m The recent move toward the metric system in the United States seems to have started in 1968 when Congress in response to what was happening in the rest of the world passed a Metric Study Act Congress continued to pro mote a voluntary switch to the metric system by passing the Metric Conver sion Act in 1975 A trade bill passed by Congress in 1988 set a September 1992 deadline for all federal agencies to convert to the metric system How ever the deadlines were relaxed later with no clear plans for the future The industries that are heavily involved in international trade such as the automotive soft drink and liquor industries have been quick to convert to the metric system for economic reasons having a single worldwide design fewer sizes smaller inventories etc Today nearly all the cars manufactured in the United States are metric Most car owners probably do not realize this until they try an English socket wrench on a metric bolt Most industries however resisted the change thus slowing down the conversion process At present the United States is a dualsystem society and it will stay that way until the transition to the metric system is completed This puts an extra burden on todays engineering students since they are expected to retain their understanding of the English system while learning thinking and working in terms of the SI Given the position of the engineers in the transition period both unit systems are used in this text with particular emphasis on SI units As pointed out the SI is based on a decimal relationship between units The prefixes used to express the multiples of the various units are listed in Table 12 They are standard for all units and the student is encouraged to memorize them because of their widespread use Fig 16 TABLE 11 The seven fundamental or primary dimensions and their units in SI Dimension Unit Length meter m Mass kilogram kg Time second s Temperature kelvin K Electric current ampere A Amount of light candela cd Amount of matter mole mol TABLE 12 Standard prefixes in SI units Multiple Prefix 1024 yotta Y 1021 zetta Z 1018 exa E 1015 peta P 1012 tera T 109 giga G 106 mega M 103 kilo k 102 hecto h 101 deka da 101 deci d 102 centi c 103 milli m 106 micro μ 109 nano n 1012 pico p 1015 femto f 1018 atto a 1021 zepto z 1024 yocto y FIGURE 16 The SI unit prefixes are used in all branches of engineering 1 kg 200 mL 02 L 103 g 1 MΩ 106 Ω Final PDF to printer 6 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 6 110317 0710 AM FIGURE 17 The definition of the force units m 1 kg m 32174 lbm a 1 ms2 a 1 fts2 F 1 lbf F 1 N Some SI and English Units In SI the units of mass length and time are the kilogram kg meter m and second s respectively The respective units in the English system are the poundmass lbm foot ft and second s The pound symbol lb is actually the abbreviation of libra which was the ancient Roman unit of weight The English retained this symbol even after the end of the Roman occupation of Britain in 410 The mass and length units in the two systems are related to each other by 1 lbm 045356 kg 1 ft 03048 m In the English system force is usually considered to be one of the primary dimensions and is assigned a nonderived unit This is a source of confusion and error that necessitates the use of a dimensional constant gc in many for mulas To avoid this nuisance we consider force to be a secondary dimension whose unit is derived from Newtons second law that is Force Mass Acceleration or F ma 11 In SI the force unit is the newton N and it is defined as the force required to accelerate a mass of 1 kg at a rate of 1 ms2 In the English system the force unit is the poundforce lbf and is defined as the force required to accelerate a mass of 1 slug 32174 lbm at a rate of 1 fts2 Fig 17 That is 1 N 1 kgm s 2 1 lbf 32174 lbmft s 2 A force of 1 N is roughly equivalent to the weight of a small apple m 102 g whereas a force of 1 lbf is roughly equivalent to the weight of four medium apples mtotal 454 g as shown in Fig 18 Another force unit in common use in many European countries is the kilogramforce kgf which is the weight of 1 kg mass at sea level 1 kgf 9807 N The term weight is often incorrectly used to express mass particularly by the weight watchers Unlike mass weight W is a force It is the gravitational force applied to a body and its magnitude is determined from Newtons sec ond law W mg N 12 where m is the mass of the body and g is the local gravitational acceleration g is 9807 ms2 or 32174 fts2 at sea level and 45 latitude An ordinary bathroom scale measures the gravitational force acting on a body The mass of a body remains the same regardless of its location in the uni verse Its weight however changes with a change in gravitational accelera tion A body weighs less on top of a mountain since g decreases with altitude FIGURE 18 The relative magnitudes of the force units newton N kilogramforce kgf and poundforce lbf 1 kgf 10 apples m 1 kg 4 apples m 1 lbm 1 lbf 1 apple m 102 g 1 N Final PDF to printer 7 CHAPTER 1 cen22672ch01001050indd 7 110317 0710 AM On the surface of the moon an astronaut weighs about onesixth of what she or he normally weighs on earth Fig 19 At sea level a mass of 1 kg weighs 9807 N as illustrated in Fig 110 A mass of 1 lbm however weighs 1 lbf which misleads people into believing that poundmass and poundforce can be used interchangeably as pound lb which is a major source of error in the English system It should be noted that the gravity force acting on a mass is due to the attrac tion between the masses and thus it is proportional to the magnitudes of the masses and inversely proportional to the square of the distance between them Therefore the gravitational acceleration g at a location depends on latitude the distance to the center of the earth and to a lesser extent the positions of the moon and the sun The value of g varies with location from 9832 ms2 at the poles 9789 at the equator to 7322 ms2 at 1000 km above sea level However at altitudes up to 30 km the variation of g from the sealevel value of 9807 ms2 is less than 1 percent Therefore for most practical purposes the gravitational acceleration can be assumed to be constant at 9807 ms2 often rounded to 981 ms2 It is interesting to note that at locations below sea level the value of g increases with distance from the sea level reaches a maximum at about 4500 m and then starts decreasing What do you think the value of g is at the center of the earth The primary cause of confusion between mass and weight is that mass is usually measured indirectly by measuring the gravity force it exerts This approach also assumes that the forces exerted by other effects such as air buoyancy and fluid motion are negligible This is like measuring the distance to a star by measuring its redshift or measuring the altitude of an airplane by measuring barometric pressure Both of these are also indirect measurements The correct direct way of measuring mass is to compare it to a known mass This is cumbersome however and it is mostly used for calibration and mea suring precious metals Work which is a form of energy can simply be defined as force times distance therefore it has the unit newtonmeter Nm which is called a joule J That is 1 J 1 Nm 13 A more common unit for energy in SI is the kilojoule 1 kJ 103 J In the English system the energy unit is the Btu British thermal unit which is defined as the energy required to raise the temperature of 1 lbm of water at 68F by 1F In the metric system the amount of energy needed to raise the temperature of 1 g of water at 145C by 1C is defined as 1 calorie cal and 1 cal 41868 J The magnitudes of the kilojoule and Btu are almost identical 1 Btu 10551 kJ Here is a good way to get a feel for these units If you light a typical match and let it burn itself out it yields approximately one Btu or one kJ of energy Fig 111 The unit for time rate of energy is joule per second Js which is called a watt W In the case of work the time rate of energy is called power A com monly used unit of power is horsepower hp which is equivalent to 746 W Electrical energy typically is expressed in the unit kilowatthour kWh which is equivalent to 3600 kJ An electric appliance with a rated power of 1 kW consumes 1 kWh of electricity when running continuously for one hour FIGURE 19 A body weighing 150 lbf on earth will weigh only 25 lbf on the moon FIGURE 110 The weight of a unit mass at sea level g 9807 ms2 W 9807 kgms2 9807 N 1 kgf W 32174 lbmfts2 1 lbf g 32174 fts2 kg lbm FIGURE 111 A typical match yields about one Btu or one kJ of energy if completely burned John M Cimbala Final PDF to printer 8 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 8 110317 0710 AM When dealing with electric power generation the units kW and kWh are often confused Note that kW or kJs is a unit of power whereas kWh is a unit of energy Therefore statements like the new wind turbine will generate 50 kW of electricity per year are meaningless and incorrect A correct statement should be something like the new wind turbine with a rated power of 50 kW will generate 120000 kWh of electricity per year Dimensional Homogeneity We all know that apples and oranges do not add But we somehow man age to do it by mistake of course In engineering all equations must be dimensionally homogeneous That is every term in an equation must have the same unit If at some stage of an analysis we find ourselves in a position to add two quantities that have different units it is a clear indication that we have made an error at an earlier stage So checking dimensions can serve as a valuable tool to spot errors FIGURE 112 A wind turbine as discussed in Example 11 Bear Dancer StudiosMark Dierker RF EXAMPLE 11 Electric Power Generation by a Wind Turbine A school is paying 012kWh for electric power To reduce its power bill the school installs a wind turbine Fig 112 with a rated power of 30 kW If the turbine operates 2200 hours per year at the rated power determine the amount of electric power gener ated by the wind turbine and the money saved by the school per year SOLUTION A wind turbine is installed to generate electricity The amount of elec tric energy generated and the money saved per year are to be determined Analysis The wind turbine generates electric energy at a rate of 30 kW or 30 kJs Then the total amount of electric energy generated per year becomes Total energy Energy per unit time Time interval 30 kW 2200 h 66000 kWh The money saved per year is the monetary value of this energy determined as Money saved Total energy Unit cost of energy 66000 kWh 012 kWh 7920 Discussion The annual electric energy production also could be determined in kJ by unit manipulations as Total energy 30 kW2200 h 3600 s 1 h 1 kJ s 1 kW 238 10 8 kJ which is equivalent to 66000 kWh 1 kWh 3600 kJ We all know from experience that units can give terrible headaches if they are not used carefully in solving a problem However with some attention and skill units can be used to our advantage They can be used to check formulas sometimes they can even be used to derive formulas as explained in the fol lowing example Final PDF to printer 9 CHAPTER 1 cen22672ch01001050indd 9 110317 0710 AM You should keep in mind that a formula that is not dimensionally homo geneous is definitely wrong Fig 114 but a dimensionally homogeneous formula is not necessarily right Unity Conversion Ratios Just as all nonprimary dimensions can be formed by suitable combinations of primary dimensions all nonprimary units secondary units can be formed by combinations of primary units Force units for example can be expressed as 1 N 1 kg m s 2 and 1 lbf 32174 lbm ft s 2 They can also be expressed more conveniently as unity conversion ratios as 1 N 1 kgm s 2 1 and 1 lbf 32174 lbmft s 2 1 Unity conversion ratios are identically equal to 1 and are unitless and thus such ratios or their inverses can be inserted conveniently into any calculation to properly convert units Fig 115 You are encouraged to always use unity conversion ratios such as those given here when converting units Some text books insert the archaic gravitational constant gc defined as gc 32174 lbmft lbfs2 1 kgmNs2 1 into equations in order to force units to match This practice leads to unnecessary confusion and is strongly discouraged by the present authors We recommend that you instead use unity conversion ratios FIGURE 113 Schematic for Example 12 Oil V 2 m3 m ρ 850 kgm3 EXAMPLE 12 Obtaining Formulas from Unit Considerations A tank is filled with oil whose density is ρ 850 kgm3 If the volume of the tank is V 2 m3 determine the amount of mass m in the tank SOLUTION The volume of an oil tank is given The mass of oil is to be determined Assumptions Oil is a nearly incompressible substance and thus its density is constant Analysis A sketch of the system just described is given in Fig 113 Suppose we for got the formula that relates mass to density and volume However we know that mass has the unit of kilograms That is whatever calculations we do we should end up with the unit of kilograms Putting the given information into perspective we have ρ 850 kg m 3 and V 2 m 3 It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities Therefore the formula we are looking for should be m ρV Thus m 850 kg m 3 2 m 3 1700 kg Discussion Note that this approach may not work for more complicated formulas Nondimensional constants also may be present in the formulas and these cannot be derived from unit considerations alone FIGURE 114 Always check the units in your calculations FIGURE 115 Every unity conversion ratio as well as its inverse is exactly equal to 1 Shown here are a few commonly used unity conversion ratios each within its own set of parentheses 03048 m 1 ft 1 min 60 s 1 lbm 045359 kg 32174 lbmfts2 1 lbf 1 kgms2 1 N 1 kPa 1000 Nm2 1 kJ 1000 Nm 1 W 1 Js Final PDF to printer 10 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 10 110317 0710 AM EXAMPLE 13 The Weight of One PoundMass Using unity conversion ratios show that 100 lbm weighs 100 lbf on earth Fig 116 SOLUTION A mass of 100 lbm is subjected to standard earth gravity Its weight in lbf is to be determined Assumptions Standard sealevel conditions are assumed Properties The gravitational constant is g 32174 fts2 Analysis We apply Newtons second law to calculate the weight force that corresponds to the known mass and acceleration The weight of any object is equal to its mass times the local value of gravitational acceleration Thus W mg 100 lbm32174 ft s 2 1 lbf 32174 lbm fts 2 100 lbf Discussion The quantity in large parentheses in this equation is a unity conversion ratio Mass is the same regardless of its location However on some other planet with a different value of gravitational acceleration the weight of 1 lbm would differ from that calculated here FIGURE 116 A mass of 1 lbm weighs 1 lbf on earth lbm When you buy a box of breakfast cereal the printing may say Net weight One pound 454 grams See Fig 117 Technically this means that the cereal inside the box weighs 100 lbf on earth and has a mass of 4536 g 04536 kg Using Newtons second law the actual weight of the cereal on earth is W mg 4536 g981 m s 2 1 N 1 kg ms 2 1 kg 1000 g 449 N 13 SYSTEMS AND CONTROL VOLUMES A system is defined as a quantity of matter or a region in space chosen for study The mass or region outside the system is called the surroundings The real or imaginary surface that separates the system from its surround ings is called the boundary Fig 118 The boundary of a system can be fixed or movable Note that the boundary is the contact surface shared by both the system and the surroundings Mathematically speaking the bound ary has zero thickness and thus it can neither contain any mass nor occupy any volume in space Systems may be considered to be closed or open depending on whether a fixed mass or a fixed volume in space is chosen for study A closed system also known as a control mass or just system when the context makes it clear consists of a fixed amount of mass and no mass can cross its bound ary That is no mass can enter or leave a closed system as shown in Fig 119 But energy in the form of heat or work can cross the boundary and the volume of a closed system does not have to be fixed If as a special case even energy is not allowed to cross the boundary that system is called an isolated system FIGURE 117 A quirk in the metric system of units Net weight One pound 454 grams FIGURE 118 System surroundings and boundary Surroundings Boundary System Final PDF to printer 11 CHAPTER 1 cen22672ch01001050indd 11 110317 0710 AM Consider the pistoncylinder device shown in Fig 120 Let us say that we would like to find out what happens to the enclosed gas when it is heated Since we are focusing our attention on the gas it is our system The inner surfaces of the piston and the cylinder form the boundary and since no mass is crossing this boundary it is a closed system Notice that energy may cross the boundary and part of the boundary the inner surface of the piston in this case may move Everything outside the gas including the piston and the cylinder is the surroundings An open system or a control volume as it is often called is a properly selected region in space It usually encloses a device that involves mass flow such as a compressor turbine or nozzle Flow through these devices is best studied by selecting the region within the device as the control volume Both mass and energy can cross the boundary of a control volume A large number of engineering problems involve mass flow in and out of a system and therefore are modeled as control volumes A water heater a car radiator a turbine and a compressor all involve mass flow and should be ana lyzed as control volumes open systems instead of as control masses closed systems In general any arbitrary region in space can be selected as a con trol volume There are no concrete rules for the selection of control volumes but the proper choice certainly makes the analysis much easier If we were to analyze the flow of air through a nozzle for example a good choice for the control volume would be the region within the nozzle The boundaries of a control volume are called a control surface and they can be real or imaginary In the case of a nozzle the inner surface of the nozzle forms the real part of the boundary and the entrance and exit areas form the imaginary part since there are no physical surfaces there Fig 121a A control volume can be fixed in size and shape as in the case of a nozzle or it may involve a moving boundary as shown in Fig 121b Most control volumes however have fixed boundaries and thus do not involve any moving boundaries A control volume can also involve heat and work interactions just as a closed system in addition to mass interaction FIGURE 119 Mass cannot cross the boundaries of a closed system but energy can Closed system Yes m constant Energy No Mass FIGURE 120 A closed system with a moving boundary Gas 2 kg 15 m3 Gas 2 kg 1 m3 Moving boundary Fixed boundary FIGURE 121 A control volume can involve fixed moving real and imaginary boundaries CV Moving boundary Fixed boundary Real boundary b A control volume CV with fixed and moving boundaries as well as real and imaginary boundaries a A control volume CV with real and imaginary boundaries Imaginary boundary CV a nozzle CV Moving boundary Fixed boundary Real boundary b A control volume CV with fixed and moving boundaries as well as real and imaginary boundaries a A control volume CV with real and imaginary boundaries Imaginary boundary CV a nozzle Final PDF to printer 12 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 12 110317 0710 AM As an example of an open system consider the water heater shown in Fig 122 Let us say that we would like to determine how much heat we must transfer to the water in the tank in order to supply a steady stream of hot water Since hot water will leave the tank and be replaced by cold water it is not convenient to choose a fixed mass as our system for the analysis Instead we can concentrate our attention on the volume formed by the interior surfaces of the tank and consider the hot and cold water streams as mass leaving and entering the control volume The interior surfaces of the tank form the control surface for this case and mass is crossing the control surface at two locations In an engineering analysis the system under study must be defined care fully In most cases the system investigated is quite simple and obvious and defining the system may seem like a tedious and unnecessary task In other cases however the system under study may be rather involved and a proper choice of the system may greatly simplify the analysis 14 PROPERTIES OF A SYSTEM Any characteristic of a system is called a property Some familiar properties are pressure P temperature T volume V and mass m The list can be extended to include less familiar ones such as viscosity thermal conductivity modu lus of elasticity thermal expansion coefficient electric resistivity and even velocity and elevation Properties are considered to be either intensive or extensive Intensive properties are those that are independent of the mass of a system such as tem perature pressure and density Extensive properties are those whose values depend on the sizeor extentof the system Total mass total volume and total momentum are some examples of extensive properties An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition as shown in Fig 123 Each part will have the same value of intensive properties as the original system but half the value of the extensive properties Generally uppercase letters are used to denote extensive properties with mass m being a major exception and lowercase letters are used for intensive properties with pressure P and temperature T being the obvious exceptions Extensive properties per unit mass are called specific properties Some examples of specific properties are specific volume v Vm and specific total energy e Em Continuum Matter is made up of atoms that are widely spaced in the gas phase Yet it is very convenient to disregard the atomic nature of a substance and view it as a continuous homogeneous matter with no holes that is a continuum The continuum idealization allows us to treat properties as point functions and to assume the properties vary continually in space with no jump discontinuities This idealization is valid as long as the size of the system we deal with is large relative to the space between the molecules This is the case in practically all problems except some specialized ones The continuum idealization is implicit in many statements we make such as the density of water in a glass is the same at any point FIGURE 122 An open system a control volume with one inlet and one exit McGrawHill EducationChristopher Kerrigan FIGURE 123 Criterion to differentiate intensive and extensive properties V V V Final PDF to printer 13 CHAPTER 1 cen22672ch01001050indd 13 110317 0710 AM To have a sense of the distance involved at the molecular level consider a container filled with oxygen at atmospheric conditions The diameter of the oxygen molecule is about 3 1010 m and its mass is 53 1026 kg Also the mean free path of oxygen at 1 atm pressure and 20C is 63 108 m That is an oxygen molecule travels on average a distance of 63 108 m about 200 times its diameter before it collides with another molecule Also there are about 3 1016 molecules of oxygen in the tiny volume of 1 mm3 at 1 atm pressure and 20C Fig 124 The continuum model is appli cable as long as the characteristic length of the system such as its diameter is much larger than the mean free path of the molecules At very high vacuums or very high elevations the mean free path may become large for example it is about 01 m for atmospheric air at an elevation of 100 km For such cases the rarefied gas flow theory should be used and the impact of individual molecules should be considered In this text we will limit our consideration to substances that can be modeled as a continuum 15 DENSITY AND SPECIFIC GRAVITY Density is defined as mass per unit volume Fig 125 Density ρ m V kg m 3 14 The reciprocal of density is the specific volume v which is defined as volume per unit mass That is v V m 1 ρ 15 For a differential volume element of mass δm and volume δV density can be expressed as ρ δmδV The density of a substance in general depends on temperature and pres sure The density of most gases is proportional to pressure and inversely pro portional to temperature Liquids and solids on the other hand are essentially incompressible substances and the variation of their density with pressure is usually negligible At 20C for example the density of water changes from 998 kgm3 at 1 atm to 1003 kgm3 at 100 atm a change of just 05 percent The density of liquids and solids depends more strongly on temperature than it does on pressure At 1 atm for example the density of water changes from 998 kgm3 at 20C to 975 kgm3 at 75C a change of 23 percent which can still be neglected in many engineering analyses Sometimes the density of a substance is given relative to the density of a wellknown substance Then it is called specific gravity or relative density and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature usually water at 4C for which ρ H 2 O 1000 kgm3 That is Specific gravity SG ρ ρ H 2 O 16 Note that the specific gravity of a substance is a dimensionless quantity How ever in SI units the numerical value of the specific gravity of a substance is FIGURE 124 Despite the relatively large gaps between molecules a gas can usually be treated as a continuum because of the very large number of molecules even in an extremely small volume Void 1 atm 20C O2 3 1016 moleculesmm3 FIGURE 125 Density is mass per unit volume specific volume is volume per unit mass 12 m3 m 3 kg 3 3kg 025 kgm 4 m 1 v V ρ ρ Final PDF to printer 14 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 14 110317 0710 AM exactly equal to its density in gcm3 or kgL or 0001 times the density in kgm3 since the density of water at 4C is 1 gcm3 1 kgL 1000 kgm3 The spe cific gravity of mercury at 0C for example is 136 Therefore its density at 0C is 136 gcm3 136 kgL 13600 kgm3 The specific gravities of some substances at 0C are given in Table 13 Note that substances with specific gravities less than 1 are lighter than water and thus they would float on water The weight of a unit volume of a substance is called specific weight and is expressed as Specific weight γ s ρg N m 3 17 where g is the gravitational acceleration The densities of liquids are essentially constant and thus they can often be approximated as being incompressible substances during most processes without sacrificing much in accuracy 16 STATE AND EQUILIBRIUM Consider a system not undergoing any change At this point all the properties can be measured or calculated throughout the entire system which gives us a set of properties that completely describes the condition or the state of the system At a given state all the properties of a system have fixed values If the value of even one property changes the state will change to a different one In Fig 126 a system is shown at two different states Thermodynamics deals with equilibrium states The word equilibrium implies a state of balance In an equilibrium state there are no unbalanced potentials or driving forces within the system A system in equilibrium experiences no changes when it is isolated from its surroundings There are many types of equilibrium and a system is not in thermodynamic equilibrium unless the conditions of all the relevant types of equilibrium are satisfied For example a system is in thermal equilibrium if the temperature is the same throughout the entire system as shown in Fig 127 That is the system involves no temperature differential which is the driving force for heat flow Mechanical equilibrium is related to pressure and a system is in mechanical equilibrium if there is no change in pressure at any point of the system with time However the pressure may vary within the system with elevation as a result of gravitational effects For example the higher pressure at a bottom layer is balanced by the extra weight it must carry and therefore there is no imbalance of forces The varia tion of pressure as a result of gravity in most thermodynamic systems is rela tively small and usually disregarded If a system involves two phases it is in phase equilibrium when the mass of each phase reaches an equilibrium level and stays there Finally a system is in chemical equilibrium if its chemical com position does not change with time that is no chemical reactions occur A system will not be in equilibrium unless all the relevant equilibrium criteria are satisfied The State Postulate As noted earlier the state of a system is described by its properties But we know from experience that we do not need to specify all the properties in order to fix a state Once a sufficient number of properties are specified FIGURE 126 A system at two different states m 2 kg T2 20C 25 m3 a State 1 m 2 kg T1 20C 15 m3 b State 2 V1 V2 FIGURE 127 A closed system reaching thermal equilibrium 20C a Before b After 23C 35C 40C 30C 42C 32C 32C 32C 32C 32C 32C TABLE 13 Specific gravities of some substances at 0C Substance SG Water 10 Blood 105 Seawater 1025 Gasoline 07 Ethyl alcohol 079 Mercury 136 Wood 0309 Gold 192 Bones 1720 Ice 092 Air at 1 atm 00013 Final PDF to printer 15 CHAPTER 1 cen22672ch01001050indd 15 110317 0710 AM the rest of the properties assume certain values automatically That is specifying a certain number of properties is sufficient to fix a state The number of properties required to fix the state of a system is given by the state postulate The state of a simple compressible system is completely specified by two independent intensive properties A system is called a simple compressible system in the absence of elec trical magnetic gravitational motion and surface tension effects These effects are due to external force fields and are negligible for most engineering problems Otherwise an additional property needs to be specified for each effect that is significant If the gravitational effects are to be considered for example the elevation z needs to be specified in addition to the two properties necessary to fix the state The state postulate requires that the two properties specified be indepen dent to fix the state Two properties are independent if one property can be varied while the other one is held constant Temperature and specific volume for example are always independent properties and together they can fix the state of a simple compressible system Fig 128 Temperature and pressure however are independent properties for singlephase systems but are depen dent properties for multiphase systems At sea level P 1 atm water boils at 100C but on a mountaintop where the pressure is lower water boils at a lower temperature That is T f P during a phasechange process thus temperature and pressure are not sufficient to fix the state of a twophase sys tem Phasechange processes are discussed in detail in Chap 3 17 PROCESSES AND CYCLES Any change that a system undergoes from one equilibrium state to another is called a process and the series of states through which a system passes during a process is called the path of the process Fig 129 To describe a process completely one should specify the initial and final states of the pro cess as well as the path it follows and the interactions with the surroundings When a process proceeds in such a manner that the system remains infini tesimally close to an equilibrium state at all times it is called a quasistatic or quasiequilibrium process A quasiequilibrium process can be viewed as a sufficiently slow process that allows the system to adjust itself internally so that properties in one part of the system do not change any faster than those at other parts This is illustrated in Fig 130 When a gas in a pistoncylinder device is compressed suddenly the molecules near the face of the piston will not have enough time to escape and they will have to pile up in a small region in front of the piston thus creating a highpressure region there Because of this pres sure difference the system can no longer be said to be in equilibrium and this makes the entire process nonquasiequilibrium However if the piston is moved slowly the molecules will have sufficient time to redistribute and there will not be a molecule pileup in front of the piston As a result the pressure inside the cylinder will always be nearly uniform and will rise at the same rate at all locations Since equilibrium is maintained at all times this is a quasiequilibrium process FIGURE 128 The state of nitrogen is fixed by two independent intensive properties Nitrogen T 25C v 09 m3kg FIGURE 129 A process between states 1 and 2 and the process path State 1 State 2 Process path Property B Property A FIGURE 130 Quasiequilibrium and nonquasi equilibrium compression processes a Slow compression quasiequilibrium b Very fast compression nonquasiequilibrium Final PDF to printer 16 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 16 110317 0710 AM It should be pointed out that a quasiequilibrium process is an idealized process and is not a true representation of an actual process But many actual processes closely approximate it and they can be modeled as quasiequilibrium with negligible error Engineers are interested in quasiequilibrium processes for two reasons First they are easy to analyze second workproducing devices deliver the most work when they operate on quasiequilibrium pro cesses Therefore quasiequilibrium processes serve as standards to which actual processes can be compared Process diagrams plotted by employing thermodynamic properties as coor dinates are very useful in visualizing the processes Some common properties that are used as coordinates are temperature T pressure P and volume V or specific volume v Figure 131 shows the PV diagram of a compression process of a gas Note that the process path indicates a series of equilibrium states through which the system passes during a process and has significance for quasi equilibrium processes only For nonquasiequilibrium processes we cannot characterize the entire system by a single state and thus we cannot speak of a process path for a system as a whole A nonquasiequilibrium process is denoted by a dashed line between the initial and final states instead of a solid line The prefix iso is often used to designate a process for which a particular property remains constant An isothermal process for example is a process during which the temperature T remains constant an isobaric process is a process during which the pressure P remains constant and an isochoric or isometric process is a process during which the specific volume v remains constant A system is said to have undergone a cycle if it returns to its initial state at the end of the process That is for a cycle the initial and final states are identical The SteadyFlow Process The terms steady and uniform are used often in engineering and thus it is important to have a clear understanding of their meanings The term steady implies no change with time The opposite of steady is unsteady or transient The term uniform however implies no change with location over a specified region These meanings are consistent with their everyday use steady girl friend uniform properties etc A large number of engineering devices operate for long periods of time under the same conditions and they are classified as steadyflow devices Processes involving such devices can be represented reasonably well by a somewhat idealized process called the steadyflow process which can be defined as a process during which a fluid flows through a control volume steadily Fig 132 That is the fluid properties can change from point to point within the control volume but at any fixed point they remain the same during the entire process Therefore the volume V the mass m and the total energy content E of the control volume remain constant during a steadyflow process Fig 133 Steadyflow conditions can be closely approximated by devices that are intended for continuous operation such as turbines pumps boilers FIGURE 131 The PV diagram of a compression process Initial state Final state Process path 2 1 P V2 V1 V 2 System 1 FIGURE 132 During a steadyflow process fluid properties within the control volume may change with position but not with time 300C 250C 200C 150C Control volume 225C Mass in Time 1 PM Mass out 300C 250C 200C 150C Control volume 225C Mass in Time 3 PM Mass out Final PDF to printer 17 CHAPTER 1 cen22672ch01001050indd 17 110317 0710 AM condensers and heat exchangers or power plants or refrigeration systems Some cyclic devices such as reciprocating engines or compressors do not satisfy any of the conditions stated above since the flow at the inlets and the exits will be pulsating and not steady However the fluid properties vary with time in a periodic manner and the flow through these devices can still be analyzed as a steadyflow process by using timeaveraged values for the properties 18 TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS Although we are familiar with temperature as a measure of hotness or coldness it is not easy to give an exact definition for it Based on our physi ological sensations we express the level of temperature qualitatively with words like freezing cold cold warm hot and redhot However we cannot assign numerical values to temperatures based on our sensations alone Fur thermore our senses may be misleading A metal chair for example will feel much colder than a wooden one even when both are at the same temperature Fortunately several properties of materials change with temperature in a repeatable and predictable way and this forms the basis for accurate tempera ture measurement The commonly used mercuryinglass thermometer for example is based on the expansion of mercury with temperature Temperature is also measured by using several other temperaturedependent properties It is a common experience that a cup of hot coffee left on the table eventu ally cools off and a cold drink eventually warms up That is when a body is brought into contact with another body that is at a different temperature heat is transferred from the body at higher temperature to the one at lower temperature until both bodies attain the same temperature Fig 134 At that point the heat transfer stops and the two bodies are said to have reached thermal equilibrium The equality of temperature is the only requirement for thermal equilibrium The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with a third body they are also in thermal equilibrium with each other It may seem silly that such an obvious fact is called one of the basic laws of thermodynamics However it cannot be concluded from the other laws of thermodynamics and it serves as a basis for the validity of tempera ture measurement By replacing the third body with a thermometer the zeroth law can be restated as two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact The zeroth law was first formulated and labeled by R H Fowler in 1931 As the name suggests its value as a fundamental physical principle was rec ognized more than half a century after the formulation of the first and the second laws of thermodynamics It was named the zeroth law since it should have preceded the first and the second laws of thermodynamics Temperature Scales Temperature scales enable us to use a common basis for temperature mea surements and several have been introduced throughout history All tempera ture scales are based on some easily reproducible states such as the freezing FIGURE 133 Under steadyflow conditions the mass and energy contents of a control volume remain constant Control volume mCV const ECV const Mass in Mass out FIGURE 134 Two bodies reaching thermal equilibrium after being brought into contact in an isolated enclosure 150C Iron 20C Copper 60C Iron 60C Copper Final PDF to printer 18 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 18 110317 0710 AM and boiling points of water which are also called the ice point and the steam point respectively A mixture of ice and water that is in equilibrium with air saturated with vapor at 1 atm pressure is said to be at the ice point and a mixture of liquid water and water vapor with no air in equilibrium at 1 atm pressure is said to be at the steam point The temperature scales used in the SI and in the English system today are the Celsius scale formerly called the centigrade scale in 1948 it was renamed after the Swedish astronomer A Celsius 17021744 who devised it and the Fahrenheit scale named after the German instrument maker G Fahrenheit 16861736 respectively On the Celsius scale the ice and steam points were originally assigned the values of 0 and 100C respectively The corresponding values on the Fahrenheit scale are 32 and 212F These are often referred to as twopoint scales since temperature values are assigned at two different points In thermodynamics it is very desirable to have a temperature scale that is independent of the properties of any substance or substances Such a tempera ture scale is called a thermodynamic temperature scale which is developed later in conjunction with the second law of thermodynamics The thermo dynamic temperature scale in the SI is the Kelvin scale named after Lord Kelvin 18241907 The temperature unit on this scale is the kelvin which is designated by K not K the degree symbol was officially dropped from kel vin in 1967 The lowest temperature on the Kelvin scale is absolute zero or 0 K Then it follows that only one nonzero reference point needs to be assigned to establish the slope of this linear scale Using nonconventional refrigeration techniques scientists have approached absolute zero kelvin they achieved 0000000002 K in 1989 The thermodynamic temperature scale in the English system is the Rankine scale named after William Rankine 18201872 The temperature unit on this scale is the rankine which is designated by R A temperature scale that turns out to be nearly identical to the Kelvin scale is the idealgas temperature scale The temperatures on this scale are mea sured using a constantvolume gas thermometer which is basically a rigid vessel filled with a gas usually hydrogen or helium at low pressure This thermometer is based on the principle that at low pressures the temperature of a gas is proportional to its pressure at constant volume That is the temper ature of a gas of fixed volume varies linearly with pressure at sufficiently low pressures Then the relationship between the temperature and the pressure of the gas in the vessel can be expressed as T a bP 18 where the values of the constants a and b for a gas thermometer are deter mined experimentally Once a and b are known the temperature of a medium can be calculated from this relation by immersing the rigid vessel of the gas thermometer into the medium and measuring the gas pressure when ther mal equilibrium is established between the medium and the gas in the vessel whose volume is held constant An idealgas temperature scale can be developed by measuring the pres sures of the gas in the vessel at two reproducible points such as the ice and the steam points and assigning suitable values to temperatures at those two Final PDF to printer 19 CHAPTER 1 cen22672ch01001050indd 19 110317 0710 AM points Considering that only one straight line passes through two fixed points on a plane these two measurements are sufficient to determine the constants a and b in Eq 18 Then the unknown temperature T of a medium correspond ing to a pressure reading P can be determined from that equation by a simple calculation The values of the constants will be different for each thermom eter depending on the type and the amount of the gas in the vessel and the temperature values assigned at the two reference points If the ice and steam points are assigned the values 0C and 100C respectively then the gas tem perature scale will be identical to the Celsius scale In this case the value of the constant a which corresponds to an absolute pressure of zero is determined to be 27315C regardless of the type and the amount of the gas in the vessel of the gas thermometer That is on a PT diagram all the straight lines pass ing through the data points in this case will intersect the temperature axis at 27315C when extrapolated as shown in Fig 135 This is the lowest tem perature that can be obtained by a gas thermometer and thus we can obtain an absolute gas temperature scale by assigning a value of zero to the constant a in Eq 18 In that case Eq 18 reduces to T bP and thus we need to specify the temperature at only one point to define an absolute gas temperature scale It should be noted that the absolute gas temperature scale is not a thermo dynamic temperature scale since it cannot be used at very low temperatures due to condensation and at very high temperatures due to dissociation and ionization However absolute gas temperature is identical to the thermody namic temperature in the temperature range in which the gas thermometer can be used Thus we can view the thermodynamic temperature scale at this point as an absolute gas temperature scale that utilizes an ideal or imaginary gas that always acts as a lowpressure gas regardless of the temperature If such a gas thermometer existed it would read zero kelvin at absolute zero pressure which corresponds to 27315C on the Celsius scale Fig 136 The Kelvin scale is related to the Celsius scale by T K T C 27315 19 The Rankine scale is related to the Fahrenheit scale by T R T F 45967 110 It is common practice to round the constant in Eq 19 to 273 and that in Eq 110 to 460 The temperature scales in the two unit systems are related by T R 18T K 111 T F 18T C 32 112 A comparison of various temperature scales is given in Fig 137 The reference temperature chosen in the original Kelvin scale was 27315 K or 0C which is the temperature at which water freezes or ice melts and water exists as a solidliquid mixture in equilibrium under standard atmo spheric pressure the ice point At the Tenth General Conference on Weights and Measures in 1954 the reference point was changed to a much more pre cisely reproducible point the triple point of water the state at which all three FIGURE 135 P versus T plots of the experimental data obtained from a constantvolume gas thermometer using four different gases at different but low pressures Measured data points P Gas A Gas B Gas C Gas D 0 27315 Extrapolation T C FIGURE 136 A constantvolume gas thermometer would read 27315C at absolute zero pressure Absolute vacuum V constant T C T K 0 0 27315 P kPa 275 250 225 200 0 25 50 75 0 40 80 120 Final PDF to printer 20 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 20 110317 0710 AM phases of water coexist in equilibrium which is assigned the value 27316 K The Celsius scale was also redefined at this conference in terms of the ideal gas temperature scale and a single fixed point which is again the triple point of water with an assigned value of 001C The boiling temperature of water the steam point was experimentally determined to be again 10000C and thus the new and old Celsius scales were in good agreement The International Temperature Scale of 1990 ITS90 The International Temperature Scale of 1990 which supersedes the Interna tional Practical Temperature Scale of 1968 IPTS68 1948 ITPS48 and 1927 ITS27 was adopted by the International Committee of Weights and Measures at its meeting in 1989 at the request of the Eighteenth General Con ference on Weights and Measures The ITS90 is similar to its predecessors except that it is more refined with updated values of fixed temperatures has an extended range and conforms more closely to the thermodynamic temper ature scale On this scale the unit of thermodynamic temperature T is again the kelvin K defined as the fraction 127316 of the thermodynamic tem perature of the triple point of water which is the sole defining fixed point of both the ITS90 and the Kelvin scale and is the most important thermometric fixed point used in the calibration of thermometers to ITS90 The unit of Celsius temperature is the degree Celsius C which is by defi nition equal in magnitude to the kelvin K A temperature difference may be expressed in kelvins or degrees Celsius The ice point remains the same at 0C 27315 K in both ITS90 and ITPS68 but the steam point is 99975C in ITS90 with an uncertainty of 0005C whereas it was 100000C in IPTS 68 The change is due to precise measurements made by gas thermometry by paying particular attention to the effect of sorption the impurities in a gas absorbed by the walls of the bulb at the reference temperature being desorbed at higher temperatures causing the measured gas pressure to increase The ITS90 extends upward from 065 K to the highest temperature practically measurable in terms of the Planck radiation law using monochromatic radiation It is based on specifying definite temperature values on a number of fixed and easily reproducible points to serve as benchmarks and expressing the variation of temperature in a number of ranges and subranges in functional form In ITS90 the temperature scale is considered in four ranges In the range of 065 to 5 K the temperature scale is defined in terms of the vapor pressure temperature relations for 3He and 4He Between 3 and 245561 K the triple point of neon it is defined by means of a properly calibrated helium gas ther mometer From 138033 K the triple point of hydrogen to 123493 K the freezing point of silver it is defined by means of platinum resistance ther mometers calibrated at specified sets of defining fixed points Above 123493 K it is defined in terms of the Planck radiation law and a suitable defining fixed point such as the freezing point of gold 133733 K We emphasize that the magnitudes of each division of 1 K and 1C are identical Fig 138 Therefore when we are dealing with temperature dif ferences ΔT the temperature interval on both scales is the same Raising the temperature of a substance by 10C is the same as raising it by 10 K That is ΔT K ΔT C 113 FIGURE 138 Comparison of magnitudes of various temperature units 1C 1 K 18F 18 R FIGURE 137 Comparison of temperature scales 27315 C 0 27316 001 K 45967 F 0 49169 3202 R Triple point of water Absolute zero Final PDF to printer 21 CHAPTER 1 cen22672ch01001050indd 21 110317 0710 AM ΔT R ΔT F 114 Some thermodynamic relations involve the temperature T and often the question arises of whether it is in K or C If the relation involves temperature differences such as a bΔT it makes no difference and either can be used However if the relation involves temperatures only instead of temperature differences such as a bT then K must be used When in doubt it is always safe to use K because there are virtually no situations in which the use of K is incorrect but there are many thermodynamic relations that will yield an erroneous result if C is used EXAMPLE 14 Expressing Temperatures in Different Units Humans are most comfortable when the temperature is between 65F and 75F Express these temperature limits in C Convert the size of this temperature range 10F to K C and R Is there any difference in the size of this range as measured in relative or absolute units SOLUTION A temperature range given in F unit is to be converted to C unit and the temperature difference in F is to be expressed in K C and R Analysis The lower and upper limits of comfort in C are T C T F 32 18 65 32 18 183 C T C T F 32 18 75 32 18 239 C A temperature change of 10F in various units are ΔT R ΔT F 10 R ΔT C ΔT F 18 10 18 56 C ΔT K ΔT C 56 K Therefore the units C and K in the SI system and F and R in the English system are interchangeable when dealing with temperature differences Discussion Students should be careful when making temperature unit conversions They should identify first whether the conversion involves a temperature value or a tem perature change value 19 PRESSURE Pressure is defined as a normal force exerted by a fluid per unit area Nor mally we speak of pressure when we deal with a gas or a liquid The coun terpart of pressure in solids is normal stress Note however that pressure is a scalar quantity while stress is a tensor Since pressure is defined as force per unit area it has the unit of newtons per square meter Nm2 which is called a pascal Pa That is 1 Pa 1 N m 2 Final PDF to printer 22 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 22 110317 0710 AM FIGURE 139 The normal stress or pressure on the feet of a chubby person is much greater than on the feet of a slim person 150 pounds Afeet 50 in2 P 3 psi P 6 psi 300 pounds W Afeet 150 lbf 50 in2 P n 3 psi σ The pressure unit pascal is too small for most pressures encountered in practice Therefore its multiples kilopascal 1 kPa 103 Pa and megapascal 1 MPa 106 Pa are commonly used Three other pressure units commonly used in practice especially in Europe are bar standard atmosphere and kilogramforce per square centimeter 1 bar 10 5 Pa 01 MPa 100 kPa 1 atm 101325 Pa 101325 kPa 101325 bars 1 kgf cm 2 9807 N cm 2 9807 10 4 N m 2 9807 10 4 Pa 09807 bar 09679 atm Note the pressure units bar atm and kgfcm2 are almost equivalent to each other In the English system the pressure unit is poundforce per square inch lbfin2 or psi and 1 atm 14696 psi The pressure units kgfcm2 and lbfin2 are also denoted by kgcm2 and lbin2 respectively and they are commonly used in tire gages It can be shown that 1 kgfcm2 14223 psi Pressure is also used on solid surfaces as synonymous with normal stress which is the force acting perpendicular to the surface per unit area For exam ple a 150pound person with a total foot imprint area of 50 in2 exerts a pressure of 150 lbf50 in2 30 psi on the floor Fig 139 If the person stands on one foot the pressure doubles If the person gains excessive weight he or she is likely to encounter foot discomfort because of the increased pressure on the foot the size of the bottom of the foot does not change with weight gain This also explains how a person can walk on fresh snow without sinking by wearing large snowshoes and how a person cuts with little effort when using a sharp knife The actual pressure at a given position is called the absolute pressure and it is measured relative to absolute vacuum ie absolute zero pressure Most pressuremeasuring devices however are calibrated to read zero in the atmosphere Fig 140 and so they indicate the difference between the absolute pressure and the local atmospheric pressure This difference is called the gage pressure Pgage can be positive or negative but pressures below atmospheric pressure are sometimes called vacuum pressures and are measured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pres sure Absolute gage and vacuum pressures are related to each other by P gage P abs P atm 115 P vac P atm P abs 116 This is illustrated in Fig 141 Like other pressure gages the gage used to measure the air pressure in an automobile tire reads the gage pressure Therefore the common reading of 320 psi 225 kgfcm2 indicates a pressure of 320 psi above the atmospheric pressure At a location where the atmospheric pressure is 143 psi for exam ple the absolute pressure in the tire is 320 143 463 psi In thermodynamic relations and tables absolute pressure is almost always used Throughout this text the pressure P will denote absolute pressure unless specified otherwise Often the letters a for absolute pressure and g for gage pressure are added to pressure units such as psia and psig to clarify what is meant FIGURE 140 Some basic pressure gages Dresser Instruments Dresser Inc Used by permission Final PDF to printer 23 CHAPTER 1 cen22672ch01001050indd 23 110317 0710 AM FIGURE 141 Absolute gage and vacuum pressures Absolute vacuum Absolute vacuum Pabs Pvac Patm Patm Patm Pgage Pabs Pabs 0 FIGURE 142 The pressure of a fluid at rest increases with depth as a result of added weight Pgage EXAMPLE 15 Absolute Pressure of a Vacuum Chamber A vacuum gage connected to a chamber reads 58 psi at a location where the atmospheric pressure is 145 psi Determine the absolute pressure in the chamber SOLUTION The gage pressure of a vacuum chamber is given The absolute pressure in the chamber is to be determined Analysis The absolute pressure is easily determined from Eq 116 to be P abs P atm P vac 145 58 87 psi Discussion Note that the local value of the atmospheric pressure is used when determining the absolute pressure Variation of Pressure with Depth It will come as no surprise to you that pressure in a fluid at rest does not change in the horizontal direction This can be shown easily by considering a thin horizontal layer of fluid and doing a force balance in any horizontal direction However this is not the case in the vertical direction in a gravity field Pressure in a fluid increases with depth because more fluid rests on deeper layers and the effect of this extra weight on a deeper layer is bal anced by an increase in pressure Fig 142 To obtain a relation for the variation of pressure with depth consider a rectangular fluid element of height Δz length Δx and unit depth Δy 1 into the page in equilibrium as shown in Fig 143 Assuming the density of the fluid ρ to be constant a force balance in the vertical zdirection gives F z m a z 0 P 1 Δx Δy P 2 Δx Δy ρg Δx Δy Δz 0 where W mg ρg Δx Δy Δz is the weight of the fluid element and Δz z2 z1 Dividing by Δx Δy and rearranging gives ΔP P 2 P 1 ρg Δz γ s Δz 117 where γs ρg is the specific weight of the fluid Thus we conclude that the pressure difference between two points in a constant density fluid is FIGURE 143 Freebody diagram of a rectangular fluid element in equilibrium P1 W P2 x 0 z z z2 z1 x Δ Δ g Final PDF to printer 24 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 24 110317 0710 AM proportional to the vertical distance Δz between the points and the density ρ of the fluid Noting the negative sign pressure in a static fluid increases linearly with depth This is what a diver experiences when diving deeper in a lake An easier equation to remember and apply between any two points in the same fluid under hydrostatic conditions is P below P above ρg Δz P above γ s Δz 118 where below refers to the point at lower elevation deeper in the fluid and above refers to the point at higher elevation If you use this equation consis tently you should avoid sign errors For a given fluid the vertical distance Δz is sometimes used as a measure of pressure and it is called the pressure head We also conclude from Eq 117 that for small to moderate distances the variation of pressure with height is negligible for gases because of their low density The pressure in a tank containing a gas for example can be consid ered to be uniform since the weight of the gas is too small to make a significant difference Also the pressure in a room filled with air can be approximated as a constant Fig 144 If we take the above point to be at the free surface of a liquid open to the atmosphere Fig 145 where the pressure is the atmospheric pressure Patm then from Eq 118 the pressure at a depth h below the free surface becomes P P atm ρgh or P gage ρgh 119 Liquids are essentially incompressible substances and thus the variation of density with depth is negligible This is also the case for gases when the eleva tion change is not very large The variation of density of liquids or gases with temperature can be significant however and may need to be considered when high accuracy is desired Also at great depths such as those encountered in oceans the change in the density of a liquid can be significant because of the compression by the tremendous amount of liquid weight above The gravitational acceleration g varies from 9807 ms2 at sea level to 9764 ms2 at an elevation of 14000 m where large passenger planes cruise This is a change of just 04 percent in this extreme case Therefore g can be approxi mated as a constant with negligible error For fluids whose density changes significantly with elevation a relation for the variation of pressure with elevation can be obtained by dividing Eq 117 by Δz and taking the limit as Δz 0 This yields dP dz ρg 120 Note that dP is negative when dz is positive since pressure decreases in an upward direction When the variation of density with elevation is known the pressure difference between any two points 1 and 2 can be determined by integration to be ΔP P 2 P 1 1 2 ρgdz 121 For constant density and constant gravitational acceleration this relation reduces to Eq 117 as expected FIGURE 144 In a room filled with a gas the variation of pressure with height is negligible Ptop 1 atm Air A 5mhigh room Pbottom 1006 atm FIGURE 145 Pressure in a liquid at rest increases linearly with distance from the free surface Pabove Patm Pbelow Patm ρgh h Final PDF to printer 25 CHAPTER 1 cen22672ch01001050indd 25 110317 0710 AM Pressure in a fluid at rest is independent of the shape or cross section of the container It changes with the vertical distance but remains constant in other directions Therefore the pressure is the same at all points on a horizontal plane in a given fluid The Dutch mathematician Simon Stevin 15481620 published in 1586 the principle illustrated in Fig 146 Note that the pres sures at points A B C D E F and G are the same since they are at the same depth and they are interconnected by the same static fluid However the pressures at points H and I are not the same since these two points cannot be interconnected by the same fluid ie we cannot draw a curve from point I to point H while remaining in the same fluid at all times although they are at the same depth Can you tell at which point the pressure is higher Also notice that the pressure force exerted by the fluid is always normal to the sur face at the specified points A consequence of the pressure in a fluid remaining constant in the hori zontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount This is called Pascals law after Blaise Pascal 16231662 Pascal also knew that the force applied by a fluid is proportional to the surface area He realized that two hydraulic cylinders of different areas could be connected and the larger could be used to exert a pro portionally greater force than that applied to the smaller Pascals machine has been the source of many inventions that are a part of our daily lives such as hydraulic brakes and lifts This is what enables us to lift a car easily with one arm as shown in Fig 147 Noting that P1 P2 since both pistons are at the same level the effect of small height differences is negligible especially at high pressures the ratio of output force to input force is determined to be P 1 P 2 F 1 A 1 F 2 A 2 F 2 F 1 A 2 A 1 122 FIGURE 146 Under hydrostatic conditions the pressure is the same at all points on a horizontal plane in a given fluid regardless of geometry provided that the points are interconnected by the same fluid h A B C D E Water Mercury F G I H Patm PA PB PC PD PE PF PG Patm ρgh FIGURE 147 Lifting of a large weight by a small force by the application of Pascals law A common example is a hydraulic jack Top StockbyteGetty Images RF F1 P1A1 1 2 A1 P1 F2 P2A2 A2 P2 Final PDF to printer 26 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 26 110317 0710 AM The area ratio A2 A1 is called the ideal mechanical advantage of the hydrau lic lift Using a hydraulic car jack with a piston area ratio of A2 A1 100 for example a person can lift a 1000kg car by applying a force of just 10 kgf 908 N 110 PRESSURE MEASUREMENT DEVICES The Barometer Atmospheric pressure is measured by a device called a barometer thus the atmospheric pressure is often referred to as the barometric pressure The Italian Evangelista Torricelli 16081647 was the first to conclusively prove that the atmospheric pressure can be measured by inverting a mercury filled tube into a mercury container that is open to the atmosphere as shown in Fig 148 The pressure at point B is equal to the atmospheric pressure and the pressure at point C can be taken to be zero since there is only mercury vapor above point C and the pressure is very low relative to Patm and can be neglected to an excellent approximation Writing a force balance in the verti cal direction gives P atm ρgh 123 where ρ is the density of mercury g is the local gravitational acceleration and h is the height of the mercury column above the free surface Note that the length and the crosssectional area of the tube have no effect on the height of the fluid column of a barometer Fig 149 A frequently used pressure unit is the standard atmosphere which is defined as the pressure produced by a column of mercury 760 mm in height at 0C ρHg 13595 kgm3 under standard gravitational acceleration g 9807 ms2 If water instead of mercury were used to measure the standard atmospheric pressure a water column of about 103 m would be needed Pres sure is sometimes expressed especially by weather forecasters in terms of the height of the mercury column The standard atmospheric pressure for example is 760 mmHg 2992 inHg at 0C The unit mmHg is also called the torr in honor of Torricelli Therefore 1 atm 760 torr and 1 torr 1333 Pa Atmospheric pressure Patm changes from 101325 kPa at sea level to 8988 7950 5405 265 and 553 kPa at altitudes of 1000 2000 5000 10000 and 20000 meters respectively The typical atmospheric pressure in Denver ele vation 1610 m for example is 834 kPa Remember that the atmospheric pressure at a location is simply the weight of the air above that location per unit surface area Therefore it changes not only with elevation but also with weather conditions The decline of atmospheric pressure with elevation has farreaching ram ifications in daily life For example cooking takes longer at high altitudes since water boils at a lower temperature at lower atmospheric pressures Nose bleeding is a common experience at high altitudes since the difference between the blood pressure and the atmospheric pressure is larger in this case and the delicate walls of veins in the nose are often unable to withstand this extra stress For a given temperature the density of air is lower at high altitudes and thus a given volume contains less air and less oxygen So it is no surprise FIGURE 148 The basic barometer h A h B Vacuum Mercury C Patm W ρghA FIGURE 149 The length and the crosssectional area of the tube have no effect on the height of the fluid column of a barometer provided that the tube diameter is large enough to avoid surface tension capillary effects A2 A1 A3 Final PDF to printer 27 CHAPTER 1 cen22672ch01001050indd 27 110317 0710 AM that we tire more easily and experience breathing problems at high altitudes To compensate for this effect people living at higher altitudes develop more efficient lungs Similarly a 20L car engine will act like a 17L car engine at 1500 m altitude unless it is turbocharged because of the 15 percent drop in pressure and thus 15 percent drop in the density of air Fig 150 A fan or compressor will displace 15 percent less air at that altitude for the same vol ume displacement rate Therefore larger cooling fans may need to be selected for operation at high altitudes to ensure the specified mass flow rate The lower pressure and thus lower density also affects lift and drag airplanes need a longer runway at high altitudes to develop the required lift and they climb to very high altitudes for cruising in order to reduce drag and thus achieve better fuel efficiency FIGURE 150 At high altitudes a car engine generates less power and a person gets less oxygen because of the lower density of air Engine Lungs EXAMPLE 16 Measuring Atmospheric Pressure with a Barometer Determine the atmospheric pressure at a location where the barometric reading is 740 mmHg and the gravitational acceleration is g 9805 ms2 Assume the tempera ture of mercury to be 10C at which its density is 13570 kgm3 SOLUTION The barometric reading at a location in height of mercury column is given The atmospheric pressure is to be determined Assumptions The temperature of mercury is assumed to be 10C Properties The density of mercury is given to be 13570 kgm3 Analysis From Eq 123 the atmospheric pressure is determined to be P atm ρgh 13570 kg m 3 9805 m s 2 0740 m 1 N 1 kgm s 2 1 kPa 1000 N m 2 985 kPa Discussion Note that density changes with temperature and thus this effect should be considered in calculations FIGURE 151 Schematic for Example 17 12 m Patm IV bottle EXAMPLE 17 GravityDriven Flow from an IV Bottle Intravenous infusions usually are driven by gravity by hanging the fluid bottle at suf ficient height to counteract the blood pressure in the vein and to force the fluid into the body Fig 151 The higher the bottle is raised the higher the flow rate of the fluid will be a If it is observed that the fluid and the blood pressures balance each other when the bottle is 12 m above the arm level determine the gage pressure of the blood b If the gage pressure of the fluid at the arm level needs to be 20 kPa for sufficient flow rate determine how high the bottle must be placed Take the density of the fluid to be 1020 kgm3 SOLUTION It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined Assumptions 1 The IV fluid is incompressible 2 The IV bottle is open to the atmosphere Final PDF to printer 28 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 28 110317 0710 AM Properties The density of the IV fluid is given to be ρ 1020 kgm3 Analysis a Noting that the IV fluid and the blood pressures balance each other when the bottle is 12 m above the arm level the gage pressure of the blood in the arm is simply equal to the gage pressure of the IV fluid at a depth of 12 m P gagearm P abs P atm ρg h armbottle 1020 kg m 3 981 m s 2 120 m 1 kN 1000 kgm s 2 1 kPa 1 kN m 2 120 kPa b To provide a gage pressure of 20 kPa at the arm level the height of the sur face of the IV fluid in the bottle from the arm level is again determined from Pgagearm ρgharmbottle to be h armbottle P gagearm ρg 20 kPa 1020 kg m 3 981 m s 2 1000 kgm s 2 1 kN 1 kN m 2 1 kPa 200 m Discussion Note that the height of the reservoir can be used to control flow rates in gravitydriven flows When there is flow the pressure drop in the tube due to frictional effects also should be considered For a specified flow rate this requires raising the bottle a little higher to overcome the pressure drop EXAMPLE 18 Hydrostatic Pressure in a Solar Pond with Variable Density Solar ponds are small artificial lakes a few meters deep that are used to store solar energy The rise of heated and thus less dense water to the surface is prevented by adding salt at the pond bottom In a typical salt gradient solar pond the density of water increases in the gradient zone as shown in Fig 152 and the density can be expressed as ρ ρ 0 1 tan 2 π 4 s H where ρ0 is the density on the water surface s is the vertical distance measured down ward from the top of the gradient zone s z and H is the thickness of the gradient zone For H 4 m ρ0 1040 kgm3 and a thickness of 08 m for the surface zone calculate the gage pressure at the bottom of the gradient zone SOLUTION The variation of density of saline water in the gradient zone of a solar pond with depth is given The gage pressure at the bottom of the gradient zone is to be determined Assumptions The density in the surface zone of the pond is constant Properties The density of brine on the surface is given to be 1040 kgm3 FIGURE 152 Schematic for Example 18 Increasing salinity and density Surface zone Sun H 4 m s Gradient zone Storage zone 1 2 ρ0 1040 kgm3 Final PDF to printer 29 CHAPTER 1 cen22672ch01001050indd 29 110317 0710 AM Analysis We label the top and the bottom of the gradient zone as 1 and 2 respectively Noting that the density of the surface zone is constant the gage pressure at the bottom of the surface zone which is the top of the gradient zone is P 1 ρg h 1 1040 kg m 3 981 m s 2 08 m 1 kN 1000 kgm s 2 816 kPa since 1 kNm2 1 kPa Since s z the differential change in hydrostatic pressure across a vertical distance of ds is given by dP ρg ds Integrating from the top of the gradient zone point 1 where s 0 to any location s in the gradient zone no subscript gives P P 1 0 s ρg ds P P 1 0 s ρ 0 1 tan 2 π 4 s H g ds Performing the integration gives the variation of gage pressure in the gradient zone to be P P 1 ρ 0 g 4H π sinh 1 tan π 4 s H Then the pressure at the bottom of the gradient zone s H 4 m becomes P 2 816 kPa 1040 kg m 3 981 m s 2 4 4 m π sinh 1 tan π 4 4 4 1 kN 1000 kgm s 2 540 kPa gage Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig 153 The dashed line indicates the hydrostatic pressure for the case of constant density at 1040 kgm3 and is given for reference Note that the variation of pressure with depth is not linear when density varies with depth That is why integration was required FIGURE 153 The variation of gage pressure with depth in the gradient zone of the solar pond 4 3 Constant density Variable density 2 35 25 15 1 05 0 0 10 20 30 P kPa s m 40 50 60 The Manometer We notice from Eq 117 that an elevation change of Δz in a fluid at rest corresponds to ΔPρg which suggests that a fluid column can be used to measure pressure differences A device based on this principle is called a manometer and it is commonly used to measure small and moderate pres sure differences A manometer consists of a glass or plastic Utube containing one or more fluids such as mercury water alcohol or oil Fig 154 To keep the size of the manometer to a manageable level heavy fluids such as mer cury are used if large pressure differences are anticipated Consider the manometer shown in Fig 155 that is used to measure the pressure in the tank Since the gravitational effects of gases are negligible the pressure anywhere in the tank and at position 1 has the same value Furthermore since pressure in a fluid does not vary in the horizontal direction within a fluid the pressure at point 2 is the same as the pressure at point 1 P2 P1 FIGURE 154 A simple Utube manometer with high pressure applied to the right side John M Cimbala Final PDF to printer 30 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 30 110317 0710 AM The differential fluid column of height h is in static equilibrium and it is open to the atmosphere Then the pressure at point 2 is determined directly from Eq 118 to be P 2 P atm ρgh 124 where ρ is the density of the manometer fluid in the tube Note that the cross sectional area of the tube has no effect on the differential height h and thus on the pressure exerted by the fluid However the diameter of the tube should be large enough more than several millimeters to ensure that the surface ten sion effect and thus the capillary rise is negligible FIGURE 156 Schematic for Example 19 P h 55 cm SG 085 Patm 96 kPa FIGURE 155 The basic manometer Gas h 1 2 EXAMPLE 19 Measuring Pressure with a Manometer A manometer is used to measure the pressure of a gas in a tank The fluid used has a specific gravity of 085 and the manometer column height is 55 cm as shown in Fig 156 If the local atmospheric pressure is 96 kPa determine the absolute pressure within the tank SOLUTION The reading of a manometer attached to a tank and the atmospheric pressure are given The absolute pressure in the tank is to be determined Assumptions The density of the gas in the tank is much lower than the density of the manometer fluid Properties The specific gravity of the manometer fluid is given to be 085 We take the standard density of water to be 1000 kgm3 Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water ρ SG ρ water 0851000 kg m 3 850 kg m 3 Then from Eq 124 P P atm ρgh 96 kPa 850 kg m 3 981 m s 2 055 m 1 N 1 kgm s 2 1 kPa 1000 N m 2 1006 kPa Discussion Note that the gage pressure in the tank is 46 kPa Some manometers use a slanted or inclined tube in order to increase the resolution precision when reading the fluid height Such devices are called inclined manometers Many engineering problems and some manometers involve multiple immis cible fluids of different densities stacked on top of each other Such systems can be analyzed easily by remembering that 1 the pressure change across a fluid column of height h is ΔP ρgh 2 pressure increases downward in a given fluid and decreases upward ie Pbottom Ptop and 3 two points at the same elevation in a continuous fluid at rest are at the same pressure The last principle which is a result of Pascals law allows us to jump from one fluid column to the next in manometers without worrying about pressure change as long as we stay in the same continuous fluid and the fluid Final PDF to printer 31 CHAPTER 1 cen22672ch01001050indd 31 110317 0710 AM is at rest Then the pressure at any point can be determined by starting with a point of known pressure and adding or subtracting ρgh terms as we advance toward the point of interest For example the pressure at the bottom of the tank in Fig 157 can be determined by starting at the free surface where the pressure is Patm moving downward until we reach point 1 at the bottom and setting the result equal to P1 It gives P atm ρ 1 g h 1 ρ 2 g h 2 ρ 3 g h 3 P 1 In the special case of all fluids having the same density this relation reduces to Patm ρgh1 h2 h3 P1 Manometers are particularly wellsuited to measure pressure drops across a horizontal flow section between two specified points due to the presence of a device such as a valve or heat exchanger or any resistance to flow This is done by connecting the two legs of the manometer to these two points as shown in Fig 158 The working fluid can be either a gas or a liquid whose density is ρ1 The density of the manometer fluid is ρ2 and the differential fluid height is h The two fluids must be immiscible and ρ2 must be greater than ρ1 A relation for the pressure difference P1 P2 can be obtained by starting at point 1 with P1 moving along the tube by adding or subtracting the ρgh terms until we reach point 2 and setting the result equal to P2 P 1 ρ 1 g a h ρ 2 gh ρ 1 ga P 2 125 Note that we jumped from point A horizontally to point B and ignored the part underneath since the pressure at both points is the same Simplifying P 1 P 2 ρ 2 ρ 1 gh 126 Note that the distance a must be included in the analysis even though it has no effect on the result Also when the fluid flowing in the pipe is a gas then ρ1 ρ2 and the relation in Eq 126 simplifies to P1 P2 ρ2gh FIGURE 157 In stackedup fluid layers at rest the pressure change across each fluid layer of density ρ and height h is ρgh Patm 1 h3 h2 h1 Fluid 2 Fluid 1 Fluid 3 FIGURE 158 Measuring the pressure drop across a flow section or a flow device by a differential manometer a h ρ1 A B Fluid A flow section or flow device 1 2 ρ2 EXAMPLE 110 Measuring Pressure with a Multifluid Manometer The water in a tank is pressurized by air and the pressure is measured by a multifluid manometer as shown in Fig 159 The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 856 kPa Determine the air pressure in the tank if h1 01 m h2 02 m and h3 035 m Take the densities of water oil and mercury to be 1000 kgm3 850 kgm3 and 13600 kgm3 respectively SOLUTION The pressure in a pressurized water tank is measured by a multifluid manometer The air pressure in the tank is to be determined Assumptions The air pressure in the tank is uniform ie its variation with eleva tion is negligible due to its low density and thus we can determine the pressure at the airwater interface Properties The densities of water oil and mercury are given to be 1000 kgm3 850 kgm3 and 13600 kgm3 respectively FIGURE 159 Schematic for Example 110 drawing not to scale h1 h2 h3 Oil Mercury Water Air 1 2 Final PDF to printer 32 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 32 110317 0710 AM Other Pressure Measurement Devices Another type of commonly used mechanical pressure measurement device is the Bourdon tube named after the French engineer and inventor Eugene Bourdon 18081884 which consists of a bent coiled or twisted hollow metal tube whose end is closed and connected to a dial indicator needle Fig 160 When the tube is open to the atmosphere the tube is undeflected and the needle on the dial at this state is calibrated to read zero gage pres sure When the fluid inside the tube is pressurized the tube stretches and moves the needle in proportion to the applied pressure Electronics have made their way into every aspect of life including pressure measurement devices Modern pressure sensors called pressure transducers use various techniques to convert the pressure effect to an electrical effect such as a change in voltage resistance or capacitance Pressure transducers are smaller and faster and they can be more sensitive reliable and precise than their mechanical counterparts They can measure pressures from less than a millionth of 1 atm to several thousands of atm A wide variety of pressure transducers is available to measure gage abso lute and differential pressures in a wide range of applications Gage pres sure transducers use the atmospheric pressure as a reference by venting the back side of the pressuresensing diaphragm to the atmosphere and they give a zero signal output at atmospheric pressure regardless of altitude Abso lute pressure transducers are calibrated to have a zero signal output at full vacuum Differential pressure transducers measure the pressure difference between two locations directly instead of using two pressure transducers and taking their difference Straingage pressure transducers work by having a diaphragm deflect between two chambers open to the pressure inputs As the diaphragm Analysis Starting with the pressure at point 1 at the airwater interface moving along the tube by adding or subtracting the ρgh terms until we reach point 2 and setting the result equal to Patm since the tube is open to the atmosphere gives P 1 ρ water g h 1 ρ oil g h 2 ρ mercury g h 3 P 2 P atm Solving for P1 and substituting P 1 P atm ρ water g h 1 ρ oil g h 2 ρ mercury g h 3 P atm g ρ mercury h 3 ρ water h 1 ρ oil h 2 856 kPa 981 m s 2 13600 kg m 3 035 m 1000 kg m 3 01 m 850 kg m 3 02 m 1 N 1 kgm s 2 1 kPa 1000 N m 2 130 kPa Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis considerably Also note that mercury is a toxic fluid and mercury manometers and thermometers are being replaced by ones with safer fluids because of the risk of exposure to mercury vapor during an accident FIGURE 160 Various types of Bourdon tubes used to measure pressure They work on the same principle as party noise makers bottom photo due to the flat tube cross section Photo John M Cimbala Ctype Spiral Twisted tube Tube cross section Helical Final PDF to printer 33 CHAPTER 1 cen22672ch01001050indd 33 110317 0710 AM stretches in response to a change in pressure difference across it the strain gage stretches and a Wheatstone bridge circuit amplifies the output A capaci tance transducer works similarly but capacitance change is measured instead of resistance change as the diaphragm stretches Piezoelectric transducers also called solidstate pressure transducers work on the principle that an electric potential is generated in a crystalline substance when it is subjected to mechanical pressure This phenomenon first discovered by brothers Pierre and Jacques Curie in 1880 is called the piezoelectric or presselectric effect Piezoelectric pressure transducers have a much faster frequency response compared to diaphragm units and are very suitable for highpressure applications but they are generally not as sensitive as diaphragmtype transducers especially at low pressures Another type of mechanical pressure gage called a deadweight tester is used primarily for calibration and can measure extremely high pressures Fig 161 As its name implies a deadweight tester measures pressure directly through application of a weight that provides a force per unit area the fundamental definition of pressure It is constructed with an internal chamber filled with a fluid usually oil along with a tightfitting piston cyl inder and plunger Weights are applied to the top of the piston which exerts a force on the oil in the chamber The total force F acting on the oil at the pistonoil interface is the sum of the weight of the piston plus the applied weights Since the piston crosssectional area Ae is known the pressure is calculated as P FAe The only significant source of error is that due to static friction along the interface between the piston and cylinder but even this error is usually negligibly small The reference pressure port is connected to either an unknown pressure that is to be measured or to a pressure sensor that is to be calibrated 111 PROBLEMSOLVING TECHNIQUE The first step in learning any science is to grasp the fundamentals and to gain a sound knowledge of it The next step is to master the fundamentals by test ing this knowledge This is done by solving significant realworld problems Solving such problems especially complicated ones requires a systematic approach By using a stepbystep approach an engineer can reduce the solu tion of a complicated problem into the solution of a series of simple problems Fig 162 When you are solving a problem we recommend that you use the following steps zealously as applicable This will help you avoid some of the common pitfalls associated with problem solving Step 1 Problem Statement In your own words briefly state the problem the key information given and the quantities to be found This is to make sure that you understand the prob lem and the objectives before you try to solve the problem Step 2 Schematic Draw a realistic sketch of the physical system involved and list the relevant information on the figure The sketch does not have to be something elab orate but it should resemble the actual system and show the key features FIGURE 161 A deadweight tester can measure extremely high pressures up to 10000 psi in some applications Ae F Oil reservoir Adjustable plunger Crank Oil Internal chamber Reference pressure port Piston Weights FIGURE 162 A stepbystep approach can greatly simplify problem solving Solution Hard way Easy way Problem Final PDF to printer 34 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 34 110317 0710 AM Indicate any energy and mass interactions with the surroundings Listing the given information on the sketch helps one to see the entire problem at once Also check for properties that remain constant during a process such as tem perature during an isothermal process and indicate them on the sketch Step 3 Assumptions and Approximations State any appropriate assumptions and approximations made to simplify the problem to make it possible to obtain a solution Justify the questionable assumptions Assume reasonable values for missing quantities that are neces sary For example in the absence of specific data for atmospheric pressure it can be taken to be 1 atm However it should be noted in the analysis that the atmospheric pressure decreases with increasing elevation For example it drops to 083 atm in Denver elevation 1610 m Fig 163 Step 4 Physical Laws Apply all the relevant basic physical laws and principles such as the conserva tion of mass and reduce them to their simplest form by utilizing the assump tions made However the region to which a physical law is applied must be clearly identified first For example the increase in speed of water flowing through a nozzle is analyzed by applying conservation of mass between the inlet and outlet of the nozzle Step 5 Properties Determine the unknown properties at known states necessary to solve the problem from property relations or tables List the properties separately and indicate their source if applicable Step 6 Calculations Substitute the known quantities into the simplified relations and perform the calculations to determine the unknowns Pay particular attention to the units and unit cancellations and remember that a dimensional quantity without a unit is meaningless Also dont give a false implication of high precision by copying all the digits from the screen of the calculatorround the results to an appropriate number of significant digits see the subsection on significant digits at the end of this section Step 7 Reasoning Verification and Discussion Check to make sure that the results obtained are reasonable and intuitive and verify the validity of the questionable assumptions Repeat the calculations that resulted in unreasonable values For example insulating a water heater that uses 80 worth of natural gas a year cannot result in savings of 200 a year Fig 164 Also point out the significance of the results and discuss their implications State the conclusions that can be drawn from the results and any recommen dations that can be made from them Emphasize the limitations under which the results are applicable and caution against any possible misunderstandings and using the results in situations where the underlying assumptions do not apply For example if you determined that wrapping a water heater with a 30 insulation jacket will reduce the energy cost by 40 a year indicate that FIGURE 163 The assumptions made while solving an engineering problem must be reasonable and justifiable Given Air temperature in Denver To be found Density of air Missing information Atmospheric pressure Assumption 1 Take P 1 atm Inappropriate Ignores effect of altitude Will cause more than 15 error Assumption 2 Take P 083 atm Appropriate Ignores only minor effects such as weather FIGURE 164 The results obtained from an engineering analysis must be checked for reasonableness Energy use Energy saved by insulation IMPOSSIBLE 80yr 200yr Final PDF to printer 35 CHAPTER 1 cen22672ch01001050indd 35 110317 0710 AM the insulation will pay for itself from the energy it saves in less than a year However also indicate that the analysis does not consider labor costs and that this will be the case if you install the insulation yourself Keep in mind that the solutions you present to your instructors and any engineering analysis presented to others is a form of communication There fore neatness organization completeness and visual appearance are of utmost importance for maximum effectiveness Fig 165 Besides neatness also serves as a great checking tool since it is very easy to spot errors and inconsistencies in neat work Carelessness and skipping steps to save time often end up costing more time and unnecessary anxiety The approach described here is used in the solved example problems with out explicitly stating each step as well as in the Solutions Manual of this text For some problems some of the steps may not be applicable or necessary For example often it is not practical to list the properties separately However we cannot overemphasize the importance of a logical and orderly approach to problem solving Most difficulties encountered while solving a problem are not due to a lack of knowledge rather they are due to a lack of organization You are strongly encouraged to follow these steps in problem solving until you develop your own approach that works best for you Engineering Software Packages You may be wondering why we are about to undertake an indepth study of the fundamentals of another engineering science After all almost all such problems we are likely to encounter in practice can be solved using one of several sophisticated software packages readily available in the market today These software packages not only give the desired numerical results but also supply the outputs in colorful graphical form for impressive presentations It is unthinkable to practice engineering today without using some of these packages This tremendous computing power available to us at the touch of a button is both a blessing and a curse It certainly enables engineers to solve problems easily and quickly but it also opens the door for abuses and misin formation In the hands of poorly educated people these software packages are as dangerous as sophisticated powerful weapons in the hands of poorly trained soldiers Thinking that a person who can use the engineering software packages with out proper training on fundamentals can practice engineering is like thinking that a person who can use a wrench can work as a car mechanic If it were true that the engineering students do not need all these fundamental courses they are taking because practically everything can be done by computers quickly and easily then it would also be true that the employers would no longer need highsalaried engineers since any person who knows how to use a word processing program can also learn how to use those software packages How ever the statistics show that the need for engineers is on the rise not on the decline despite the availability of these powerful packages We should always remember that all the computing power and the engi neering software packages available today are just tools and tools have meaning only in the hands of masters Having the best wordprocessing program does not make a person a good writer but it certainly makes the job of a good writer much easier and makes the writer more productive Fig 166 Hand calculators did not eliminate the need to teach our children FIGURE 165 Neatness and organization are highly valued by employers FIGURE 166 An excellent wordprocessing program does not make a person a good writer it simply makes a good writer a more efficient writer Caia ImagesGlow Images RF Final PDF to printer 36 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 36 110317 0710 AM how to add or subtract and the sophisticated medical software packages did not take the place of medical school training Neither will engineering software packages replace the traditional engineering education They will simply cause a shift in emphasis in the courses from mathematics to phys ics That is more time will be spent in the classroom discussing the physical aspects of the problems in greater detail and less time on the mechanics of solution procedures All these marvelous and powerful tools available today put an extra burden on todays engineers They must still have a thorough understanding of the fundamentals develop a feel of the physical phenomena be able to put the data into proper perspective and make sound engineering judgments just like their predecessors However they must do it much better and much faster using more realistic models because of the powerful tools available today The engineers in the past had to rely on hand calculations slide rules and later hand calculators and computers Today they rely on software packages The easy access to such power and the possibility of a simple misunderstanding or misinterpretation causing great damage make it more important today than ever to have solid training in the fundamentals of engineering In this text we make an extra effort to put the emphasis on developing an intuitive and physical understanding of natural phenomena instead of on the mathematical details of solution procedures Equation Solvers You are probably familiar with the equation solving capabilities of spread sheets such as Microsoft Excel Despite its simplicity Excel is commonly used in solving systems of equations in engineering as well as finance It enables the user to conduct parametric studies plot the results and ask what if questions It can also solve simultaneous equations if properly set up There are also many sophisticated equation solvers that are commonly used in engineering practice such as the Engineering Equation Solver EES which is a program that easily solves systems of linear or nonlinear algebraic or differential equations numerically It has a large library of builtin thermody namic property functions as well as mathematical functions and it allows the user to supply additional property data Unlike some software packages equation solvers do not solve engineer ing problems they only solve the equations supplied by the user Therefore the user must understand the problem and formulate it by applying any rel evant physical laws and relations Equation solvers save the user considerable time and effort by simply solving the resulting mathematical equations This makes it possible to attempt significant engineering problems not suitable for hand calculations and to conduct parametric studies quickly and conveniently EXAMPLE 111 Solving a System of Equations Numerically The difference of two numbers is 4 and the sum of the squares of these two numbers is equal to the sum of the numbers plus 20 Determine these two numbers SOLUTION Relations are given for the difference and the sum of the squares of two numbers The two numbers are to be determined Final PDF to printer 37 CHAPTER 1 cen22672ch01001050indd 37 110317 0710 AM Analysis We first solve the problem using EES We start the EES program by double clicking on its icon open a new file and type the following on the blank screen that appears x y 4 x 2 y 2 x y 20 which is an exact mathematical expression of the problem statement with x and y denoting the unknown numbers The solution to this system of equations one linear and one nonlinear with two unknowns is obtained by a single click on the calcula tor icon on the taskbar It gives Fig 167 x 5 and y 1 We now solve the same problem using Excel Start Excel FileOptionsAddIns Solver AddInOK where the underline means to click on that option and the slash separates each sequential option Choose a cell for x and a cell for y and enter initial guesses there we chose cells C25 and D25 and guessed 05 and 05 We must rewrite the two equations so that no variables are on the righthand side RHS x y 4 and x 2 y 2 x y 20 Choose a cell for the RHS of each equation and enter the formula there we chose cells D20 and D21 see the equations in Fig 168a Data Solver Set the cell for the RHS of the first equation D20 as the Objective with a value of 4 set the cells for x and y C25D25 as those subject to constraints and set the constraint such that the cell for the RHS of the second equation D21 must equal 20 SolveOK The solution iterates to the correct final values of x 5 and y 1 respectively Fig 168b Note For better convergence the precision number of allowed iterations etc can be changed in DataSolverOptions Discussion Note that all we did was formulate the problem as we would on paper EES or Excel took care of all the mathematical details of the solution Also note that equations can be linear or nonlinear and they can be entered in any order with unknowns on either side Friendly equation solvers such as EES allow the user to concentrate on the physics of the problem without worrying about the mathematical complexities associated with the solution of the resulting system of equations FIGURE 168 Excel screen images for Example 111 a Equations with initial guesses highlighted b Final results after using Excels Solver with converged values highlighted a b FIGURE 167 EES screen images for Example 111 A Remark on Significant Digits In engineering calculations the information given is not known to more than a certain number of significant digits usually three digits Consequently the results obtained cannot possibly be accurate to more significant digits Reporting results in more significant digits falsely implies greater accuracy than exists and it should be avoided For example consider a 375L container filled with gasoline whose density is 0845 kgL and try to determine its mass Probably the first thought that comes to your mind is to multiply the volume and density to obtain 316875 kg for the mass which falsely implies that the mass determined is accurate to six significant digits In reality however the mass cannot be more accurate than three significant digits since both the volume and the density are accurate to three significant digits only Therefore the result should be rounded to three significant digits and the mass should be reported to be 317 kg instead of what appears in the screen of the calculator The result 316875 kg would Final PDF to printer 38 INTRODUCTION AND BASIC CONCEPTS cen22672ch01001050indd 38 110317 0710 AM be correct only if the volume and density were given to be 375000 L and 0845000 kgL respectively The value 375 L implies that we are fairly con fident that the volume is accurate within 001 L and it cannot be 374 or 376 L However the volume can be 3746 3750 3753 etc since they all round to 375 L Fig 169 It is more appropriate to retain all the digits during intermediate calculations and to do the rounding in the final step since this is what a computer will normally do When solving problems we will assume the given information to be accu rate to at least three significant digits Therefore if the length of a pipe is given to be 40 m we will assume it to be 400 m in order to justify using three significant digits in the final results You should also keep in mind that all experimentally determined values are subject to measurement errors and such errors will reflect in the results obtained For example if the density of a substance has an uncertainty of 2 percent then the mass determined using this density value will also have an uncertainty of 2 percent You should also be aware that we sometimes knowingly introduce small errors in order to avoid the trouble of searching for more accurate data For example when dealing with liquid water we just use the value of 1000 kgm3 for density which is the density value of pure water at 0C Using this value at 75C will result in an error of 25 percent since the density at this temperature is 975 kgm3 The minerals and impurities in the water will introduce addi tional error This being the case you should have no reservation in rounding the final results to a reasonable number of significant digits Besides having a few percent uncertainty in the results of engineering analysis is usually the norm not the exception FIGURE 169 A result with more significant digits than that of given data falsely implies more precision Given Also 375 0845 316875 Volume Density Find Mass m V 316875 kg Rounding to 3 significant digits m 317 kg 3 significant digits V 375 L ρ 0845 kgL ρ SUMMARY In this chapter the basic concepts of thermodynamics are introduced and discussed Thermodynamics is the science that primarily deals with energy The first law of thermodynamics is simply an expression of the conservation of energy prin ciple and it asserts that energy is a thermodynamic property The second law of thermodynamics asserts that energy has quality as well as quantity and actual processes occur in the direction of decreasing quality of energy A system of fixed mass is called a closed system or control mass and a system that involves mass transfer across its boundaries is called an open system or control volume The massdependent properties of a system are called extensive properties and the others intensive properties Density is mass per unit volume and specific volume is volume per unit mass A system is said to be in thermodynamic equilibrium if it maintains thermal mechanical phase and chemical equilib rium Any change from one state to another is called a process A process with identical end states is called a cycle During a quasistatic or quasiequilibrium process the system remains practically in equilibrium at all times The state of a simple compressible system is completely specified by two indepen dent intensive properties The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact The temperature scales used in the SI and the English sys tem today are the Celsius scale and the Fahrenheit scale respectively They are related to absolute temperature scales by T K T C 27315 T R T F 45967 The magnitudes of each division of 1 K and 1C are identical and so are the magnitudes of each division of 1 R and 1F Therefore ΔT K ΔT C and ΔT R ΔT F The normal force exerted by a fluid per unit area is called pres sure and its unit is the pascal 1 Pa 1 Nm2 The pressure relative to absolute vacuum is called the absolute pressure and the difference between the absolute pressure and the local atmospheric pressure is called the gage pressure Pressures Final PDF to printer cen22672ch01001050indd 39 110317 0710 AM 39 CHAPTER 1 below atmospheric pressure are called vacuum pressures The absolute gage and vacuum pressures are related by P gage P abs P atm for pressure above P atm P vac P atm P abs for pressure below P atm The pressure at a point in a fluid has the same magnitude in all directions The variation of pressure with elevation is given by dP dz ρg where the positive z direction is taken to be upward When the density of the fluid is constant the pressure difference across a fluid layer of thickness Δz is ΔP P 2 P 1 ρg Δz The absolute and gage pressures in a liquid open to the atmosphere at a depth h from the free surface are P P atm ρgh or P gage ρgh Small to moderate pressure differences are measured by a manometer The pressure in a stationary fluid remains constant in the horizontal direction Pascals principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount The atmospheric pressure is measured by a barometer and is given by P atm ρgh where h is the height of the liquid column REFERENCES AND SUGGESTED READINGS 1 American Society for Testing and Materials Standards for Metric Practice ASTM E 38079 January 1980 2 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley 2006 3 J A Schooley Thermometry Boca Raton FL CRC Press 1986 PROBLEMS Thermodynamics 11C Why does a bicyclist pick up speed on a downhill road even when he is not pedaling Does this violate the conserva tion of energy principle 12C One of the most amusing things a person can experience is when a car in neutral appears to go uphill when its brakes are released Can this really happen or is it an optical illusion How can you verify if a road is pitched uphill or downhill 13C An office worker claims that a cup of cold coffee on his table warmed up to 80C by picking up energy from the surrounding air which is at 25C Is there any truth to his claim Does this process violate any thermodynamic laws 14C What is the difference between the classical and the statistical approaches to thermodynamics Mass Force and Units 15C Explain why the lightyear has the dimension of length 16C What is the difference between poundmass and poundforce 17C What is the net force acting on a car cruising at a con stant velocity of 70 kmh a on a level road and b on an uphill road 18 What is the weight in N of an object with a mass of 200 kg at a location where g 96 ms2 19E If the mass of an object is 10 lbm what is its weight in lbf at a location where g 320 fts2 110 The acceleration of highspeed aircraft is sometimes expressed in gs in multiples of the standard acceleration of gravity Determine the upward force in N that a 90kg man would experience in an aircraft whose acceleration is 6 gs 111 The value of the gravitational acceleration g decreases with elevation from 9807 ms2 at sea level to 9767 ms2 at an altitude of 13000 m where large passenger planes cruise Determine the percent reduction in the weight of an airplane cruising at 13000 m relative to its weight at sea level Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch01001050indd 40 110317 0710 AM 40 INTRODUCTION AND BASIC CONCEPTS 112 A 3kg plastic tank that has a volume of 02 m3 is filled with liquid water Assuming the density of water is 1000 kgm3 determine the weight of the combined system 113 A 2kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 979 ms2 Determine the acceleration of the rock in ms2 114 Solve Prob 113 using appropriate software Print out the entire solution including the numer ical results with proper units 115 A 4kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level Determine the amount of electric energy used in both kWh and kJ 116E A 150lbm astronaut took his bathroom scale a spring scale and a beam scale compares masses to the moon where the local gravity is g 548 fts2 Determine how much he will weigh a on the spring scale and b on the beam scale Answers a 255 lbf b 150 lbf 117 The gas tank of a car is filled with a nozzle that discharges gasoline at a constant flow rate Based on unit considerations of quantities obtain a relation for the filling time in terms of the volume V of the tank in L and the discharge rate of gasoline V in Ls Systems Properties State and Processes 118C How would you define a system to determine the rate at which an automobile adds carbon dioxide to the atmosphere 119C A large fraction of the thermal energy generated in the engine of a car is rejected to the air by the radiator through the circulating water Should the radiator be analyzed as a closed system or as an open system Explain 121C How would you define a system to determine the temperature rise created in a lake when a portion of its water is used to cool a nearby electrical power plant 122C How would you describe the state of the air in the atmosphere What kind of process does this air undergo from a cool morning to a warm afternoon 123C What is the difference between intensive and exten sive properties 124C The specific weight of a system is defined as the weight per unit volume note that this definition violates the normal specific propertynaming convention Is the specific weight an extensive or intensive property 125C Is the number of moles of a substance contained in a system an extensive or intensive property 126C Is the state of the air in an isolated room completely specified by the temperature and the pressure Explain 127C What is a quasiequilibrium process What is its importance in engineering 128C Define the isothermal isobaric and isochoric processes 129C What is specific gravity How is it related to density 130 The density of atmospheric air varies with eleva tion decreasing with increasing altitude a Using the data given in the table obtain a relation for the variation of density with elevation and calculate the density at an elevation of 7000 m b Calculate the mass of the atmo sphere using the correlation you obtained Assume the earth to be a perfect sphere with a radius of 6377 km and take the thickness of the atmosphere to be 25 km FIGURE P119C McGrawHill EducationChristopher Kerrigan 120C A can of soft drink at room temperature is put into the refrigerator so that it will cool Would you model the can of soft drink as a closed system or as an open system Explain z km ρ kgm3 6377 1225 6378 1112 6379 1007 6380 09093 6381 08194 6382 07364 6383 06601 6385 05258 6387 04135 6392 01948 6397 008891 6402 004008 Temperature 131C What are the ordinary and absolute temperature scales in the SI and the English system Final PDF to printer cen22672ch01001050indd 41 110317 0710 AM 41 CHAPTER 1 132C Consider an alcohol and a mercury thermometer that read exactly 0C at the ice point and 100C at the steam point The distance between the two points is divided into 100 equal parts in both thermometers Do you think these thermometers will give exactly the same reading at a temperature of say 60C Explain 133C Consider two closed systems A and B System A con tains 3000 kJ of thermal energy at 20C whereas system B contains 200 kJ of thermal energy at 50C Now the systems are brought into contact with each other Determine the direc tion of any heat transfer between the two systems 134E Consider a system whose temperature is 18C Express this temperature in R K and F 135E Steam enters a heat exchanger at 300 K What is the temperature of this steam in F 136 The temperature of a system rises by 130C during a heating process Express this rise in temperature in kelvins 137E The temperature of a system drops by 45F during a cooling process Express this drop in temperature in K R and C 138E The temperature of the lubricating oil in an automo bile engine is measured as 150F What is the temperature of this oil in C 139E Heated air is at 150C What is the temperature of this air in F Pressure Manometer and Barometer 140C What is the difference between gage pressure and absolute pressure 141C Explain why some people experience nose bleed ing and some others experience shortness of breath at high elevations 142C A health magazine reported that physicians measured 100 adults blood pressure using two different arm positions parallel to the body along the side and perpendicular to the body straight out Readings in the parallel position were up to 10 percent higher than those in the perpendicular posi tion regardless of whether the patient was standing sitting or lying down Explain the possible cause for the difference 143C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled Do you agree Explain 144C Consider two identical fans one at sea level and the other on top of a high mountain running at identical speeds How would you compare a the volume flow rates and b the mass flow rates of these two fans 145E The absolute pressure in a compressed air tank is 200 kPa What is this pressure in psia 146E A manometer measures a pressure difference as 40 inches of water What is this pressure difference in pound force per square inch psi Answer 144 psi 147 A vacuum gage connected to a chamber reads 35 kPa at a location where the atmospheric pressure is 92 kPa Deter mine the absolute pressure in the chamber 148E The maximum safe air pressure of a tire is typically written on the tire itself The label on a tire indicates that the maximum pressure is 35 psi gage Express this maximum pressure in kPa FIGURE P148E 149E A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 291 in Hg Determine the absolute pressure in the tank Take ρHg 8484 lbmft3 Answer 643 psia 150 A pressure gage connected to a tank reads 500 kPa at a location where the atmospheric pressure is 94 kPa Determine the absolute pressure in the tank 151E A 200pound man has a total foot imprint area of 72 in2 Determine the pressure this man exerts on the ground if a he stands on both feet and b he stands on one foot 152 The gage pressure in a liquid at a depth of 3 m is read to be 42 kPa Determine the gage pressure in the same liquid at a depth of 9 m 153 The absolute pressure in water at a depth of 9 m is read to be 185 kPa Determine a the local atmospheric pres sure and b the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 085 at the same location 154 Consider a 175mtall man standing vertically in water and completely submerged in a pool Determine the difference between the pressures acting at the head and at the toes of the man in kPa 155 The barometer of a mountain hiker reads 750 mbars at the beginning of a hiking trip and 650 mbars at the end Neglecting the effect of altitude on local gravitational accel eration determine the vertical distance climbed Assume an average air density of 120 kgm3 Answer 850 m 156 The basic barometer can be used to measure the height of a building If the barometric readings at the top and at the bottom of a building are 675 and 695 mmHg respectively Final PDF to printer cen22672ch01001050indd 42 110317 0710 AM 42 INTRODUCTION AND BASIC CONCEPTS determine the height of the building Take the densities of air and mercury to be 118 kgm3 and 13600 kgm3 respectively 159 Reconsider Prob 158 Using appropriate soft ware investigate the effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder Plot the pressure against the spring force and discuss the results 160 The piston of a vertical pistoncylinder device contain ing a gas has a mass of 60 kg and a crosssectional area of 004 m2 as shown in Fig P160 The local atmospheric pres sure is 097 bar and the gravitational acceleration is 981 ms2 a Determine the pressure inside the cylinder b If some heat is transferred to the gas and its volume is doubled do you expect the pressure inside the cylinder to change FIGURE P160 A 004 m2 P Patm 097 bar m 60 kg FIGURE P156 McGrawHill Education 157 Solve Prob 156 using appropriate software Print out the entire solution including the numer ical results with proper units 158 A gas is contained in a vertical frictionless piston cylinder device The piston has a mass of 32 kg and a cross sectional area of 35 cm2 A compressed spring above the piston exerts a force of 150 N on the piston If the atmospheric pres sure is 95 kPa determine the pressure inside the cylinder Answer 147 kPa FIGURE P158 A 35 cm2 Patm 95 kPa mp 32 kg 150 N 161 Both a gage and a manometer are attached to a gas tank to measure its pressure If the reading on the pressure gage is 80 kPa determine the distance between the two fluid levels of the manometer if the fluid is a mercury ρ 13600 kgm3 or b water ρ 1000 kgm3 FIGURE P161 Pg 80 kPa Gas h 162 Reconsider Prob 161 Using appropriate soft ware investigate the effect of the manometer fluid density in the range of 800 to 13000 kgm3 on the dif ferential fluid height of the manometer Plot the differential fluid height against the density and discuss the results 163 A manometer containing oil ρ 850 kgm3 is attached to a tank filled with air If the oillevel difference between the two columns is 80 cm and the atmospheric pressure is 98 kPa determine the absolute pressure of the air in the tank Answer 105 kPa Final PDF to printer cen22672ch01001050indd 43 110317 0710 AM 43 CHAPTER 1 FIGURE P171 70 cm Water Oil 164E A manometer is used to measure the air pressure in a tank The fluid used has a specific gravity of 125 and the differential height between the two arms of the manometer is 28 in If the local atmospheric pressure is 127 psia determine the absolute pressure in the tank for the cases of the manom eter arm with the a higher and b lower fluid level being attached to the tank 165 A mercury manometer ρ 13600 kgm3 is connected to an air duct to measure the pressure inside The difference in the manometer levels is 30 mm and the atmospheric pressure is 100 kPa a Judging from Fig P165 determine if the pres sure in the duct is above or below the atmospheric pressure b Determine the absolute pressure in the duct 169E Blood pressure is usually measured by wrapping a closed airfilled jacket equipped with a pressure gage around the upper arm of a person at the level of the heart Using a mercury manometer and a stethoscope the systolic pressure the maximum pressure when the heart is pumping and the diastolic pressure the minimum pressure when the heart is resting are measured in mmHg The systolic and diastolic pressures of a healthy person are about 120 mmHg and 80 mmHg respectively and are indicated as 12080 Express both of these gage pressures in kPa psi and meter water column 170 The maximum blood pressure in the upper arm of a healthy person is about 120 mmHg If a vertical tube open to the atmosphere is connected to the vein in the arm of the per son determine how high the blood will rise in the tube Take the density of the blood to be 1050 kgm3 FIGURE P165 Air P h 30 mm Mercury 166 Repeat Prob 165 for a differential mercury height of 45 mm 167E The pressure in a natural gas pipeline is measured by the manometer shown in Fig P167E with one of the arms open to the atmosphere where the local atmospheric pressure is 142 psia Determine the absolute pressure in the pipeline FIGURE P167E 10 in 6 in 2 in 25 in Natural Gas Water Air Mercury SG 136 168E Repeat Prob 167E by replacing air with oil with a specific gravity of 069 FIGURE P170 h 171 Consider a Utube whose arms are open to the atmo sphere Now water is poured into the Utube from one arm and light oil ρ 790 kgm3 from the other One arm contains 70cmhigh water while the other arm contains both fluids with an oiltowater height ratio of 4 Determine the height of each fluid in that arm 172 Consider a doublefluid manometer attached to an air pipe shown in Fig P172 If the specific gravity of one fluid Final PDF to printer cen22672ch01001050indd 44 110317 0710 AM 44 INTRODUCTION AND BASIC CONCEPTS 173 Calculate the absolute pressure P1 of the manometer shown in Fig P173 in kPa The local atmospheric pressure is 758 mmHg kPa indicated by the manometer when the local atmospheric pressure is 720 mmHg 176 The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2500 kg Deter mine the fluid gage pressure that must be maintained in the reservoir 177 Consider the system shown in Fig P177 If a change of 07 kPa in the pressure of air causes the brinemercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant determine the ratio of A2 A1 FIGURE P177 Brine pipe SG 11 Area A2 Mercury SG 1356 Water Air Area A1 FIGURE P172 SG2 Air P 69 kPa 28 cm 40 cm SG1 1355 is 1355 determine the specific gravity of the other fluid for the indicated absolute pressure of air Take the atmospheric pressure to be 100 kPa Answer 159 FIGURE P173 P1 12 cm 5 cm 30 cm 15 cm Atmospheric pressure Fluid B 8 kNm3 Fluid A 10 kNm3 174 Consider the manometer in Fig P173 If the specific weight of fluid A is 100 kNm3 what is the absolute pressure in kPa indicated by the manometer when the local atmo spheric pressure is 90 kPa 175 Consider the manometer in Fig P173 If the specific weight of fluid B is 20 kNm3 what is the absolute pressure in 178 The gage pressure of the air in the tank shown in Fig P178 is measured to be 80 kPa Determine the differen tial height h of the mercury column FIGURE P178 Air 80 kPa 30 cm 75 cm h Mercury SG 136 Water Oil SG 072 179 Repeat Prob 178 for a gage pressure of 40 kPa Solving Engineering Problems and Equation Solvers 180C What is the value of the engineering software packages in a engineering education and b engineering practice Final PDF to printer cen22672ch01001050indd 45 110317 0710 AM 45 CHAPTER 1 181 Determine a positive real root of this equation using appropriate software 2 x 3 10 x 05 3x 3 182 Solve this system of two equations with two unknowns using appropriate software x 3 y 2 59 3xy y 35 183 Solve this system of three equations with three unknowns using appropriate software 2x y z 7 3 x 2 3y z 3 xy 2z 4 184 Solve this system of three equations with three unknowns using appropriate software x 2 y z 1 x 3 y 05 xz 2 x y z 2 Review Problems 185E The reactive force developed by a jet engine to push an airplane forward is called thrust and the thrust developed by the engine of a Boeing 777 is about 85000 lbf Express this thrust in N and kgf 186 The weight of bodies may change somewhat from one location to another as a result of the variation of the gravitational acceleration g with elevation Accounting for this variation using the relation g a bz where a 9807 ms2 and b 332 106 s2 determine the weight of an 80kg person at sea level z 0 in Denver z 1610 m and on the top of Mount Everest z 8848 m 187E A man goes to a traditional market to buy a steak for dinner He finds a 12oz steak 1 lbm 16 oz for 550 He then goes to the adjacent international market and finds a 300g steak of identical quality for 520 Which steak is the better buy 188E What is the weight of a 1kg substance in N kN kgms2 kgf lbmfts2 and lbf 189E The pressure in a steam boiler is given to be 92 kgfcm2 Express this pressure in psi kPa atm and bars 191 The average atmospheric pressure on earth is approxi mated as a function of altitude by the relation Patm 101325 1 002256z5256 where Patm is the atmospheric pressure in kPa and z is the altitude in km with z 0 at sea level Determine the approximate atmospheric pressures at Atlanta z 306 m Denver z 1610 m Mexico City z 2309 m and the top of Mount Everest z 8848 m 192E Hyperthermia of 5C ie 5C rise above the normal body temperature is considered fatal Express this fatal level of hyperthermia in a K b F and c R 193E The boiling temperature of water decreases by about 3C for each 1000m rise in altitude What is the decrease in the boiling temperature in a K b F and c R for each 1000m rise in altitude 194E A house is losing heat at a rate of 1800 kJh per C temperature difference between the indoor and the outdoor temperatures Express the rate of heat loss from this house per a K b F and c R difference between the indoor and the outdoor temperature 195E The average body temperature of a person rises by about 2C during strenuous exercise What is the rise in the body temperature in a K b F and c R during strenuous exercise 196 The average temperature of the atmosphere in the world is approximated as a function of altitude by the relation T atm 28815 65z where Tatm is the temperature of the atmosphere in K and z is the altitude in km with z 0 at sea level Determine the average temperature of the atmosphere outside an airplane that is cruising at an altitude of 12000 m FIGURE P190 F1 D2 F2 25 kg 10 cm Weight 1900 kg 190 A hydraulic lift is to be used to lift a 1900kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm Determine the diameter of the piston on which the weight is to be placed Final PDF to printer cen22672ch01001050indd 46 110317 0710 AM 46 INTRODUCTION AND BASIC CONCEPTS 197 A vertical frictionless pistoncylinder device contains a gas at 180 kPa absolute pressure The atmospheric pressure outside is 100 kPa and the piston area is 25 cm2 Determine the mass of the piston 198 A vertical pistoncylinder device contains a gas at a pressure of 100 kPa The piston has a mass of 10 kg and a diameter of 14 cm Pressure of the gas is to be increased by placing some weights on the piston Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas inside the cylinder Answers 936 kPa 157 kg Determine the upward force the water will exert on the duct Take the densities of air and water to be 13 kgm3 and 1000 kgm3 respectively 1101 Balloons are often filled with helium gas because it weighs only about oneseventh of what air weighs under identical conditions The buoyancy force which can be expressed as Fb ρairgVballoon will push the balloon upward If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released Assume the density of air is ρ 116 kgm3 and neglect the weight of the ropes and the cage Answer 224 ms2 FIGURE P1101 m 170 kg Helium D 12 m ρHe ρair 1 7 FIGURE P199 P1 P2 P3 D1 D2 Spring 199 The force generated by a spring is given by F kx where k is the spring constant and x is the deflection of the spring The spring of Fig P199 has a spring constant of 8 kNcm The pressures are P1 5000 kPa P2 10000 kPa and P3 1000 kPa If the piston diameters are D1 8 cm and D2 3 cm how far will the spring be deflected Answer 172 cm FIGURE P198 Weights Gas 1100 An airconditioning system requires a 35mlong section of 15cmdiameter ductwork to be laid underwater 1102 Reconsider Prob 1101 Using appropriate software investigate the effect of the number of people carried in the balloon on acceleration Plot the acceler ation against the number of people and discuss the results 1103 Determine the maximum amount of load in kg the balloon described in Prob 1101 can carry Answer 900 kg 1104 The lower half of a 6mhigh cylindrical container is filled with water ρ 1000 kgm3 and the upper half with oil that has a specific gravity of 085 Determine the pressure difference between the top and bottom of the cylinder Answer 544 kPa FIGURE P1104 Water ρ 1000 kgm3 Oil SG 085 h 6 m Final PDF to printer cen22672ch01001050indd 47 110317 0710 AM 47 CHAPTER 1 1105 A pressure cooker cooks a lot faster than an ordi nary pan by maintaining a higher pressure and temperature inside The lid of a pressure cooker is well sealed and steam can escape only through an opening in the middle of the lid A separate metal piece the petcock sits on top of this open ing and prevents steam from escaping until the pressure force overcomes the weight of the petcock The periodic escape of the steam in this manner prevents any potentially dangerous pressure buildup and keeps the pressure inside at a constant value Determine the mass of the petcock of a pressure cooker whose operation pressure is 100 kPa gage and has an open ing crosssectional area of 4 mm2 Assume an atmospheric pressure of 101 kPa and draw the freebody diagram of the petcock Answer 408 g 1108E Consider a Utube whose arms are open to the atmo sphere Now equal volumes of water and light oil ρ 493 lbmft3 are poured from different arms A person blows from the oil side of the Utube until the contact surface of the two fluids moves to the bottom of the Utube and thus the liquid levels in the two arms are the same If the fluid height in each arm is 30 in deter mine the gage pressure the person exerts on the oil by blowing FIGURE P1105 Patm 101 kPa Pressure cooker Petcock A 4 mm2 1106 The pilot of an airplane reads the altitude 6400 m and the absolute pressure 45 kPa when flying over a city Calcu late the local atmospheric pressure in that city in kPa and in mmHg Take the densities of air and mercury to be 0828 kgm3 and 13600 kgm3 respectively FIGURE P1106 Altitude 64 km P 45 kPa Michał KrakowiakGetty Images RF determine how high the water will rise in the tube in m Take the density of water to be 1000 kgm3 FIGURE P1107 Patm 99 kPa h Water FIGURE P1108E Air Water 30 in Oil 1109E A water pipe is connected to a doubleU manometer as shown in Fig P1109E at a location where the local atmo spheric pressure is 142 psia Determine the absolute pressure at the center of the pipe FIGURE P1109E Mercury SG 136 Oil SG 080 Oil SG 080 Water pipe 60 in 25 in 30 in 20 in 1107 A glass tube is attached to a water pipe as shown in Fig P1107 If the water pressure at the bottom of the tube is 107 kPa and the local atmospheric pressure is 99 kPa Final PDF to printer cen22672ch01001050indd 48 110317 0710 AM 48 INTRODUCTION AND BASIC CONCEPTS 1110 A gasoline line is connected to a pressure gage through a doubleU manometer as shown in Fig P1110 If the reading of the pressure gage is 370 kPa determine the gage pressure of the gasoline line P aI b where a and b are constants and calculate the pres sure that corresponds to a signal of 10 mA FIGURE P1110 45 cm 10 cm 22 cm 50 cm Mercury SG 136 Gasoline SG 070 Water Air Oil SG 079 Pgage 370 kPa Pipe FIGURE P1112 Duct 45 12 cm L Air 1111 Repeat Prob 1110 for a pressure gage reading of 180 kPa 1112 When measuring small pressure differences with a manometer often one arm of the manometer is inclined to improve the accuracy of reading The pressure difference is still proportional to the vertical distance and not the actual length of the fluid along the tube The air pressure in a circular duct is to be measured using a manometer whose open arm is inclined 45 from the horizontal as shown in Fig P1112 The density of the liquid in the manometer is 081 kgL and the vertical distance between the fluid levels in the two arms of the manometer is 12 cm Determine the gage pressure of air in the duct and the length of the fluid column in the inclined arm above the fluid level in the vertical arm 1113 Pressure transducers are commonly used to measure pressure by generating analog signals usually in the range of 4 mA to 20 mA or 0 Vdc to 10 Vdc in response to applied pressure The system whose schematic is shown in Fig P1113 can be used to calibrate pressure trans ducers A rigid container is filled with pressurized air and pressure is measured by the manometer attached A valve is used to regulate the pressure in the container Both the pres sure and the electric signal are measured simultaneously for various settings and the results are tabulated For the given set of measurements obtain the calibration curve in the form of 1114 Consider the flow of air through a wind turbine whose blades sweep an area of diameter D in m The average air velocity through the swept area is V in ms On the bases of the units of the quantities involved show that the mass flow rate of air in kgs through the swept area is proportional to air density the wind velocity and the square of the diameter of the swept area 1115 The drag force exerted on a car by air depends on a dimensionless drag coefficient the density of air the car velocity and the frontal area of the car That is FD FDCdrag Afront ρ V Based on unit considerations alone obtain a rela tion for the drag force Δh mm 280 1815 2978 4131 7659 I mA 421 578 697 815 1176 Δh mm 1027 1149 1362 1458 1536 I mA 1443 1568 1786 1884 1964 FIGURE P1115 Air V Pressure transducer Pressurized air P Rigid container Manometer Mercury SG 1356 Δh Multimeter Valve FIGURE P1113 Final PDF to printer cen22672ch01001050indd 49 110317 0710 AM 49 CHAPTER 1 1116E It is well known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the chilling effect of the wind This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities The equivalent wind chill tempera ture in F is given by ASHRAE Handbook of Fundamentals Atlanta GA 1993 p 815 T equiv 914 914 T ambient 0475 00203V 0304 V where V is the wind velocity in mih and Tambient is the ambient air temperature in F in calm air which is taken to be air with light winds at speeds up to 4 mih The constant 914F in the given equation is the mean skin temperature of a resting person in a comfortable environment Windy air at temperature Tambient and velocity V will feel as cold as the calm air at temperature Tequiv Using proper conversion factors obtain an equivalent relation in SI units where V is the wind velocity in kmh and Tambient is the ambient air temperature in C Answer T equiv 330 330 T ambient 0475 00126V 0240 V 1117E Reconsider Prob 1116E Using appropriate software plot the equivalent wind chill tem peratures in F as a function of wind velocity in the range of 4 to 40 mph for the ambient temperatures of 20 40 and 60F Discuss the results Fundamentals of Engineering FE Exam Problems 1118 During a heating process the temperature of an object rises by 10C This temperature rise is equivalent to a tempera ture rise of a 10F b 42F c 18 K d 18 R e 283 K 1119 An apple loses 36 kJ of heat as it cools per C drop in its temperature The amount of heat loss from the apple per F drop in its temperature is a 05 kJ b 18 kJ c 20 kJ d 36 kJ e 65 kJ 1120 At sea level the weight of 1 kg mass in SI units is 981 N The weight of 1 lbm mass in English units is a 1 lbf b 981 lbf c 322 lbf d 01 lbf e 0031 lbf 1121 Consider a fish swimming 5 m below the free surface of water The increase in the pressure exerted on the fish when it dives to a depth of 25 m below the free surface is a 196 Pa b 5400 Pa c 30000 Pa d 196000 Pa e 294000 Pa 1122 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 960 and 980 kPa If the density of air is 10 kgm3 the height of the building is a 17 m b 20 m c 170 m d 204 m e 252 m 1123 Consider a 25mdeep swimming pool The pressure difference between the top and bottom of the pool is a 25 kPa b 120 kPa c 196 kPa d 245 kPa e 250 kPa Design and Essay Problems 1124 Write an essay on different temperature measurement devices Explain the operational principle of each device its advantages and disadvantages its cost and its range of appli cability Which device would you recommend for use in the following cases taking the temperatures of patients in a doc tors office monitoring the variations of temperature of a car engine block at several locations and monitoring the tempera tures in the furnace of a power plant 1125 Write an essay on the various mass and volume measurement devices used throughout history Also explain the development of the modern units for mass and volume Final PDF to printer cen22672ch01001050indd 50 110317 0710 AM Final PDF to printer cen22672ch02051108indd 51 110317 0749 AM 51 OBJECTIVES The objectives of Chapter 2 are to Introduce the concept of energy and define its various forms Discuss the nature of internal energy Define the concept of heat and the terminology associated with energy transfer by heat Define the concept of work including electrical work and several forms of mechanical work Introduce the first law of thermodynamics energy balances and mechanisms of energy transfer to or from a system Determine that a fluid flowing across a control surface of a control volume carries energy across the control surface in addition to any energy transfer across the control surface that may be in the form of heat andor work Define energy conversion efficiencies Discuss the implications of energy conversion on the environment EN E R GY E N ERGY T R A NS F E R A N D G EN E R A L E N E R GY AN ALYS I S W hether we realize it or not energy is an important part of most aspects of daily life The quality of life and even its sustenance depends on the availability of energy Therefore it is important to have a good understanding of the sources of energy the conversion of energy from one form to another and the ramifications of these conversions Energy exists in numerous forms such as thermal mechanical electric chemical and nuclear Even mass can be considered a form of energy Energy can be transferred to or from a closed system a fixed mass in two distinct forms heat and work For control volumes energy can also be transferred by mass flow An energy transfer to or from a closed system is heat if it is caused by a temperature difference Otherwise it is work and it is caused by a force acting through a distance We start this chapter with a discussion of various forms of energy and energy transfer by heat We then introduce various forms of work and discuss energy transfer by work We continue with developing a general intuitive expression for the first law of thermodynamics also known as the conservation of energy principle which is one of the most fundamental principles in nature and we then demonstrate its use Finally we discuss the efficiencies of some familiar energy conversion processes and examine the impact on energy conversion on the environment Detailed treatments of the first law of thermodynamics for closed systems and control volumes are given in Chaps 4 and 5 respectively It should be kept in mind that physical laws or the laws of nature such as the first law of thermodynamics are universally accepted statements related to observed phenomena They are conclusions drawn on the basis of numerous scientific experiments and observations over a long period of time A physi cal law dictates that a particular phenomenon always occurs when specified conditions are present Everything in the universe obeys them with no excep tion and there can be no violations As such physical laws serve as powerful predictive tools enabling scientists to predict how things will happen in the physical universe before they happen They remain unchanged since first dis covered and nothing in nature seems to affect the laws of nature 2 CHAPTER Final PDF to printer 52 ENERGY ENERGY TRANSFER cen22672ch02051108indd 52 110317 0749 AM 21 INTRODUCTION We are familiar with the conservation of energy principle which is an expres sion of the first law of thermodynamics back from our high school years We are told repeatedly that energy cannot be created or destroyed during a pro cess it can only change from one form to another This seems simple enough but lets test ourselves to see how well we understand and truly believe in this principle Consider a room whose door and windows are tightly closed and whose walls are wellinsulated so that heat loss or gain through the walls is negli gible Now lets place a refrigerator in the middle of the room with its door open and plug it into a wall outlet Fig 21 You may even use a small fan to circulate the air in order to maintain temperature uniformity in the room Now what do you think will happen to the average temperature of air in the room Will it be increasing or decreasing Or will it remain constant Probably the first thought that comes to mind is that the average air tem perature in the room will decrease as the warmer room air mixes with the air cooled by the refrigerator Some may draw our attention to the heat generated by the motor of the refrigerator and may argue that the average air tempera ture may rise if this heating effect is greater than the cooling effect But they will get confused if it is stated that the motor is made of superconducting materials and thus there is hardly any heat generation in the motor Heated discussion may continue with no end in sight until we remember the conservation of energy principle that we take for granted If we take the entire roomincluding the air and the refrigeratoras the system which is an adiabatic closed system since the room is wellsealed and wellinsulated the only energy interaction involved is the electrical energy crossing the sys tem boundary and entering the room The conservation of energy requires the energy content of the room to increase by an amount equal to the amount of the electrical energy drawn by the refrigerator which can be measured by an ordinary electric meter The refrigerator or its motor does not store this energy Therefore this energy must now be in the room air and it will manifest itself as a rise in the air temperature The temperature rise of air can be calculated on the basis of the conservation of energy principle using the properties of air and the amount of electrical energy consumed What do you think would happen if we had a window air conditioning unit instead of a refrigerator placed in the middle of this room What if we operated a fan in this room instead Fig 22 Note that energy is conserved during the process of operating the refrig erator placed in a roomthe electrical energy is converted into an equiva lent amount of thermal energy stored in the room air If energy is already conserved then what are all those speeches on energy conservation and the measures taken to conserve energy Actually by energy conservation what is meant is the conservation of the quality of energy not the quantity Elec tricity which is of the highest quality of energy for example can always be converted to an equal amount of thermal energy also called heat But only a small fraction of thermal energy which is the lowest quality of energy can be converted back to electricity as we discuss in Chap 6 Think about the things that you can do with the electrical energy that the refrigerator has consumed and the air in the room that is now at a higher temperature FIGURE 21 A refrigerator operating with its door open in a wellsealed and well insulated room Wellsealed and wellinsulated room FIGURE 22 A fan running in a wellsealed and wellinsulated room will raise the temperature of air in the room Fan Wellsealed and wellinsulated room Final PDF to printer 53 CHAPTER 2 cen22672ch02051108indd 53 110317 0749 AM Now if asked to name the energy transformations associated with the opera tion of a refrigerator we may still have a hard time answering because all we see is electrical energy entering the refrigerator and heat dissipated from the refrigerator to the room air Obviously there is need to study the various forms of energy first and this is exactly what we do next followed by a study of the mechanisms of energy transfer 22 FORMS OF ENERGY Energy can exist in numerous forms such as thermal mechanical kinetic potential electric magnetic chemical and nuclear Fig 23 and their sum constitutes the total energy E of a system The total energy of a system on a unit mass basis is denoted by e and is expressed as e E m kJkg 21 Thermodynamics provides no information about the absolute value of the total energy It deals only with the change of the total energy which is what matters in engineering problems Thus the total energy of a system can be assigned a value of zero E 0 at some convenient reference point The change in total energy of a system is independent of the reference point selected The decrease in the potential energy of a falling rock for example depends on only the elevation difference and not the reference level selected In thermodynamic analysis it is often helpful to consider the various forms of energy that make up the total energy of a system in two groups macro scopic and microscopic The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame such as kinetic and potential energies Fig 24 The microscopic forms of energy are those related to the molecular structure of a system and the degree of the molecular activity and they are independent of outside reference frames The sum of all the microscopic forms of energy is called the internal energy of a system and is denoted by U The term energy was coined in 1807 by Thomas Young and its use in ther modynamics was proposed in 1852 by Lord Kelvin The term internal energy and its symbol U first appeared in the works of Rudolph Clausius and William Rankine in the second half of the 19th century and it eventually replaced the alternative terms inner work internal work and intrinsic energy commonly used at the time The macroscopic energy of a system is related to motion and the influence of some external effects such as gravity magnetism electricity and surface tension The energy that a system possesses as a result of its motion relative to some reference frame is called kinetic energy KE When all parts of a system move with the same velocity the kinetic energy is expressed as KE m V 2 2 kJ 22 or on a unit mass basis ke V 2 2 kJkg 23 FIGURE 24 The macroscopic energy of an object changes with velocity and elevation a FIGURE 23 At least six different forms of energy are encountered in bringing power from a nuclear plant to your home nuclear thermal mechanical kinetic magnetic and electrical a Gary GladstoneGetty Images RF b Tetra ImagesGetty Images RF b Final PDF to printer 54 ENERGY ENERGY TRANSFER cen22672ch02051108indd 54 110317 0749 AM where V denotes the velocity of the system relative to some fixed reference frame The kinetic energy of a rotating solid body is given by 1 2 I ω 2 where I is the moment of inertia of the body and ω is the angular velocity The energy that a system possesses as a result of its elevation in a gravita tional field is called potential energy PE and is expressed as PE mgz kJ 24 or on a unit mass basis pe gz kJkg 25 where g is the gravitational acceleration and z is the elevation of the center of gravity of a system relative to some arbitrarily selected reference level The magnetic electric and surface tension effects are significant in some specialized cases only and are usually ignored In the absence of such effects the total energy of a system consists of the kinetic potential and internal energies and is expressed as E U KE PE U m V 2 2 mgz kJ 26 or on a unit mass basis e u ke pe u V 2 2 gz kJkg 27 Most closed systems remain stationary during a process and thus experi ence no change in their kinetic and potential energies Closed systems whose velocity and elevation of the center of gravity remain constant during a pro cess are often referred to as stationary systems The change in the total energy ΔE of a stationary system is identical to the change in its internal energy ΔU In this text a closed system is assumed to be stationary unless stated otherwise Control volumes typically involve fluid flow for long periods of time and it is convenient to express the energy flow associated with a fluid stream in the rate form This is done by incorporating the mass flow rate m which is the amount of mass flowing through a cross section per unit time It is related to the volume flow rate V which is the volume of a fluid flowing through a cross section per unit time by Mass flow rate m ρ V ρ A c V avg kgs 28 which is analogous to m ρ V Here ρ is the fluid density Ac is the cross sectional area of flow and Vavg is the average flow velocity normal to Ac The dot over a symbol is used to indicate time rate throughout the book Then the energy flow rate associated with a fluid flowing at a rate of m is Fig 25 Energy flow rate E m e kJs or kW 29 which is analogous to E me FIGURE 25 Mass and energy flow rates associated with the flow of steam in a pipe of inner diameter D with an average velocity of Vavg D Steam Vavg Ac πD24 m ρAcVavg E me Final PDF to printer 55 CHAPTER 2 cen22672ch02051108indd 55 110317 0749 AM Some Physical Insight to Internal Energy Internal energy was defined earlier as the sum of all the microscopic forms of energy of a system It is related to the molecular structure and the degree of molecular activity and can be viewed as the sum of the kinetic and potential energies of the molecules To have a better understanding of internal energy let us examine a system at the molecular level The molecules of a gas move through space with some velocity and thus they possess some kinetic energy This is known as the translational energy The atoms of polyatomic molecules rotate about an axis and the energy associated with this rotation is the rotational kinetic energy The atoms of a polyatomic molecule may also vibrate about their common center of mass and the energy associated with this backandforth motion is the vibrational kinetic energy For gases the kinetic energy is mostly due to translational and rotational motions with vibrational motion becoming signif icant at higher temperatures The electrons in an atom rotate about the nucleus and thus possess rotational kinetic energy Electrons at outer orbits have larger kinetic energies Electrons also spin about their axes and the energy associated with this motion is the spin energy Other particles in the nucleus of an atom also possess spin energy The portion of the internal energy of a system associated with the kinetic energies of the molecules is called the sensible energy Fig 26 The average velocity and the degree of activity of the molecules are proportional to the temperature of the gas Therefore at higher temperatures the molecules possess higher kinetic energies and as a result the system has a higher internal energy The internal energy is also associated with various binding forces between the molecules of a substance between the atoms within a molecule and between the particles within an atom and its nucleus The forces that bind the molecules to each other are as one would expect strongest in solids and weakest in gases If sufficient energy is added to the molecules of a solid or liquid the molecules overcome these molecular forces and break away turn ing the substance into a gas This is a phasechange process Because of this added energy a system in the gas phase is at a higher internal energy level than it is in the solid or the liquid phase The internal energy associated with the phase of a system is called the latent energy The phasechange process can occur without a change in the chemical composition of a system Most practical problems fall into this category and one does not need to pay any attention to the forces binding the atoms in a molecule to each other An atom consists of neutrons and positively charged protons bound together by very strong nuclear forces in the nucleus and negatively charged electrons orbiting around it The internal energy associated with the atomic bonds in a molecule is called chemical energy During a chemical reaction such as a combustion process some chemical bonds are destroyed while others are formed As a result the internal energy changes The nuclear forces are much larger than the forces that bind the electrons to the nucleus The tremendous amount of energy associated with the strong bonds within the nucleus of the atom itself is called nuclear energy Fig 27 Obviously we need not be concerned with nuclear energy in thermodynamics unless of course we deal with fusion or fission reactions A chemical reaction involves changes in the structure of the electrons of the atoms but a nuclear reaction involves changes FIGURE 26 The various forms of microscopic energies that make up sensible energy Molecular translation Molecular rotation Electron translation Molecular vibration Electron spin Nuclear spin FIGURE 27 The internal energy of a system is the sum of all forms of the microscopic energies Nuclear energy Chemical energy Sensible and latent energy Final PDF to printer 56 ENERGY ENERGY TRANSFER cen22672ch02051108indd 56 110317 0749 AM in the core or nucleus Therefore an atom preserves its identity during a chemical reaction but loses it during a nuclear reaction Atoms may also pos sess electric and magnetic dipolemoment energies when subjected to exter nal electric and magnetic fields due to the twisting of the magnetic dipoles produced by the small electric currents associated with the orbiting electrons The forms of energy already discussed which constitute the total energy of a system can be contained or stored in a system and thus can be viewed as the static forms of energy The forms of energy not stored in a system can be viewed as the dynamic forms of energy or as energy interactions The dynamic forms of energy are recognized at the system boundary as they cross it and they represent the energy gained or lost by a system during a process The only two forms of energy interactions associated with a closed system are heat transfer and work An energy interaction is heat transfer if its driving force is a temperature difference Otherwise it is work as explained in the next section A control volume can also exchange energy via mass transfer since any time mass is transferred into or out of a system the energy content of the mass is also transferred with it In daily life we often refer to the sensible and latent forms of internal energy as heat and we talk about heat content of bodies In thermodynam ics however we usually refer to those forms of energy as thermal energy to prevent any confusion with heat transfer A distinction should be made between the macroscopic kinetic energy of an object as a whole and the microscopic kinetic energies of its molecules that constitute the sensible internal energy of the object Fig 28 The kinetic energy of an object is an organized form of energy associated with the orderly motion of all molecules in one direction in a straight path or around an axis In contrast the kinetic energies of the molecules are com pletely random and highly disorganized As you will see in later chapters the organized energy is much more valuable than the disorganized energy and a major application area of thermodynamics is the conversion of disor ganized energy heat into organized energy work You will also see that the organized energy can be converted to disorganized energy completely but only a fraction of disorganized energy can be converted to organized energy by specially built devices called heat engines like car engines and power plants A similar argument can be given for the macroscopic poten tial energy of an object as a whole and the microscopic potential energies of the molecules More on Nuclear Energy The bestknown fission reaction involves the splitting of the uranium atom the U235 isotope into other elements It is commonly used to generate elec tricity in nuclear power plants 450 reactors in 2016 with 392000 MW capac ity to power nuclear submarines and aircraft carriers and even to power spacecraft in addition to its use in nuclear bombs The percentage of electric ity produced by nuclear power is 76 percent in France 19 percent in Russia and the UK 14 percent in Germany and 20 percent in the United States The first nuclear chain reaction was achieved by Enrico Fermi in 1942 and the first largescale nuclear reactors were built in 1944 for the purpose of producing material for nuclear weapons When a uranium235 atom absorbs FIGURE 28 The macroscopic kinetic energy is an organized form of energy and is much more useful than the disorganized microscopic kinetic energies of the molecules Water Dam Macroscopic kinetic energy turns the wheel Microscopic kinetic energy of molecules does not turn the wheel Final PDF to printer 57 CHAPTER 2 cen22672ch02051108indd 57 110317 0749 AM a neutron and splits during a fission process it produces a cesium140 atom a rubidium93 atom three neutrons and 32 1011 J of energy In practical terms the complete fission of 1 kg of uranium235 releases 8314 1010 kJ of heat which is more than the heat released when 3700 tons of coal are burned Therefore for the same amount of fuel a nuclear fission reaction releases several million times more energy than a chemical reaction The safe disposal of used nuclear fuel however remains a concern Nuclear energy by fusion is released when two small nuclei combine into a larger one The huge amount of energy radiated by the sun and the other stars originates from such a fusion process which involves the combination of two hydrogen nuclei into a helium nucleus When two heavy hydrogen deute rium nuclei combine during a fusion process they produce a helium3 atom a free neutron and 51 1013 J of energy Fig 29 Fusion reactions are much more difficult to achieve in practice because of the strong repulsion between the positively charged nuclei called the Cou lomb repulsion To overcome this repulsive force and to enable the two nuclei to fuse together the energy level of the nuclei must be raised by heating them to about 100 million C But such high temperatures are found only in the center of stars or in exploding atomic bombs the Abomb In fact the uncontrolled fusion reaction in a hydrogen bomb the Hbomb is initiated by a small atomic bomb The uncontrolled fusion reaction was achieved in the early 1950s but all the efforts since then to achieve controlled fusion by mas sive lasers powerful magnetic fields and electric currents to generate power have failed FIGURE 29 The fission of uranium and the fusion of hydrogen during nuclear reactions and the release of nuclear energy a Fission of uranium 51 1013 J neutron b Fusion of hydrogen He3 n H2 H2 U235 32 1011 J 3 neutrons neutron Uranium Ce140 Rb93 n n n n FIGURE 210 Schematic for Example 21 Nuclear fuel EXAMPLE 21 A Car Powered by Nuclear Fuel An average car consumes about 5 L of gasoline a day and the capacity of the fuel tank of a car is about 50 L Therefore a car needs to be refueled once every 10 days Also the density of gasoline ranges from 068 to 078 kgL and its lower heating value is about 44000 kJkg that is 44000 kJ of heat is released when 1 kg of gasoline is completely burned Suppose all the problems associated with the radioactivity and waste disposal of nuclear fuels are resolved and a car is to be powered by U235 If a new car comes equipped with 01 kg of the nuclear fuel U235 determine if this car will ever need refueling under average driving conditions Fig 210 SOLUTION A car powered by nuclear energy comes equipped with nuclear fuel It is to be determined if this car will ever need refueling Assumptions 1 Gasoline is an incompressible substance with an average density of 075 kgL 2 Nuclear fuel is completely converted to thermal energy Analysis The mass of gasoline used per day by the car is m gasoline ρV gasoline 075 kgL 5 L day 375 kgday Noting that the heating value of gasoline is 44000 kJkg the energy supplied to the car per day is E m gasoline Heating value 375 kgday 44000 kJkg 165000 kJday Final PDF to printer 58 ENERGY ENERGY TRANSFER cen22672ch02051108indd 58 110317 0749 AM Mechanical Energy Many engineering systems are designed to transport a fluid from one loca tion to another at a specified flow rate velocity and elevation difference and the system may generate mechanical work in a turbine or it may consume mechanical work in a pump or fan during this process Fig 211 These systems do not involve the conversion of nuclear chemical or thermal energy to mechanical energy Also they do not involve any heat transfer in any sig nificant amount and they operate essentially at constant temperature Such systems can be analyzed conveniently by considering the mechanical forms of energy only and the frictional effects that cause the mechanical energy to be lost ie to be converted to thermal energy that usually cannot be used for any useful purpose The mechanical energy can be defined as the form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine Kinetic and potential energies are the familiar forms of mechanical energy Thermal energy is not mechanical energy how ever since it cannot be converted to work directly and completely the second law of thermodynamics A pump transfers mechanical energy to a fluid by raising its pres sure and a turbine extracts mechanical energy from a fluid by dropping its pressure Therefore the pressure of a flowing fluid is also associated with its mechanical energy In fact the pressure unit Pa is equivalent to Pa Nm2 Nmm3 Jm3 which is energy per unit volume and the prod uct Pv or its equivalent Pρ has the unit Jkg which is energy per unit mass Note that pressure itself is not a form of energy but a pressure force acting on a fluid through a distance produces work called flow work in the amount of Pρ per unit mass Flow work is expressed in terms of fluid properties and it is convenient to view it as part of the energy of a flowing fluid and call it flow energy Therefore the mechanical energy of a flowing fluid can be expressed on a unit mass basis as e mech P ρ V 2 2 gz 210 FIGURE 211 Mechanical energy is a useful concept for flows that do not involve significant heat transfer or energy conversion such as the flow of gasoline from an underground tank into a car altrendo imagesGetty Images RF The complete fission of 01 kg of uranium235 releases 8314 10 10 kJkg 01 kg 8314 10 9 kJ of heat which is sufficient to meet the energy needs of the car for No of days Energy content of fuel Daily energy use 8314 10 9 kJ 165000 kJday 50390 days which is equivalent to about 138 years Considering that no car will last more than 100 years this car will never need refueling It appears that nuclear fuel of the size of a cherry is sufficient to power a car during its lifetime Discussion Note that this problem is not quite realistic since the necessary critical mass cannot be achieved with such a small amount of fuel Further all of the uranium cannot be converted in fission again because of the critical mass problems after par tial conversion Final PDF to printer 59 CHAPTER 2 cen22672ch02051108indd 59 110317 0749 AM where Pρ is the flow energy V 22 is the kinetic energy and gz is the potential energy of the fluid all per unit mass It can also be expressed in rate form as E mech m e mech m P ρ V 2 2 gz 211 where m is the mass flow rate of the fluid Then the mechanical energy change of a fluid during incompressible ρ constant flow becomes Δ e mech P 2 P 1 ρ V 2 2 V 1 2 2 g z 2 z 1 kJkg 212 and Δ E mech m Δ e mech m P 2 P 1 ρ V 2 2 V 1 2 2 g z 2 z 1 kW 213 Therefore the mechanical energy of a fluid does not change during flow if its pressure density velocity and elevation remain constant In the absence of any irreversible losses the mechanical energy change represents the mechan ical work supplied to the fluid if Δemech 0 or extracted from the fluid if Δemech 0 The maximum ideal power generated by a turbine for example is W max m Δ e mech as shown in Fig 212 W max m Δ e mech m g z 1 z 4 m gh since P 1 P 4 P atm and V 1 V 4 0 a 4 1 Generator Turbine h W Generator Turbine W 2 3 W max m Δ e mech m P 2 P 3 ρ m ΔP ρ since V 1 V 3 and z 2 z 3 b FIGURE 212 Mechanical energy is illustrated by an ideal hydraulic turbine coupled with an ideal generator In the absence of irreversible losses the maximum produced power is proportional to a the change in water surface elevation from the upstream to the downstream reservoir or b closeup view the drop in water pressure from just upstream to just downstream of the turbine EXAMPLE 22 Wind Energy A site evaluated for a wind farm is observed to have steady winds at a speed of 85 ms Fig 213 Determine the wind energy a per unit mass b for a mass of 10 kg and c for a flow rate of 1154 kgs for air SOLUTION A site with a specified wind speed is considered Wind energy per unit mass for a specified mass and for a given mass flow rate of air are to be determined Assumptions Wind flows steadily at the specified speed Analysis The only harvestable form of energy of atmospheric air is the kinetic energy which is captured by a wind turbine a Wind energy per unit mass of air is e ke V 2 2 85 ms 2 2 1 Jkg 1 m 2 s 2 361 J kg b Wind energy for an air mass of 10 kg is E me 10 kg 361 Jkg 361 J c Wind energy for a mass flow rate of 1154 kgs is E m e 1154 kgs361 Jkg 1 kW 1000 Js 417 kW Discussion It can be shown that the specified mass flow rate corresponds to a 12mdiameter flow section when the air density is 12 kgm3 Therefore a wind tur bine with a wind span diameter of 12 m has a power generation potential of 417 kW Real wind turbines convert about onethird of this potential to electric power FIGURE 213 A site for a wind farm as discussed in Example 22 Image SourceGetty Images RF Final PDF to printer 60 ENERGY ENERGY TRANSFER cen22672ch02051108indd 60 110317 0749 AM 23 ENERGY TRANSFER BY HEAT Energy can cross the boundary of a closed system in two distinct forms heat and work Fig 214 It is important to distinguish between these two forms of energy Therefore they will be discussed first to form a sound basis for the development of the laws of thermodynamics We know from experience that a can of cold soda left on a table eventually warms up and that a hot baked potato on the same table cools down When a body is left in a medium that is at a different temperature energy trans fer takes place between the body and the surrounding medium until thermal equilibrium is established that is the body and the medium reach the same temperature The direction of energy transfer is always from the higher tem perature body to the lower temperature one Once the temperature equality is established energy transfer stops In the processes described above energy is said to be transferred in the form of heat Heat is defined as the form of energy that is transferred between two sys tems or a system and its surroundings by virtue of a temperature differ ence Fig 215 That is an energy interaction is heat only if it takes place because of a temperature difference Then it follows that there cannot be any heat transfer between two systems that are at the same temperature Several phrases in common use todaysuch as heat flow heat addition heat rejection heat absorption heat removal heat gain heat loss heat storage heat generation electrical heating resistance heating frictional heating gas heating heat of reaction liberation of heat specific heat sen sible heat latent heat waste heat body heat process heat heat sink and heat sourceare not consistent with the strict thermodynamic meaning of the term heat which limits its use to the transfer of thermal energy during a process However these phrases are deeply rooted in our vocabulary and they are used by both ordinary people and scientists without causing any misunderstanding since they are usually interpreted properly instead of being taken literally Besides no acceptable alternatives exist for some of these phrases For example the phrase body heat is understood to mean the thermal energy content of a body Likewise heat flow is understood to mean the transfer of thermal energy not the flow of a fluidlike sub stance called heat although the latter incorrect interpretation which is based on the caloric theory is the origin of this phrase Also the transfer of heat into a system is often referred to as heat addition and the trans fer of heat out of a system as heat rejection Perhaps there are thermody namic reasons for being so reluctant to replace heat with thermal energy It takes less time and energy to say write and comprehend heat than it does thermal energy Heat is energy in transition It is recognized only as it crosses the boundary of a system Consider the hot baked potato one more time The potato con tains energy but this energy is heat transfer only as it passes through the skin of the potato the system boundary to reach the air as shown in Fig 216 Once in the surroundings the transferred heat becomes part of the internal energy of the surroundings Thus in thermodynamics the term heat simply means heat transfer A process during which there is no heat transfer is called an adiabatic process Fig 217 The word adiabatic comes from the Greek word adiabatos FIGURE 214 Energy can cross the boundaries of a closed system in the form of heat and work Work Closed system m constant Heat System boundary FIGURE 215 Temperature difference is the driving force for heat transfer The larger the temperature difference the higher is the rate of heat transfer Room air 25C No heat transfer Heat Heat 8 Js 16 Js 25C 15C 5C FIGURE 216 Energy is recognized as heat transfer only as it crosses the system boundary Surrounding air Baked potato System boundary Heat 2 kJ thermal energy 2 kJ thermal energy 2 kJ heat Final PDF to printer 61 CHAPTER 2 cen22672ch02051108indd 61 110317 0749 AM which means not to be passed There are two ways a process can be adiabatic Either the system is well insulated so that only a negligible amount of heat can pass through the boundary or both the system and the surroundings are at the same temperature and therefore there is no driving force temperature dif ference for heat transfer An adiabatic process should not be confused with an isothermal process Even though there is no heat transfer during an adiabatic process the energy content and thus the temperature of a system can still be changed by other means such as work As a form of energy heat has energy units kJ or Btu being the most com mon one The amount of heat transferred during the process between two states states 1 and 2 is denoted by Q12 or just Q Heat transfer per unit mass of a system is denoted q and is determined from q Q m kJkg 214 Sometimes it is desirable to know the rate of heat transfer the amount of heat transferred per unit time instead of the total heat transferred over some time interval Fig 218 The heat transfer rate is denoted Q where the over dot stands for the time derivative or per unit time The heat transfer rate Q has the unit kJs which is equivalent to kW When Q varies with time the amount of heat transfer during a process is determined by integrating Q over the time interval of the process Q t 1 t 2 Q dt kJ 215 When Q remains constant during a process this relation reduces to Q Q Δt kJ 216 where Δt t2 t1 is the time interval during which the process takes place Historical Background on Heat Heat has always been perceived to be something that produces in us a sen sation of warmth and one would think that the nature of heat is one of the first things understood by mankind However it was only in the middle of the 19th century that we had a true physical understanding of the nature of heat thanks to the development at that time of the kinetic theory which treats molecules as tiny balls that are in motion and thus possess kinetic energy Heat is then defined as the energy associated with the random motion of atoms and molecules Although it was suggested in the 18th and early 19th centuries that heat is the manifestation of motion at the molecular level called the live force the prevailing view of heat until the middle of the 19th century was based on the caloric theory proposed by the French chemist Antoine Lavoisier 17441794 in 1789 The caloric theory asserts that heat is a fluidlike substance called the caloric that is a massless colorless odorless and tasteless substance that can be poured from one body into another Fig 219 When caloric was added to a body its temperature increased when caloric was removed from a body its temperature decreased When a body could not contain any more caloric FIGURE 217 During an adiabatic process a system exchanges no heat with its surroundings Q 0 Insulation Adiabatic system FIGURE 218 The relationships among q Q and Q Q 30 kJ m 2 kg t 5 s Δ Q 6 kW q 15 kJkg 30 kJ heat FIGURE 219 In the early 19th century heat was thought to be an invisible fluid called the caloric that flowed from warmer bodies to cooler ones Hot body Cold body Contact surface Caloric Final PDF to printer 62 ENERGY ENERGY TRANSFER cen22672ch02051108indd 62 110317 0749 AM much the same as when a glass of water could not dissolve any more salt or sugar the body was said to be saturated with caloric This interpretation gave rise to the terms saturated liquid and saturated vapor that are still in use today The caloric theory came under attack soon after its introduction It main tained that heat is a substance that could not be created or destroyed Yet it was known that heat can be generated indefinitely by rubbing ones hands together or rubbing two pieces of wood together In 1798 the American Benjamin Thompson Count Rumford 17541814 showed in his papers that heat can be generated continuously through friction The validity of the caloric theory was also challenged by several others But it was the careful experiments of the Englishman James P Joule 18181889 published in 1843 that finally convinced the skeptics that heat was not a substance after all and thus put the caloric theory to rest Although the caloric theory was totally abandoned in the middle of the 19th century it contributed greatly to the development of thermodynamics and heat transfer Heat is transferred by three mechanisms conduction convection and radi ation Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles Convection is the transfer of energy between a solid sur face and the adjacent fluid that is in motion and it involves the combined effects of conduction and fluid motion Radiation is the transfer of energy due to the emission of electromagnetic waves or photons An overview of the three mechanisms of heat transfer is given at the end of this chapter as a Topic of Special Interest 24 ENERGY TRANSFER BY WORK Work like heat is an energy interaction between a system and its sur roundings As mentioned earlier energy can cross the boundary of a closed system in the form of heat or work Therefore if the energy crossing the boundary of a closed system is not heat it must be work Heat is easy to recognize Its driving force is a temperature difference between the system and its surroundings Then we can simply say that an energy interaction that is not caused by a temperature difference between a system and its sur roundings is work More specifically work is the energy transfer associated with a force acting through a distance A rising piston a rotating shaft and an electric wire crossing the system boundaries are all associated with work interactions Work is also a form of energy transferred like heat and therefore has energy units such as kJ The work done during a process between states 1 and 2 is denoted by W12 or simply W The work done per unit mass of a system is denoted by w and is expressed as w W m kJkg 217 The work done per unit time is called power and is denoted W Fig 220 The unit of power is kJs or kW FIGURE 220 The relationships among w W and W W 30 kJ m 2 kg t 5 s Δ W 6 kW w 15 kJkg 30 kJ work Final PDF to printer 63 CHAPTER 2 cen22672ch02051108indd 63 110317 0749 AM Heat and work are directional quantities and thus the complete description of a heat or work interaction requires the specification of both the magnitude and direction One way of doing that is to adopt a sign convention The gen erally accepted formal sign convention for heat and work interactions is as follows heat transfer to a system and work done by a system are positive heat transfer from a system and work done on a system are negative Another way is to use the subscripts in and out to indicate direction Fig 221 For example a work input of 5 kJ can be expressed as Win 5 kJ while a heat loss of 3 kJ can be expressed as Qout 3 kJ When the direction of a heat or work interaction is not known we can simply assume a direction for the interaction using the subscript in or out and solve for it A positive result indicates the assumed direction is right A negative result on the other hand indicates that the direction of the interaction is the opposite of the assumed direction This is just like assuming a direction for an unknown force when solving a statics problem and reversing the direction when a negative result is obtained for the force We will use this intuitive approach in this book as it eliminates the need to adopt a formal sign convention and the need to carefully assign negative values to some interactions Note that a quantity that is transferred to or from a system during an inter action is not a property since the amount of such a quantity depends on more than just the state of the system Heat and work are energy transfer mecha nisms between a system and its surroundings and there are many similarities between them 1 Both are recognized at the boundaries of a system as they cross the boundaries That is both heat and work are boundary phenomena 2 Systems possess energy but not heat or work 3 Both are associated with a process not a state Unlike properties heat or work has no meaning at a state 4 Both are path functions ie their magnitudes depend on the path fol lowed during a process as well as the end states Path functions have inexact differentials designated by the symbol δ There fore a differential amount of heat or work is represented by δQ or δW respec tively instead of dQ or dW Properties however are point functions ie they depend on the state only and not on how a system reaches that state and they have exact differentials designated by the symbol d A small change in volume for example is represented by dV and the total volume change dur ing a process between states 1 and 2 is 1 2 dV V 2 V 1 ΔV That is the volume change during process 12 is always the volume at state 2 minus the volume at state 1 regardless of the path followed Fig 222 The total work done during process 12 however is 1 2 δW W 12 not ΔW FIGURE 221 Specifying the directions of heat and work System Surroundings Qin Qout Win Wout FIGURE 222 Properties are point functions but heat and work are path functions their magnitudes depend on the path followed P 5 m3 2 m3 2 1 Process B Process A V ΔVA 3 m3 WA 8 kJ ΔVB 3 m3 WB 12 kJ Final PDF to printer 64 ENERGY ENERGY TRANSFER cen22672ch02051108indd 64 110317 0749 AM That is the total work is obtained by following the process path and adding the differential amounts of work δW done along the way The integral of δW is not W2 W1 ie the work at state 2 minus work at state 1 which is meaningless since work is not a property and systems do not possess work at a state EXAMPLE 23 Burning of a Candle in an Insulated Room A candle is burning in a wellinsulated room Taking the room the air plus the can dle as the system determine a if there is any heat transfer during this burning pro cess and b if there is any change in the internal energy of the system SOLUTION A candle burning in a wellinsulated room is considered It is to be determined whether there is any heat transfer and any change in internal energy Analysis a The interior surfaces of the room form the system boundary as indi cated by the dashed lines in Fig 223 As pointed out earlier heat is recognized as it crosses the boundaries Since the room is well insulated we have an adiabatic system and no heat will pass through the boundaries Therefore Q 0 for this process b The internal energy involves energies that exist in various forms sensible latent chemical nuclear During the process just described part of the chemical energy is converted to sensible energy Since there is no increase or decrease in the total inter nal energy of the system ΔU 0 for this process EXAMPLE 24 Heating of a Potato in an Oven A potato initially at room temperature 25C is being baked in an oven that is main tained at 200C as shown in Fig 224 Is there any heat transfer during this baking process SOLUTION A potato is being baked in an oven It is to be determined whether there is any heat transfer during this process Analysis This is not a welldefined problem since the system is not specified Let us assume that we are observing the potato which will be our system Then the outer surface of the skin of the potato can be viewed as the system boundary Part of the energy in the oven will pass through the skin to the potato Since the driving force for this energy transfer is a temperature difference this is a heat transfer process EXAMPLE 25 Heating of an Oven by Work Transfer A wellinsulated electric oven is being heated through its heating element If the entire oven including the heating element is taken to be the system determine whether this is a heat or work interaction SOLUTION A wellinsulated electric oven is being heated by its heating element It is to be determined whether this is a heat or work interaction Analysis For this problem the interior surfaces of the oven form the system bound ary as shown in Fig 225 The energy content of the oven obviously increases during FIGURE 223 Schematic for Example 23 Room Insulation FIGURE 224 Schematic for Example 24 Oven 200C Heat Potato 25C FIGURE 225 Schematic for Example 25 System boundary Electric oven Heating element Final PDF to printer 65 CHAPTER 2 cen22672ch02051108indd 65 110317 0749 AM Electrical Work It was pointed out in Example 25 that electrons crossing the system bound ary do electrical work on the system In an electric field electrons in a wire move under the effect of electromotive forces doing work When N coulombs of electrical charge move through a potential difference V the electrical work done is W e VN which can also be expressed in the rate form as W e VI W 218 where W e is the electrical power and I is the number of electrical charges flowing per unit time that is the current Fig 227 In general both V and I vary with time and the electrical work done during a time interval Δt is expressed as W e 1 2 V I dt kJ 219 When both V and I remain constant during the time interval Δt it reduces to W e VI Δt kJ 220 EXAMPLE 26 Heating of an Oven by Heat Transfer Answer the question in Example 25 if the system is taken as only the air in the oven without the heating element SOLUTION The question in Example 25 is to be reconsidered by taking the sys tem to be only the air in the oven Analysis This time the system boundary will include the outer surface of the heat ing element and will not cut through it as shown in Fig 226 Therefore no electrons will be crossing the system boundary at any point Instead the energy generated in the interior of the heating element will be transferred to the air around it as a result of the temperature difference between the heating element and the air in the oven Therefore this is a heat transfer process Discussion For both cases the amount of energy transfer to the air is the same These two examples show that an energy transfer can be heat or work depending on how the system is selected FIGURE 226 Schematic for Example 26 System boundary Electric oven Heating element this process as evidenced by a rise in temperature This energy transfer to the oven is not caused by a temperature difference between the oven and the surrounding air Instead it is caused by electrons crossing the system boundary and thus doing work Therefore this is a work interaction FIGURE 227 Electrical power in terms of resistance R current I and potential difference V V I R We VI I 2R V2R Final PDF to printer 66 ENERGY ENERGY TRANSFER cen22672ch02051108indd 66 110317 0749 AM 25 MECHANICAL FORMS OF WORK There are several different ways of doing work each in some way related to a force acting through a distance Fig 228 In elementary mechanics the work done by a constant force F on a body displaced a distance s in the direc tion of the force is given by W Fs kJ 221 If the force F is not constant the work done is obtained by adding ie inte grating the differential amounts of work W 1 2 F ds kJ 222 Obviously one needs to know how the force varies with displacement to per form this integration Equations 221 and 222 give only the magnitude of the work The sign is easily determined from physical considerations The work done on a system by an external force acting in the direction of motion is negative and work done by a system against an external force acting in the opposite direction to motion is positive There are two requirements for a work interaction between a system and its surroundings to exist 1 there must be a force acting on the boundary and 2 the boundary must move Therefore the presence of forces on the bound ary without any displacement of the boundary does not constitute a work interaction Likewise the displacement of the boundary without any force to oppose or drive this motion such as the expansion of a gas into an evacuated space is not a work interaction since no energy is transferred In many thermodynamic problems mechanical work is the only form of work involved It is associated with the movement of the boundary of a sys tem or with the movement of the entire system as a whole Some common forms of mechanical work are discussed next Shaft Work Energy transmission with a rotating shaft is very common in engineering practice Fig 229 Often the torque T applied to the shaft is constant which means that the force F applied is also constant For a specified constant torque the work done during n revolutions is determined as fol lows A force F acting through a moment arm r generates a torque T of Fig 230 T Fr F T r 223 This force acts through a distance s which is related to the radius r by s 2πrn 224 Then the shaft work is determined from W sh Fs T r 2πrn 2πnT kJ 225 FIGURE 228 The work done is proportional to the force applied F and the distance traveled s s F F FIGURE 229 Energy transmission through rotating shafts is commonly encountered in practice Engine Boat FIGURE 230 Shaft work is proportional to the torque applied and the number of revolutions of the shaft Wsh 2πnT r F n Torque Fr Final PDF to printer 67 CHAPTER 2 cen22672ch02051108indd 67 110317 0749 AM The power transmitted through the shaft is the shaft work done per unit time which can be expressed as W sh 2π n T kW 226 where n is the number of revolutions per unit time Spring Work It is common knowledge that when a force is applied on a spring the length of the spring changes Fig 232 When the length of the spring changes by a differential amount dx under the influence of a force F the work done is δ W spring F dx 227 To determine the total spring work we need to know a functional relationship between F and x For linear elastic springs the displacement x is proportional to the force applied Fig 233 That is F kx kN 228 where k is the spring constant and has the unit kNm The displacement x is measured from the undisturbed position of the spring that is x 0 when F 0 Substituting Eq 228 into Eq 227 and integrating yield W spring 1 2 k x 2 2 x 1 2 kJ 229 where x1 and x2 are the initial and the final displacements of the spring respectively measured from the undisturbed position of the spring There are many other forms of mechanical work Next we introduce some of them briefly Work Done on Elastic Solid Bars Solids are often modeled as linear springs because under the action of a force they contract or elongate as shown in Fig 234 and when the force is lifted they return to their original lengths like a spring This is true as long as the EXAMPLE 27 Power Transmission by the Shaft of a Car Determine the power transmitted through the shaft of a car when the torque applied is 200 Nm and the shaft rotates at a rate of 4000 revolutions per minute rpm SOLUTION The torque and the rpm for a car engine are given The power trans mitted is to be determined Analysis A sketch of the car is given in Fig 231 The shaft power is determined directly from W sh 2π n T 2π 4000 1 min 200 Nm 1 min 60 s 1 kJ 1000 Nm 838 kW or 112 hp Discussion Note that power transmitted by a shaft is proportional to torque and the rotational speed FIGURE 231 Schematic for Example 27 T 200 Nm n 4000 rpm FIGURE 232 Elongation of a spring under the influence of a force Rest position dx x F FIGURE 233 The displacement of a linear spring doubles when the force is doubled x1 1 mm Rest position F1 300 N x2 2 mm F2 600 N Final PDF to printer 68 ENERGY ENERGY TRANSFER cen22672ch02051108indd 68 110317 0749 AM force is in the elastic range that is not large enough to cause permanent plas tic deformations Therefore the equations given for a linear spring can also be used for elastic solid bars Alternately we can determine the work associated with the expansion or contraction of an elastic solid bar by replacing pressure P with its counterpart in solids normal stress σn FA in the work expression W elastic 1 2 F dx 1 2 σ n A dx kJ 230 where A is the crosssectional area of the bar Note that the normal stress has pressure units Work Associated with the Stretching of a Liquid Film Consider a liquid film such as soap film suspended on a wire frame Fig 235 We know from experience that it will take some force to stretch this film by the movable portion of the wire frame This force is used to over come the microscopic forces between molecules at the liquidair interfaces These microscopic forces are perpendicular to any line in the surface and the force generated by these forces per unit length is called the surface tension σs whose unit is Nm Therefore the work associated with the stretching of a film is also called surface tension work It is determined from W surface 1 2 σ s dA kJ 231 where dA 2b dx is the change in the surface area of the film The factor 2 is due to the fact that the film has two surfaces in contact with air The force acting on the movable wire as a result of surface tension effects is F 2bσs where σs is the surface tension force per unit length Work Done to Raise or to Accelerate a Body When a body is raised in a gravitational field its potential energy increases Likewise when a body is accelerated its kinetic energy increases The con servation of energy principle requires that an equivalent amount of energy must be transferred to the body being raised or accelerated Remember that energy can be transferred to a given mass by heat and work and the energy transferred in this case obviously is not heat since it is not driven by a tem perature difference Therefore it must be work Then we conclude that 1 the work transfer needed to raise a body is equal to the change in the potential energy of the body and 2 the work transfer needed to accelerate a body is equal to the change in the kinetic energy of the body Fig 236 Simi larly the potential or kinetic energy of a body represents the work that can be obtained from the body as it is lowered to the reference level or decelerated to zero velocity This discussion together with the consideration for friction and other losses form the basis for determining the required power rating of motors used to drive devices such as elevators escalators conveyor belts and ski lifts It also plays a primary role in the design of automotive and aircraft engines and in the determination of the amount of hydroelectric power that can be produced FIGURE 234 Solid bars behave as springs under the influence of a force x F FIGURE 235 Stretching a liquid film with a Ushaped wire and the forces acting on the movable wire of length b δx F F Movable wire Rigid wire frame Liquid film Wire Surface of film b x σs σs FIGURE 236 The energy transferred to a body while being raised is equal to the change in its potential energy Motor Elevator car Final PDF to printer 69 CHAPTER 2 cen22672ch02051108indd 69 110317 0749 AM from a given water reservoir which is simply the potential energy of the water relative to the location of the hydraulic turbine FIGURE 237 Schematic for Example 28 McGrawHill EducationLars A Niki EXAMPLE 29 Power Needs of a Car to Accelerate Determine the power required to accelerate a 900kg car shown in Fig 238 from rest to a velocity of 80 kmh in 20 s on a level road SOLUTION The power required to accelerate a car to a specified velocity is to be determined Analysis The work needed to accelerate a body is simply the change in the kinetic energy of the body W a 1 2 m V 2 2 V 1 2 1 2 900 kg 80000 m 3600 s 2 0 2 1 kJkg 1000 m 2 s 2 222 kJ The average power is determined from W a W a Δt 222 kJ 20 s 111 kW or 149 hp Discussion This is in addition to the power required to overcome friction rolling resistance and other imperfections FIGURE 238 Schematic for Example 29 m 900 kg 0 80 kmh EXAMPLE 28 Power Needs of a Car to Climb a Hill A man whose mass is 100 kg pushes a cart whose mass including its contents is 100 kg up a ramp that is inclined at an angle of 20 from the horizontal Fig 237 The local gravitational acceleration is 98 ms2 Determine the work in kJ needed to move along this ramp a distance of 100 m considering a the man and b the cart and its contents as the system SOLUTION A man is pushing a cart with its contents up a ramp that is inclined at an angle of 20 from the horizontal The work needed to move along this ramp is to be determined considering a the man and b the cart and its contents as the system Analysis a Considering the man as the system letting l be the displacement along the ramp and letting θ be the inclination angle of the ramp W Fl sin θ mgl sin θ 100 100 kg98 m s 2 100 msin 20 1 kJ kg 1000 m 2 s 2 670 kJ This is work that the man must do to raise the weight of the cart and contents plus his own weight a distance of l sin θ b Applying the same logic to the cart and its contents gives W Fl sin θ mgl sin θ 100 kg98 m s 2 100 msin 20 1 kJ kg 1000 m 2 s 2 335 kJ Discussion The result in part a is more realistic since the man has to move him self in addition to the cart Final PDF to printer 70 ENERGY ENERGY TRANSFER cen22672ch02051108indd 70 110317 0749 AM Nonmechanical Forms of Work The treatment in Section 25 represents a fairly comprehensive cover age of mechanical forms of work except the moving boundary work that is covered in Chap 4 Some work modes encountered in practice are not mechanical in nature However these nonmechanical work modes can be treated in a similar manner by identifying a generalized force F acting in the direction of a generalized displacement x Then the work associated with the differential displacement under the influence of this force is deter mined from δW Fdx Some examples of nonmechanical work modes are electrical work where the generalized force is the voltage the electrical potential and the generalized displacement is the electrical charge as discussed earlier magnetic work where the generalized force is the magnetic field strength and the generalized displacement is the total magnetic dipole moment and electrical polarization work where the generalized force is the electric field strength and the generalized displacement is the polarization of the medium the sum of the electric dipole rotation moments of the molecules Detailed consideration of these and other nonmechanical work modes can be found in specialized books on these topics 26 THE FIRST LAW OF THERMODYNAMICS So far we have considered various forms of energy such as heat Q work W and total energy E individually and no attempt is made to relate them to each other during a process The first law of thermodynamics also known as the conservation of energy principle provides a sound basis for studying the relationships among the various forms of energy and energy interactions Based on experimental observations the first law of thermodynamics states that energy can be neither created nor destroyed during a process it can only change forms Therefore every bit of energy should be accounted for during a process We all know that a rock at some elevation possesses some potential energy and part of this potential energy is converted to kinetic energy as the rock falls Fig 239 Experimental data show that the decrease in potential energy mg Δz exactly equals the increase in kinetic energy m V 2 2 V 1 2 2 when the air resistance is negligible thus confirming the conservation of energy principle for mechanical energy Consider a system undergoing a series of adiabatic processes from a spec ified state 1 to another specified state 2 Being adiabatic these processes obviously cannot involve any heat transfer but they may involve several kinds of work interactions Careful measurements during these experiments indicate the following For all adiabatic processes between two specified states of a closed system the net work done is the same regardless of the nature of the closed system and the details of the process Considering that there are an infinite number of ways to perform work interactions under adi abatic conditions this statement appears to be very powerful with a poten tial for farreaching implications This statement which is largely based on the experiments of Joule in the first half of the 19th century cannot be drawn FIGURE 239 Energy cannot be created or destroyed it can only change forms PE1 10 kJ m KE1 0 PE2 7 kJ m KE2 3 kJ Δz FIGURE 240 The increase in the energy of a potato in an oven is equal to the amount of heat transferred to it Qin 5 kJ Potato ΔE 5 kJ FIGURE 241 In the absence of any work interactions the energy change of a system is equal to the net heat transfer ΔE Qnet 12 kJ Qin 15 kJ Qout 3 kJ Final PDF to printer 71 CHAPTER 2 cen22672ch02051108indd 71 110317 0749 AM from any other known physical principle and is recognized as a fundamental principle This principle is called the first law of thermodynamics or just the first law A major consequence of the first law is the existence and the defini tion of the property total energy E Considering that the net work is the same for all adiabatic processes of a closed system between two specified states the value of the net work must depend on the end states of the system only and thus it must correspond to a change in a property of the system This property is the total energy Note that the first law makes no reference to the value of the total energy of a closed system at a state It simply states that the change in the total energy during an adiabatic process must be equal to the net work done Therefore any convenient arbitrary value can be assigned to total energy at a specified state to serve as a reference point Implicit in the first law statement is the conservation of energy Although the essence of the first law is the existence of the property total energy the first law is often viewed as a statement of the conservation of energy princi ple Next we develop the first law or the conservation of energy relation with the help of some familiar examples using intuitive arguments First we consider some processes that involve heat transfer but no work interactions The potato baked in the oven is a good example for this case Fig 240 As a result of heat transfer to the potato the energy of the potato will increase If we disregard any mass transfer moisture loss from the potato the increase in the total energy of the potato becomes equal to the amount of heat transfer That is if 5 kJ of heat is transferred to the potato the energy increase of the potato will also be 5 kJ As another example consider the heating of water in a pan on top of a range Fig 241 If 15 kJ of heat is transferred to the water from the heat ing element and 3 kJ of it is lost from the water to the surrounding air the increase in energy of the water will be equal to the net heat transfer to water which is 12 kJ Now consider a wellinsulated ie adiabatic room heated by an electric heater as our system Fig 242 As a result of electrical work done the energy of the system will increase Since the system is adiabatic and cannot have any heat transfer to or from the surroundings Q 0 the conservation of energy principle dictates that the electrical work done on the system must equal the increase in energy of the system Next let us replace the electric heater with a paddle wheel Fig 243 As a result of the stirring process the energy of the system will increase Again since there is no heat interaction between the system and its surroundings Q 0 the shaft work done on the system must show up as an increase in the energy of the system Many of you have probably noticed that the temperature of air rises when it is compressed Fig 244 This is because energy is transferred to the air in the form of boundary work In the absence of any heat transfer Q 0 the entire boundary work will be stored in the air as part of its total energy The conservation of energy principle again requires that the increase in the energy of the system be equal to the boundary work done on the system FIGURE 242 The work electrical done on an adiabatic system is equal to the increase in the energy of the system Win 5 kJ Adiabatic Battery ΔE 5 kJ FIGURE 243 The work shaft done on an adiabatic system is equal to the increase in the energy of the system Adiabatic ΔE 8 kJ Wshin 8 kJ FIGURE 244 The work boundary done on an adiabatic system is equal to the increase in the energy of the system Wbin 10 kJ Adiabatic ΔE 10 kJ Final PDF to printer 72 ENERGY ENERGY TRANSFER cen22672ch02051108indd 72 110317 0749 AM We can extend these discussions to systems that involve various heat and work interactions simultaneously For example if a system gains 12 kJ of heat during a process while 6 kJ of work is done on it the increase in the energy of the system during that process is 18 kJ Fig 245 That is the change in the energy of a system during a process is simply equal to the net energy transfer to or from the system Energy Balance In the light of the preceding discussions the conservation of energy principle can be expressed as follows The net change increase or decrease in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process That is Total energy entering the system Total energy leaving the system Change in the total energy of the system or E in E out Δ E system This relation is often referred to as the energy balance and is applicable to any kind of system undergoing any kind of process The successful use of this relation to solve engineering problems depends on understanding the various forms of energy and recognizing the forms of energy transfer Energy Change of a System ΔEsystem The determination of the energy change of a system during a process involves the evaluation of the energy of the system at the beginning and at the end of the process and taking their difference That is Energy change Energy at final state Energy at initial state or Δ E system E final E initial E 2 E 1 232 Note that energy is a property and the value of a property does not change unless the state of the system changes Therefore the energy change of a system is zero if the state of the system does not change during the process Also energy can exist in numerous forms such as internal sensible latent chemical and nuclear kinetic potential electric and magnetic and their sum constitutes the total energy E of a system In the absence of electric magnetic and surface tension effects ie for simple compressible systems the change in the total energy of a system during a process is the sum of the changes in its internal kinetic and potential energies and can be expressed as ΔE ΔU ΔKE ΔPE 233 where ΔU m u 2 u 1 ΔKE 1 2 m V 2 2 V 1 2 ΔPE mg z 2 z 1 FIGURE 245 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings Wshin 6 kJ ΔE 15 3 6 18 kJ Qin 15 kJ Qout 3 kJ Final PDF to printer 73 CHAPTER 2 cen22672ch02051108indd 73 110317 0749 AM When the initial and final states are specified the values of the specific inter nal energies u1 and u2 can be determined directly from the property tables or thermodynamic property relations Most systems encountered in practice are stationary that is they do not involve any changes in their velocity or elevation during a process Fig 246 Thus for stationary systems the changes in kinetic and potential energies are zero that is ΔKE ΔPE 0 and the total energy change relation in Eq 233 reduces to ΔE ΔU for such systems Also the energy of a system during a process will change even if only one form of its energy changes while the other forms of energy remain unchanged Mechanisms of Energy Transfer Ein and Eout Energy can be transferred to or from a system in three forms heat work and mass flow Energy interactions are recognized at the system boundary as they cross it and they represent the energy gained or lost by a system during a pro cess The only two forms of energy interactions associated with a fixed mass or closed system are heat transfer and work 1 Heat Transfer Q Heat transfer to a system heat gain increases the energy of the molecules and thus the internal energy of the system and heat trans fer from a system heat loss decreases it since the energy transferred out as heat comes from the energy of the molecules of the system 2 Work Transfer W An energy interaction that is not caused by a temperature difference between a system and its surroundings is work A rising piston a rotating shaft and an electrical wire crossing the system boundaries are all associated with work interactions Work transfer to a system ie work done on a system increases the energy of the system and work transfer from a system ie work done by the system decreases it since the energy transferred out as work comes from the energy con tained in the system Car engines and hydraulic steam or gas turbines produce work while compressors pumps and mixers consume work 3 Mass Flow m Mass flow in and out of the system serves as an addi tional mechanism of energy transfer When mass enters a system the energy of the system increases because mass carries energy with it in fact mass is energy Likewise when some mass leaves the system the energy contained within the system decreases because the departing mass takes out some energy with it For example when some hot water is taken out of a water heater and is replaced by the same amount of cold water the energy content of the hotwater tank the control volume decreases as a result of this mass interaction Fig 247 Noting that energy can be transferred in the forms of heat work and mass and that the net transfer of a quantity is equal to the difference between the amounts transferred in and out the energy balance can be written more explicitly as E in E out Q in Q out W in W out E massin E massout Δ E system 234 where the subscripts in and out denote quantities that enter and leave the sys tem respectively All six quantities on the right side of the equation represent amounts and thus they are positive quantities The direction of any energy transfer is described by the subscripts in and out FIGURE 246 For stationary systems ΔKE ΔPE 0 thus ΔE ΔU Stationary Systems z1 z2 ΔPE 0 V1 V2 ΔKE 0 ΔE ΔU FIGURE 247 The energy content of a control volume can be changed by mass flow as well as by heat and work interactions Mass in Mass out Control volume Q W Final PDF to printer 74 ENERGY ENERGY TRANSFER cen22672ch02051108indd 74 110317 0749 AM The heat transfer Q is zero for adiabatic systems the work transfer W is zero for systems that involve no work interactions and the energy transport with mass Emass is zero for systems that involve no mass flow across their boundar ies ie closed systems Energy balance for any system undergoing any kind of process can be expressed more compactly as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies kJ 235 or in the rate form as E in E out Rate of net energy transfer by heat work and mass dE system dt Rate of change in internal kinetic potential etc energies kW 236 For constant rates the total quantities during a time interval Δt are related to the quantities per unit time as Q Q Δt W W Δt and ΔE dE dt Δt kJ 237 The energy balance can be expressed on a per unit mass basis as e in e out Δ e system kJkg 238 which is obtained by dividing all the quantities in Eq 235 by the mass m of the system Energy balance can also be expressed in the differential form as δ E in δ E out d E system or δ e in δ e out d e system 239 For a closed system undergoing a cycle the initial and final states are iden tical and thus ΔEsystem E2 E1 0 Then the energy balance for a cycle simplifies to Ein Eout 0 or Ein Eout Noting that a closed system does not involve any mass flow across its boundaries the energy balance for a cycle can be expressed in terms of heat and work interactions as W netout Q netin or W netout Q netin for a cycle 240 That is the net work output during a cycle is equal to net heat input Fig 248 FIGURE 248 For a cycle ΔE 0 thus Q W P Qnet Wnet V EXAMPLE 210 Cooling of a Hot Fluid in a Tank A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel Initially the internal energy of the fluid is 800 kJ During the cooling process the fluid loses 500 kJ of heat and the paddle wheel does 100 kJ of work on the fluid Determine the final internal energy of the fluid Neglect the energy stored in the paddle wheel SOLUTION A fluid in a rigid tank loses heat while being stirred The final internal energy of the fluid is to be determined Final PDF to printer 75 CHAPTER 2 cen22672ch02051108indd 75 110317 0749 AM Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 Therefore ΔE ΔU and internal energy is the only form of the systems energy that may change during this process 2 Energy stored in the paddle wheel is negligible Analysis Take the contents of the tank as the system Fig 249 This is a closed system since no mass crosses the boundary during the process We observe that the volume of a rigid tank is constant and thus there is no mov ing boundary work Also heat is lost from the system and shaft work is done on the system Applying the energy balance on the system gives E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W shin Q out ΔU U 2 U 1 100 kJ 500 kJ U 2 800 kJ U 2 400 kJ Therefore the final internal energy of the system is 400 kJ FIGURE 249 Schematic for Example 210 Wshin 100 kJ Qout 500 kJ U1 800 kJ U2 Fluid EXAMPLE 211 Acceleration of Air by a Fan A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 10 kgs at a discharge velocity of 8 ms Fig 250 Determine if this claim is reasonable SOLUTION A fan is claimed to increase the velocity of air to a specified value while consuming electric power at a specified rate The validity of this claim is to be investigated Assumptions The ventilating room is relatively calm and air velocity in it is negligible Analysis First lets examine the energy conversions involved The motor of the fan converts part of the electrical power it consumes to mechanical shaft power which is used to rotate the fan blades in air The blades are shaped such that they impart a large fraction of the mechanical power of the shaft to air by mobilizing it In the limiting ideal case of no losses no conversion of electrical and mechanical energy to thermal energy in steady operation the electric power input will be equal to the rate of increase of the kinetic energy of air Therefore for a control volume that encloses the fanmotor unit the energy balance can be written as E in E out Rate of net energy transfer by heat work and mass d E system dt 0 steady Rate of change in internal kinetic potential etc energies 0 E in E out W electin m air ke out m air V out 2 2 Solving for Vout and substituting gives the maximum air outlet velocity to be V out 2 W electin m air 2 20 Js 10 kgs 1 m 2 s 2 1 Jkg 63 ms which is less than 8 ms Therefore the claim is false FIGURE 250 Schematic for Example 211 Design PicsPunchStock RF Air 8 ms Final PDF to printer 76 ENERGY ENERGY TRANSFER cen22672ch02051108indd 76 110317 0749 AM Discussion The conservation of energy principle requires the energy to be pre served as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process From the first law point of view there is nothing wrong with the conversion of the entire electrical energy into kinetic energy Therefore the first law has no objection to air velocity reaching 63 msbut this is the upper limit Any claim of higher velocity is in violation of the first law and thus impossible In reality the air velocity will be considerably lower than 63 ms because of the losses associated with the conversion of electrical energy to mechanical shaft energy and the conversion of mechanical shaft energy to kinetic energy or air EXAMPLE 212 Heating Effect of a Fan A room is initially at the outdoor temperature of 25C Now a large fan that consumes 200 W of electricity when running is turned on Fig 251 The heat transfer rate between the room and the outdoor air is given as Q UA T i T o where U 6 Wm2C is the overall heat transfer coefficient A 30 m2 is the exposed surface area of the room and Ti and To are the indoor and outdoor air temperatures respectively Deter mine the indoor air temperature when steady operating conditions are established SOLUTION A large fan is turned on and kept on in a room that loses heat to the out doors The indoor air temperature is to be determined when steady operation is reached Assumptions 1 Heat transfer through the floor is negligible 2 There are no other energy interactions involved Analysis The electricity consumed by the fan is energy input for the room and thus the room gains energy at a rate of 200 W As a result the room air temperature tends to rise But as the room air temperature rises the rate of heat loss from the room increases until the rate of heat loss equals the electric power consumption At that point the temperature of the room air and thus the energy content of the room remains constant and the conservation of energy for the room becomes E in E out Rate of net energy transfer by heat work and mass d E system dt 0 steady Rate of change in internal kinetic potential etc energies 0 E in E out W electin Q out UA T i T o Substituting 200 W 6 Wm 2 C30 m 2 T i 25 C It gives T i 261 C Therefore the room air temperature will remain constant after it reaches 261C Discussion Note that a 200W fan heats a room just like a 200W resistance heater In the case of a fan the motor converts part of the electric energy it draws to mechani cal energy in the form of a rotating shaft while the remaining part is dissipated as heat to the room air because of the motor inefficiency no motor converts 100 percent FIGURE 251 Schematic for Example 212 Fan Room Qout Welectin Final PDF to printer 77 CHAPTER 2 cen22672ch02051108indd 77 110317 0749 AM of the electric energy it receives to mechanical energy although some large motors come close with a conversion efficiency of over 97 percent Part of the mechanical energy of the shaft is converted to kinetic energy of air through the blades which is then converted to thermal energy as air molecules slow down because of friction At the end the entire electric energy drawn by the fan motor is converted to thermal energy of air which manifests itself as a rise in temperature EXAMPLE 213 Annual Lighting Cost of a Classroom The lighting needs of a classroom are met by 30 fluorescent lamps each consuming 80 W of electricity Fig 252 The lights in the classroom are kept on for 12 hours a day and 250 days a year For a unit electricity cost of 11 cents per kWh determine the annual energy cost of lighting for this classroom Also discuss the effect of lighting on the heating and airconditioning requirements of the room SOLUTION The lighting of a classroom by fluorescent lamps is considered The annual electricity cost of lighting for this classroom is to be determined and the light ings effect on the heating and airconditioning requirements is to be discussed Assumptions The effect of voltage fluctuations is negligible so each fluorescent lamp consumes its rated power Analysis The electric power consumed by the lamps when all are on and the num ber of hours they are kept on per year are Lighting power Power consumed per lamp No of lamps 80 Wlamp30 lamps 2400 W 24 kW Operating hours 12 hday250 daysyear 3000 hyear Then the amount and cost of electricity used per year become Lighting energy Lighting powerOperating hours 24 kW3000 hyear 7200 kWhyear Lighting cost Lighting energyUnit cost 7200 kWhyear011 kWh 792 year Light is absorbed by the surfaces it strikes and is converted to thermal energy Disregarding the light that escapes through the windows the entire 24 kW of elec tric power consumed by the lamps eventually becomes part of thermal energy of the classroom Therefore the lighting system in this room reduces the heating require ments by 24 kW but increases the airconditioning load by 24 kW Discussion Note that the annual lighting cost of this classroom alone is close to 800 This shows the importance of energy conservation measures If incandescent lightbulbs were used instead of fluorescent tubes the lighting costs would be four times as much since incandescent lamps use four times as much power for the same amount of light produced FIGURE 252 Fluorescent lamps lighting a classroom as discussed in Example 213 PhotoLinkGetty Images RF Final PDF to printer 78 ENERGY ENERGY TRANSFER cen22672ch02051108indd 78 110317 0749 AM 27 ENERGY CONVERSION EFFICIENCIES Efficiency is one of the most often used terms in thermodynamics and it indicates how well an energy conversion or transfer process is accomplished Efficiency is also one of the most often misused terms in thermodynamics and a source of misunderstandings This is because efficiency is often used without being properly defined first Next we will clarify this further and define some efficiencies commonly used in practice Efficiency in general can be expressed in terms of the desired output and the required input as Efficiency Desired output Required input 241 If you are shopping for a water heater a knowledgeable salesperson will tell you that the efficiency of a conventional electric water heater is about 90 percent Fig 253 You may find this confusing since the heating ele ments of electric water heaters are resistance heaters and the efficiency of all resistance heaters is 100 percent as they convert all the electrical energy they consume into thermal energy A knowledgeable salesperson will clarify this by explaining that the heat losses from the hotwater tank to the sur rounding air amount to 10 percent of the electrical energy consumed and the efficiency of a water heater is defined as the ratio of the energy deliv ered to the house by hot water to the energy supplied to the water heater A clever salesperson may even talk you into buying a more expensive water heater with thicker insulation that has an efficiency of 94 percent If you are a knowledgeable consumer and have access to natural gas you will probably purchase a gas water heater whose efficiency is only 55 percent since a gas unit costs about the same as an electric unit to purchase and install but the annual energy cost of a gas unit will be much less than that of an electric unit Perhaps you are wondering how the efficiency for a gas water heater is defined and why it is much lower than the efficiency of an electric heater As a general rule the efficiency of equipment that involves the combustion of a fuel is based on the heating value of the fuel which is the amount of heat released when a unit amount of fuel at room temperature is completely burned and the combustion products are cooled to the room temperature Fig 254 Then the performance of combustion equipment can be charac terized by combustion equipment efficiency ηcomb equip defined as η comb equip Q useful HV Useful heat delivered by the combustion equipment Heating value of the fuel burned 242 This efficiency can take different names depending on the type of the combustion unit such as furnace efficiency ηfurnace boiler efficiency ηboiler or heater efficiency ηheater For example an efficiency of 70 percent for a coal burning heater used to heat a building in winter indicates that 70 percent of the heating value of the coal is transferred to the building as useful heat while the remaining 30 percent is lost mostly by the hot stack gases leaving the heater Most fuels contain hydrogen which forms water when burned and the heating value of a fuel will be different depending on whether the water in FIGURE 253 Typical efficiencies of conventional and highefficiency electric and natural gas water heaters McGrawHill EducationChristopher Kerrigan Water heater Type Efficiency Gas conventional Gas highefficiency Electric conventional Electric highefficiency 55 62 90 94 FIGURE 254 The definition of the heating value of gasoline Air 25C Combustion gases 25C CO2 H2O N2 etc 1 kg Gasoline 25C LHV 44000 kJkg Combustion chamber Final PDF to printer 79 CHAPTER 2 cen22672ch02051108indd 79 110317 0749 AM combustion products is in the liquid or vapor form The heating value is called the lower heating value or LHV when the water leaves as a vapor and the higher heating value or HHV when the water in the combustion gases is completely condensed and thus the heat of vaporization is also recovered The difference between these two heating values is equal to the product of the amount of water and the enthalpy of vaporization of water at room tem perature For example the lower and higher heating values of gasoline are 44000 kJkg and 47300 kJkg respectively An efficiency definition should make it clear whether it is based on the higher or lower heating value of the fuel Efficiencies of cars and jet engines are normally based on lower heating values since water normally leaves as a vapor in the exhaust gases and it is not practical to try to recover the heat of vaporization Efficiencies of fur naces on the other hand are based on higher heating values The efficiency of space heating systems of residential and commercial build ings is usually expressed in terms of the annual fuel utilization efficiency or AFUE which accounts for the combustion equipment efficiency as well as other losses such as heat losses to unheated areas and startup and cooldown losses The AFUE of most new heating systems is about 85 percent although the AFUE of some old heating systems is under 60 percent The AFUE of some new highefficiency furnaces exceeds 96 percent but the high cost of such furnaces cannot be justified for locations with mild to moderate winters Such high efficiencies are achieved by reclaiming most of the heat in the flue gases condensing the water vapor and discharging the flue gases at tempera tures as low as 38C or 100F instead of about 200C or 400F for the conventional models For car engines the work output is understood to be the power delivered by the crankshaft But for power plants the work output can be the mechanical power at the turbine exit or the electrical power output of the generator A generator is a device that converts mechanical energy to electrical energy and the effectiveness of a generator is characterized by the generator efficiency which is the ratio of the electrical power output to the mechani cal power input The thermal efficiency of a power plant which is of primary interest in thermodynamics is usually defined as the ratio of the net shaft work output of the turbine to the heat input to the working fluid The effects of other factors are incorporated by defining an overall efficiency for the power plant as the ratio of the net electrical power output to the rate of fuel energy input That is η overall η comb equip η thermal η generator W netelectric HHV m fuel 243 The overall efficiencies are about 2530 percent for gasoline automotive engines 3540 percent for diesel engines and up to 60 percent for large power plants We are all familiar with the conversion of electrical energy to light by incan descent lightbulbs fluorescent tubes and highintensity discharge lamps The efficiency for the conversion of electricity to light can be defined as the ratio of the energy converted to light to the electrical energy consumed For exam ple common incandescent lightbulbs convert about 5 percent of the electrical energy they consume to light the rest of the energy consumed is dissipated as Final PDF to printer 80 ENERGY ENERGY TRANSFER cen22672ch02051108indd 80 110317 0749 AM heat which adds to the cooling load of the air conditioner in summer How ever it is more common to express the effectiveness of this conversion pro cess by lighting efficacy which is defined as the amount of light output in lumens per W of electricity consumed The efficacy of different lighting systems is given in Table 21 Note that a compact fluorescent lightbulb produces about four times as much light as an incandescent lightbulb per W and thus a 15W fluorescent bulb can replace a 60W incandescent lightbulb Fig 255 Also a compact fluorescent bulb lasts about 10000 h which is 10 times as long as an incandescent bulb and it plugs directly into the socket of an incandescent lamp Therefore despite their higher initial cost compact fluorescents reduce the lighting costs consider ably through reduced electricity consumption Sodiumfilled highintensity discharge lamps provide the most efficient lighting but their use is limited to outdoor use because of their yellowish light We can also define efficiency for cooking appliances since they convert electrical or chemical energy to heat for cooking The efficiency of a cooking appliance can be defined as the ratio of the useful energy transferred to the food to the energy consumed by the appliance Fig 256 Electric ranges are more efficient than gas ranges but it is much cheaper to cook with natural gas than with electricity because of the lower unit cost of natural gas Table 22 The cooking efficiency depends on user habits as well as the individual appliances Convection and microwave ovens are inherently more efficient than conventional ovens On average convection ovens save about onethird and microwave ovens save about twothirds of the energy used by conven tional ovens The cooking efficiency can be increased by using the smallest oven for baking using a pressure cooker using an electric slow cooker for stews and soups using the smallest pan that will do the job using the smaller heating element for small pans on electric ranges using flatbottomed pans on electric burners to assure good contact keeping burner drip pans clean and shiny defrosting frozen foods in the refrigerator before cooking avoiding preheating unless it is necessary keeping the pans covered during cooking using timers and thermometers to avoid overcooking using the selfcleaning feature of ovens right after cooking and keeping inside surfaces of microwave ovens clean This value depends on the spectral distribution of the assumed ideal light source For white light sources the upper limit is about 300 lmW for metal halide 350 lmW for fluorescents and 400 lmW for LEDs Spectral maximum occurs at a wavelength of 555 nm green with a light output of 683 lmW TABLE 21 The efficacy of different lighting systems Type of lighting Efficacy lumensW Combustion Candle Kerosene lamp 03 12 Incandescent Ordinary Halogen 620 1535 Fluorescent Compact Tube 4087 60120 Highintensity discharge Mercury vapor Metal halide Highpressure sodium Lowpressure sodium 4060 65118 85140 70200 SolidState LED OLED 20160 1560 Theoretical limit 300 Assumes a unit cost of 0095kWh for electricity and 120therm for gas From J T Amann A Wilson and K Ackerly Consumer Guide to Home Energy Savings 9th ed American Council for an EnergyEfficient Economy Washington DC 2007 p 163 TABLE 22 Energy costs of cooking a casserole with different appliances Cooking appliance Cooking temperature Cooking time Energy used Cost of energy Electric oven 350F 177C 1 h 20 kWh 019 Convection oven elect 325F 163C 45 min 139 kWh 013 Gas oven 350F 177C 1 h 0112 therm 013 Frying pan 420F 216C 1 h 09 kWh 009 Toaster oven 425F 218C 50 min 095 kWh 009 Crockpot 200F 93C 7 h 07 kWh 007 Microwave oven High 15 min 036 kWh 003 FIGURE 255 A 15W compact fluorescent lamp provides as much light as a 60W incandescent lamp 15 W 60 W Final PDF to printer 81 CHAPTER 2 cen22672ch02051108indd 81 110317 0749 AM Using energyefficient appliances and practicing energy conservation mea sures help our pocketbooks by reducing our utility bills This also helps the environment by reducing the amount of pollutants emitted to the atmosphere during the combustion of fuel at home or at the power plants where electricity is generated The combustion of each therm of natural gas produces 64 kg of carbon dioxide which causes global climate change 47 g of nitrogen oxides and 054 g of hydrocarbons which cause smog 20 g of carbon monoxide which is toxic and 0030 g of sulfur dioxide which causes acid rain Each therm of natural gas saved eliminates the emission of these pollutants while saving 060 for the average consumer in the United States Each kWh of electricity conserved saves 04 kg of coal and 10 kg of CO2 and 15 g of SO2 from a coal power plant FIGURE 256 The efficiency of a cooking appliance represents the fraction of the energy supplied to the appliance that is transferred to the food 5 kW 3 kW 2 kW Efficiency Energy utilized Energy supplied to appliance 3 kWh 5 kWh 060 EXAMPLE 214 Cost of Cooking with Electric and Gas Ranges The efficiency of cooking appliances affects the internal heat gain from them since an inefficient appliance consumes a greater amount of energy for the same task and the excess energy consumed shows up as heat in the living space The efficiency of open burners is determined to be 73 percent for electric units and 38 percent for gas units Fig 257 Consider a 2kW electric burner at a location where the unit costs of electricity and natural gas are 012kWh and 120therm respectively Determine the rate of energy consumption by the burner and the unit cost of utilized energy for both electric and gas burners SOLUTION The operation of electric and gas ranges is considered The rate of energy consumption and the unit cost of utilized energy are to be determined Analysis The efficiency of the electric heater is given to be 73 percent Therefore a burner that consumes 2 kW of electrical energy will supply Q utilized Energy input Efficiency 2 kW 073 146 kW of useful energy The unit cost of utilized energy is inversely proportional to the effi ciency and is determined from Cost of utilized energy Cost of energy input Efficiency 012 kWh 073 0164 kWh Noting that the efficiency of a gas burner is 38 percent the energy input to a gas burner that supplies utilized energy at the same rate 146 kW is Q inputgas Q utilized Efficiency 146 kW 038 384 kW 13100 Btuh since 1 kW 3412 Btuh Therefore a gas burner should have a rating of at least 13100 Btuh to perform as well as the electric unit Noting that 1 therm 293 kWh the unit cost of utilized energy in the case of a gas burner is determined to be Cost of utilized energy Cost of energy input Efficiency 120 293 kWh 038 0108 kWh FIGURE 257 Schematic of the 73 percent efficient electric heating unit and 38 percent efficient gas burner discussed in Example 214 Gas Range Electric Range 73 38 Final PDF to printer 82 ENERGY ENERGY TRANSFER cen22672ch02051108indd 82 110317 0749 AM Efficiencies of Mechanical and Electrical Devices The transfer of mechanical energy is usually accomplished by a rotating shaft and thus mechanical work is often referred to as shaft work A pump or a fan receives shaft work usually from an electric motor and transfers it to the fluid as mechanical energy less frictional losses A turbine on the other hand converts the mechanical energy of a fluid to shaft work In the absence of any irreversibilities such as friction mechanical energy can be converted entirely from one mechanical form to another and the mechanical efficiency of a device or process can be defined as Fig 258 η mech Mechanical energy output Mechanical energy input E mechout E mechin 1 E mechloss E mechin 244 A conversion efficiency of less than 100 percent indicates that conversion is less than perfect and some losses have occurred during conversion A mechanical efficiency of 97 percent indicates that 3 percent of the mechanical energy input is converted to thermal energy as a result of frictional heating and this will manifest itself as a slight rise in the temperature of the fluid In fluid systems we are usually interested in increasing the pressure veloc ity andor elevation of a fluid This is done by supplying mechanical energy to the fluid by a pump a fan or a compressor we will refer to all of them as pumps Or we are interested in the reverse process of extracting mechanical energy from a fluid by a turbine and producing mechanical power in the form of a rotating shaft that can drive a generator or any other rotary device The degree of perfection of the conversion process between the mechanical work supplied or extracted and the mechanical energy of the fluid is expressed by the pump efficiency and turbine efficiency defined as η pump Mechanical energy increase of the fluid Mechanical energy input Δ E mechfluid W shaftin W pumpu W pump 245 where Δ E mechfluid E mechout E mechin is the rate of increase in the mechanical energy of the fluid which is equivalent to the useful pumping power W pumpu supplied to the fluid and η turbine Mechanical energy output Mechanical energy decrease of the fluid W shaftout Δ E mechfluid W turbine W turbinee 246 where Δ E mechfluid E mechin E mechout is the rate of decrease in the mechani cal energy of the fluid which is equivalent to the mechanical power extracted from the fluid by the turbine W turbinee and we use the absolute value sign to avoid negative values for efficiencies A pump or turbine efficiency of 100 percent indicates perfect conversion between the shaft work and the mechanical energy of the fluid and this value can be approached but never attained as the frictional effects are minimized Discussion The cost of utilized gas is less than that of utilized electricity There fore despite its higher efficiency cooking with an electric burner will cost about 52 percent more compared to a gas burner in this case This explains why cost conscious consumers always ask for gas appliances and it is not wise to use electric ity for heating purposes FIGURE 258 The mechanical efficiency of a fan is the ratio of the rate of increase of the mechanical energy of air to the mechanical power input m 0506 kgs Fan 500 W 1 2 V1 0 V2 121 ms z1 z2 P1 Patm and P2 Patm 0741 ηmechfan ΔEmechfluid Wshaftin 0506 kgs121 ms22 500 W mV 2 22 Wshaftin Final PDF to printer 83 CHAPTER 2 cen22672ch02051108indd 83 110317 0749 AM Electrical energy is commonly converted to rotating mechanical energy by electric motors to drive fans compressors robot arms car starters and so forth The effectiveness of this conversion process is characterized by the motor efficiency ηmotor which is the ratio of the mechanical energy output of the motor to the electrical energy input The fullload motor efficiencies range from about 35 percent for small motors to over 97 percent for large highefficiency motors The difference between the electrical energy con sumed and the mechanical energy delivered is dissipated as waste heat The mechanical efficiency should not be confused with the motor efficiency and the generator efficiency which are defined as Motor η motor Mechanical power output Electric power input W shaftout W electin 247 and Generator η generator Electric power output Mechanical power input W electout W shaftin 248 A pump is usually packaged together with its motor and a turbine with its generator Therefore we are usually interested in the combined or overall efficiency of pumpmotor and turbinegenerator combinations Fig 259 which are defined as η pumpmotor η pump η motor W pumpu W electin Δ E mechfluid W electin 249 and η turbinegen η turbine η generator W electout W turbinee W electout Δ E mechfluid 250 All the efficiencies just defined range between 0 and 100 percent The lower limit of 0 percent corresponds to the conversion of the entire mechanical or electric energy input to thermal energy and the device in this case functions like a resistance heater The upper limit of 100 percent corresponds to the case of perfect conversion with no friction or other irreversibilities and thus no conversion of mechanical or electric energy to thermal energy FIGURE 260 Schematic for Example 215 2 1 h 70 m h 70 m m 1500 kgs m 1500 kgs Generator Generator Turbine Turbine EXAMPLE 215 Power Generation from a Hydroelectric Plant Electric power is to be generated by installing a hydraulic turbinegenerator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kgs steadily Fig 260 If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW determine the turbine efficiency and the combined turbinegenerator efficiency of this plant Neglect losses in the pipes SOLUTION A hydraulic turbinegenerator installed at a large reservoir is to gener ate electricity The combined turbinegenerator efficiency and the turbine efficiency are to be determined Assumptions 1 The water elevation in the reservoir remains constant 2 The mechanical energy of water at the turbine exit is negligible FIGURE 259 The overall efficiency of a turbine generator is the product of the efficiency of the turbine and the efficiency of the generator and it represents the fraction of the mechanical power of the fluid converted to electrical power Welect out Turbine Generator ηturbine 075 ηgenerator 097 075 097 073 ηturbinegen ηturbineηgenerator Final PDF to printer 84 ENERGY ENERGY TRANSFER cen22672ch02051108indd 84 110317 0749 AM Analysis We take the free surface of water in the reservoir to be point 1 and the turbine exit to be point 2 We also take the turbine exit as the reference level z2 0 so that the potential energies at 1 and 2 are pe1 gz1 and pe2 0 The flow energy Pρ at both points is zero since both 1 and 2 are open to the atmosphere P1 P2 Patm Further the kinetic energy at both points is zero ke1 ke2 0 since the water at point 1 is essentially motionless and the kinetic energy of water at turbine exit is assumed to be negligible The potential energy of water at point 1 is pe 1 g z 1 981 ms 2 70 m 1 kJkg 1000 m 2 s 2 0687 kJkg Then the rate at which the mechanical energy of water is supplied to the turbine becomes Δ E mechfluid m e mechin e mechout m pe 1 0 m pe 1 1500 kgs0687 kJkg 1031 kW The combined turbinegenerator and the turbine efficiency are determined from their definitions to be η turbine gen W electout Δ E mechfluid 750 kW 1031 kW 0727 or 727 η turbine W shaftout E mechfluid 800 kW 1031 kW 0776 or 776 Therefore the reservoir supplies 1031 kW of mechanical energy to the turbine which converts 800 kW of it to shaft work that drives the generator which then generates 750 kW of electric power Discussion This problem can also be solved by taking point 1 to be at the turbine inlet and using flow energy instead of potential energy It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir FIGURE 261 Schematic for Example 216 Standard motor 60 hp η 890 Highefficiency motor 60 hp η 932 EXAMPLE 216 Cost Savings Associated with HighEfficiency Motors A 60hp electric motor a motor that delivers 60 hp of shaft power at full load that has an efficiency of 890 percent is worn out and is to be replaced by a 932 percent efficient highefficiency motor Fig 261 The motor operates 3500 hours a year at full load Taking the unit cost of electricity to be 008kWh determine the amount of energy and money saved as a result of installing the highefficiency motor instead of the standard motor Also determine the simple payback period if the purchase prices of the standard and highefficiency motors are 4520 and 5160 respectively SOLUTION A wornout standard motor is to be replaced by a highefficiency one The amount of electrical energy and money saved as well as the simple payback period are to be determined Assumptions The load factor of the motor remains constant at 1 full load when operating Final PDF to printer 85 CHAPTER 2 cen22672ch02051108indd 85 110317 0749 AM 28 ENERGY AND ENVIRONMENT The conversion of energy from one form to another often affects the environ ment and the air we breathe in many ways and thus the study of energy is not complete without considering its impact on the environment Fig 262 Fos sil fuels such as coal oil and natural gas have been powering the industrial development and the amenities of modern life that we enjoy since the 1700s but this has not been without any undesirable side effects From the soil we farm and the water we drink to the air we breathe the environment has been paying a heavy toll for it Pollutants emitted during the combustion of fossil fuels are responsible for smog acid rain global warming and climate change The environmental pollution has reached such high levels that it has become a serious threat to vegetation wildlife and human health Air pollution has been the cause of numerous health problems including asthma and cancer Analysis The electric power drawn by each motor and their difference can be expressed as W electric in standard W shaft η st Rated power Load factor η st W electric in efficient W shaft η eff Rated power Load factor η eff Power savings W electric in standard W electric in efficient Rated power Load factor 1 η st 1 η eff where ηst is the efficiency of the standard motor and ηeff is the efficiency of the com parable highefficiency motor Then the annual energy and cost savings associated with the installation of the highefficiency motor become Energy savings Power savings Operating hours Rated powerOperating hoursLoad factor1 η st 1 η eff 60 hp07457 kWhp3500 hyear11 089 1 0932 7929 kWhyear Cost savings Energy savings Unit cost of energy 7929 kWhyear 008 kWh 634 year Also Excess initial cost Purchase price differential 5160 4520 640 This gives a simple payback period of Simple payback period Excess initial cost Annual cost savings 640 634 year 101 year Discussion Note that the highefficiency motor pays for its price differential within about one year from the electrical energy it saves Considering that the service life of electric motors is several years the purchase of the higher efficiency motor is definitely indicated in this case FIGURE 262 Energy conversion processes are often accompanied by environmental pollution Comstock ImagesAlamy RF Final PDF to printer 86 ENERGY ENERGY TRANSFER cen22672ch02051108indd 86 110317 0749 AM It is estimated that over 60000 people in the United States alone die each year due to heart and lung diseases related to air pollution Hundreds of elements and compounds such as benzene and formaldehyde are known to be emitted during the combustion of coal oil natural gas and wood in electric power plants engines of vehicles furnaces and even fire places Some compounds are added to liquid fuels for various reasons such as MTBE to raise the octane number of the fuel and also to oxygenate the fuel in winter months to reduce urban smog The largest source of air pol lution is the motor vehicles and the pollutants released by the vehicles are usually grouped as hydrocarbons HC nitrogen oxides NOx and carbon monoxide CO Fig 263 The HC emissions make up a large portion of volatile organic compound VOC emissions and the two terms are generally used interchangeably for motor vehicle emissions A significant portion of the VOC or HC emissions is caused by the evaporation of fuels during refueling or spillage during spitback or by evaporation from gas tanks with faulty caps that do not close tightly The solvents propellants and household cleaning products that contain benzene butane or other HC products are also signifi cant sources of HC emissions The increase of environmental pollution at alarming rates and the rising awareness of its dangers made it necessary to control it by legislation and international treaties In the United States the Clean Air Act of 1970 whose passage was aided by the 14day smog alert in Washington that year set lim its on pollutants emitted by large plants and vehicles These early standards focused on emissions of hydrocarbons nitrogen oxides and carbon monox ide New cars were required to have catalytic converters in their exhaust sys tems to reduce HC and CO emissions As a side benefit the removal of lead from gasoline to permit the use of catalytic converters led to a significant reduction in toxic lead emissions Emission limits for HC NOx and CO from cars have been declining steadily since 1970 The Clean Air Act of 1990 made the requirements on emissions even tougher primarily for ozone CO nitrogen dioxide and particulate mat ter PM As a result todays industrial facilities and vehicles emit a fraction of the pollutants they used to emit a few decades ago The HC emissions of cars for example decreased from about 8 gpm grams per mile in 1970 to 033 gpm in 2010 This is a significant reduction since many of the gaseous toxics from motor vehicles and liquid fuels are hydrocarbons Children are most susceptible to the damages caused by air pollutants since their organs are still developing They are also exposed to more pollution since they are more active and thus they breathe faster People with heart and lung problems especially those with asthma are most affected by air pollut ants This becomes apparent when the air pollution levels in their neighbor hoods rise to high levels Ozone and Smog If you live in a metropolitan area such as Los Angeles you are probably famil iar with urban smogthe dark yellow or brown haze that builds up in a large stagnant air mass and hangs over populated areas on calm hot summer days Smog is made up mostly of groundlevel ozone O3 but it also contains many other chemicals including carbon monoxide CO particulate matter such FIGURE 263 Motor vehicles are the largest source of air pollution NOx CO HC Final PDF to printer 87 CHAPTER 2 cen22672ch02051108indd 87 110317 0749 AM as soot and dust and volatile organic compounds VOCs such as benzene butane and other hydrocarbons The harmful groundlevel ozone should not be confused with the useful ozone layer high in the stratosphere that protects the earth from the suns harmful ultraviolet rays Ozone at ground level is a pollutant with several adverse health effects The primary source of both nitrogen oxides and hydrocarbons is motor vehicles Hydrocarbons and nitrogen oxides react in the presence of sunlight on hot calm days to form groundlevel ozone Fig 264 Smog formation usually peaks in late afternoons when the temperatures are highest and there is plenty of sunlight Although groundlevel smog and ozone form in urban areas with heavy traffic or industry the prevailing winds can transport them several hundred miles to other cities This shows that pollution knows no boundaries and it is a global problem Ozone irritates eyes and damages the air sacs in the lungs where oxygen and carbon dioxide are exchanged causing eventual hardening of this soft and spongy tissue It also causes shortness of breath wheezing fatigue head aches and nausea and it aggravates respiratory problems such as asthma Every exposure to ozone does a little damage to the lungs just like cigarette smoke eventually reducing a persons lung capacity Staying indoors and minimizing physical activity during heavy smog minimizes damage Ozone also harms vegetation by damaging leaf tissues To improve the air quality in areas with the worst ozone problems reformulated gasoline RFG that con tains at least 2 percent oxygen was introduced The use of RFG has resulted in a significant reduction in the emission of ozone and other pollutants and its use is mandatory in many smogprone areas The other serious pollutant in smog is carbon monoxide which is a color less odorless poisonous gas It is mostly emitted by motor vehicles and it can build to dangerous levels in areas with heavy congested traffic It deprives the bodys organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen At low levels carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles slows body reactions and reflexes and impairs judgment It poses a serious threat to people with heart disease because of the fragile condition of the circulatory sys tem and to fetuses because of the oxygen needs of the developing brain At high levels it can be fatal as evidenced by the many deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars Smog also contains suspended particulate matter such as dust and soot emitted by vehicles and industrial facilities Such particles irritate the eyes and the lungs since they may carry compounds such as acids and metals Acid Rain Fossil fuels are mixtures of various chemicals including small amounts of sulfur The sulfur in the fuel reacts with oxygen to form sulfur dioxide SO2 which is an air pollutant The main source of SO2 is the electric power plants that burn highsulfur coal The Clean Air Act of 1970 has limited the SO2 emissions severely which forced the plants to install SO2 scrubbers to switch to lowsulfur coal or to gasify the coal and recover the sulfur Motor vehicles also contribute to SO2 emissions since gasoline and diesel fuel also contain small amounts of sulfur Volcanic eruptions and hot springs also release sul fur oxides the cause of the rotten egg smell FIGURE 264 Groundlevel ozone which is the primary component of smog forms when HC and NOx react in the presence of sunlight on hot calm days O3 NOx HC Smog Sun Final PDF to printer 88 ENERGY ENERGY TRANSFER cen22672ch02051108indd 88 110317 0749 AM The sulfur oxides and nitric oxides react with water vapor and other chemi cals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids Fig 265 The acids formed usually dissolve in the suspended water droplets in clouds or fog These acidladen droplets which can be as acidic as lemon juice are washed from the air onto the soil by rain or snow This is known as acid rain Soil can neutralize a certain amount of acid but the amounts produced by burning highsulfur coal are more than the soil can handle As a result many lakes and rivers in industrial areas such as New York Pennsylvania and Michigan have become too acidic for fish to grow Forests in those areas also experience a slow death because they absorb the acids through their leaves needles and roots Even marble structures deterio rate due to acid rain The magnitude of the problem was not recognized until the early 1970s and measures have been taken since then to reduce the sulfur dioxide emissions drastically by installing scrubbers in power plants and by desulfurizing coal before combustion The Greenhouse Effect Global Warming and Climate Change You have probably noticed that when you leave your car under direct sun light on a sunny day the interior of the car gets much warmer than the air outside and you may have wondered why the car acts like a heat trap This is because glass at thicknesses encountered in practice transmits over 90 percent of radiation in the visible range and is practically opaque non transparent to radiation in the longer wavelength infrared regions There fore glass allows the solar radiation to enter freely but blocks the infrared radiation emitted by the interior surfaces This causes a rise in the interior temperature as a result of the thermal energy buildup in the car This heating effect is known as the greenhouse effect because it is exploited primarily in greenhouses The greenhouse effect is also experienced on a larger scale on earth The surface of the earth which warms up during the day as a result of the absorp tion of solar energy cools down at night by radiating part of its energy into deep space as infrared radiation Carbon dioxide CO2 water vapor and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth Fig 266 Therefore these gases are called greenhouse gases with CO2 being the primary component Water vapor is usually taken out of this list since it comes down as rain or snow as part of the water cycle and because human activities that produce water such as the burning of fossil fuels do not have much impact on its concentration in the atmosphere which is mostly due to evaporation from rivers lakes and oceans CO2 is different however in that peoples activities do make a difference in CO2 concentration in the atmosphere The greenhouse effect makes life on earth possible by keeping the earth warm about 30C warmer However excessive amounts of these gases disturb the delicate balance by trapping too much energy which causes the average temperature of the earth to rise and the climate at some localities to change These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change FIGURE 265 Sulfuric acid and nitric acid are formed when sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight Sun FIGURE 266 The greenhouse effect on earth Solar radiation passes through and is mostly absorbed by earths surface Some infrared radiation emitted by earth is absorbed by greenhouse gases and emitted back Greenhouse gases Sun Final PDF to printer 89 CHAPTER 2 cen22672ch02051108indd 89 110317 0749 AM Global climate change is due to the excessive use of fossil fuels such as coal petroleum products and natural gas in electric power generation transporta tion buildings and manufacturing and it has been a concern in recent decades In 2016 a total of 99 billion tons of carbon was released to the atmosphere as CO2 The current concentration of CO2 in the atmosphere is about 400 ppm or 004 percent This is 30 percent higher than the level a century ago and it is projected to increase to over 700 ppm by the year 2100 Under normal conditions vegetation consumes CO2 and releases O2 during photosynthesis and thus it keeps the CO2 concentration in the atmosphere in check A mature growing tree consumes about 12 kg of CO2 a year and exhales enough oxygen to support a family of four However deforestation and the huge increase in CO2 production in recent decades has disturbed this balance In a 1995 report the worlds leading climate scientists concluded that the earth has already warmed about 05C during the last century and they esti mate that the earths temperature will rise another 2C by the year 2100 It is feared that a rise of this magnitude could cause severe changes in weather patterns with storms heavy rains and flooding in some places and drought in others major floods due to the melting of ice at the poles a loss of wet lands and coastal areas due to rising sea levels variations in water supply changes in the ecosystem due to the inability of some animal and plant spe cies to adjust to the changes increases in epidemic diseases due to the warmer temperatures and adverse side effects on human health and socioeconomic conditions in some areas The seriousness of these threats has moved the United Nations to establish a committee on climate change A world summit in 1992 in Rio de Janeiro Brazil attracted world attention to the problem The agreement prepared by the committee in 1992 to control greenhouse gas emissions was signed by 162 nations In the 1997 meeting in Kyoto Japan the worlds industrialized countries adopted the Kyoto Protocol and committed to reduce their CO2 and other greenhouse gas emissions by 5 percent below the 1990 levels by 2008 to 2012 In December 2011 countries agreed in Durban South Africa to forge a new deal forcing the biggest polluting countries to limit greenhouse gas emis sions The Kyoto Protocol was extended to allow five more years to finalize a wider agreement The goal was to produce a new legally binding accord to cut greenhouse gas emissions that would be completed by 2015 and would come into force by 2020 In 2015 the United Nations Climate Change Conference was held in Paris France resulting in the Paris Agreement on the reduction of climate change The conference included participants from 196 nations The main result of the conference was the establishment of a goal to limit global warming to less than 2C compared to preindustrial times According to the agreement humanmade greenhouse emissions should be eliminated during the second half of the 21st century Greenhouse gas emissions can be reduced by increasing conservation efforts and improving conversion efficiencies while new energy demands should be met by the use of renewable energy such as hydroelectric solar wind and geothermal energy rather than by fossil fuels The United States is the largest contributor of greenhouse gases with over 5 tons of carbon emissions per person per year Major sources of greenhouse gas emissions are the industrial sector and transportation Each kilowatthour Final PDF to printer 90 ENERGY ENERGY TRANSFER cen22672ch02051108indd 90 110317 0749 AM of electricity produced by a fossilfueled power plant produces 06 to 10 kg 13 to 22 lbm of carbon dioxide Each liter of gasoline burned by a vehicle produces about 25 kg of CO2 or each gallon of gasoline burned produces about 20 lbm of CO2 An average car in the United States is driven about 13500 miles a year and it consumes about 600 gallons of gasoline Therefore a car emits about 12000 lbm of CO2 to the atmosphere a year which is about four times the weight of a typical car Fig 267 This and other emissions can be reduced significantly by buying an energyefficient car that burns less fuel over the same distance and by driving sensibly Saving fuel also saves money and the environment For example choosing a vehicle that gets 30 rather than 20 miles per gallon will prevent 2 tons of CO2 from being released to the atmo sphere every year while reducing the fuel cost by 500 per year under average driving conditions of 13500 miles a year and at a fuel cost of 22gal Considerable amounts of pollutants are emitted as the chemical energy in fossil fuels is converted to thermal mechanical or electrical energy by combustion and thus power plants motor vehicles and even stoves take the blame for air pollution In contrast no pollution is emitted as electric ity is converted to thermal chemical or mechanical energy and thus elec tric cars are often touted as zero emission vehicles and their widespread use is seen by some as the ultimate solution to the air pollution problem It should be remembered however that the electricity used by electric cars is generated somewhere else mostly by burning fuel and thus emitting pollu tion Therefore each time an electric car consumes 1 kWh of electricity it bears the responsibility for the pollution emitted as 1 kWh of electricity plus the conversion and transmission losses is generated elsewhere Electric cars can truly be zeroemission vehicles only when the electricity they consume is generated by emissionfree renewable resources such as hydroelectric solar wind and geothermal energy Fig 268 Therefore the use of renewable energy should be encouraged worldwide with incentives as necessary to make the earth a better place to live in The advancements in thermodynamics have contributed greatly in recent decades to improve conversion efficiencies in some cases doubling them and thus to reduce pollution As individuals we can also help by practicing energy conservation measures and by making energy efficiency a high priority in our purchases EXAMPLE 217 Reducing Air Pollution by Geothermal Heating A geothermal power plant in Nevada is generating electricity using geothermal water extracted at 180C and injected back into the ground at 85C It is proposed to use the injected brine to heat the residential and commercial buildings in the area and calcu lations show that the geothermal heating system can save 18 million therms of natural gas a year Determine the amount of NOx and CO2 emissions the geothermal sys tem will save each year Take the average NOx and CO2 emissions of gas furnaces to be 00047 kgtherm and 64 kgtherm respectively SOLUTION The gas heating systems in an area are being replaced by a geothermal district heating system The amounts of NOx and CO2 emissions saved per year are to be determined FIGURE 267 The average car produces several times its weight in CO2 every year it is driven 13500 miles a year consumes 600 gallons of gasoline and produces 20 lbm of CO2 per gallon milehightraveleriStockphotoGetty Images RF FIGURE 268 Renewable energies such as wind are called green energy since they emit no pollutants or greenhouse gases Bear Dancer StudiosMark Dierker RF Final PDF to printer 91 CHAPTER 2 cen22672ch02051108indd 91 110317 0749 AM Analysis The amounts of emissions saved per year are equivalent to the amounts emitted by furnaces when 18 million therms of natural gas are burned NO x savings NO x emission per therm No of therms per year 00047 kgtherm18 10 6 thermyear 85 10 4 kgyear C O 2 savings C O 2 emission per thermNo of therms per year 64 kgtherm18 10 6 thermyear 12 10 8 kgyear Discussion A typical car on the road generates about 85 kg of NOx and 6000 kg of CO2 a year Therefore the environmental impact of replacing the gas heating systems in the area with the geothermal heating system is equivalent to taking 10000 cars off the road for NOx emission and taking 20000 cars off the road for CO2 emission The proposed system should have a significant effect on reducing smog in the area FIGURE 269 Heat conduction from warm air to a cold canned drink through the wall of the aluminum can Cola T1 Heat Wall of aluminum can Δx Air T2 Cola Heat Air ΔT Heat can be transferred in three different ways conduction convection and radiation We will give a brief description of each mode to familiarize you with the basic mechanisms of heat transfer All modes of heat transfer require the existence of a temperature difference and all modes of heat transfer are from the hightemperature medium to a lower temperature one Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles Conduction can take place in solids liquids or gases In gases and liquids conduction is due to the collisions of the molecules during their random motion In solids it is due to the combination of vibrations of mol ecules in a lattice and energy transport by free electrons A cold canned drink in a warm room for example eventually warms up to the room temperature as a result of heat transfer from the room to the drink through the aluminum can by conduction Fig 269 It is observed that the rate of heat conduction Q cond through a layer of con stant thickness Δx is proportional to the temperature difference ΔT across the layer and the area A normal to the direction of heat transfer and is inversely proportional to the thickness of the layer Therefore Q cond kA ΔT Δx W 251 where the constant of proportionality k is the thermal conductivity of the material which is a measure of the ability of a material to conduct heat Table 23 Materials such as copper and silver which are good electric conductors are also good heat conductors and therefore have high k values TOPIC OF SPECIAL INTEREST Mechanisms of Heat Transfer This section can be skipped without a loss in continuity Final PDF to printer 92 ENERGY ENERGY TRANSFER cen22672ch02051108indd 92 110317 0749 AM Materials such as rubber wood and styrofoam are poor conductors of heat and therefore have low k values In the limiting case of Δx 0 the preceding equation reduces to the dif ferential form Q cond kA dT dx W 252 which is known as Fouriers law of heat conduction It indicates that the rate of heat conduction in a direction is proportional to the temperature gra dient in that direction Heat is conducted in the direction of decreasing tem perature and the temperature gradient becomes negative when temperature decreases with increasing x Therefore a negative sign is added in Eq 252 to make heat transfer in the positive x direction a positive quantity Temperature is a measure of the kinetic energies of the molecules In a liquid or gas the kinetic energy of the molecules is due to the random motion of the molecules as well as the vibrational and rotational motions When two mol ecules possessing different kinetic energies collide part of the kinetic energy of the more energetic higher temperature molecule is transferred to the less energetic lower temperature particle in much the same way as when two elastic balls of the same mass at different velocities collide part of the kinetic energy of the faster ball is transferred to the slower one In solids heat conduction is due to two effects the lattice vibrational waves induced by the vibrational motions of the molecules positioned at relatively fixed position in a periodic manner called a lattice and the energy trans ported via the free flow of electrons in the solid The thermal conductivity of a solid is obtained by adding the lattice and the electronic components The thermal conductivity of pure metals is primarily due to the electronic component whereas the thermal conductivity of nonmetals is primarily due to the lattice component The lattice component of thermal conductivity strongly depends on the way the molecules are arranged For example the thermal conductivity of diamond which is a highly ordered crystalline solid is much higher than the thermal conductivities of pure metals as can be seen from Table 23 Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion and it involves the combined effects of conduction and fluid motion The faster the fluid motion the greater the convection heat transfer In the absence of any bulk fluid motion heat transfer between a solid surface and the adjacent fluid is by pure conduction The pres ence of bulk motion of the fluid enhances the heat transfer between the solid surface and the fluid but it also complicates the determination of heat transfer rates Consider the cooling of a hot block by blowing cool air over its top surface Fig 270 Energy is first transferred to the air layer adjacent to the surface of the block by conduction This energy is then carried away from the surface by convectionthat is by the combined effects of conduction within the air which is due to random motion of air molecules and the bulk or macroscopic motion of the air which removes the heated air near the surface and replaces it with the cooler air TABLE 23 Thermal conductivities of some materials at room conditions Material Thermal conductivity WmK Diamond 2300 Silver 429 Copper 401 Gold 317 Aluminum 237 Iron 802 Mercury l 854 Glass 14 Brick 072 Water l 0613 Human skin 037 Wood oak 017 Helium g 0152 Soft rubber 013 Glass fiber 0043 Air g 0026 Urethane rigid foam 0026 FIGURE 270 Heat transfer from a hot surface to air by convection Velocity variation of air V A Hot block Temperature variation of air Air flow T Ts Tf Qconv Final PDF to printer 93 CHAPTER 2 cen22672ch02051108indd 93 110317 0749 AM Convection is called forced convection if the fluid is forced to flow in a tube or over a surface by external means such as a fan pump or the wind In contrast convection is called free or natural convection if the fluid motion is caused by buoyancy forces induced by density differences due to the varia tion of temperature in the fluid Fig 271 For example in the absence of a fan heat transfer from the surface of the hot block in Fig 270 will be by natu ral convection since any motion in the air in this case will be due to the rise of the warmer and thus lighter air near the surface and the fall of the cooler and thus heavier air to fill its place Heat transfer between the block and sur rounding air will be by conduction if the temperature difference between the air and the block is not large enough to overcome the resistance of air to move and thus to initiate natural convection currents Heat transfer processes that involve change of phase of a fluid are also con sidered to be convection because of the fluid motion induced during the pro cess such as the rise of the vapor bubbles during boiling or the fall of the liquid droplets during condensation The rate of heat transfer by convection Q conv is determined from Newtons law of cooling expressed as Q conv hA T s T f W 253 where h is the convection heat transfer coefficient A is the surface area through which heat transfer takes place Ts is the surface temperature and Tf is bulk fluid temperature away from the surface At the surface the fluid tem perature equals the surface temperature of the solid The convection heat transfer coefficient h is not a property of the fluid It is an experimentally determined parameter whose value depends on all the vari ables that influence convection such as the surface geometry the nature of fluid motion the properties of the fluid and the bulk fluid velocity Typical values of h in Wm2K are in the range of 225 for the free convection of gases 501000 for the free convection of liquids 25250 for the forced convection of gases 5020000 for the forced convection of liquids and 2500100000 for convection in boiling and condensation processes Radiation is the energy emitted by matter in the form of electromagnetic waves or photons as a result of the changes in the electronic configurations of the atoms or molecules Unlike conduction and convection the transfer of energy by radiation does not require the presence of an intervening medium Fig 272 In fact energy transfer by radiation is fastest at the speed of light and it suffers no attenuation in a vacuum This is exactly how the energy of the sun reaches the earth In heat transfer studies we are interested in thermal radiation which is the form of radiation emitted by bodies because of their temperature It dif fers from other forms of electromagnetic radiation such as Xrays gamma rays microwaves radio waves and television waves that are not related to temperature All bodies at a temperature above absolute zero emit thermal radiation Radiation is a volumetric phenomenon and all solids liquids and gases emit absorb or transmit radiation of varying degrees However radiation is usually considered to be a surface phenomenon for solids that are opaque to FIGURE 271 The cooling of a boiled egg by forced and natural convection Forced convection Air hot egg Natural convection Air hot egg FIGURE 272 Unlike conduction and convection heat transfer by radiation can occur between two bodies even when they are separated by a medium colder than both of them Fire 900C Air 5C Person 30C Radiation Final PDF to printer 94 ENERGY ENERGY TRANSFER cen22672ch02051108indd 94 110317 0749 AM FIGURE 273 The absorption of radiation incident on an opaque surface of absorptivity α Qincident Qabs Qincident Qref 1 Qincident α α thermal radiation such as metals wood and rocks since the radiation emitted by the interior regions of such material can never reach the surface and the radiation incident on such bodies is usually absorbed within a few microns from the surface The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts is given by the StefanBoltzmann law as Q emitmax σA T s 4 W 254 where A is the surface area and σ 567 108 Wm2K4 is the Stefan Boltzmann constant The idealized surface that emits radiation at this maxi mum rate is called a blackbody and the radiation emitted by a blackbody is called blackbody radiation The radiation emitted by all real surfaces is less than the radiation emitted by a blackbody at the same temperatures and is expressed as Q emit ɛσA T s 4 W 255 where ɛ is the emissivity of the surface The property emissivity whose value is in the range 0 ɛ 1 is a measure of how closely a surface approximates a blackbody for which ɛ 1 The emissivities of some surfaces are given in Table 24 Another important radiation property of a surface is its absorptivity α which is the fraction of the radiation energy incident on a surface that is absorbed by the surface Like emissivity its value is in the range 0 α 1 A blackbody absorbs the entire radiation incident on it That is a blackbody is a perfect absorber α 1 as well as a perfect emitter In general both ɛ and α of a surface depend on the temperature and the wavelength of the radiation Kirchhoffs law of radiation states that the emis sivity and the absorptivity of a surface are equal at the same temperature and wavelength In most practical applications the dependence of ɛ and α on the temperature and wavelength is ignored and the average absorptivity of a sur face is taken to be equal to its average emissivity The rate at which a surface absorbs radiation is determined from Fig 273 Q abs α Q incident W 256 where Q incident is the rate at which radiation is incident on the surface and α is the absorptivity of the surface For opaque nontransparent surfaces the por tion of incident radiation that is not absorbed by the surface is reflected back The difference between the rates of radiation emitted by the surface and the radiation absorbed is the net radiation heat transfer If the rate of radiation absorption is greater than the rate of radiation emission the surface is said to be gaining energy by radiation Otherwise the surface is said to be losing energy by radiation In general the determination of the net rate of heat trans fer by radiation between two surfaces is a complicated matter since it depends on the properties of the surfaces their orientation relative to each other and the interaction of the medium between the surfaces with radiation However in the special case of a relatively small surface of emissivity ɛ and surface TABLE 24 Emissivity of some materials at 300 K Material Emissivity Aluminum foil 007 Anodized aluminum 082 Polished copper 003 Polished gold 003 Polished silver 002 Polished stainless steel 017 Black paint 098 White paint 090 White paper 092097 Asphalt pavement 085093 Red brick 093096 Human skin 095 Wood 082092 Soil 093096 Water 096 Vegetation 092096 Final PDF to printer 95 CHAPTER 2 cen22672ch02051108indd 95 110317 0749 AM area A at absolute temperature Ts that is completely enclosed by a much larger surface at absolute temperature Tsurr separated by a gas such as air that does not intervene with radiation ie the amount of radiation emitted absorbed or scattered by the medium is negligible the net rate of radiation heat transfer between these two surfaces is determined from Fig 274 Q rad εσA T s 4 T surr 4 W 257 In this special case the emissivity and the surface area of the surrounding sur face do not have any effect on the net radiation heat transfer FIGURE 275 Heat transfer from the person described in Example 218 Qcond Room air 29C 20C Qrad Qconv EXAMPLE 218 Heat Transfer from a Person Consider a person standing in a breezy room at 20C Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 16 m2 and 29C respectively and the convection heat transfer coefficient is 6 Wm2C Fig 275 SOLUTION A person is standing in a breezy room The total rate of heat loss from the person is to be determined Assumptions 1 The emissivity and heat transfer coefficient are constant and uni form 2 Heat conduction through the feet is negligible 3 Heat loss by evaporation is disregarded Analysis The heat transfer between the person and the air in the room will be by convection instead of conduction since it is conceivable that the air in the vicinity of the skin or clothing will warm up and rise as a result of heat transfer from the body initiating natural convection currents It appears that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area m2 per unit temperature difference in K or C between the person and the air away from the person Thus the rate of convection heat transfer from the person to the air in the room is from Eq 253 Q conv hA T s T f 6 Wm 2 C16 m 2 29 20 C 864 W The person will also lose heat by radiation to the surrounding wall surfaces We take the temperature of the surfaces of the walls ceiling and the floor to be equal to the air temperature in this case for simplicity but we recognize that this does not need to be the case These surfaces may be at a higher or lower temperature than the average temperature of the room air depending on the outdoor conditions and the structure of the walls Considering that air does not intervene with radiation and the person is completely enclosed by the surrounding surfaces the net rate of radiation heat transfer from the person to the surrounding walls ceiling and floor is from Eq 257 Q rad εσA T s 4 T surr 4 095567 10 8 Wm 2 K 4 16 m 2 29 273 4 20 273 4 K 4 817 W FIGURE 274 Radiation heat transfer between a body and the inner surfaces of a much larger enclosure that completely surrounds it Qrad Tsurr Large enclosure ε A T Small body s Final PDF to printer 96 ENERGY ENERGY TRANSFER cen22672ch02051108indd 96 110317 0749 AM Note that we must use absolute temperatures in radiation calculations Also note that we used the emissivity value for the skin and clothing at room temperature since the emissivity is not expected to change significantly at a slightly higher temperature Then the rate of total heat transfer from the body is determined by adding these two quantities to be Q total Q conv Q rad 864 817 1681 W The heat transfer would be much higher if the person were not dressed since the exposed surface temperature would be higher Thus an important function of the clothes is to serve as a barrier against heat transfer Discussion In the preceding calculations heat transfer through the feet to the floor by conduction which is usually very small is neglected Heat transfer from the skin by perspiration which is the dominant mode of heat transfer in hot environments is not considered here SUMMARY The sum of all forms of energy of a system is called total energy which consists of internal kinetic and potential energy for simple compressible systems Internal energy rep resents the molecular energy of a system and may exist in sen sible latent chemical and nuclear forms Mass flow rate m is defined as the amount of mass flow ing through a cross section per unit time It is related to the volume flow rate V which is the volume of a fluid flowing through a cross section per unit time by m ρ V ρ A c V avg The energy flow rate associated with a fluid flowing at a rate of m is E m e which is analogous to E me The mechanical energy is defined as the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as an ideal turbine It is expressed on a unit mass basis and rate form as e mech P ρ V 2 2 gz and E mech m e mech m P ρ V 2 2 gz where Pρ is the flow energy V 22 is the kinetic energy and gz is the potential energy of the fluid per unit mass Energy can cross the boundaries of a closed system in the form of heat or work For control volumes energy can also be transported by mass If the energy transfer is due to a tempera ture difference between a closed system and its surroundings it is heat otherwise it is work Work is the energy transferred as a force acts on a system through a distance Various forms of work are expressed as follows Electrical work W e VI Δt Shaft work W sh 2πnT Spring work W spring 1 2 k x 2 2 x 1 2 The first law of thermodynamics is essentially an expres sion of the conservation of energy principle also called the energy balance The general energy balance for any system undergoing any process can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies kJ It can also be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass dE system dt Rate of change in internal kinetic potential etc energies kW The efficiencies of various devices are defined as η pump Δ E mechfluid W shaftin W pumpu W pump Final PDF to printer 97 CHAPTER 2 cen22672ch02051108indd 97 110317 0749 AM REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Fundamentals SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 2 Y A Çengel An Intuitive and Unified Approach to Teach ing Thermodynamics ASME International Mechanical Engineering Congress and Exposition Atlanta Georgia AESVol 36 pp 251260 November 1722 1996 PROBLEMS Forms of Energy 21C What is the difference between the macroscopic and microscopic forms of energy 22C What is total energy Identify the different forms of energy that constitute the total energy 23C List the forms of energy that contribute to the internal energy of a system 24C How are heat internal energy and thermal energy related to each other 25C What is mechanical energy How does it differ from thermal energy What are the forms of mechanical energy of a fluid stream 26C Portable electric heaters are commonly used to heat small rooms Explain the energy transformation involved dur ing this heating process 27C Natural gas which is mostly methane CH4 is a fuel and a major energy source Can we say the same about hydro gen gas H2 28C Consider the falling of a rock off a cliff into seawater and eventually settling at the bottom of the sea Starting with the potential energy of the rock identify the energy transfers and transformations involved during this process 29 Electric power is to be generated by installing a hydraulic turbinegenerator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kgs steadily Determine the power generation potential 210E The specific kinetic energy of a moving mass is given by ke V22 where V is the velocity of the mass Determine the specific kinetic energy of a mass whose velocity is 100 fts in Btulbm Answer 02 Btulbm 211 Determine the specific kinetic energy of a mass whose velocity is 30 ms in kJkg 212E Calculate the total potential energy in Btu of an object that is 20 ft below a datum level at a location where g 317 fts2 and which has a mass of 100 lbm 213 Determine the specific potential energy in kJkg of an object 50 m above a datum in a location where g 98 ms2 214 An object whose mass is 100 kg is located 20 m above a datum level in a location where standard gravitational acceleration exists Determine the total potential energy in kJ of this object 215 A water jet that leaves a nozzle at 60 ms at a flow rate of 120 kgs is to be used to generate power by striking the buckets located on the perimeter of a wheel Determine the power generation potential of this water jet η turbine W shaftout Δ E mechfluid W turbine W turbinee η motor Mechanical power output Electric power input W shaftout W electin η generator Electric power output Mechanical power input W electout W shaftin η pumpmotor η pump η motor Δ E mechfluid W electin η turbinegen η turbine η generator W electout Δ E mechfluid The conversion of energy from one form to another is often associated with adverse effects on the environment and envi ronmental impact should be an important consideration in the conversion and utilization of energy Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch02051108indd 98 110317 0749 AM 98 ENERGY ENERGY TRANSFER 216 Consider a river flowing toward a lake at an average velocity of 3 ms at a rate of 500 m3s at a location 90 m above the lake surface Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location 228E A construction crane lifts a prestressed concrete beam weighing 3 short tons from the ground to the top of piers that are 24 ft above the ground Determine the amount of work done considering a the beam and b the crane as the system Express your answers in both lbfft and Btu 229E Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm 230E A spring whose spring constant is 200 lbfin has an initial force of 100 lbf acting on it Determine the work in Btu required to compress it another 1 in 231 How much work in kJ can a spring whose spring con stant is 3 kNcm produce after it has been compressed 3 cm from its unloaded length 232 A ski lift has a oneway length of 1 km and a vertical rise of 200 m The chairs are spaced 20 m apart and each chair can seat three people The lift is operating at a steady speed of 10 kmh Neglecting friction and air drag and assuming that the average mass of each loaded chair is 250 kg determine the power required to operate this ski lift Also estimate the power required to accelerate this ski lift in 5 s to its operating speed when it is first turned on 233 The engine of a 1500kg automobile has a power rating of 75 kW Determine the time required to accelerate this car from rest to a speed of 100 kmh at full power on a level road Is your answer realistic 234 A damaged 1200kg car is being towed by a truck Neglecting the friction air drag and rolling resistance determine the extra power required a for constant velocity on a level road b for constant velocity of 50 kmh on a 30 from horizontal uphill road and c to accelerate on a level road from stop to 90 kmh in 12 s Answers a 0 b 817 kW c 313 kW 235 As a spherical ammonia vapor bubble rises in liquid ammonia its diameter changes from 1 cm to 3 cm Calculate the amount of work produced by this bubble in kJ if the sur face tension of ammonia is 002 Nm Answer 503 108 kJ 236 A steel rod of 05 cm diameter and 10 m length is stretched 3 cm Youngs modulus for this steel is 21 kNcm2 How much work in kJ is required to stretch this rod The First Law of Thermodynamics 237C What are the different mechanisms for transferring energy to or from a control volume 238C For a cycle is the net work necessarily zero For what kinds of systems will this be the case 239C On a hot summer day a student turns his fan on when he leaves his room in the morning When he returns in the eve ning will the room be warmer or cooler than the neighbor ing rooms Why Assume all the doors and windows are kept closed FIGURE P216 River 3 ms 90 m 217 At a certain location wind is blowing steadily at 10 ms Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60mdiameter blades at that location Take the air density to be 125 kgm3 Energy Transfer by Heat and Work 218C What is the caloric theory When and why was it abandoned 219C In what forms can energy cross the boundaries of a closed system 220C What is an adiabatic process What is an adiabatic system 221C When is the energy crossing the boundaries of a closed system heat and when is it work 222C Consider an automobile traveling at a constant speed along a road Determine the direction of the heat and work interactions taking the following as the system a the car radiator b the car engine c the car wheels d the road and e the air surrounding the car 223C A room is heated by an iron that is left plugged in Is this a heat or work interaction Take the entire room including the iron as the system 224C A room is heated as a result of solar radiation coming in through the windows Is this a heat or work interaction for the room 225C A gas in a pistoncylinder device is compressed and as a result its temperature rises Is this a heat or work interaction 226 A small electrical motor produces 5 W of mechani cal power What is this power in a N m and s units and b kg m and s units Answers a 5 Nms b 5 kgm2s3 Mechanical Forms of Work 227C A car is accelerated from rest to 85 kmh in 10 s Would the energy transferred to the car be different if it were accelerated to the same speed in 5 s Final PDF to printer cen22672ch02051108indd 99 110317 0749 AM 99 CHAPTER 2 240 Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel During the process 30 kJ of heat is transferred to the water and 5 kJ of heat is lost to the surrounding air The paddlewheel work amounts to 500 Nm Determine the final energy of the system if its initial energy is 125 kJ Answer 380 kJ 246 The lighting needs of a storage room are being met by six fluorescent light fixtures each fixture containing four lamps rated at 60 W each All the lamps are on during operat ing hours of the facility which are 6 am to 6 pm 365 days a year The storage room is actually used for an average of 3 h a day If the price of electricity is 011kWh determine the amount of energy and money that will be saved as a result of installing motion sensors Also determine the simple payback period if the purchase price of the sensor is 32 and it takes 1 h to install it at a cost of 40 247 A university campus has 200 classrooms and 400 fac ulty offices The classrooms are equipped with 12 fluorescent tubes each consuming 110 W including the electricity used by the ballasts The faculty offices on average have half as many tubes The campus is open 240 days a year The class rooms and faculty offices are not occupied an average of 4 h a day but the lights are kept on If the unit cost of electricity is 011kWh determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods 248 Consider a room that is initially at the outdoor temper ature of 20C The room contains a 40W lightbulb a 110W TV set a 300W refrigerator and a 1200W iron Assuming no heat transfer through the walls determine the rate of increase of the energy content of the room when all of these electric devices are on 249 An escalator in a shopping center is designed to move 50 people 75 kg each at a constant speed of 06 ms at 45 slope Determine the minimum power input needed to drive this escalator What would your answer be if the escala tor velocity were to be doubled 250 Consider a 2100kg car cruising at constant speed of 70 kmh Now the car starts to pass another car by accelerating to 110 kmh in 5 s Determine the additional power needed to achieve this acceleration What would your answer be if the total mass of the car were only 700 kg Answers 117 kW 389 kW 251E One way to improve the fuel efficiency of a car is to use tires that have a lower rolling resistancetires that roll with less resistance Highway tests at 65 mph showed that tires with the lowest rolling resistance can improve fuel efficiency by nearly 2 mpg miles per gallon Consider a car that gets 35 mpg on highrollingresistance tires and is driven 15000 miles per year For a fuel cost of 35gal determine how much money will be saved per year by switching to lowrollingresistance tires Energy Conversion Efficiencies 252C What is mechanical efficiency What does a mechan ical efficiency of 100 percent mean for a hydraulic turbine 253C How is the combined pumpmotor efficiency of a pump and motor system defined Can the combined pumpmotor effi ciency be greater than either the pump or the motor efficiency FIGURE P240 500 Nm 30 kJ 5 kJ 241 An adiabatic closed system is accelerated from 0 ms to 30 ms Determine the specific energy change of this system in kJkg 242 A fan is to accelerate quiescent air to a velocity of 8 ms at a rate of 9 m3s Determine the minimum power that must be supplied to the fan Take the density of air to be 118 kgm3 Answer 340 W 243E A vertical pistoncylinder device contains water and is being heated on top of a range During the process 65 Btu of heat is transferred to the water and heat losses from the side walls amount to 8 Btu The piston rises as a result of evaporation and 5 Btu of work is done by the vapor Determine the change in the energy of the water for this process Answer 52 Btu 244E At winter design conditions a house is projected to lose heat at a rate of 60000 Btuh The internal heat gain from people lights and appliances is estimated to be 6000 Btuh If this house is to be heated by electric resistance heaters deter mine the required rated power of these heaters in kW to main tain the house at constant temperature 245E A water pump increases the water pressure from 15 psia to 70 psia Determine the power input required in hp to pump 08 ft3s of water Does the water temperature at the inlet have any significant effect on the required flow power Answer 115 hp Final PDF to printer cen22672ch02051108indd 100 110317 0749 AM 100 ENERGY ENERGY TRANSFER 254C Can the combined turbinegenerator efficiency be greater than either the turbine efficiency or the generator effi ciency Explain 255 Consider a 24kW hooded electric open burner in an area where the unit costs of electricity and natural gas are 010kWh and 120therm 1 therm 105500 kJ respec tively The efficiency of open burners can be taken to be 73 percent for electric burners and 38 percent for gas burners Determine the rate of energy consumption and the unit cost of utilized energy for both electric and gas burners 256E The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 55 106 Btuh The combustion efficiency of the boiler is measured to be 07 by a handheld flue gas analyzer After tuning up the boiler the combustion efficiency rises to 08 The boiler operates 4200 h a year intermittently Taking the unit cost of energy to be 13106 Btu determine the annual energy and cost savings as a result of tuning up the boiler 257E Reconsider Prob 256E Using appropriate soft ware study the effects of the unit cost of energy the new combustion efficiency on the annual energy and cost savings Let the efficiency vary from 07 to 09 and let the unit cost vary from 12 to 14 per million Btu Plot the annual energy and cost savings against the efficiency for unit costs of 12 13 and 14 per million Btu and discuss the results 258 A 75hp shaft output motor that has an efficiency of 910 percent is worn out and is replaced by a highefficiency 75hp motor that has an efficiency of 954 percent Determine the reduction in the heat gain of the room due to higher effi ciency under fullload conditions 259 A 90hp shaft output electric car is powered by an elec tric motor mounted in the engine compartment If the motor has an average efficiency of 91 percent determine the rate of heat supply by the motor to the engine compartment at full load 260 An exercise room has six weightlifting machines that have no motors and seven treadmills each equipped with a 25hp shaft output motor The motors operate at an average load factor of 07 at which their efficiency is 077 During peak evening hours all 13 pieces of exercising equipment are used continuously and there are also two people doing light exercises while waiting in line for one piece of the equipment Assuming the average rate of heat dissipation from people in an exercise room is 600 W determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions 261 A room is cooled by circulating chilled water through a heat exchanger located in the room The air is circulated through the heat exchanger by a 025hp shaft output fan Typical efficiency of small electric motors driving 025hp equipment is 60 percent Determine the rate of heat supply by the fanmotor assembly to the room 262 The water in a large lake is to be used to generate elec tricity by the installation of a hydraulic turbinegenerator at a location where the depth of the water is 50 m Water is to be supplied at a rate of 5000 kgs If the electric power gener ated is measured to be 1862 kW and the generator efficiency is 95 percent determine a the overall efficiency of the turbine generator b the mechanical efficiency of the turbine and c the shaft power supplied by the turbine to the generator 263 A 7hp shaft pump is used to raise water to an ele vation of 15 m If the mechanical efficiency of the pump is 82 percent determine the maximum volume flow rate of water 264 A geothermal pump is used to pump brine whose den sity is 1050 kgm3 at a rate of 03 m3s from a depth of 200 m For a pump efficiency of 74 percent determine the required power input to the pump Disregard frictional losses in the pipes and assume the geothermal water at 200 m depth to be exposed to the atmosphere 265 At a certain location wind is blowing steadily at 7 ms Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 80mdiameter blades at that location Also determine the actual electric power generation assuming an overall efficiency of 30 percent Take the air density to be 125 kgm3 266 Reconsider Prob 265 Using appropriate soft ware investigate the effect of wind velocity and the blade span diameter on wind power generation Let the velocity vary from 5 to 20 ms in increments of 5 ms and let the diameter vary from 20 to 120 m in increments of 20 m Tabulate the results and discuss their significance 267 Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of shaft power The free surface of the upper reservoir is 45 m higher than that of the lower reservoir If the flow rate of water is measured to be 003 m3s determine mechanical power that is converted to thermal energy during this process due to frictional effects FIGURE P267 1 2 45 m Pump z1 0 003 m3s 20 kW 268E An 80percentefficient pump with a power input of 20 hp is pumping water from a lake to a nearby pool at a Final PDF to printer cen22672ch02051108indd 101 110317 0749 AM 101 CHAPTER 2 rate of 15 ft3s through a constantdiameter pipe The free surface of the pool is 80 ft above that of the lake Determine the mechanical power used to overcome frictional effects in piping Answer 237 hp 269 Water is pumped from a lake to a storage tank 15 m above at a rate of 70 Ls while consuming 154 kW of elec tric power Disregarding any frictional losses in the pipes and any changes in kinetic energy determine a the overall effi ciency of the pumpmotor unit and b the pressure difference between the inlet and the exit of the pump 273 An oil pump is drawing 44 kW of electric power while pumping oil with ρ 860 kgm3 at a rate of 01 m3s The inlet and outlet diameters of the pipe are 8 cm and 12 cm respec tively If the pressure rise of oil in the pump is measured to be 500 kPa and the motor efficiency is 90 percent determine the mechanical efficiency of the pump FIGURE P269 Pump Storage tank 15 m 270 Large wind turbines with a power capacity of 8 MW and blade span diameters of over 160 m are available for electric power generation Consider a wind turbine with a blade span diameter of 100 m installed at a site subjected to steady winds at 8 ms Taking the overall efficiency of the wind turbine to be 32 percent and the air density to be 125 kgm3 determine the electric power generated by this wind turbine Also assum ing steady winds of 8 ms during a 24h period determine the amount of electric energy and the revenue generated per day for a unit price of 009kWh for electricity 271 A hydraulic turbine has 85 m of elevation difference available at a flow rate of 025 m3s and its overall turbine generator efficiency is 91 percent Determine the electric power output of this turbine 272 The water behind Hoover Dam in Nevada is 206 m higher than the Colorado River below it At what rate must water pass through the hydraulic turbines of this dam to pro duce 50 MW of power if the turbines are 100 percent efficient FIGURE P272 Photo by Lynn Betts USDA Natural Resources Conservation Service FIGURE P273 12 cm Pump ΔP 500 kPa Motor 44 kW 8 cm 01 m3s Oil 274 A wind turbine is rotating at 15 rpm under steady winds flowing through the turbine at a rate of 42000 kgs The tip velocity of the turbine blade is measured to be 250 kmh If 180 kW power is produced by the turbine determine a the average velocity of the air and b the conversion efficiency of the turbine Take the density of air to be 131 kgm3 Energy and Environment 275C How does energy conversion affect the environment What are the primary chemicals that pollute the air What is the primary source of these pollutants 276C What is acid rain Why is it called a rain How do the acids form in the atmosphere What are the adverse effects of acid rain on the environment 277C Why is carbon monoxide a dangerous air pollutant How does it affect human health at low levels and at high levels 278C What is the greenhouse effect How does the excess CO2 gas in the atmosphere cause the greenhouse effect What are the potential longterm consequences of the greenhouse effect How can we combat this problem 279C What is smog What does it consist of How does groundlevel ozone form What are the adverse effects of ozone on human health 280E Consider a household that uses 14000 kWh of elec tricity per year and 900 gal of fuel oil during a heating season The average amount of CO2 produced is 264 lbmgal of fuel oil and 154 lbmkWh of electricity If this household reduces its oil and electricity usage by 15 percent by implementing some energy conservation measures determine the reduction in the amount of CO2 emissions by that household per year 281 When a hydrocarbon fuel is burned almost all of the carbon in the fuel burns completely to form CO2 carbon dioxide which is the principal gas causing the greenhouse Final PDF to printer cen22672ch02051108indd 102 110317 0749 AM 102 ENERGY ENERGY TRANSFER effect and thus global climate change On average 059 kg of CO2 is produced for each kWh of electricity generated from a power plant that burns natural gas A typical new household refrigerator uses about 700 kWh of electricity per year Deter mine the amount of CO2 production that is due to the refrigera tors in a city with 300000 households 282 Repeat Prob 281 assuming the electricity is pro duced by a power plant that burns coal The average produc tion of CO2 in this case is 11 kg per kWh 283 A typical car driven 20000 km a year emits to the atmosphere about 11 kg per year of NOx nitrogen oxides which cause smog in major population areas Natural gas burned in the furnace emits about 43 g of NOx per therm 1 therm 105500 kJ and the electric power plants emit about 71 g of NOx per kWh of electricity produced Consider a household that has two cars and consumes 9000 kWh of elec tricity and 1200 therms of natural gas per year Determine the amount of NOx emission to the atmosphere per year for which this household is responsible 291 The inner and outer surfaces of a 5m 6m brick wall of thickness 30 cm and thermal conductivity 069 WmC are maintained at temperatures of 20C and 5C respectively Determine the rate of heat transfer through the wall in W FIGURE P283 11 kg NOx per year 284E A Ford Taurus driven 12000 miles a year will use about 650 gal of gasoline compared to a Ford Explorer that would use 850 gal About 197 lbm of CO2 which causes global warming is released to the atmosphere when a gallon of gasoline is burned Determine the extra amount of CO2 pro duction a man is responsible for during a 5year period if he trades his Taurus for an Explorer Special Topic Mechanisms of Heat Transfer 285C What are the mechanisms of heat transfer 286C Which is a better heat conductor diamond or silver 287C How does forced convection differ from natural convection 288C What is a blackbody How do real bodies differ from a blackbody 289C Define emissivity and absorptivity What is Kirchhoffs law of radiation 290C Does any of the energy of the sun reach the earth by conduction or convection FIGURE P291 Brick wall 30 cm 20C 5C 292 The inner and outer surfaces of a 05cmthick 2m 2m window glass in winter are 15C and 6C respec tively If the thermal conductivity of the glass is 078 WmC determine the amount of heat loss in kJ that occurs through the glass over a period of 10 h What would your answer be if the glass were 1 cm thick 293 Reconsider Prob 292 Using appropriate soft ware investigate the effect of glass thickness on heat loss for the specified glass surface temperatures Let the glass thickness vary from 02 to 2 cm Plot the heat loss versus the glass thickness and discuss the results 294 An aluminum pan whose thermal conductivity is 237 WmC has a flat bottom whose diameter is 20 cm and thickness 06 cm Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 700 W If the inner surface of the bottom of the pan is 105C determine the tem perature of the outer surface of the bottom of the pan 295 The inner and outer glasses of a 2m 2m double pane window are at 18C and 6C respectively If the 1cm space between the two glasses is filled with still air determine the rate of heat transfer through the air layer by conduction in kW 296 Two surfaces of a 2cmthick plate are maintained at 0C and 100C respectively If it is determined that heat is transferred through the plate at a rate of 500 Wm2 determine its thermal conductivity 297 Hot air at 80C is blown over a 2m 4m flat sur face at 30C If the convection heat transfer coefficient is 55 Wm2C determine the rate of heat transfer from the air to the plate in kW 298 For heat transfer purposes a standing man can be mod eled as a 30cmdiameter 175cmlong vertical cylinder with both the top and bottom surfaces insulated and with the side Final PDF to printer cen22672ch02051108indd 103 110317 0749 AM 103 CHAPTER 2 surface at an average temperature of 34C For a convection heat transfer coefficient of 10 Wm2C determine the rate of heat loss from this man by convection in an environment at 20C Answer 231 W 299 A 9cmdiameter spherical ball whose surface is main tained at a temperature of 110C is suspended in the middle of a room at 20C If the convection heat transfer coefficient is 15 Wm2C and the emissivity of the surface is 08 determine the total rate of heat transfer from the ball 2100 Reconsider Prob 299 Using appropriate soft ware investigate the effect of the convection heat transfer coefficient and surface emissivity on the heat transfer rate from the ball Let the heat transfer coefficient vary from 5 to 30 Wm2C Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 01 05 08 and 1 and discuss the results 2101 A 1000W iron is left on the ironing board with its base exposed to the air at 23C The convection heat transfer coefficient between the base surface and the surrounding air is 20 Wm2C If the base has an emissivity of 04 and a surface area of 002 m2 determine the temperature of the base of the iron 2104 Reconsider Prob 2103 Using appropriate soft ware investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate Let the heat transfer coefficient vary from 10 to 90 Wm2C Plot the surface temperature against the convection heat transfer coeffi cient and discuss the results 2105 The outer surface of a spacecraft in space has an emis sivity of 06 and an absorptivity of 02 for solar radiation If solar radiation is incident on the spacecraft at a rate of 1000 Wm2 determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed 2106 Reconsider Prob 2105 Using appropri ate software investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilib rium surface temperature Plot the surface temperature against emissivity for solar absorptivities of 01 05 08 and 1 and discuss the results 2107 A hollow spherical iron container whose outer diam eter is 40 cm and thickness is 04 cm is filled with iced water at 0C If the outer surface temperature is 3C determine the approximate rate of heat loss from the sphere and the rate at which ice melts in the container FIGURE P2103 α 08 25C 450 Wm2 FIGURE P2101 1000W iron Air 23C 2102 A 7cmexternaldiameter 18mlong hotwater pipe at 80C is losing heat to the surrounding air at 5C by natu ral convection with a heat transfer coefficient of 25 Wm2C Determine the rate of heat loss from the pipe by natural con vection in kW 2103 A thin metal plate is insulated on the back and exposed to solar radiation on the front surface The exposed surface of the plate has an absorptivity of 08 for solar radiation If solar radiation is incident on the plate at a rate of 450 Wm2 and the surrounding air temperature is 25C determine the sur face temperature of the plate when the heat loss by convec tion equals the solar energy absorbed by the plate Assume the FIGURE P2107 Iced water 04 cm 3C convection heat transfer coefficient to be 50 Wm2C and dis regard heat loss by radiation Final PDF to printer cen22672ch02051108indd 104 110317 0749 AM 104 ENERGY ENERGY TRANSFER Review Problems 2108 Some engineers have developed a device that pro vides lighting to rural areas with no access to grid electricity The device is intended for indoor use It is driven by gravity and it works as follows A bag of rock or sand is raised by human power to a higher location As the bag descends very slowly it powers a sprocketwheel which also rotates slowly A gear train mechanism converts this slow motion to high speed which drives a DC generator The electric output from the gen erator is used to power an LED bulb Consider a gravitydriven LED bulb that provides 16 lumens of lighting The device uses a 10kg sandbag that is raised by human power to a 2m height For continuous lighting the bag needs to be raised every 20 minutes Using an efficacy of 150 lumens per watt for the LED bulb determine a the velocity of the sandbag as it descends and b the overall efficiency of the device 2109 Consider a classroom for 55 students and one instruc tor each generating heat at a rate of 100 W Lighting is pro vided by 18 fluorescent lightbulbs 40 W each and the ballasts consume an additional 10 percent Determine the rate of inter nal heat generation in this classroom when it is fully occupied 2110 Consider a homeowner who is replacing his 25year old natural gas furnace that has an efficiency of 55 percent The homeowner is considering a conventional furnace that has an efficiency of 82 percent and costs 1600 and a highefficiency furnace that has an efficiency of 95 percent and costs 2700 The homeowner would like to buy the highefficiency furnace if the savings from the natural gas pay for the additional cost in less than 8 years If the homeowner now pays 1200 a year for heating determine if he should buy the conventional or the high efficiency model 2111 A homeowner is considering these heating systems for heating his house Electric resistance heating with 012kWh and 1 kWh 3600 kJ gas heating with 124therm and 1 therm 105500 kJ and oil heating with 23gal and 1 gal of oil 138500 kJ Assuming efficiencies of 100 percent for the electric furnace and 87 percent for the gas and oil furnaces determine the heating system with the lowest energy cost 2112 The US Department of Energy estimates that 570000 barrels of oil would be saved per day if every house hold in the United States lowered the thermostat setting in win ter by 6F 33C Assuming the average heating season to be 180 days and the cost of oil to be 55barrel determine how much money would be saved per year 2113 A typical household pays about 1200 a year on energy bills and the US Department of Energy estimates that 46 percent of this energy is used for heating and cooling 15 percent for heating water 15 percent for refrigerating and freezing and the remaining 24 percent for lighting cooking and running other appliances The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation If the cost of insulation is 200 determine how long it will take for the insulation to pay for itself from the energy it saves 2114 A diesel engine with an engine volume of 40 L and an engine speed of 2500 rpm operates on an airfuel ratio of 18 kg airkg fuel The engine uses light diesel fuel that con tains 500 ppm parts per million of sulfur by mass All of this sulfur is exhausted to the environment where the sulfur is converted to sulfurous acid H2SO3 If the rate of the air entering the engine is 336 kgh determine the mass flow rate of sulfur in the exhaust Also determine the mass flow rate of sulfurous acid added to the environment if for each kmol of sulfur in the exhaust 1 kmol sulfurous acid will be added to the environment 2115 The force F required to compress a spring a distance x is given by F F0 kx where k is the spring constant and F0 is the preload Determine the work in kJ required to compress a spring a distance of 1 cm when its spring constant is 300 Ncm and the spring is initially compressed by a force of 100 N 2116 The force required to expand the gas in a gas spring a distance x is given by F Constant x k where the constant is determined by the geometry of this device and k is determined by the gas used in the device Such a gas spring is arranged to have a constant of 1000 Nm13 and k 13 Determine the work in kJ required to compress this spring from 01 m to 03 m Answer 187 kJ 2117 Consider a TV set that consumes 120 W of electric power when it is on and is kept on for an average of 6 h per day For a unit electricity cost of 12 cents per kWh determine the cost of electricity this TV consumes per month 30 days 2118E Water is pumped from a 200ftdeep well into a 100fthigh storage tank Determine the power in kW that would be required to pump 200 galmin 2119 Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty The weight of the elevator cabin is partially balanced by a 400kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the eleva tor well Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless determine a the power required while the fully loaded cabin is rising at a con stant speed of 12 ms and b the power required while the empty cabin is descending at a constant speed of 12 ms What would your answer be to a if no counterweight were used What would your answer be to b if a friction force of 800 N has developed between the cabin and the guide rails 2120 A grist mill of the 1800s employed a waterwheel that was 14 m high 480 Lmin of water flowed onto the wheel near the top How much power in kW could this waterwheel have produced Answer 110 kW Final PDF to printer cen22672ch02051108indd 105 110317 0749 AM 105 CHAPTER 2 2122 The demand for electric power is usually much higher during the day than it is at night and utility companies often sell power at night at much lower prices to encourage consum ers to use the available power generation capacity and to avoid building new expensive power plants that will be used only a short time during peak periods Utilities are also willing to purchase power produced during the day from private parties at a high price Suppose a utility company is selling electric power for 005kWh at night and is willing to pay 012kWh for power produced during the day To take advantage of this opportunity an entrepreneur is considering building a large reservoir 40 m above the lake level pumping water from the lake to the res ervoir at night using cheap power and letting the water flow from the reservoir back to the lake during the day producing power as the pumpmotor operates as a turbinegenerator dur ing reverse flow Preliminary analysis shows that a water flow rate of 2 m3s can be used in either direction The combined pumpmotor and turbinegenerator efficiencies are expected to be 75 percent each Disregarding the frictional losses in pip ing and assuming the system operates for 10 h each in the pump and turbine modes during a typical day determine the potential revenue this pumpturbine system can generate per year 2123 The pump of a water distribution system is powered by a 15kW electric motor whose efficiency is 90 percent The water flow rate through the pump is 50 Ls The diameters of the inlet and outlet pipes are the same and the elevation dif ference across the pump is negligible If the pressures at the inlet and outlet of the pump are measured to be 100 kPa and 300 kPa absolute respectively determine the mechanical efficiency of the pump Answer 741 percent FIGURE P2121 2 1 90 m Generator Turbine 65 m3s η turbinegen 84 FIGURE P2122 40 m Pump turbine Reservoir Lake FIGURE P2123 50 Ls η motor 90 100 kPa Water 300 kPa W pump Motor 15 kW FIGURE P2124 Flow channel 2124 An automobile moving through the air causes the air velocity measured with respect to the car to decrease and fill a larger flow channel An automobile has an effective flow chan nel area of 3 m2 The car is traveling at 90 kmh on a day when the barometric pressure is 70 cm of mercury and the tempera ture is 20C Behind the car the air velocity with respect to the car is measured to be 82 kmh and the temperature is 20C Determine the power required to move this car through the air and the area of the effective flow channel behind the car 2121 In a hydroelectric power plant 65 m3s of water flows from an elevation of 90 m to a turbine where electric power is generated The overall efficiency of the turbinegenerator is 84 percent Disregarding frictional losses in piping estimate the electric power output of this plant Answer 482 MW Final PDF to printer cen22672ch02051108indd 106 110317 0749 AM 106 ENERGY ENERGY TRANSFER Fundamentals of Engineering FE Exam Problems 2125 A 2kW electric resistance heater in a room is turned on and kept on for 50 min The amount of energy transferred to the room by the heater is a 2 kJ b 100 kJ c 3000 kJ d 6000 kJ e 12000 kJ 2126 Consider a refrigerator that consumes 320 W of elec tric power when it is running If the refrigerator runs only onequarter of the time and the unit cost of electricity is 013kWh the electricity cost of this refrigerator per month 30 days is a 49 b 58 c 75 d 83 e 97 2127 A 75hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an efficiency of 93 percent If the unit cost of electricity is 011kWh the annual electricity cost of this compressor is a 14300 b 15380 c 16540 d 19180 e 22180 2128 On a hot summer day the air in a wellsealed room is circulated by a 050hp fan driven by a 65 percent efficient motor Note that the motor delivers 050 hp of net shaft power to the fan The rate of energy supply from the fanmotor assembly to the room is a 0769 kJs b 0325 kJs c 0574 kJs d 0373 kJs e 0242 kJs 2129 A fan is to accelerate quiescent air to a velocity of 9 ms at a rate of 3 m3s If the density of air is 115 kgm3 the minimum power that must be supplied to the fan is a 41 W b 122 W c 140 W d 206 W e 280 W 2130 A 900kg car cruising at a constant speed of 60 kmh is to accelerate to 100 kmh in 4 s The additional power needed to achieve this acceleration is a 56 kW b 222 kW c 25 kW d 62 kW e 90 kW 2131 The elevator of a large building is to raise a net mass of 550 kg at a constant speed of 12 ms using an electric motor The minimum power rating of the motor should be a 0 kW b 48 kW c 12 kW d 45 kW e 65 kW 2132 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3s from an elevation of 65 m using a turbinegenerator with an efficiency of 85 percent When frictional losses in piping are disregarded the electric power output of this plant is a 39 MW b 38 MW c 45 MW d 53 MW e 65 MW 2133 A 2kW pump is used to pump kerosene ρ 0820 kgL from a tank on the ground to a tank at a higher elevation Both tanks are open to the atmosphere and the elevation difference between the free surfaces of the tanks is 30 m The maximum volume flow rate of kerosene is a 83 Ls b 72 Ls c 68 Ls d 121 Ls e 178 Ls 2134 A glycerin pump is powered by a 5kW electric motor The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa If the flow rate through the pump is 18 Ls and the changes in elevation and the flow velocity across the pump are negligible the overall efficiency of the pump is a 69 percent b 72 percent c 76 percent d 79 percent e 82 percent The Following Problems Are Based on the Optional Special Topic of Heat Transfer 2135 A 10cmhigh and 20cmwide circuit board houses on its surface 100 closely spaced chips each generating heat at a rate of 008 W and transferring it by convection to the surround ing air at 25C Heat transfer from the back surface of the board is negligible If the convection heat transfer coefficient on the surface of the board is 10 Wm2C and radiation heat transfer is negligible the average surface temperature of the chips is a 26C b 45C c 15C d 80C e 65C 2136 A 50cmlong 02cmdiameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally The sur face temperature of the wire is measured to be 130C when a wattmeter indicates the electric power consumption to be 41 kW Then the heat transfer coefficient is a 43500 Wm2C b 137 Wm2C c 68330 Wm2C d 10038 Wm2C e 37540 Wm2C 2137 A 3m2 hot black surface at 80C is losing heat to the sur rounding air at 25C by convection with a convection heat transfer coefficient of 12 Wm2C and by radiation to the surrounding surfaces at 15C The total rate of heat loss from the surface is a 1987 W b 2239 W c 2348 W d 3451 W e 3811 W 2138 Heat is transferred steadily through a 02mthick 8 m 4 m wall at a rate of 24 kW The inner and outer surface temperatures of the wall are measured to be 15C and 5C The average thermal conductivity of the wall is a 0002 WmC b 075 WmC c 10 WmC d 15 WmC e 30 WmC 2139 The roof of an electrically heated house is 7 m long 10 m wide and 025 m thick It is made of a flat layer of concrete whose thermal conductivity is 092 WmC During a certain winter night the temperatures of the inner and outer surfaces of the roof were measured to be 15C and 4C respectively The average rate of heat loss through the roof that night was a 41 W b 177 W c 4894 W d 5567 W e 2834 W Final PDF to printer cen22672ch02051108indd 107 110317 0749 AM 107 CHAPTER 2 Design and Essay Problems 2140 An average vehicle puts out nearly 20 lbm of carbon dioxide into the atmosphere for every gallon of gasoline it burns and thus one thing we can do to reduce global warming is to buy a vehicle with higher fuel economy A US govern ment publication states that a vehicle that gets 25 rather than 20 miles per gallon will prevent 10 tons of carbon dioxide from being released over the lifetime of the vehicle Making reason able assumptions evaluate if this is a reasonable claim or a gross exaggeration 2141 Your neighbor lives in a 2500squarefoot about 250 m2 older house heated by natural gas The current gas heater was installed in the early 1980s and has an efficiency called the Annual Fuel Utilization Efficiency rating or AFUE of 65 percent It is time to replace the furnace and the neighbor is trying to decide between a conventional fur nace that has an efficiency of 80 percent and costs 1500 and a highefficiency furnace that has an efficiency of 95 percent and costs 2500 Your neighbor offered to pay you 100 if you help him make the right decision Considering the weather data typical heating loads and the price of natural gas in your area make a recommendation to your neighbor based on a convincing economic analysis 2142 Find out the prices of heating oil natural gas and electricity in your area and determine the cost of each per kWh of energy supplied to the house as heat Go through your utility bills and determine how much money you spent for heating last January Also determine how much your January heating bill would be for each of the heating systems if you had the latest and most efficient system installed 2143 Prepare a report on the heating systems available in your area for residential buildings Discuss the advantages and disadvantages of each system and compare their initial and operating costs What are the important factors in the selec tion of a heating system Give some guidelines Identify the conditions under which each heating system would be the best choice in your area 2144 The roofs of many homes in the United States are covered with photovoltaic PV solar cells that resemble roof tiles generating electricity quietly from solar energy An arti cle stated that over its projected 30year service life a 4kW roof PV system in California will reduce the production of CO2 that causes global warming by 433000 lbm sulfates that cause acid rain by 2900 lbm and nitrates that cause smog by 1660 lbm The article also claims that a PV roof will save 253000 lbm of coal 21000 gal of oil and 27 million ft3 of natural gas Making reasonable assumptions for incident solar radiation efficiency and emissions evaluate these claims and make corrections if necessary 2145 Pressure changes across atmospheric weather fronts are typically a few centimeters of mercury while the tempera ture changes are typically 220C Develop a plot of front pres sure change versus front temperature change that will cause a maximum wind velocity of 10 ms or more 2146 The performance of a device is defined as the ratio of the desired output to the required input and this definition can be extended to nontechnical fields For example your per formance in this course can be viewed as the grade you earn relative to the effort you put in If you have been investing a lot of time in this course and your grades do not reflect it you are performing poorly In that case perhaps you should try to find out the underlying cause and how to correct the problem Give three other definitions of performance from nontechnical fields and discuss them 2147 Some engineers have suggested that air compressed into tanks can be used to propel personal transportation vehicles Current compressedair tank technology permits us to compress and safely hold air at up to 4000 psia Tanks made of compos ite materials require about 10 lbm of construction materials for each 1 ft3 of stored gas Approximately 001 hp is required per pound of vehicle weight to move a vehicle at a speed of 30 miles per hour What is the maximum range that this vehicle can have Account for the weight of the tanks only and assume perfect conversion of the energy in the compressed air Final PDF to printer cen22672ch02051108indd 108 110317 0749 AM Final PDF to printer cen22672ch03109160indd 109 092217 1145 AM 109 CHAPTER3 P R O P E RT I E S O F PUR E S UB STA N C E S W e start this chapter with the introduction of the concept of a pure substance and a discussion of the physics of phasechange processes We then illustrate the various property diagrams and PvT surfaces of pure substances After demonstrating the use of the property tables the hypothetical substance ideal gas and the idealgas equation of state are discussed The compressibility factor which accounts for the deviation of real gases from idealgas behavior is introduced and some of the best known equations of state such as the van der Waals BeattieBridgeman and BenedictWebbRubin equations are presented OBJECTIVES The objectives of Chapter 3 are to Introduce the concept of a pure substance Discuss the physics of phase change processes Illustrate the Pv Tv and PT property diagrams and PvT surfaces of pure substances Demonstrate the procedures for determining thermodynamic properties of pure substances from tables of property data Describe the hypothetical substance ideal gas and the idealgas equation of state Apply the idealgas equation of state in the solution of typical problems Introduce the compressibility factor which accounts for the deviation of real gases from idealgas behavior Present some of the bestknown equations of state Final PDF to printer 110 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 110 092217 1145 AM 31 PURE SUBSTANCE A substance that has a fixed chemical composition throughout is called a pure substance Water nitrogen helium and carbon dioxide for example are all pure substances A pure substance does not have to be of a single chemical element or compound however A mixture of various chemical elements or compounds also qualifies as a pure substance as long as the mixture is homogeneous Air for example is a mixture of several gases but it is often considered to be a pure substance because it has a uniform chemical composition Fig 31 However a mixture of oil and water is not a pure substance Since oil is not soluble in water it will collect on top of the water forming two chemically dissimilar regions A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same Fig 32 A mixture of ice and liquid water for example is a pure substance because both phases have the same chemical composition A mixture of liquid air and gas eous air however is not a pure substance since the composition of liquid air is different from the composition of gaseous air and thus the mixture is no longer chemically homogeneous This is due to different components in air condensing at different temperatures at a specified pressure 32 PHASES OF A PURE SUBSTANCE We all know from experience that substances exist in different phases At room temperature and pressure copper is a solid mercury is a liquid and nitrogen is a gas Under different conditions each may appear in a differ ent phase Even though there are three principal phasessolid liquid and gasa substance may have several phases within a principal phase each with a different molecular structure Carbon for example may exist as graphite or diamond in the solid phase Helium has two liquid phases iron has three solid phases Ice may exist at seven different phases at high pres sures A phase is identified as having a distinct molecular arrangement that is homogeneous throughout and separated from the others by easily identi fiable boundary surfaces The two phases of H2O in iced water represent a good example of this When studying phases or phase changes in thermodynamics one does not need to be concerned with the molecular structure and behavior of differ ent phases However it is very helpful to have some understanding of the molecular phenomena involved in each phase and a brief discussion of phase transformations follows Intermolecular bonds are strongest in solids and weakest in gases This is why molecules in solids are closely packed together whereas in gases they are separated by relatively large distances The molecules in a solid are arranged in a threedimensional pattern lattice that is repeated throughout Fig 33 Because of the small dis tances between molecules in a solid the attractive forces of molecules on each other are large and keep the molecules at fixed positions Note that the attractive forces between molecules turn to repulsive forces as the distance between the molecules approaches zero thus preventing the molecules from FIGURE 31 Nitrogen and gaseous air are pure substances Air N2 FIGURE 32 A mixture of liquid and gaseous water is a pure substance but a mixture of liquid and gaseous air is not Gas Liquid Liquid a H2O b Air Vapor FIGURE 33 The molecules in a solid are kept at their positions by the large springlike intermolecular forces Final PDF to printer 111 CHAPTER 3 cen22672ch03109160indd 111 092217 1145 AM piling up on top of each other Even though the molecules in a solid can not move relative to each other they continually oscillate about their equi librium positions The velocity of the molecules during these oscillations depends on the temperature At sufficiently high temperatures the velocity and thus the momentum of the molecules may reach a point where the intermolecular forces are partially overcome and groups of molecules break away Fig 34 This is the beginning of the melting process The molecular spacing in the liquid phase is not much different from that of the solid phase except the molecules are no longer at fixed positions relative to each other and they can rotate and translate freely In a liquid the intermolecular forces are weaker relative to solids but still relatively strong compared with gases The distances between molecules generally experi ence a slight increase as a solid turns liquid with water being a notable exception In the gas phase the molecules are far apart from each other and a molecu lar order is nonexistent Gas molecules move about at random continually colliding with each other and the walls of the container they are in Particu larly at low densities the intermolecular forces are very small and collisions are the only mode of interaction between the molecules Molecules in the gas phase are at a considerably higher energy level than they are in the liquid or solid phases Therefore the gas must release a large amount of its energy before it can condense or freeze 33 PHASECHANGE PROCESSES OF PURE SUBSTANCES There are many practical situations where two phases of a pure substance coexist in equilibrium Water exists as a mixture of liquid and vapor in the boiler and the condenser of a steam power plant The refrigerant turns from liquid to vapor in the freezer of a refrigerator Even though many homeowners consider the freezing of water in underground pipes to be the most impor tant phasechange process attention in this section is focused on the liquid FIGURE 34 The arrangement of atoms in different phases a molecules are at relatively fixed positions in a solid b groups of molecules move about each other in the liquid phase and c molecules move about at random in the gas phase a b c Final PDF to printer 112 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 112 092217 1145 AM and vapor phases and their mixture As a familiar substance water is used to demonstrate the basic principles involved Remember however that all pure substances exhibit the same general behavior Compressed Liquid and Saturated Liquid Consider a pistoncylinder device containing liquid water at 20C and 1 atm pressure state 1 Fig 35 Under these conditions water exists in the liquid phase and it is called a compressed liquid or a subcooled liquid meaning that it is not about to vaporize Heat is now transferred to the water until its temperature rises to say 40C As the temperature rises the liquid water expands slightly and so its specific volume increases To accommodate this expansion the piston moves up slightly The pressure in the cylinder remains constant at 1 atm during this process since it depends on the outside barometric pressure and the weight of the piston both of which are constant Water is still a compressed liquid at this state since it has not started to vaporize As more heat is transferred the temperature keeps rising until it reaches 100C state 2 Fig 36 At this point water is still a liquid but any heat addi tion will cause some of the liquid to vaporize That is a phasechange process from liquid to vapor is about to take place A liquid that is about to vaporize is called a saturated liquid Therefore state 2 is a saturated liquid state Saturated Vapor and Superheated Vapor Once boiling starts the temperature stops rising until the liquid is com pletely vaporized That is the temperature will remain constant during the entire phasechange process if the pressure is held constant This can easily be verified by placing a thermometer into boiling pure water on top of a stove At sea level P 1 atm the thermometer will always read 100C if the pan is uncovered or covered with a light lid During a boiling process the only change we will observe is a large increase in the volume and a steady decline in the liquid level as a result of more liquid turning to vapor Midway about the vaporization line state 3 Fig 37 the cylinder con tains equal amounts of liquid and vapor As we continue transferring heat the vaporization process continues until the last drop of liquid is vaporized state 4 Fig 38 At this point the entire cylinder is filled with vapor that is on the borderline of the liquid phase Any heat loss from this vapor will cause some of the vapor to condense phase change from vapor to liquid A vapor that is about to condense is called a saturated vapor Therefore state 4 is a saturated vapor state A substance at states between 2 and 4 is referred to as a saturated liquidvapor mixture since the liquid and vapor phases coexist in equilibrium at these states Once the phasechange process is completed we are back to a single phase region again this time vapor and further transfer of heat results in an increase in both the temperature and the specific volume Fig 39 At state 5 the temperature of the vapor is let us say 300C and if we transfer some heat from the vapor the temperature may drop somewhat but no con densation will take place as long as the temperature remains above 100C for FIGURE 35 At 1 atm and 20C water exists in the liquid phase compressed liquid FIGURE 36 At 1 atm pressure and 100C water exists as a liquid that is ready to vaporize saturated liquid Heat State 2 P 1 atm T 100C FIGURE 37 As more heat is transferred part of the saturated liquid vaporizes saturated liquidvapor mixture Heat State 3 Saturated vapor Saturated liquid P 1 atm T 100C Final PDF to printer 113 CHAPTER 3 cen22672ch03109160indd 113 092217 1145 AM P 1 atm A vapor that is not about to condense ie not a saturated vapor is called a superheated vapor Therefore water at state 5 is a superheated vapor This constantpressure phasechange process is illustrated on a Tv diagram in Fig 310 If the entire process described here is reversed by cooling the water while maintaining the pressure at the same value the water will go back to state 1 retracing the same path and in so doing the amount of heat released will exactly match the amount of heat added during the heating process In our daily life water implies liquid water and steam implies water vapor In thermodynamics however both water and steam usually mean only one thing H2O Saturation Temperature and Saturation Pressure It probably came as no surprise to you that water started to boil at 100C Strictly speaking the statement water boils at 100C is incorrect The cor rect statement is water boils at 100C at 1 atm pressure The only reason water started boiling at 100C was because we held the pressure constant at 1 atm 101325 kPa If the pressure inside the cylinder were raised to 500 kPa by adding weights on top of the piston water would start boiling at 1518C That is the temperature at which water starts boiling depends on the pres sure therefore if the pressure is fixed so is the boiling temperature At a given pressure the temperature at which a pure substance changes phase is called the saturation temperature Tsat Likewise at a given tem perature the pressure at which a pure substance changes phase is called the saturation pressure Psat At a pressure of 101325 kPa Tsat is 9997C Con versely at a temperature of 9997C Psat is 101325 kPa At 10000C Psat is 10142 kPa in the ITS90 discussed in Chap 1 Saturation tables that list the saturation pressure against the temperature or the saturation temperature against the pressure are available for practically FIGURE 38 At 1 atm pressure the temperature remains constant at 100C until the last drop of liquid is vaporized saturated vapor Heat State 4 P 1 atm T 100C FIGURE 39 As more heat is transferred the temperature of the vapor starts to rise superheated vapor Heat State 5 P 1 atm T 300C FIGURE 310 Tv diagram for the heating process of water at constant pressure 2 Saturated mixture 4 5 Superheated T C 300 100 20 1 vapor 3 Compressed liquid P 1 atm v Final PDF to printer 114 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 114 092217 1145 AM all substances A partial listing of such a table is given in Table 31 for water This table indicates that the pressure of water changing phase boiling or con densing at 25C must be 317 kPa and the pressure of water must be main tained at 3976 kPa about 40 atm to have it boil at 250C Also water can be frozen by dropping its pressure below 061 kPa It takes a large amount of energy to melt a solid or vaporize a liquid The amount of energy absorbed or released during a phasechange process is called the latent heat More specifically the amount of energy absorbed during melting is called the latent heat of fusion and is equivalent to the amount of energy released during freezing Similarly the amount of energy absorbed during vaporization is called the latent heat of vaporization and is equivalent to the energy released during condensation The magnitudes of the latent heats depend on the temperature or pressure at which the phase change occurs At 1 atm pressure the latent heat of fusion of water is 3337 kJkg and the latent heat of vaporization is 22565 kJkg During a phasechange process pressure and temperature are obviously dependent properties and there is a definite relation between them that is Psat fTsat A plot of Psat versus Tsat such as the one given for water in Fig 311 is called a liquidvapor saturation curve A curve of this kind is characteristic of all pure substances It is clear from Fig 311 that Tsat increases with Psat Thus a substance at higher pressures boils at higher temperatures In the kitchen higher boiling temperatures mean shorter cooking times and energy savings A beef stew for example may take 1 to 2 h to cook in a regular pan that operates at 1 atm pressure but only 20 min in a pressure cooker operating at 3 atm absolute pressure corresponding boiling temperature 134C The atmospheric pressure and thus the boiling temperature of water decreases with elevation Therefore it takes longer to cook at higher altitudes than it does at sea level unless a pressure cooker is used For example the standard atmospheric pressure at an elevation of 2000 m is 7950 kPa which corresponds to a boiling temperature of 933C as opposed to 100C at sea level zero elevation The variation of the boiling temperature of water with altitude at standard atmospheric conditions is given in Table 32 For each 1000 m increase in elevation the boiling temperature drops by a little over 3C Note that the atmospheric pressure at a location and thus the boiling temperature changes slightly with the weather conditions But the corre sponding change in the boiling temperature is no more than about 1C Some Consequences of Tsat and Psat Dependence We mentioned earlier that a substance at a specified pressure boils at the satu ration temperature corresponding to that pressure This phenomenon allows us to control the boiling temperature of a substance by simply controlling the pressure and it has numerous applications in practice In this section we give some examples The natural drive to achieve phase equilibrium by allowing some liquid to evaporate is at work behind the scenes Consider a sealed can of liquid refrigerant134a in a room at 25C If the can has been in the room long enough the temperature of the refrigerant in the can is also 25C Now if the lid is opened slowly and some refrigerant is FIGURE 311 The liquidvapor saturation curve of a pure substance numerical values are for water 400 200 0 0 200 150 100 50 600 TsatC Psat kPa TABLE 31 Saturation or vapor pressure of water at various temperatures Temperature T C Saturation pressure Psat kPa 10 0260 5 0403 0 0611 5 0872 10 123 15 171 20 234 25 317 30 425 40 738 50 1235 100 1013 1 atm 150 4758 200 1554 250 3973 300 8581 Final PDF to printer 115 CHAPTER 3 cen22672ch03109160indd 115 092217 1145 AM allowed to escape the pressure in the can will start dropping until it reaches the atmospheric pressure If you are holding the can you will notice its tem perature dropping rapidly and even ice forming outside the can if the air is humid A thermometer inserted in the can will register 26C when the pres sure drops to 1 atm which is the saturation temperature of refrigerant134a at that pressure The temperature of the liquid refrigerant will remain at 26C until the last drop of it vaporizes Another aspect of this interesting physical phenomenon is that a liquid cannot vaporize unless it absorbs energy in the amount of the latent heat of vaporization which is 217 kJkg for refrigerant134a at 1 atm Therefore the rate of vaporization of the refrigerant depends on the rate of heat transfer to the can the larger the rate of heat transfer the higher the rate of vaporization The rate of heat transfer to the can and thus the rate of vaporization of the refrigerant can be minimized by insulating the can heavily In the limiting case of no heat transfer the refrigerant will remain in the can as a liquid at 26C indefinitely The boiling temperature of nitrogen at atmospheric pressure is 196C see Table A3a This means the temperature of liquid nitrogen exposed to the atmosphere must be 196C since some nitrogen will be evaporat ing The temperature of liquid nitrogen remains constant at 196C until it is depleted For this reason nitrogen is commonly used in lowtemperature scientific studies such as superconductivity and cryogenic applications to maintain a test chamber at a constant temperature of 196C This is done by placing the test chamber into a liquid nitrogen bath that is open to the atmosphere Any heat transfer from the environment to the test section is absorbed by the nitrogen which evaporates isothermally and keeps the test chamber temperature constant at 196C Fig 312 The entire test section must be insulated heavily to minimize heat transfer and thus liquid nitrogen consumption Liquid nitrogen is also used for medical purposes to burn off unsightly spots on the skin This is done by soaking a cotton swab in liquid nitrogen and wetting the target area with it As the nitrogen evaporates it freezes the affected skin by rapidly absorbing heat from it A practical way of cooling leafy vegetables is vacuum cooling which is based on reducing the pressure of the sealed cooling chamber to the satu ration pressure at the desired low temperature and evaporating some water from the products to be cooled The heat of vaporization during evaporation is absorbed from the products which lowers the product temperature The satu ration pressure of water at 0C is 061 kPa and the products can be cooled to 0C by lowering the pressure to this level The cooling rate can be increased by lowering the pressure below 061 kPa but this is not desirable because of the danger of freezing and the added cost In vacuum cooling there are two distinct stages In the first stage the prod ucts at ambient temperature say at 25C are loaded into the chamber and the operation begins The temperature in the chamber remains constant until the saturation pressure is reached which is 317 kPa at 25C In the second stage that follows saturation conditions are maintained inside at progressively lower pressures and the corresponding lower temperatures until the desired temperature is reached Fig 313 Vacuum cooling is usually more expensive than the conventional refriger ated cooling and its use is limited to applications that result in much faster FIGURE 312 The temperature of liquid nitrogen exposed to the atmosphere remains constant at 196C and thus it maintains the test chamber at 196C TABLE 32 Variation of the standard atmospheric pressure and the boiling saturation temperature of water with altitude Elevation m Atmospheric pressure kPa Boiling tempera ture C 0 1000 2000 5000 10000 20000 10133 8955 7950 5405 2650 553 1000 965 933 833 663 347 FIGURE 313 The variation of the temperature of fruits and vegetables with pressure during vacuum cooling from 25C to 0C Pressure kPa End of cooling 0C 061 kPa Start of cooling 25C 100 kPa Temperature C 25 0 100 10 317 061 0 1 Final PDF to printer 116 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 116 092217 1145 AM cooling Products with large surface area per unit mass and a high tendency to release moisture such as lettuce and spinach are well suited for vacuum cool ing Products with a low surface areatomass ratio are not suitable especially those that have relatively impervious peels such as tomatoes and cucumbers Some products such as mushrooms and green peas can be vacuum cooled suc cessfully by wetting them first The vacuum cooling just described becomes vacuum freezing if the vapor pressure in the vacuum chamber is dropped below 061 kPa the saturation pressure of water at 0C The idea of making ice by using a vacuum pump is nothing new Dr William Cullen actually made ice in Scotland in 1775 by evacuating the air in a water tank Fig 314 Package icing is commonly used in smallscale cooling applications to remove heat and keep the products cool during transit by taking advantage of the large latent heat of fusion of water but its use is limited to products that are not harmed by contact with ice Also ice provides moisture as well as refrigeration 34 PROPERTY DIAGRAMS FOR PHASECHANGE PROCESSES The variations of properties during phasechange processes are best studied and understood with the help of property diagrams Next we develop and discuss the Tv Pv and PT diagrams for pure substances 1 The Tv Diagram The phasechange process of water at 1 atm pressure was described in detail in the last section and plotted on a Tv diagram in Fig 310 Now we repeat this process at different pressures to develop the Tv diagram Let us add weights on top of the piston until the pressure inside the cylinder reaches 1 MPa At this pressure water has a somewhat smaller specific volume than it does at 1 atm pressure As heat is transferred to the water at this new pressure the process follows a path that looks very much like the process path at 1 atm pressure as shown in Fig 315 but there are some noticeable differences First water starts boiling at a much higher temperature 1799C at this pressure Second the specific volume of the saturated liquid is larger and the specific volume of the saturated vapor is smaller than the corresponding values at 1 atm pressure That is the horizontal line that connects the saturated liquid and saturated vapor states is much shorter As the pressure is increased further this saturation line continues to shrink as shown in Fig 315 and it becomes a point when the pressure reaches 2206 MPa for the case of water This point is called the critical point and it is defined as the point at which the saturated liquid and saturated vapor states are identical The temperature pressure and specific volume of a substance at the critical point are called respectively the critical temperature Tcr critical pressure Pcr and critical specific volume vcr The criticalpoint properties of water are Pcr 2206 MPa Tcr 37395C and vcr 0003106 m3kg For helium FIGURE 314 In 1775 ice was made by evacuating the airspace in a water tank Final PDF to printer 117 CHAPTER 3 cen22672ch03109160indd 117 092217 1145 AM they are 023 MPa 26785C and 001444 m3kg The critical properties for various substances are given in Table A1 in the appendix At pressures above the critical pressure there is not a distinct phasechange process Fig 316 Instead the specific volume of the substance continu ally increases and at all times there is only one phase present Eventually it resembles a vapor but we can never tell when the change has occurred Above the critical state there is no line that separates the compressed liquid region and the superheated vapor region However it is customary to refer to the substance as superheated vapor at temperatures above the critical temperature and as compressed liquid at temperatures below the critical temperature The saturated liquid states in Fig 315 can be connected by a line called the saturated liquid line and saturated vapor states in the same figure can be connected by another line called the saturated vapor line These two lines meet at the critical point forming a dome as shown in Fig 317a All the compressed liquid states are located in the region to the left of the saturated liquid line called the compressed liquid region All the superheated vapor states are located to the right of the saturated vapor line called the superheated vapor region In these two regions the substance exists in a single phase a liquid or a vapor All the states that involve both phases in equilibrium are located under the dome called the saturated liquidvapor mixture region or the wet region FIGURE 315 Tv diagram of constantpressure phasechange processes of a pure substance at various pressures numerical values are for water 0003106 37395 Saturated vapor Saturated liquid Critical point T C P 25 MPa P 2206 MPa P 15 MPa P 8 MPa P 1 MPa P 01 MPa P 001 MPa m3kg v FIGURE 316 At supercritical pressures P Pcr there is no distinct phasechange boiling process T Liquid Critical point P Pcr P Pcr Pcr Phase change Tcr Vapor vcr v Final PDF to printer 118 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 118 092217 1145 AM 2 The Pv Diagram The general shape of the Pv diagram of a pure substance is very much like the Tv diagram but the T constant lines on this diagram have a downward trend as shown in Fig 317b Consider again a pistoncylinder device that contains liquid water at 1 MPa and 150C Water at this state exists as a compressed liquid Now the weights on top of the piston are removed one by one so that the pressure inside the cylinder decreases gradually Fig 318 The water is allowed to exchange heat with the surroundings so its temperature remains constant As the pres sure decreases the volume of the water increases slightly When the pressure reaches the saturationpressure value at the specified temperature 04762 MPa the water starts to boil During this vaporization process both the temperature and the pressure remain constant but the specific volume increases Once the last drop of liquid is vaporized further reduction in pressure results in a further increase in specific volume Notice that during the phasechange process we did not remove any weights Doing so would cause the pressure and therefore the temperature to drop since Tsat fPsat and the process would no longer be isothermal When the process is repeated for other temperatures similar paths are obtained for the phasechange processes Connecting the saturated liquid and the saturated vapor states by a curve we obtain the Pv diagram of a pure substance as shown in Fig 317b Extending the Diagrams to Include the Solid Phase The two equilibrium diagrams developed so far represent the equilibrium states involving the liquid and the vapor phases only However these diagrams can easily be extended to include the solid phase as well as the solidliquid FIGURE 317 Property diagrams of a pure substance T Critical point line liquid Saturated line vapor Saturated P2 const P1 P1 const Compressed liquid region Saturated liquidvapor region Superheated vapor region P Critical point line liquid Saturated line vapor Saturated T2 const T1 T1 const Compressed liquid region Saturated liquidvapor region Superheated vapor region v v a Tv diagram of a pure substance b Pv diagram of a pure substance FIGURE 318 The pressure in a pistoncylinder device can be reduced by reducing the weight of the piston Heat P 1 MPa T 150C Final PDF to printer 119 CHAPTER 3 cen22672ch03109160indd 119 092217 1145 AM and the solidvapor saturation regions The basic principles discussed in conjunction with the liquidvapor phasechange process apply equally to the solidliquid and solidvapor phasechange processes Most substances con tract during a solidification ie freezing process Others like water expand as they freeze The Pv diagrams for both groups of substances are given in Figs 319a and 319b These two diagrams differ only in the solidliquid saturation region The Tv diagrams look very much like the Pv diagrams especially for substances that contract on freezing The fact that water expands upon freezing has vital consequences in nature If water contracted on freezing as most other substances do the ice formed would be heavier than the liquid water and it would settle to the bottom of rivers lakes and oceans instead of floating at the top The suns rays would never reach these ice layers and the bottoms of many rivers lakes and oceans would be covered with ice at times seriously disrupting marine life We are all familiar with two phases being in equilibrium but under some conditions all three phases of a pure substance coexist in equilibrium Fig 320 On Pv or Tv diagrams these triplephase states form a line called the triple line The states on the triple line of a substance have the same pres sure and temperature but different specific volumes The triple line appears as a point on the PT diagrams and therefore is often called the triple point The triplepoint temperatures and pressures of various substances are given in Table 33 For water the triplepoint temperature and pressure are 001C and 06117 kPa respectively That is all three phases of water coexist in equilibrium only if the temperature and pressure have precisely these values No substance can exist in the liquid phase in stable equilibrium at pressures below the triplepoint pressure The same can be said for temperature for sub stances that contract on freezing However substances at high pressures can exist in the liquid phase at temperatures below the triplepoint temperature For example water cannot exist in liquid form in equilibrium at atmospheric FIGURE 319 Pv diagrams of different substances P Critical point Liquid vapor Vapor Liquid Solid liquid Solid Triple line Solid vapor P Critical point Liquid vapor Vapor Liquid Solid liquid Solid Triple line Solid vapor v v a Pv diagram of a substance that contracts on freezing b Pv diagram of a substance that expands on freezing such as water FIGURE 320 At triplepoint pressure and temperature a substance exists in three phases in equilibrium Vapor Liquid Solid Final PDF to printer 120 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 120 092217 1145 AM pressure at temperatures below 0C but it can exist as a liquid at 20C at 200 MPa pressure Also ice exists at seven different solid phases at pressures above 100 MPa There are two ways a substance can pass from the solid to the vapor phase either it melts first into a liquid and subsequently evaporates or it evaporates directly without melting first The latter occurs at pressures below the triple point value since a pure substance cannot exist in the liquid phase at those pressures Fig 321 Passing from the solid phase directly into the vapor phase is called sublimation For substances that have a triplepoint pres sure above the atmospheric pressure such as solid CO2 dry ice sublimation is the only way to change from the solid to the vapor phase at atmospheric conditions 3 The PT Diagram Figure 322 shows the PT diagram of a pure substance This diagram is often called the phase diagram since all three phases are separated from each other by three lines The sublimation line separates the solid and vapor FIGURE 321 At low pressures below the triple point value solids evaporate without melting first sublimation Solid Vapor TABLE 33 Triplepoint temperatures and pressures of various substances Substance Formula Ttp K Ptp kPa Acetylene Ammonia Argon Carbon graphite Carbon dioxide Carbon monoxide Deuterium Ethane Ethylene Helium 4 λ point Hydrogen Hydrogen chloride Mercury Methane Neon Nitric oxide Nitrogen Nitrous oxide Oxygen Palladium Platinum Sulfur dioxide Titanium Uranium hexafluoride Water Xenon Zinc C2H2 NH3 A C CO2 CO D2 C2H6 C2H4 He H2 HCl Hg CH4 Ne NO N2 N2O O2 Pd Pt SO2 Ti UF6 H2O Xe Zn 1924 19540 8381 3900 21655 6810 1863 8989 1040 219 1384 15896 2342 9068 2457 10950 6318 18234 5436 1825 2045 19769 1941 33717 27316 1613 69265 120 6076 689 10100 517 1537 171 8 104 012 51 704 139 165 107 117 432 2192 126 8785 0152 35 103 20 104 167 53 103 1517 061 815 0065 Source Data from National Bureau of Standards US Circ 500 1952 Final PDF to printer 121 CHAPTER 3 cen22672ch03109160indd 121 092217 1145 AM FIGURE 322 PT diagram of pure substances T P Critical point Triple point Sublimation Melting Melting Vaporization Liquid Vapor Substances that expand on freezing Substances that contract on freezing Solid FIGURE 323 PvT surface of a substance that contracts on freezing Critical point Solid Solidliquid Solidvapor Temperature Specific volume Triple line Vapor Gas Liquid Pressure Liquid vapor FIGURE 324 PvT surface of a substance that expands on freezing like water Triple line Critical point Solid Solidvapor Temperature Liquid vapor Vapor Liquid Pressure Gas Specific volume regions the vaporization line separates the liquid and vapor regions and the melting or fusion line separates the solid and liquid regions These three lines meet at the triple point where all three phases coexist in equilibrium The vaporization line ends at the critical point because no distinction can be made between liquid and vapor phases above the critical point Substances that expand and contract on freezing differ only in the melting line on the PT diagram The PvT Surface The state of a simple compressible substance is fixed by any two indepen dent intensive properties Once the two appropriate properties are fixed all the other properties become dependent properties Remembering that any equation with two independent variables in the form z zx y repre sents a surface in space we can represent the PvT behavior of a substance as a surface in space as shown in Figs 323 and 324 Here T and v may be viewed as the independent variables the base and P as the dependent variable the height All the points on the surface represent equilibrium states All states along the path of a quasiequilibrium process lie on the PvT surface since such a process must pass through equilibrium states The singlephase regions appear as curved surfaces on the PvT surface and the twophase regions as surfaces perpendicular to the PT plane This is expected since the projections of twophase regions on the PT plane are lines All the twodimensional diagrams we have discussed so far are merely projections of this threedimensional surface onto the appropriate planes A Pv diagram is just a projection of the PvT surface on the Pv plane and a Tv diagram is nothing more than the birdseye view of this surface The Final PDF to printer 122 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 122 092217 1145 AM PvT surfaces present a great deal of information at once but in a thermody namic analysis it is more convenient to work with twodimensional diagrams such as the Pv and Tv diagrams 35 PROPERTY TABLES For most substances the relationships among thermodynamic properties are too complex to be expressed by simple equations Therefore properties are frequently presented in the form of tables Some thermodynamic properties can be measured easily but others cannot and the latter are calculated by using the relations between them and measurable properties The results of these measurements and calculations are presented in tables in a convenient format In the following discussion the steam tables are used to demonstrate the use of thermodynamic property tables Property tables of other substances are used in the same manner For each substance the thermodynamic properties are listed in more than one table In fact a separate table is prepared for each region of interest such as the superheated vapor compressed liquid and saturated mixture regions Property tables are given in the Appendix in both SI and English units The tables in English units carry the same number as the corresponding tables in SI followed by an identifier E Tables A6 and A6E for example list prop erties of superheated water vapor the former in SI and the latter in English units Before we get into the discussion of property tables we define a new property called enthalpy EnthalpyA Combination Property A person looking at the tables will notice two new properties enthalpy h and entropy s Entropy is a property associated with the second law of thermody namics and we will not use it until it is properly defined in Chap 7 However it is appropriate to introduce enthalpy at this point In the analysis of certain types of processes particularly in power gener ation and refrigeration Fig 325 we often encounter the combination of properties u Pv For the sake of simplicity and convenience this combina tion is defined as a new property enthalpy and given the symbol h h u Pv kJ kg 31 or H U PV kJ 32 Both the total enthalpy H and specific enthalpy h are simply referred to as enthalpy since the context clarifies which one is meant Notice that the equations given above are dimensionally homogeneous That is the unit of the pressurevolume product may differ from the unit of the internal energy by only a factor Fig 326 For example it can be easily shown that 1 kPam3 1 kJ In some tables encountered in practice the internal energy u is frequently not listed but it can always be determined from u h Pv The widespread use of the property enthalpy is due to Professor Richard Mollier who recognized the importance of the group u Pv in the analysis of steam turbines and in the representation of the properties of steam in tabular FIGURE 325 The combination u Pv is often encountered in the analysis of control volumes u1 P1v1 u2 P2v2 Control volume FIGURE 326 The product pressure volume has energy units kPam3 kPam3kg barm3 MPam3 psift3 kJ kJkg 100 kJ 1000 kJ 018505 Btu Final PDF to printer 123 CHAPTER 3 cen22672ch03109160indd 123 092217 1145 AM and graphical form as in the famous Mollier chart Mollier referred to the group u Pv as heat content and total heat These terms were not quite con sistent with the modern thermodynamic terminology and were replaced in the 1930s by the term enthalpy from the Greek word enthalpien which means to heat 1a Saturated Liquid and Saturated Vapor States The properties of saturated liquid and saturated vapor for water are listed in Tables A4 and A5 Both tables give the same information The only dif ference is that in Table A4 properties are listed under temperature and in Table A5 under pressure Therefore it is more convenient to use Table A4 when temperature is given and Table A5 when pressure is given The use of Table A4 is illustrated in Fig 327 The subscript f is used to denote properties of a saturated liquid and the subscript g to denote the properties of saturated vapor These symbols are commonly used in thermodynamics and originated from German Another subscript commonly used is fg which denotes the difference between the saturated vapor and saturated liquid values of the same property For example v f specific volume of saturated liquid v g specific volume of saturated vapor v fg difference between v g and v f that is v fg v g v f The quantity hfg is called the enthalpy of vaporization or latent heat of vaporization It represents the amount of energy needed to vaporize a unit mass of saturated liquid at a given temperature or pressure It decreases as the temperature or pressure increases and becomes zero at the critical point FIGURE 327 A partial list of Table A4 Temperature Specific volume of saturated liquid Corresponding saturation pressure Specific volume of saturated vapor 85 57868 90 70183 95 84609 0001032 28261 0001036 23593 0001040 19808 Sat Temp C T Specific volume m3kg Sat liquid vf Sat vapor vg press kPa Psat EXAMPLE 31 Pressure of Saturated Liquid in a Tank A rigid tank contains 50 kg of saturated liquid water at 90C Determine the pressure in the tank and the volume of the tank SOLUTION A rigid tank contains saturated liquid water The pressure and volume of the tank are to be determined Analysis The state of the saturated liquid water is shown on a Tv diagram in Fig 328 Since saturation conditions exist in the tank the pressure must be the saturation pressure at 90C P P sat 90C 70183 kPa Table A4 The specific volume of the saturated liquid at 90C is v v f 90C 0001036 m 3 kg Table A4 Then the total volume of the tank becomes V mv 50 kg 0001036 m 3 kg 00518 m 3 FIGURE 328 Schematic and Tv diagram for Example 31 P 7 0 1 8 3 kPa T C vf v 90 Sat liquid T 90C Final PDF to printer 124 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 124 092217 1145 AM EXAMPLE 32 Temperature of Saturated Vapor in a Cylinder A pistoncylinder device contains 2 ft3 of saturated water vapor at 50psia pressure Determine the temperature and the mass of the vapor inside the cylinder SOLUTION A cylinder contains saturated water vapor The temperature and the mass of vapor are to be determined Analysis The state of the saturated water vapor is shown on a Pv diagram in Fig 329 Since the cylinder contains saturated vapor at 50 psia the temperature inside must be the saturation temperature at this pressure T T sat 50 psia 2899F Table A5E The specific volume of the saturated vapor at 50 psia is v v g 50 psia 85175 ft 3 lbm Table A5E Then the mass of water vapor inside the cylinder becomes m V v 2 ft 3 85175 ft 3 lbm 0235 lbm FIGURE 329 Schematic and Pv diagram for Example 32 P psia vg v 50 Saturated vapor P 50 psia V 2 ft3 T 28099F EXAMPLE 33 Volume and Energy Change during Evaporation A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa Determine a the volume change and b the amount of energy transferred to the water SOLUTION Saturated liquid water is vaporized at constant pressure The volume change and the energy transferred are to be determined Analysis a The process described is illustrated on a Pv diagram in Fig 330 The volume change per unit mass during a vaporization process is vfg which is the difference between vg and vf Reading these values from Table A5 at 100 kPa and substituting yield v fg v g v f 16941 0001043 16931 m 3 kg Thus ΔV m v fg 02 kg 16931 m 3 kg 03386 m 3 b The amount of energy needed to vaporize a unit mass of a substance at a given pressure is the enthalpy of vaporization at that pressure which is hfg 22575 kJkg for water at 100 kPa Thus the amount of energy transferred is m h fg 02 kg 225755 kJ kg 4515 kJ Discussion Note that we have considered the first four decimal digits of vfg and disregarded the rest This is because vg has significant numbers to the first four decimal places only and we do not know the numbers in the other decimal places Copying all the digits from the calculator would mean that we are assuming vg 1694100 which is not necessarily the case It could very well be that vg 1694138 since this number too would truncate to 16941 All the digits in our result 16931 are significant But if we did not truncate the result we would obtain vfg 1693057 which falsely implies that our result is accurate to the sixth decimal place FIGURE 330 Schematic and Pv diagram for Example 33 P kPa 100 vf vg v Sat vapor P 100 kPa Sat liquid P 100 kPa Final PDF to printer 125 CHAPTER 3 cen22672ch03109160indd 125 092217 1145 AM 1b Saturated LiquidVapor Mixture During a vaporization process a substance exists as part liquid and part vapor That is it is a mixture of saturated liquid and saturated vapor Fig 331 To analyze this mixture properly we need to know the proportions of the liquid and vapor phases in the mixture This is done by defining a new property called the quality x as the ratio of the mass of vapor to the total mass of the mixture x m vapor m total 33 where m total m liquid m vapor m f m g Quality has significance for saturated mixtures only It has no meaning in the compressed liquid or superheated vapor regions Its value is between 0 and 1 The quality of a system that consists of saturated liquid is 0 or 0 percent and the quality of a system consisting of saturated vapor is 1 or 100 percent In saturated mixtures quality can serve as one of the two independent intensive properties needed to describe a state Note that the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor During the vaporization process only the amount of saturated liquid changes not its properties The same can be said about a saturated vapor A saturated mixture can be treated as a combination of two subsystems the saturated liquid and the saturated vapor However the amount of mass for each phase is usually not known Therefore it is often more convenient to imagine that the two phases are mixed well forming a homogeneous mixture Fig 332 Then the properties of this mixture will simply be the average properties of the saturated liquidvapor mixture under consideration Here is how it is done Consider a tank that contains a saturated liquidvapor mixture The volume occupied by saturated liquid is Vf and the volume occupied by saturated vapor is Vg The total volume V is the sum of the two V V f V g V mv m t v avg m f v f m g v g m f m t m g m t v avg m t m g v f m g v g Dividing by mt yields v avg 1 x v f x v g since x mg mt This relation can also be expressed as v avg v f x v fg m 3 kg 34 where vfg vg vf Solving for quality we obtain x v avg v f v fg 35 FIGURE 332 A twophase system can be treated as a homogeneous mixture for convenience vg vf Saturated liquid Saturated vapor Saturated liquidvapor mixture vavg FIGURE 331 The relative amounts of liquid and vapor phases in a saturated mixture are specified by the quality x Saturated li quid st ate s Saturated v ap or s tat es P or T Critical point Sat liquid Sat vapor v Final PDF to printer 126 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 126 092217 1145 AM Based on this equation quality can be related to the horizontal distances on a Pv or Tv diagram Fig 333 At a given temperature or pressure the numerator of Eq 35 is the distance between the actual state and the saturated liquid state and the denominator is the length of the entire horizontal line that connects the saturated liquid and saturated vapor states A state of 50 percent quality lies in the middle of this horizontal line The analysis given above can be repeated for internal energy and enthalpy with the following results u avg u f x u fg kJ kg 36 h avg h f x h fg kJ kg 37 All the results are of the same format and they can be summarized in a single equation as y avg y f x y fg where y is v u or h The subscript avg for average is usually dropped for simplicity The values of the average properties of the mixtures are always between the values of the saturated liquid and the saturated vapor properties Fig 334 That is y f y avg y g Finally all the saturatedmixture states are located under the saturation curve and to analyze saturated mixtures all we need are saturated liquid and saturated vapor data Tables A4 and A5 in the case of water EXAMPLE 34 Pressure and Volume of a Saturated Mixture A rigid tank contains 10 kg of water at 90C If 8 kg of the water is in the liquid form and the rest is in the vapor form determine a the pressure in the tank and b the volume of the tank SOLUTION A rigid tank contains saturated mixture The pressure and the volume of the tank are to be determined Analysis a The state of the saturated liquidvapor mixture is shown in Fig 335 Since the two phases coexist in equilibrium we have a saturated mixture and the pressure must be the saturation pressure at the given temperature P P sat 90C 70183 kPa Table A4 b At 90C we have vf 0001036 m3kg and vg 23593 m3kg Table A4 One way of finding the volume of the tank is to determine the volume occupied by each phase and then add them V V f V g m f v f m g v g 8 kg0001036 m 3 kg 2 kg 23593 m 3 kg 473 m 3 Another way is to first determine the quality x then the average specific volume v and finally the total volume x m g m t 2 kg 10 kg 02 FIGURE 333 Quality is related to the horizontal distances on Pv and Tv diagrams P or T v avg v f A B AB AC x C vfg vf v vg vavg FIGURE 334 The v value of a saturated liquid vapor mixture lies between the vf and vg values at the specified T or P Sat liquid vf Sat vapor vg P or T vf vg vg v v vf Final PDF to printer 127 CHAPTER 3 cen22672ch03109160indd 127 092217 1145 AM v v f x v fg 0001036 m 3 kg 0223593 0001036 m 3 kg 0473 m 3 kg and V mv 10 kg0473 m 3 kg 473 m 3 Discussion The first method appears to be easier in this case since the masses of each phase are given In most cases however the masses of each phase are not available and the second method becomes more convenient EXAMPLE 35 Properties of Saturated LiquidVapor Mixture An 80L vessel contains 4 kg of refrigerant134a at a pressure of 160 kPa Determine a the temperature b the quality c the enthalpy of the refrigerant and d the volume occupied by the vapor phase SOLUTION A vessel is filled with refrigerant134a Some properties of the refrigerant are to be determined Analysis a The state of the saturated liquidvapor mixture is shown in Fig 336 At this point we do not know whether the refrigerant is in the compressed liquid superheated vapor or saturated mixture region This can be determined by comparing a suitable prop erty to the saturated liquid and saturated vapor values From the information given we can determine the specific volume v V m 0080 m 3 4 kg 002 m 3 kg At 160 kPa we read v f 00007435 m 3 kg v g 012355 m 3 kg Table A12 Obviously vf v vg and the refrigerant is in the saturated mixture region Thus the temperature must be the saturation temperature at the specified pressure T T sat 160 kPa 1560C b Quality can be determined from x v v f v fg 002 00007435 012355 00007435 0157 c At 160 kPa we also read from Table A12 that hf 3118 kJkg and hfg 20996 kJkg Then h h f x h fg 3118 kJ kg 641 kJ kg 0157 20996 kJ kg FIGURE 336 Schematic and Pv diagram for Example 35 P kPa vf 00007435 P 160 kPa T 1560C 160 vg 012355 hf 3118 hg 24114 h kJkg R134a v m3kg m 4 kg FIGURE 335 Schematic and Tv diagram for Example 34 T C 90 T mg 90C 2 kg mf 8 kg vf 0001036 vg 23593 v m3kg P 7 0 1 83 kPa Final PDF to printer 128 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 128 092217 1145 AM Property tables are also available for saturated solidvapor mixtures Properties of saturated icewater vapor mixtures for example are listed in Table A8 Saturated solidvapor mixtures can be handled just as saturated liquidvapor mixtures 2 Superheated Vapor In the region to the right of the saturated vapor line and at temperatures above the critical point temperature a substance exists as superheated vapor Since the superheated region is a singlephase region vapor phase only temperature and pressure are no longer dependent properties and they can conveniently be used as the two independent properties in the tables The format of the super heated vapor tables is illustrated in Fig 337 In these tables the properties are listed against temperature for selected pressures starting with the saturated vapor data The saturation temperature is given in parentheses following the pressure value Compared to saturated vapor superheated vapor is characterized by Lower pressures P P sat at a given T Higher temperatures T T sat at a given P Higher specific volumes v v g at a given P or T Higher internal energies u u g at a given P or T Higher enthalpies h h g at a given P or T EXAMPLE 36 Cooling of Superheated Water Vapor One poundmass of water fills a 229ft3 rigid container at an initial pressure of 250 psia The container is then cooled to 100F Determine the initial temperature and final pressure of the water SOLUTION A rigid container that is filled with water is cooled The initial temperature and final pressure are to be determined Analysis The initial specific volume is v 1 V m 229 ft 3 1 lbm 229 ft 3 lbm At 250 psia the specific volume of saturated vapor is vg 18440 ft3lbm Table A5E Since v1 vg the water is initially in the superheated vapor region The temperature is determined to be d The mass of the vapor is m g x m t 01574 kg 0628 kg and the volume occupied by the vapor phase is V g m g v g 0628 kg012355 m 3 kg 00776 m 3 or 776 L The rest of the volume 24 L is occupied by the liquid FIGURE 337 A partial listing of Table A6 m3kg P 05 MPa 15183C P 01 MPa 9961C TC kJkg kJkg h u Sat 100 150 1300 Sat 200 250 16941 16959 19367 72605 037483 042503 047443 25056 25062 25829 46872 25607 26433 27238 26750 26758 27766 54133 27481 28558 29610 v Final PDF to printer 129 CHAPTER 3 cen22672ch03109160indd 129 092217 1145 AM P 1 250 psia v 1 229 ft 3 lbm T 1 550 F Table A6E This is a constant volume cooling process v Vm constant as shown in Fig 338 The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature T 2 100 F v 2 v 1 229 ft 3 lbm P 2 P sat 100F 09505 psia Table A4E Discussion When a substance undergoes a process in a closed rigid tank the specific volume remains constant and the process appears as a vertical line in the Pv diagram FIGURE 338 Schematic and Pv diagram for Example 36 H2O 250 psia 1 lbm 229 ft3 Q P 1 2 v EXAMPLE 37 Temperature of Superheated Vapor Determine the temperature of water at a state of P 05 MPa and h 2890 kJkg SOLUTION The temperature of water at a specified state is to be determined Analysis At 05 MPa the enthalpy of saturated water vapor is hg 27481 kJkg Since h hg as shown in Fig 339 we again have superheated vapor Under 05 MPa in Table A6 we read T C h kJkg 200 28558 250 29610 Obviously the temperature is between 200 and 250C By linear interpolation it is determined to be T 2163C FIGURE 339 At a specified P superheated vapor exists at a higher h than the saturated vapor Example 37 h T 05 MPa h hg hg 3 Compressed Liquid Compressed liquid tables are not as commonly available and Table A7 is the only compressed liquid table in this text The format of Table A7 is very much like the format of the superheated vapor tables One reason for the lack of compressed liquid data is the relative independence of compressed liquid properties from pressure Variation of properties of compressed liquid with pressure is very mild Increasing the pressure 100 times often causes proper ties to change less than 1 percent In the absence of compressed liquid data a general approximation is to treat compressed liquid as saturated liquid at the given temperature Fig 340 This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure Thus y y f T 38 for compressed liquids where y is v u or h Of these three properties the property whose value is most sensitive to variations in the pressure is the enthalpy h Although the preceding approximation results in negligible error in v and u the error in h may reach undesirable levels However the error in FIGURE 340 A compressed liquid may be approximated as a saturated liquid at the given temperature Given P and T vf T u uf T h hf T v Final PDF to printer 130 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 130 092217 1145 AM h at low to moderate pressures and temperatures can be reduced significantly by evaluating it from h h f T v f T P P sat T 39 instead of taking it to be just hf Note however that the approximation in Eq 39 does not yield any significant improvement at moderate to high tem peratures and pressures and it may even backfire and result in greater error due to overcorrection at very high temperatures and pressures see Kostic 2006 In general a compressed liquid is characterized by Higher pressures P P sat at a given T Lower temperatures T T sat at a given P Lower specific volumes v v f at a given P or T Lower internal energies u u f at a given P or T Lower enthalpies h h f at a given P or T But unlike superheated vapor the compressed liquid properties are not much different from the corresponding saturated liquid values EXAMPLE 38 Approximating Compressed Liquid as Saturated Liquid Determine the internal energy of compressed liquid water at 80C and 5 MPa using a data from the compressed liquid table and b saturated liquid data What is the error involved in the second case SOLUTION The exact and approximate values of the internal energy of liquid water are to be determined Analysis At 80C the saturation pressure of water is 47416 kPa and since 5 MPa Psat we obviously have compressed liquid as shown in Fig 341 a From the compressed liquid table Table A7 P 5 MPa T 80C u 33382 kJ kg b From the saturation table Table A4 we read u u f 80C 33497 kJ kg The error involved is 33497 33382 33382 100 034 which is less than 1 percent FIGURE 341 Schematic and Tu diagram for Example 38 u T C 5 MPa T 80C P 5 MPa 80 u uf 80C Reference State and Reference Values The values of u h and s cannot be measured directly and they are calculated from measurable properties using the relations between thermodynamic prop erties However those relations give the changes in properties not the values of properties at specified states Therefore we need to choose a convenient reference state and assign a value of zero for a convenient property or prop erties at that state For water the state of saturated liquid at 001C is taken Final PDF to printer 131 CHAPTER 3 cen22672ch03109160indd 131 092217 1145 AM as the reference state and the internal energy and entropy are assigned zero values at that state For refrigerant134a the state of saturated liquid at 40C is taken as the reference state and the enthalpy and entropy are assigned zero values at that state Note that some properties may have negative values as a result of the reference state chosen It should be mentioned that sometimes different tables list different values for some properties at the same state as a result of using a different reference state However in thermodynamics we are concerned with the changes in properties and the reference state chosen is of no consequence in calculations as long as we use values from a single consistent set of tables or charts EXAMPLE 39 The Use of Steam Tables to Determine Properties Determine the missing properties and the phase descriptions in the following table for water T C P kPa u kJkg x Phase description a 200 06 b 125 1600 c 1000 2950 d 75 500 e 850 00 SOLUTION Properties and phase descriptions of water are to be determined at various states Analysis a The quality is given to be x 06 which implies that 60 percent of the mass is in the vapor phase and the remaining 40 percent is in the liquid phase Therefore we have saturated liquidvapor mixture at a pressure of 200 kPa Then the temperature must be the saturation temperature at the given pressure T T sat 200 kPa 12021C Table A5 At 200 kPa we also read from Table A5 that uf 50450 kJkg and ufg 20246 kJkg Then the average internal energy of the mixture is u u f x u fg 50450 kJ kg 06 20246 kJ kg 171926 kJ kg b This time the temperature and the internal energy are given but we do not know which table to use to determine the missing properties because we have no clue as to whether we have saturated mixture compressed liquid or superheated vapor To determine the region we are in we first go to the saturation table Table A4 and determine the uf and ug values at the given temperature At 125C we read uf 52483 kJkg and ug 25343 kJkg Next we compare the given u value to these uf and ug values keeping in mind that if u u f we have compressed liquid if u f u u g we have saturated mixture if u u g we have superheated vapor Final PDF to printer 132 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 132 092217 1145 AM In our case the given u value is 1600 kJkg which falls between the uf and ug values at 125C Therefore we have saturated liquidvapor mixture Then the pressure must be the saturation pressure at the given temperature P P sat 125C 23223 kPa Table A4 The quality is determined from x u u f u fg 1600 52483 20095 0535 The preceding criteria for determining whether we have compressed liquid saturated mixture or superheated vapor can also be used when enthalpy h or specific volume v is given instead of internal energy u or when pressure is given instead of temperature c This is similar to case b except pressure is given instead of temperature Following the argument given above we read the uf and ug values at the specified pressure At 1 MPa we have uf 76139 kJkg and ug 25828 kJkg The specified u value is 2950 kJkg which is greater than the ug value at 1 MPa Therefore we have superheated vapor and the temperature at this state is determined from the superheated vapor table by interpolation to be T 3952C Table A6 We would leave the quality column blank in this case since quality has no meaning for a superheated vapor d In this case the temperature and pressure are given but again we cannot tell which table to use to determine the missing properties because we do not know whether we have saturated mixture compressed liquid or superheated vapor To determine the region we are in we go to the saturation table Table A5 and determine the saturation temperature value at the given pressure At 500 kPa we have Tsat 15183C We then compare the given T value to this Tsat value keeping in mind that if T T sat given P we have compressed liquid if T T sat given P we have saturated mixture if T T sat given P we have superheated vapor In our case the given T value is 75C which is less than the Tsat value at the specified pressure Therefore we have compressed liquid Fig 342 and normally we would determine the internal energy value from the compressed liquid table But in this case the given pressure is much lower than the lowest pressure value in the com pressed liquid table which is 5 MPa and therefore we are justified to treat the compressed liquid as saturated liquid at the given temperature not pressure u u f 75C 31399 kJ kg Table A4 We would leave the quality column blank in this case since quality has no meaning in the compressed liquid region e The quality is given to be x 0 and thus we have saturated liquid at the specified pressure of 850 kPa Then the temperature must be the saturation temperature at the given pressure and the internal energy must have the saturated liquid value T T sat 850 kPa 17294C u u f 850 kPa 73100 kJ kg Table A5 FIGURE 342 At a given P and T a pure substance will exist as a compressed liquid if T Tsat P u T C P 500 kPa 75 15183 u uf 75C Final PDF to printer 133 CHAPTER 3 cen22672ch03109160indd 133 092217 1145 AM 36 THE IDEALGAS EQUATION OF STATE Property tables provide very accurate information about the properties but they are bulky and vulnerable to typographical errors A more practical and desirable approach would be to have some simple relations among the proper ties that are sufficiently general and accurate Any equation that relates the pressure temperature and specific volume of a substance is called an equation of state Property relations that involve other properties of a substance at equilibrium states are also referred to as equations of state There are several equations of state some simple and others very complex The simplest and bestknown equation of state for substances in the gas phase is the idealgas equation of state This equation predicts the PvT behavior of a gas quite accurately within some properly selected region Gas and vapor are often used as synonymous words The vapor phase of a substance is customarily called a gas when it is above the critical temperature Vapor usually implies a gas that is not far from a state of condensation In 1662 Robert Boyle an Englishman observed during his experiments with a vacuum chamber that the pressure of gases is inversely proportional to their volume In 1802 J Charles and J GayLussac Frenchmen experimen tally determined that at low pressures the volume of a gas is proportional to its temperature That is P R T v or Pv RT 310 where the constant of proportionality R is called the gas constant Equation 310 is called the idealgas equation of state or simply the idealgas relation and a gas that obeys this relation is called an ideal gas In this equation P is the absolute pressure T is the absolute temperature and v is the specific volume The gas constant R is different for each gas Fig 343 and is determined from R R u M kJ kgK or kPam 3 kgK where Ru is the universal gas constant and M is the molar mass also called molecular weight of the gas The constant Ru is the same for all substances and its value is R u 831447 kJ kmolK 831447 kPam 3 kmolK 00831447 barm 3 kmolK 198588 Btu lbmolR 107316 psiaft 3 lbmolR 154537 ftlbf lbmolR 311 The molar mass M can simply be defined as the mass of one mole also called a grammole abbreviated gmol of a substance in grams or the mass of one kmol also called a kilogrammole abbreviated kgmol in kilograms In English units it is the mass of 1 lbmol in lbm Notice that the molar mass of a substance has the same numerical value in both unit systems because of the way it is defined When we say the molar mass of nitrogen is 28 it simply FIGURE 343 Different substances have different gas constants Substance 02870 20769 02081 02968 Air Helium Argon Nitrogen R kJkgK Final PDF to printer 134 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 134 092217 1145 AM means the mass of 1 kmol of nitrogen is 28 kg or the mass of 1 lbmol of nitrogen is 28 lbm That is M 28 kgkmol 28 lbmlbmol The mass of a system is equal to the product of its molar mass M and the mole number N m MN kg 312 The values of R and M for several substances are given in Table A1 The idealgas equation of state can be written in several different forms V mv PV mRT 313 mR MN R N R u PV N R u T 314 V N v P v R u T 315 where v is the molar specific volume that is the volume per unit mole in m3kmol or ft3lbmol A bar above a property denotes values on a unitmole basis throughout this text Fig 344 By writing Eq 313 twice for a fixed mass and simplifying the properties of an ideal gas at two different states are related to each other by P 1 V 1 T 1 P 2 V 2 T 2 316 An ideal gas is an imaginary substance that obeys the relation Pv RT It has been experimentally observed that the idealgas relation given closely approximates the PvT behavior of real gases at low densities At low pressures and high tempera tures the density of a gas decreases and the gas behaves as an ideal gas under these conditions What constitutes low pressure and high temperature is explained later In the range of practical interest many familiar gases such as air nitrogen oxygen hydrogen helium argon neon and carbon dioxide and even heavier gases such as krypton can be treated as ideal gases with negligible error often less than 1 percent Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators however should not be treated as ideal gases Instead the property tables should be used for these substances FIGURE 344 Properties per unit mole are denoted with a bar on the top Per unit mass v m3kg kJkg kJkg u h Per unit mole v m3kmol kJkmol kJkmol u h EXAMPLE 310 Temperature Rise of Air in a Tire During a Trip The gage pressure of an automobile tire is measured to be 210 kPa before a trip and 220 kPa after the trip at a location where the atmospheric pressure is 95 kPa Fig 345 Assuming the volume of the tire remains constant and the air temperature before the trip is 25C determine air temperature in the tire after the trip SOLUTION The pressure in an automobile tire is measured before and after a trip The temperature of air in the tire after the trip is to be determined Assumptions 1 The volume of the tire remains constant 2 Air is an ideal gas Properties The local atmospheric pressure is 95 kPa Analysis The absolute pressures in the tire before and after the trip are P 1 P gage1 P atm 210 95 305 kPa P 2 P gage2 P atm 220 95 315 kPa FIGURE 345 StockbyteGetty Images RF Final PDF to printer 135 CHAPTER 3 cen22672ch03109160indd 135 092217 1145 AM Is Water Vapor an Ideal Gas This question cannot be answered with a simple yes or no The error involved in treating water vapor as an ideal gas is calculated and plotted in Fig 346 It is clear from this figure that at pressures below 10 kPa water vapor can be treated as an ideal gas regardless of its temperature with negligible error Note that air is an ideal gas and the volume is constant The air temperature after the trip is determined to be P 1 V 1 T 1 P 2 V 2 T 2 T 2 P 2 P 1 T 1 315 kPa 305 kPa 25 273 K 3078 K 348C Therefore the absolute temperature of air in the tire will increase by 33 percent during this trip Discussion Note that the air temperature has risen nearly 10C during this trip This shows the importance of measuring the tire pressures before long trips to avoid errors due to the temperature increase of air in tires Also note that the unit kelvin is used for temperature in the idealgas relation FIGURE 346 Percentage of error vtable videalvtable 100 involved in assuming steam to be an ideal gas and the region where steam can be treated as an ideal gas with less than 1 percent error v m3kg T C 0 100 200 300 400 500 600 100 10 1 01 001 0001 00 00 16 00 00 05 00 00 02 00 00 01 00 00 01 00 00 05 24 108 50 173 08 41 208 371 88 13 74 176 2710 562 26 167 1527 495 257 60 00 20 MPa 10 MPa 5 MPa 1 MPa 100 kPa 10 kPa 08 kPa 01 76 30 MPa Ideal Gas 00 Final PDF to printer 136 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 136 092217 1145 AM less than 01 percent At higher pressures however the idealgas assump tion yields unacceptable errors particularly in the vicinity of the critical point over 100 percent and the saturated vapor line Therefore in airconditioning applications the water vapor in the air can be treated as an ideal gas with essentially no error since the pressure of the water vapor is very low In steam power plant applications however the pressures involved are usually very high therefore the idealgas relation should not be used 37 COMPRESSIBILITY FACTORA MEASURE OF DEVIATION FROM IDEALGAS BEHAVIOR The idealgas equation is very simple and thus very convenient to use How ever as illustrated in Fig 347 gases deviate from idealgas behavior sig nificantly at states near the saturation region and the critical point This deviation from idealgas behavior at a given temperature and pressure can accurately be accounted for by the introduction of a correction factor called the compressibility factor Z defined as Z Pv RT 317 or Pv ZRT 318 It can also be expressed as Z v actual v ideal 319 where videal RTP Obviously Z 1 for ideal gases For real gases Z can be greater than or less than unity Fig 347 The farther away Z is from unity the more the gas deviates from idealgas behavior We have said that gases follow the idealgas equation closely at low pres sures and high temperatures But what exactly constitutes low pressure or high temperature Is 100C a low temperature It definitely is for most sub stances but not for air Air or nitrogen can be treated as an ideal gas at this temperature and atmospheric pressure with an error under 1 percent This is because nitrogen is well over its critical temperature 147C and away from the saturation region At this temperature and pressure however most sub stances would exist in the solid phase Therefore the pressure or temperature of a substance is high or low relative to its critical temperature or pressure Gases behave differently at a given temperature and pressure but they behave very much the same at temperatures and pressures normalized with respect to their critical temperatures and pressures The normalization is done as P R P P cr and T R T T cr 320 Here PR is called the reduced pressure and TR the reduced temperature The Z factor for all gases is approximately the same at the same reduced pres sure and temperature This is called the principle of corresponding states FIGURE 347 The compressibility factor is unity for ideal gases Z 1 1 1 Real gases Z 1 Ideal gas Final PDF to printer 137 CHAPTER 3 cen22672ch03109160indd 137 092217 1145 AM In Fig 348 the experimentally determined Z values are plotted against PR and TR for several gases The gases seem to obey the principle of corre sponding states reasonably well By curvefitting all the data we obtain the generalized compressibility chart that can be used for all gases Fig A15 The following observations can be made from the generalized compress ibility chart 1 At very low pressures PR 1 gases behave as ideal gases regardless of temperature Fig 349 2 At high temperatures TR 2 idealgas behavior can be assumed with good accuracy regardless of pressure except when PR 1 3 The deviation of a gas from idealgas behavior is greatest in the vicinity of the critical point Fig 350 FIGURE 349 At very low pressures all gases approach idealgas behavior regardless of their temperature P 0 as Ideal gas Real gas FIGURE 348 Comparison of Z factors for various gases Source GourJen Su Modified Law of Corresponding States Ind Eng Chem international ed 38 1946 p 803 01 11 70 0 10 09 08 07 06 05 04 03 02 Z Pv RT Legend Methane Ethylene Ethane Propane nButane Isopentane nHeptane Nitrogen Carbon dioxide Water Average curve based on data on hydrocarbons 65 60 55 50 45 40 35 30 25 20 15 10 05 TR 200 TR 150 TR 130 TR 120 TR 110 TR 100 Reduced pressure PR Final PDF to printer 138 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 138 092217 1145 AM When P and v or T and v are given instead of P and T the generalized compressibility chart can still be used to determine the third property but it would involve tedious trial and error Therefore it is necessary to define one more reduced properties called the pseudoreduced specific volume vR as v R v actual R T cr P cr 321 Note that vR is defined differently from PR and TR It is related to Tcr and Pcr instead of vcr Lines of constant vR are also added to the compressibility charts and this enables one to determine T or P without having to resort to timeconsuming iterations Fig 351 FIGURE 350 Gases deviate from the idealgas behavior the most in the neighborhood of the critical point T Nonidealgas behavior Idealgas behavior Idealgas behavior v FIGURE 351 The compressibility factor can also be determined from a knowledge of PR and vR Fig A15 P PR Pcr vR RTcrPcr Z v EXAMPLE 311 The Use of Generalized Charts Determine the specific volume of refrigerant134a at 1 MPa and 50C using a the idealgas equation of state and b the generalized compressibility chart Compare the values obtained to the actual value of 0021796 m3kg and determine the error involved in each case SOLUTION The specific volume of refrigerant134a is to be determined assuming ideal and nonidealgas behavior Analysis The gas constant the critical pressure and the critical temperature of refrigerant134a are determined from Table A1 to be R 00815 kPam 3 kgK P cr 4059 MPa T cr 3742 K a The specific volume of refrigerant134a under the idealgas assumption is v RT P 00815 kPa m 3 kgK 323 K 1000 kPa 0026325 m 3 kg Therefore treating the refrigerant134a vapor as an ideal gas would result in an error of 0026325 00217960021796 0208 or 208 percent in this case b To determine the correction factor Z from the compressibility chart we first need to calculate the reduced pressure and temperature P R P P cr 1 MPa 4059 MPa 0246 T R T T cr 323 K 3742 K 0863 Z 084 Thus v Z v ideal 0840026325 m 3 kg 0022113 m 3 kg Discussion The error in this result is less than 2 percent Therefore in the absence of tabulated data the generalized compressibility chart can be used with confidence Final PDF to printer 139 CHAPTER 3 cen22672ch03109160indd 139 092217 1145 AM 38 OTHER EQUATIONS OF STATE The idealgas equation of state is very simple but its range of applicability is limited It is desirable to have equations of state that represent the PvT behavior of substances accurately over a larger region with no limitations Such equations are naturally more complicated Several equations have been proposed for this purpose Fig 354 but we shall discuss only three the van der Waals equation because it is one of the earliest the BeattieBridgeman FIGURE 353 Results obtained by using the compressibility chart are usually within a few percent of actual values Z chart 1056 P psia Exact 1000 Ideal gas 1228 from Example 312 EXAMPLE 312 Using Generalized Charts to Determine Pressure Determine the pressure of water vapor at 600F and 051431 ft3lbm using a the steam tables b the idealgas equation and c the generalized compressibility chart SOLUTION The pressure of water vapor is to be determined in three different ways Analysis A sketch of the system is given in Fig 352 The gas constant the critical pressure and the critical temperature of steam are determined from Table A1E to be R 05956 psiaft 3 lbmR P cr 3200 psia T cr 11648 R a The pressure at the specified state is determined from Table A6E to be v 051431 ft 3 lbm T 600F P 1000 psia This is the experimentally determined value and thus it is the most accurate b The pressure of steam under the idealgas assumption is determined from the idealgas relation to be P RT v 05956 psia ft 3 lbmR1060 R 051431 ft 3 lbm 1228 psia Therefore treating the steam as an ideal gas would result in an error of 1228 10001000 0228 or 228 percent in this case c To determine the correction factor Z from the compressibility chart Fig A15 we first need to calculate the pseudoreduced specific volume and the reduced temperature v R v actual R T cr P cr 051431 ft 3 lbm3200 psia 05956 psia ft 3 lbmR11648 R 2372 T R T T cr 1060 R 11648 R 091 P R 033 Thus P P R P cr 033 3200 psia 1056 psia Discussion Using the compressibility chart reduced the error from 228 to 56 percent which is acceptable for most engineering purposes Fig 353 A bigger chart of course would give better resolution and reduce the reading errors Notice that we did not have to determine Z in this problem since we could read PR directly from the chart FIGURE 352 Schematic for Example 312 H2O T 600F v 051431 ft3lbm P FIGURE 354 Several equations of state have been proposed throughout history van der Waals Berthelet RedlichKwang BeattieBridgeman BenedictWebbRubin Strobridge Virial Final PDF to printer 140 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 140 092217 1145 AM equation of state because it is one of the best known and is reasonably accu rate and the BenedictWebbRubin equation because it is one of the more recent and is very accurate van der Waals Equation of State The van der Waals equation of state was proposed in 1873 and it has two constants that are determined from the behavior of a substance at the critical point It is given by P a v 2 v b RT 322 Van der Waals intended to improve the idealgas equation of state by includ ing two of the effects not considered in the idealgas model the intermolecular attraction forces and the volume occupied by the molecules themselves The term av2 accounts for the intermolecular forces and b accounts for the vol ume occupied by the gas molecules In a room at atmospheric pressure and temperature the volume actually occupied by molecules is only about one thousandth of the volume of the room As the pressure increases the volume occupied by the molecules becomes an increasingly significant part of the total volume Van der Waals proposed to correct this by replacing v in the idealgas relation with the quantity v b where b represents the volume occupied by the gas molecules per unit mass The determination of the two constants appearing in this equation is based on the observation that the critical isotherm on a Pv diagram has a horizontal inflection point at the critical point Fig 355 Thus the first and the second derivatives of P with respect to v at the critical point must be zero That is P v T T cr const 0 and 2 P v 2 T T cr const 0 By performing the differentiations and eliminating vcr the constants a and b are determined to be a 27 R 2 T cr 2 64 P cr and b R T cr 8 P cr 323 The constants a and b can be determined for any substance from the criticalpoint data alone Table A1 The accuracy of the van der Waals equation of state is often inadequate but it can be improved by using values of a and b that are based on the actual behavior of the gas over a wider range instead of a single point Despite its limitations the van der Waals equation of state has a historical value in that it was one of the first attempts to model the behavior of real gases The van der Waals equation of state can also be expressed on a unitmole basis by replac ing the v in Eq 322 by v and the R in Eqs 322 and 323 by Ru BeattieBridgeman Equation of State The BeattieBridgeman equation proposed in 1928 is an equation of state based on five experimentally determined constants It is expressed as P R u T v 2 1 c v T 3 v B A v 2 324 FIGURE 355 Critical isotherm of a pure substance has an inflection point at the critical state P Critical point Tcr constant v Final PDF to printer 141 CHAPTER 3 cen22672ch03109160indd 141 092217 1145 AM where A A 0 1 a v and B B 0 1 b v 325 The constants appearing in this equation are given in Table 34 for various substances The BeattieBridgeman equation is known to be reasonably accu rate for densities up to about 08ρcr where ρcr is the density of the substance at the critical point BenedictWebbRubin Equation of State Benedict Webb and Rubin extended the BeattieBridgeman equation in 1940 by raising the number of constants to eight It is expressed as P R u T v B 0 R u T A 0 C 0 T 2 1 v 2 b R u T a v 3 aα v 6 c v 3 T 2 1 γ v 2 e γ v 2 326 The values of the constants appearing in this equation are given in Table 34 This equation can handle substances at densities up to about 25ρcr In 1962 Strobridge further extended this equation by raising the number of constants to 16 Fig 356 TABLE 34 Constants that appear in the BeattieBridgeman and the BenedictWebbRubin equations of state a When P is in kPa v is in m3kmol T is in K and Ru 8314 kPam3kmolK the five constants in the BeattieBridgeman equa tion are as follows Gas A0 a B0 b c Air 1318441 001931 004611 0001101 434 104 Argon Ar 1307802 002328 003931 00 599 104 Carbon dioxide CO2 5072836 007132 010476 007235 660 105 Helium He 21886 005984 001400 00 40 Hydrogen H2 200117 000506 002096 004359 504 Nitrogen N2 1362315 002617 005046 000691 420 104 Oxygen O2 1510857 002562 004624 0004208 480 104 b When P is in kPa v is in m3kmol T is in K and Ru 8314 kPam3kmolK the eight constants in the BenedictWebbRubin equation are as follows Gas a A0 b B0 c C0 α γ nButane C4H10 19068 10216 0039998 012436 3205 107 1006 108 1101 103 00340 Carbon dioxide CO2 1386 27730 0007210 004991 1511 106 1404 107 8470 105 000539 Carbon monoxide CO 371 13587 0002632 005454 1054 105 8673 105 1350 104 00060 Methane CH4 500 18791 0003380 004260 2578 105 2286 106 1244 104 00060 Nitrogen N2 254 10673 0002328 004074 7379 104 8164 105 1272 104 00053 Source Gordon J Van Wylen and Richard E Sonntag Fundamentals of Classical Thermodynamics EnglishSI Version 3rd ed New York John Wiley Sons 1986 p 46 table 33 Source Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 p 815 table A21M Originally published in H W Cooper and J C Goldfrank Hydrocarbon Processing 46 no 12 1967 p 141 FIGURE 356 Complex equations of state represent the PvT behavior of gases more accurately over a wider range van der Waals 2 constants Accurate over a limited range Strobridge 16 constants More suitable for computer calculations Virial may vary Accuracy depends on the number of terms used BeattieBridgeman 5 constants Accurate for ρ 08ρcr BenedictWebbRubin 8 constants Accurate for ρ 25ρcr Final PDF to printer 142 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 142 092217 1145 AM Virial Equation of State The equation of state of a substance can also be expressed in a series form as P RT v a T v 2 b T v 3 c T v 4 d T v 5 327 This and similar equations are called the virial equations of state and the coefficients aT bT cT and so on that are functions of temperature alone are called virial coefficients These coefficients can be determined experimentally or theoretically from statistical mechanics Obviously as the pressure approaches zero all the virial coefficients will vanish and the equa tion will reduce to the idealgas equation of state The PvT behavior of a substance can be represented accurately with the virial equation of state over a wider range by including a sufficient number of terms The equations of state discussed here are applicable to the gas phase of the substances only and thus should not be used for liquids or liquidvapor mixtures FIGURE 357 Percentage of error involved in various equations of state for nitrogen error vtable vequationvtable 100 v m3kmol T K 300 200 100 0 100 10 1 01 001 207 141 21 116 63 12 32 01 10 04 01 04 01 00 02 00 00 00 00 00 00 00 00 00 57 593 187 152 745 510 79 07 52 52 06 37 33 04 25 16 02 13 08 01 08 04 01 03 100 100 100 09 01 01 04 01 01 01 MPa 02 MPa 1 MPa 2 MPa 4 MPa 10 MPa 20 MPa 00125 MPa 00 00 00 00 00 00 01 00 00 05 00 00 10 01 01 19 01 01 42 01 02 47 02 02 01 00 00 01 00 00 05 00 00 05 01 00 11 01 00 23 01 00 53 01 01 37 01 04 12 01 01 28 01 01 67 07 01 29 03 07 van der Waals top BeattieBridgeman middle BenedictWebbRubin bottom Final PDF to printer 143 CHAPTER 3 cen22672ch03109160indd 143 092217 1145 AM Complex equations represent the PvT behavior of substances reasonably well and are very suitable for digital computer applications For hand cal culations however it is suggested that the reader use the property tables or the simpler equations of state for convenience This is particularly true for specificvolume calculations since all the earlier equations are implicit in v and require a trialanderror approach The accuracy of the van der Waals BeattieBridgeman and BenedictWebbRubin equations of state is illustrated in Fig 357 It is apparent from this figure that the Benedict WebbRubin equation of state is usually the most accurate EXAMPLE 313 Different Methods of Evaluating Gas Pressure Predict the pressure of nitrogen gas at T 175 K and v 000375 m3kg on the basis of a the idealgas equation of state b the van der Waals equation of state c the BeattieBridgeman equation of state and d the BenedictWebbRubin equation of state Compare the values obtained to the experimentally determined value of 10000 kPa SOLUTION The pressure of nitrogen gas is to be determined using four different equations of state Properties The gas constant of nitrogen gas is 02968 kPam3kgK Table A1 Analysis a Using the idealgas equation of state the pressure is found to be P RT v 02968 kPa m 3 kgK 175 K 000375 m 3 kg 13851 kPa which is in error by 385 percent b The van der Waals constants for nitrogen are determined from Eq 323 to be a 0175 m 6 kPa kg 2 b 000138 m 3 kg From Eq 322 P RT v b a v 2 9471 kPa which is in error by 53 percent c The constants in the BeattieBridgeman equation are determined from Table 34 to be A 10229 B 005378 c 42 10 4 Also v Mv 28013 kgkmol000375 m3kg 010505 m3kmol Substituting these values into Eq 324 we obtain P R u T v 2 1 c v T 3 v B A v 2 10110 kPa which is in error by 11 percent Final PDF to printer 144 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 144 092217 1145 AM FIGURE 358 Atmospheric pressure is the sum of the dry air pressure Pa and the vapor pressure Pv Air Water vapor Patm Pa Pv d The constants in the BenedictWebbRubin equation are determined from Table 34 to be a 254 A 0 10673 b 0002328 B 0 004074 c 7379 10 4 C 0 8164 10 5 α 1272 10 4 γ 00053 Substituting these values into Eq 326 gives P R u T v B 0 R u T A 0 C 0 T 2 1 v 2 b R u T a v 3 aα v 6 c v 3 T 2 1 γ v 2 e γ v 2 10009 kPa which is in error by only 009 percent Thus the accuracy of the BenedictWebb Rubin equation of state is rather impressive in this case The pressure in a gas container is due to the individual molecules striking the wall of the container and exerting a force on it This force is propor tional to the average velocity of the molecules and the number of molecules per unit volume of the container ie molar density Therefore the pres sure exerted by a gas is a strong function of the density and the temperature of the gas For a gas mixture the pressure measured by a sensor such as a transducer is the sum of the pressures exerted by the individual gas species called the partial pressure It can be shown see Chap 13 that the partial pressure of a gas in a mixture is proportional to the number of moles or the mole fraction of that gas Atmospheric air can be viewed as a mixture of dry air air with zero mois ture content and water vapor also referred to as moisture and the atmo spheric pressure is the sum of the pressure of dry air Pa and the pressure of water vapor called the vapor pressure Pv Fig 358 That is P atm P a P v 328 Note that in some applications the phrase vapor pressure is used to indi cate saturation pressure The vapor pressure constitutes a small fraction usu ally under 3 percent of the atmospheric pressure since air is mostly nitrogen and oxygen and the water molecules constitute a small fraction usually under 3 percent of the total molecules in the air However the amount of water vapor in the air has a major impact on thermal comfort and many processes such as drying This section can be skipped without a loss in continuity TOPIC OF SPECIAL INTEREST Vapor Pressure and Phase Equilibrium Final PDF to printer 145 CHAPTER 3 cen22672ch03109160indd 145 092217 1145 AM FIGURE 359 Whenever there is a concentration difference of a physical quantity in a medium nature tends to equalize things by forcing a flow from the high to the lowconcentration region a Before Water Salt b After Salty water Air can hold a certain amount of moisture only and the ratio of the actual amount of moisture in the air at a given temperature to the maximum amount air can hold at that temperature is called the relative humidity ϕ The relative humidity ranges from 0 for dry air to 100 percent for saturated air air that cannot hold any more moisture The vapor pressure of saturated air at a given temperature is equal to the saturation pressure of water at that temperature For example the vapor pressure of saturated air at 25C is 317 kPa The amount of moisture in the air is completely specified by the tempera ture and the relative humidity and the vapor pressure is related to relative humidity ϕ by P v ϕ P sat T 329 where Psat T is the saturation pressure of water at the specified temperature For example the vapor pressure of air at 25C and 60 percent relative humidity is P v ϕ P sat 25C 06 317 kPa 190 kPa The desirable range of relative humidity for thermal comfort is 40 to 60 percent Note that the amount of moisture air can hold is proportional to the sat uration pressure which increases with temperature Therefore air can hold more moisture at higher temperatures Dropping the temperature of moist air reduces its moisture capacity and may result in the condensation of some of the moisture in the air as suspended water droplets fog or as a liquid film on cold surfaces dew So it is no surprise that fog and dew are common occurrences at humid locations especially in the early morning hours when the temperatures are the lowest Both fog and dew disappear evaporate as the air temperature rises shortly after sunrise You also may have noticed that elec tronic devices such as camcorders come with warnings against bringing them into humid indoor environments when the devices are cold to prevent moisture from condensing on their sensitive electronics It is commonly seen that whenever there is an imbalance of a commodity in a medium nature tends to redistribute it until a balance or equality is established This tendency is often referred to as the driving force which is the mechanism behind many naturally occurring transport phenomena such as heat transfer fluid flow electric current and mass transfer If we define the amount of a commodity per unit volume as the concentration of that com modity we can say that the flow of a commodity is always in the direction of decreasing concentration that is from the region of high concentration to the region of low concentration Fig 359 The commodity simply creeps away during redistribution and thus the flow is a diffusion process We know from experience that a wet Tshirt hanging in an open area eventu ally dries a small amount of water left in a glass evaporates and the aftershave in an open bottle quickly disappears These and many other similar examples suggest that there is a driving force between the two phases of a substance that forces the mass to transform from one phase to another The magnitude of this force depends on the relative concentrations of the two phases A wet Tshirt dries much faster in dry air than it would in humid air In fact it does not dry at all if the relative humidity of the environment is 100 percent and thus the air is saturated In this case there is no transformation from the liquid phase Final PDF to printer 146 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 146 092217 1145 AM FIGURE 360 When open to the atmosphere water is in phase equilibrium with the vapor in the air if the vapor pressure is equal to the saturation pressure of water Water vapor Pv Liquid water T to the vapor phase and the two phases are in phase equilibrium For liquid water that is open to the atmosphere the criterion for phase equilibrium can be expressed as follows The vapor pressure in the air must be equal to the satu ration pressure of water at the water temperature That is Fig 360 Phase equilibrium criterion for water exposed to air P v P sat T 330 Therefore if the vapor pressure in the air is less than the saturation pressure of water at the water temperature some liquid will evaporate The larger the difference between the vapor and saturation pressures the higher the rate of evaporation The evaporation has a cooling effect on water and thus reduces its temperature This in turn reduces the saturation pressure of water and thus the rate of evaporation until some kind of quasisteady operation is reached This explains why water is usually at a considerably lower temperature than the surrounding air especially in dry climates It also suggests that the rate of evaporation of water can be increased by increasing the water temperature and thus the saturation pressure of water Note that the air at the water surface is always saturated because of the direct contact with water and thus the vapor pressure Therefore the vapor pres sure at the lake surface is the saturation pressure of water at the temperature of the water at the surface If the air is not saturated then the vapor pressure decreases to the value in the air at some distance from the water surface and the difference between these two vapor pressures is the driving force for the evaporation of water The natural tendency of water to evaporate in order to achieve phase equilib rium with the water vapor in the surrounding air forms the basis for the opera tion of evaporative coolers also called the swamp coolers In such coolers hot and dry outdoor air is forced to flow through a wet cloth before entering a building Some of the water evaporates by absorbing heat from the air and thus cooling it Evaporative coolers are commonly used in dry climates and provide effective cooling They are much cheaper to run than air conditioners since they are inexpensive to buy and the fan of an evaporative cooler con sumes much less power than the compressor of an air conditioner Boiling and evaporation are often used interchangeably to indicate phase change from liquid to vapor Although they refer to the same physical process they differ in some aspects Evaporation occurs at the liquidvapor interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature Water in a lake at 20C for example evaporates to air at 20C and 60 percent relative humidity since the saturation pressure of water at 20C is 234 kPa and the vapor pressure of air at 20C and 60 percent relative humidity is 14 kPa Other examples of evaporation are the drying of clothes fruits and vegetables the evaporation of sweat to cool the human body and the rejection of waste heat in wet cooling towers Note that evapora tion involves no bubble formation or bubble motion Fig 361 Boiling on the other hand occurs at the solidliquid interface when a liquid is brought into contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid At 1 atm for example liquid water in contact with a solid surface at 110C boils since the saturation temperature of water at 1 atm is 100C The boiling process Final PDF to printer 147 CHAPTER 3 cen22672ch03109160indd 147 092217 1145 AM EXAMPLE 314 Temperature Drop of a Lake Due to Evaporation On a summer day the air temperature over a lake is measured to be 25C Determine the water temperature of the lake when phase equilibrium conditions are established between the water in the lake and the vapor in the air for relative humidities of 10 80 and 100 percent for the air Fig 362 SOLUTION Air at a specified temperature is blowing over a lake The equilibrium temperatures of water for three different cases are to be determined Analysis The saturation pressure of water at 25C from Table 31 is 317 kPa Then the vapor pressures at relative humidities of 10 80 and 100 percent are determined from Eq 329 to be Relative humidity 10 P v1 ϕ 1 P sat 25C 01 317 kPa 0317 kPa Relative humidity 80 P v2 ϕ 2 P sat 25C 08 317 kPa 2536 kPa Relative humidity 100 P v3 ϕ 3 P sat 25C 10 317 kPa 317 kPa The saturation temperatures corresponding to these pressures are determined from Table 31 or Table A5 by interpolation to be T 1 80C T 2 212C and T 3 25C FIGURE 362 Schematic for Example 314 Air 25C 10 Pv Pv Psat T T Lake ϕ is characterized by the rapid motion of vapor bubbles that form at the solidliquid interface detach from the surface when they reach a certain size and attempt to rise to the free surface of the liquid When cooking we do not say water is boiling unless we see the bubbles rising to the top FIGURE 361 A liquidtovapor phase change process is called evaporation if it occurs at a liquidvapor interface and boiling if it occurs at a solid liquid interface John A RizzoGetty Images RF David ChaseyGetty Images RF Final PDF to printer 148 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 148 092217 1145 AM Therefore water will freeze in the first case even though the surrounding air is hot In the last case the water temperature will be the same as the surrounding air temperature Discussion You are probably skeptical about the lake freezing when the air is at 25C and you are right The water temperature drops to 8C in the limiting case of no heat transfer to the water surface In practice the water temperature drops below the air temperature but it does not drop to 8C because 1 it is very unlikely for the air over the lake to be so dry a relative humidity of just 10 percent and 2 as the water temperature near the surface drops heat transfer from the air and the lower parts of the water body will tend to make up for this heat loss and prevent the water temperature from dropping too much The water temperature stabilizes when the heat gain from the surrounding air and the water body equals the heat loss by evaporation that is when a dynamic balance is established between heat and mass transfer instead of phase equilibrium If you try this experiment using a shallow layer of water in a wellinsulated pan you can actually freeze the water if the air is very dry and rela tively cool SUMMARY A substance that has a fixed chemical composition through out is called a pure substance A pure substance exists in different phases depending on its energy level In the liquid phase a substance that is not about to vaporize is called a compressed or subcooled liquid In the gas phase a substance that is not about to condense is called a superheated vapor During a phasechange process the temperature and pressure of a pure substance are dependent properties At a given pres sure a substance changes phase at a fixed temperature called the saturation temperature Likewise at a given temperature the pressure at which a substance changes phase is called the saturation pressure During a boiling process both the liquid and the vapor phases coexist in equilibrium and under this condition the liquid is called saturated liquid and the vapor saturated vapor In a saturated liquidvapor mixture the mass fraction of vapor is called the quality and is expressed as x m vapor m total Quality may have values between 0 saturated liquid and 1 saturated vapor It has no meaning in the compressed liquid or superheated vapor regions In the saturated mixture region the average value of any intensive property y is deter mined from y y f x y fg where f stands for saturated liquid and g for saturated vapor In the absence of compressed liquid data a general approxi mation is to treat a compressed liquid as a saturated liquid at the given temperature y y f T where y stands for v u or h The state beyond which there is no distinct vaporization process is called the critical point At supercritical pressures a substance gradually and uniformly expands from the liq uid to vapor phase All three phases of a substance coexist in equilibrium at states along the triple line characterized by tripleline temperature and pressure The compressed liquid has lower v u and h values than the saturated liquid at the same T or P Likewise superheated vapor has higher v u and h values than the saturated vapor at the same T or P Any relation among the pressure temperature and specific volume of a substance is called an equation of state The sim plest and bestknown equation of state is the idealgas equa tion of state given as Pv RT where R is the gas constant Caution should be exercised in using this relation since an ideal gas is a fictitious substance Real gases exhibit idealgas behavior at relatively low pres sures and high temperatures The deviation from idealgas behavior can be properly accounted for by using the compressibility factor Z defined as Z Pv RT or Z v actual v ideal Final PDF to printer 149 CHAPTER 3 cen22672ch03109160indd 149 092217 1145 AM The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure which are defined as T R T T cr and P R P P cr where Pcr and Tcr are the critical pressure and temperature respectively This is known as the principle of corresponding states When either P or T is unknown it can be determined from the compressibility chart with the help of the pseudo reduced specific volume defined as v R v actual R T cr P cr The PvT behavior of substances can be represented more accurately by more complex equations of state Three of the best known are van der Waals P a v 2 v b RT where a 27 R 2 T cr 2 64 P cr and b R T cr 8 P cr BeattieBridgeman P R u T v 2 1 c v T 3 v B A v 2 where A A 0 1 a v and B B 0 1 b v BenedictWebbRubin P R u T v B 0 R u T A 0 C 0 T 2 1 v 2 b R u T a v 3 aα v 6 c v 3 T 2 1 γ v 2 e γ v 2 where Ru is the universal gas constant and v is the molar spe cific volume REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Fundamentals SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 2 ASHRAE Handbook of Refrigeration SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1994 3 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley 2006 4 M Kostic Analysis of Enthalpy Approximation for Compressed Liquid Water ASME J Heat Transfer Vol 128 pp 421426 2006 PROBLEMS Pure Substances PhaseChange Processes Property Diagrams 31C A propane tank is filled with a mixture of liquid and vapor propane Can the contents of this tank be considered a pure substance Explain 32C Is iced water a pure substance Why 33C What is the difference between saturated vapor and superheated vapor 34C What is the difference between saturated liquid and compressed liquid 35C If the pressure of a substance is increased during a boiling process will the temperature also increase or will it remain constant Why 36C Is it true that water boils at higher temperature at higher pressure Explain 37C What is the difference between the critical point and the triple point 38C A househusband is cooking beef stew for his family in a pan that is a uncovered b covered with a light lid and c covered with a heavy lid For which case will the cooking time be the shortest Why 39C How does a boiling process at supercritical pressures differ from the boiling process at subcritical pressures Problems designated by a C are concept questions and stu dents are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer 150 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 150 092217 1145 AM 323E Complete this table for H2O T F P psia u Btulbm Phase description 300 782 40 Saturated liquid 500 120 400 400 Property Tables 310C What is quality Does it have any meaning in the superheated vapor region 311C Does the amount of heat absorbed as 1 kg of saturated liquid water boils at 100C have to be equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100C 312C Does the reference point selected for the properties of a substance have any effect on thermodynamic analysis Why 313C What is the physical significance of hfg Can it be obtained from a knowledge of hf and hg How 314C Does hfg change with pressure How 315C Is it true that it takes more energy to vaporize 1 kg of saturated liquid water at 100C than it would at 120C 316C Which process requires more energy completely vaporizing 1 kg of saturated liquid water at 1 atm pressure or completely vaporizing 1 kg of saturated liquid water at 8 atm pressure 317C In what kind of pot will a given volume of water boil at a higher temperature a tall and narrow one or a short and wide one Explain 318C It is well known that warm air in a cooler environ ment rises Now consider a warm mixture of air and gasoline on top of an open gasoline can Do you think this gas mixture will rise in a cooler environment 319C In the absence of compressed liquid tables how is the specific volume of a compressed liquid at a given P and T determined 320C A perfectly fitting pot and its lid often stick after cooking and it becomes very difficult to open the lid when the pot cools down Explain why this happens and what you would do to open the lid 321 Complete this table for H2O T C P kPa u kJkg Phase description 400 1450 220 Saturated vapor 190 2500 4000 3040 322 Complete this table for H2O T C P kPa v m3kg Phase description 140 0035 550 Saturated liquid 125 750 300 0140 324E Reconsider Prob 323E Using appropriate software determine the missing properties of water Repeat the solution for refrigerant134a refrigerant22 and ammonia 325 Complete this table for H2O T C P kPa h kJkg x Phase description 200 07 140 1800 950 00 80 500 800 31622 326E Complete this table for refrigerant134a T F P psia h Btulbm x Phase description 80 78 15 06 10 70 180 12946 110 10 327 Complete this table for refrigerant134a T C P kPa u kJkg Phase description 20 95 12 Saturated liquid 400 300 8 600 328 A 18m3 rigid tank contains steam at 220C Onethird of the volume is in the liquid phase and the rest is in the vapor form Determine a the pressure of the steam b the quality of the saturated mixture and c the density of the mixture FIGURE P328 Steam 18 m3 220C 329E One poundmass of water fills a container whose volume is 2 ft3 The pressure in the container is 100 psia Calculate the total internal energy and enthalpy in the container Answers 661 Btu 698 Btu Final PDF to printer 151 CHAPTER 3 cen22672ch03109160indd 151 092217 1145 AM 330 A pistoncylinder device contains 085 kg of refrigerant 134a at 10C The piston that is free to move has a mass of 12 kg and a diameter of 25 cm The local atmospheric pressure is 88 kPa Now heat is transferred to refrigerant134a until the temperature is 15C Determine a the final pressure b the change in the volume of the cylinder and c the change in the enthalpy of the refrigerant134a pressure inside the cooker in psia and in atm Would you mod ify your answer if the place were at a higher elevation FIGURE P338E Pressure cooker 250F 339E How much error would one expect in determining the specific enthalpy by applying the incompressibleliquid approximation to water at 3000 psia and 400F 340 Water is to be boiled at sea level in a 30cmdiameter stainless steel pan placed on top of a 3kW electric burner If 60 percent of the heat generated by the burner is transferred to the water during boiling determine the rate of evaporation of water FIGURE P340 Vapor 3 kW 40 60 341 Repeat Prob 340 for a location at an elevation of 1500 m where the atmospheric pressure is 845 kPa and thus the boiling temperature of water is 95C 342 10 kg of R134a at 300 kPa fills a rigid container whose volume is 14 L Determine the temperature and total enthalpy in the container The container is now heated until the pressure is 600 kPa Determine the temperature and total enthalpy when the heating is completed FIGURE P337 Evacuated Water 1 kg 11989 m3 200 kPa 331 10 kg of R134a fill a 1115m3 rigid container at an initial temperature of 30C The container is then heated until the pressure is 200 kPa Determine the final temperature and the initial pressure Answers 142C 8443 kPa 332 What is the specific internal energy of water at 50 kPa and 200C 333 What is the specific volume of water at 5 MPa and 100C What would it be if the incompressible liquid approxima tion were used Determine the accuracy of this approximation 334 What is the specific volume of R134a at 20C and 700 kPa What is the internal energy at that state 335 Refrigerant134a at 200 kPa and 25C flows through a refrigeration line Determine its specific volume 336 One kilogram of R134a fills a 014m3 weighted pistoncylinder device at a temperature of 264C The con tainer is now heated until the temperature is 100C Determine the final volume of the R134a Answer 03014 m3 337 One kilogram of water vapor at 200 kPa fills the 11989m3 left chamber of a partitioned system shown in Fig P337 The right chamber has twice the volume of the left and is initially evacuated Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 3C FIGURE P342 R134a 300 kPa 10 kg 14 L Q 338E The temperature in a pressure cooker during cooking at sea level is measured to be 250F Determine the absolute FIGURE P330 Final PDF to printer 152 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 152 092217 1145 AM 343 100 kg of R134a at 200 kPa are contained in a pistoncylinder device whose volume is 12322 m3 The piston is now moved until the volume is onehalf its original size This is done such that the pressure of the R134a does not change Determine the final temperature and the change in the total internal energy of the R134a 344 Water initially at 200 kPa and 300C is contained in a pistoncylinder device fitted with stops The water is allowed to cool at constant pressure until it exists as a saturated vapor and the piston rests on the stops Then the water continues to cool until the pressure is 100 kPa On the Tv diagram sketch with respect to the saturation lines the process curves pass ing through the initial intermediate and final states of the water Label the T P and v values for end states on the process curves Find the overall change in internal energy between the initial and final states per unit mass of water as heat determine the local atmospheric pressure in that loca tion Answer 854 kPa 350 A rigid tank with a volume of 18 m3 contains 40 kg of saturated liquidvapor mixture of water at 90C Now the water is slowly heated Determine the temperature at which the liquid in the tank is completely vaporized Also show the process on a Tv diagram with respect to saturation lines Answer 256C 351 A pistoncylinder device contains 0005 m3 of liquid water and 09 m3 of water vapor in equilibrium at 600 kPa Heat is transferred at constant pressure until the temperature reaches 200C a What is the initial temperature of the water b Determine the total mass of the water c Calculate the final volume d Show the process on a Pv diagram with respect to saturation lines FIGURE P351 P 600 kPa H2O Q FIGURE P344 345 Saturated steam coming off the turbine of a steam power plant at 40C condenses on the outside of a 3cmouter diameter 35mlong tube at a rate of 70 kgh Determine the rate of heat transfer from the steam to the cooling water flow ing through the pipe 346 A person cooks a meal in a 30cmdiameter pot that is covered with a wellfitting lid and lets the food cool to the room temperature of 20C The total mass of the food and the pot is 8 kg Now the person tries to open the pan by lifting the lid up Assuming no air has leaked into the pan dur ing cooling determine if the lid will open or if the pan will move up together with the lid 347 Water is boiled at 1 atm pressure in a 25cminternal diameter stainless steel pan on an electric range If it is observed that the water level in the pan drops by 10 cm in 45 min determine the rate of heat transfer to the pan 348 Repeat Prob 347 for a location at 2000m elevation where the standard atmospheric pressure is 795 kPa 349 Water is boiled in a pan covered with a poorly fitting lid at a specified location Heat is supplied to the pan by a 2kW resistance heater The amount of water in the pan is observed to decrease by 119 kg in 30 min If it is estimated that 75 percent of electricity consumed by the heater is transferred to the water 352 Reconsider Prob 351 Using appropriate software investigate the effect of pressure on the total mass of water in the tank Let the pressure vary from 01 MPa to 1 MPa Plot the total mass of water against pressure and discuss the results Also show the process in Prob 351 on a Pv diagram using the property plot feature of the software 353E A 5ft3 rigid tank contains a saturated mixture of refrigerant134a at 50 psia If the saturated liquid occupies 20 percent of the volume determine the quality and the total mass of the refrigerant in the tank 354E Superheated water vapor at 180 psia and 500F is allowed to cool at constant volume until the temperature drops to 250F At the final state determine a the pressure b the quality and c the enthalpy Also show the process on a Tv diagram with respect to saturation lines Answers a 2984 psia b 0219 c 4260 Btulbm 355E Reconsider Prob 354E Using appropriate software investigate the effect of initial pres sure on the quality of water at the final state Let the pressure vary from 100 psia to 300 psia Plot the quality against initial pressure and discuss the results Also show the process in Prob 354E on a Tv diagram using the property plot fea ture of the software Final PDF to printer 153 CHAPTER 3 cen22672ch03109160indd 153 092217 1145 AM 356 One kilogram of water fills a 150L rigid container at an initial pressure of 2 MPa The container is then cooled to 40C Determine the initial temperature and the final pressure of the water FIGURE P356 357 10 kg of R134a fill a 07m3 weighted pistoncylinder device at a pressure of 200 kPa The container is now heated until the temperature is 30C Determine the initial tempera ture and final volume of the R134a 358 A pistoncylinder device contains 06 kg of steam at 300C and 05 MPa Steam is cooled at constant pressure until onehalf of the mass condenses a Show the process on a Tv diagram b Find the final temperature c Determine the volume change 359 A pistoncylinder device initially contains 14 kg sat urated liquid water at 200C Now heat is transferred to the water until the volume quadruples and the cylinder contains saturated vapor only Determine a the volume of the cylin der b the final temperature and pressure and c the internal energy change of the water FIGURE P359 360 Water is being heated in a vertical pistoncylinder device The piston has a mass of 40 kg and a cross sectional area of 150 cm2 If the local atmospheric pressure is 100 kPa determine the temperature at which the water starts boiling 361 A rigid tank initially contains 14 kg saturated liquid water at 200C At this state 25 percent of the volume is occu pied by water and the rest by air Now heat is supplied to the water until the tank contains saturated vapor only Determine a the volume of the tank b the final temperature and pres sure and c the internal energy change of the water 362 A pistoncylinder device initially contains 50 L of liquid water at 40C and 200 kPa Heat is transferred to the water at constant pressure until the entire liquid is vaporized a What is the mass of the water b What is the final temperature c Determine the total enthalpy change d Show the process on a Tv diagram with respect to saturation lines Answers a 4961 kg b 12021C c 125950 kJ 363 The springloaded pistoncylinder device shown in Fig P363 is filled with 05 kg of water vapor that is initially at 4 MPa and 400C Initially the spring exerts no force against the piston The spring constant in the spring force relation F kx is k 09 kNcm and the piston diameter is D 20 cm The water now undergoes a process until its volume is onehalf of the original volume Calculate the final temperature and the specific enthalpy of the water Answers 220C 1721 kJkg FIGURE P363 D Spring Steam 364 A pistoncylinder device initially contains steam at 35 MPa superheated by 5C Now steam loses heat to the surroundings and the piston moves down hitting a set of stops at which point the cylinder contains saturated liquid water The cooling continues until the cylinder contains water at 200C Determine a the initial temperature b the enthalpy change per unit mass of the steam by the time the piston first hits the stops and c the final pressure and the quality if mixture FIGURE P364 Final PDF to printer 154 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 154 092217 1145 AM until the temperature is 0C Determine the change of the vol ume of the device during this cooling 377 A mass of 01 kg of helium fills a 02 m3 rigid vessel at 350 kPa The vessel is heated until the pressure is 700 kPa Calculate the temperature change of helium in C and K as a result of this heating FIGURE P377 Helium 01 kg 02 m3 350 kPa Q 378 A rigid tank whose volume is unknown is divided into two parts by a partition One side of the tank contains an ideal gas at 927C The other side is evacuated and has a volume twice the size of the part containing the gas The par tition is now removed and the gas expands to fill the entire tank Heat is now transferred to the gas until the pressure equals the initial pressure Determine the final temperature of the gas Answer 3327C FIGURE P378 379E A rigid tank contains 20 lbm of air at 20 psia and 70F More air is added to the tank until the pressure and tem perature rise to 25 psia and 90F respectively Determine the amount of air added to the tank Answer 409 lbm 380E In an informative article in a magazine it is stated that tires lose roughly 1 psi of pressure for every 10F drop in out side temperature Investigate whether this is a valid statement Compressibility Factor 381C What is the physical significance of the compressibil ity factor Z 382 Determine the specific volume of refrigerant134a vapor at 09 MPa and 70C based on a the idealgas equation b the generalized compressibility chart and c data from tables Also determine the error involved in the first two cases 383E Refrigerant134a at 400 psia has a specific volume of 01384 ft3lbm Determine the temperature of the refrigerant Ideal Gas 365C Under what conditions is the idealgas assumption suitable for real gases 366C What is the difference between mass and molar mass How are these two related 367C Propane and methane are commonly used for heat ing in winter and the leakage of these fuels even for short periods poses a fire danger for homes Which gas leakage do you think poses a greater risk for fire Explain 368E What is the specific volume of oxygen at 25 psia and 80F 369 A 100L container is filled with 1 kg of air at a tem perature of 27C What is the pressure in the container 370E A mass of 1 lbm of argon is maintained at 200 psia and 100F in a tank What is the volume of the tank 371 A 400L rigid tank contains 5 kg of air at 25C Deter mine the reading on the pressure gage if the atmospheric pres sure is 97 kPa 372 The pressure gage on a 25m3 oxygen tank reads 500 kPa Determine the amount of oxygen in the tank if the temperature is 28C and the atmospheric pressure is 97 kPa 373 A spherical balloon with a diameter of 9 m is filled with helium at 27C and 200 kPa Determine the mole number and the mass of the helium in the balloon Answers 306 kmol 123 kg 374 Reconsider Prob 373 Using appropriate soft ware investigate the effect of the balloon diameter on the mass of helium contained in the balloon for the pressures of a 100 kPa and b 200 kPa Let the diameter vary from 5 m to 15 m Plot the mass of helium against the diameter for both cases 375 A 1m3 tank containing air at 10C and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35C and 150 kPa Now the valve is opened and the entire system is allowed to reach thermal equilibrium with the surroundings which are at 20C Determine the volume of the second tank and the final equilibrium pressure of air Answers 177 m3 222 kPa 376 A mass of 10 g of oxygen fill a weighted piston cylinder device at 20 kPa and 100C The device is now cooled FIGURE P372 V 25 m3 O2 T 28C Pg 500 kPa Final PDF to printer 155 CHAPTER 3 cen22672ch03109160indd 155 092217 1145 AM based on a the idealgas equation b the generalized com pressibility chart and c the refrigerant tables 384 Determine the specific volume of superheated water vapor at 15 MPa and 350C using a the idealgas equation b the generalized compressibility chart and c the steam tables Also determine the error involved in the first two cases Answers a 001917 m3kg 670 percent b 001246 m3kg 85 percent c 001148 m3kg 385 Reconsider Prob 384 Solve the problem using appropriate software Compare the specific vol ume of water for the three cases at 15 MPa over the tempera ture range of 350 to 600C in 25C intervals Plot the percent error involved in the idealgas approximation against tempera ture and discuss the results 386 Determine the specific volume of superheated water vapor at 35 MPa and 450C based on a the idealgas equa tion b the generalized compressibility chart and c the steam tables Determine the error involved in the first two cases 387 Determine the specific volume of nitrogen gas at 10 MPa and 150 K based on a the ideagas equation and b the generalized compressibility chart Compare these results with the experimental value of 0002388 m3kg and determine the error involved in each case Answers a 0004452 m3kg 864 percent b 0002404 m3kg 07 percent 388 Ethylene is heated at constant pressure from 5 MPa and 20C to 200C Using the compressibility chart determine the change in the ethylenes specific volume as a result of this heating Answer 00172 m3kg 389 Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kgs CO2 is cooled at constant pressure as it flows in the pipe and the temperature of the CO2 drops to 450 K at the exit Determine the volume flow rate and the density of carbon dioxide at the inlet and the volume flow rate at the exit of the pipe using a the idealgas equation and b the generalized compressibility chart Also determine c the error involved in the first case temperature using the idealgas equation of state the com pressibility charts and the steam tables 393 What is the percentage of error involved in treat ing carbon dioxide at 5 MPa and 25C as an ideal gas Answer 45 percent Other Equations of State 394C What is the physical significance of the two constants that appear in the van der Waals equation of state On what basis are they determined 395E Refrigerant134a at 400 psia has a specific volume of 01144 ft3lbm Determine the temperature of the refriger ant based on a the idealgas equation b the van der Waals equation and c the refrigerant tables 396 A 327m3 tank contains 100 kg of nitrogen at 175 K Determine the pressure in the tank using a the idealgas equation b the van der Waals equation and c the Beattie Bridgeman equation Compare your results with the actual value of 1505 kPa 397 Nitrogen at 150 K has a specific volume of 0041884 m3kg Determine the pressure of the nitrogen using a the idealgas equation and b the BeattieBridgeman equation Compare your results to the experimental value of 1000 kPa Answers a 1063 kPa b 10004 kPa 398 Reconsider Prob 397 Using appropriate soft ware compare the pressure results of the idealgas and BeattieBridgeman equations with nitrogen data supplied by the software Plot temperature versus specific volume for a pres sure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K T 150 K 399 100 grams of carbon monoxide are contained in a weighted pistoncylinder device Initially the carbon monox ide is at 1000 kPa and 200C It is then heated until its tem perature is 500C Determine the final volume of the carbon monoxide treating it as a an ideal gas and b a Benedict WebbRubin gas 3100 A 1m3 tank contains 2841 kg of steam at 06 MPa Determine the temperature of the steam using a the idealgas equation b the van der Waals equation and c the steam tables Answers a 4576 K b 4659 K c 473 K 3101 Reconsider Prob 3100 Solve the problem using appropriate software Compare the tem perature of water for the three cases at constant specific vol ume over the pressure range of 01 MPa to 1 MPa in 01MPa increments Plot the percent error involved in the idealgas approximation against pressure and discuss the results 3102E 1 lbm of carbon dioxide is heated in a constant pressure apparatus Initially the carbon dioxide is at 1000 psia and 200F and it is heated until its temperature becomes 800F Determine the final volume of the carbon dioxide treating it as a an ideal gas and b a BenedictWebbRubin gas FIGURE P389 3 MPa 500 K 2 kgs 450 K CO2 390E Ethane in a rigid vessel is to be heated from 50 psia and 100F until its temperature is 540F What is the final pres sure of the ethane as predicted by the compressibility chart 391 A 0016773m3 tank contains 1 kg of refrigerant134a at 110C Determine the pressure of the refrigerant using a the idealgas equation b the generalized compressibility chart and c the refrigerant tables Answers a 1861 MPa b 1583 MPa c 16 MPa 392E Saturated water vapor at 400F is heated at constant pressure until its volume has doubled Determine the final Final PDF to printer 156 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 156 092217 1145 AM Special Topic Vapor Pressure and Phase Equilibrium 3103 During a hot summer day at the beach when the air temperature is 30C someone claims the vapor pressure in the air to be 52 kPa Is this claim reasonable 3104 Consider a glass of water in a room that is at 20C and 40 percent relative humidity If the water temperature is 15C determine the vapor pressure a at the free surface of the water and b at a location in the room far from the glass 3105 On a certain day the temperature and relative humid ity of air over a large swimming pool are measured to be 25C and 60 percent respectively Determine the water temperature of the pool when phase equilibrium conditions are established between the water in the pool and the vapor in the air 3106 During a hot summer day when the air temperature is 35C and the relative humidity is 70 percent you buy a sup posedly cold canned drink from a store The store owner claims that the temperature of the drink is below 10C Yet the drink does not feel so cold and you are skeptical since you notice no condensation forming outside the can Can the store owner be telling the truth 3107 Consider two rooms that are identical except that one is maintained at 25C and 40 percent relative humidity while the other is maintained at 20C and 55 percent relative humidity Noting that the amount of moisture is proportional to the vapor pressure determine which room contains more moisture 3108E A thermos bottle is halffilled with water and is left open to the atmospheric air at 60F and 35 percent relative humidity If heat transfer to the water through the thermos walls and the free surface is negligible determine the tempera ture of water when phase equilibrium is established Review Problems 3109 Complete the blank cells in the following table of properties of steam In the last column describe the condition of steam as compressed liquid saturated mixture superheated vapor or insufficient information and if applicable give the quality P kPa T C v m3kg u kJkg Phase description and quality if applicable 200 30 2703 130 400 15493 300 0500 500 3084 mixture superheated vapor or insufficient information and if applicable give the quality FIGURE P3112 3 MPa 500 K 04 kmols 450 K CO2 P kPa T C v m3kg u kJkg Phase description and quality if applicable 320 12 1000 3937 40 117794 180 00700 200 249 3111 A rigid tank contains an ideal gas at 300 kPa and 600 K Now half of the gas is withdrawn from the tank and the gas is found at 100 kPa at the end of the process Determine a the final temperature of the gas and b the final pressure if no mass was withdrawn from the tank and the same final tempera ture was reached at the end of the process 3112 Carbon dioxide gas at 3 MPa and 500 K flows steadily in a pipe at a rate of 04 kmols Determine a the volume and mass flow rates and the density of carbon dioxide at this state If CO2 is cooled at constant pressure as it flows in the pipe so that the temperature of CO2 drops to 450 K at the exit of the pipe determine b the volume flow rate at the exit of the pipe 3113 The gage pressure of an automobile tire is measured to be 200 kPa before a trip and 220 kPa after the trip at a loca tion where the atmospheric pressure is 90 kPa Assuming the volume of the tire remains constant at 0035 m3 determine the percent increase in the absolute temperature of the air in the tire 3114 A tank contains argon at 600C and 200 kPa gage The argon is cooled in a process by heat transfer to the sur roundings such that the argon reaches a final equilibrium state at 300C Determine the final gage pressure of the argon Assume atmospheric pressure is 100 kPa FIGURE P3111 Ideal gas 300 kPa 600 K 3110 Complete the blank cells in the following table of properties of refrigerant134a In the last column describe the condition of refrigerant134a as compressed liquid saturated Final PDF to printer 157 CHAPTER 3 cen22672ch03109160indd 157 092217 1145 AM 3115 The combustion in a gasoline engine may be approxi mated by a constantvolume heat addition process The cylin der contains the airfuel mixture before combustion and the combustion gases after it and both may be approximated as air an ideal gas In a gasoline engine the cylinder conditions are 12 MPa and 450C before the combustion and 1900C after it Determine the pressure at the end of the combustion process Answer 361 MPa 3121 A 10kg mass of superheated refrigerant134a at 12 MPa and 70C is cooled at constant pressure until it exists as a compressed liquid at 20C a Show the process on a Tv diagram with respect to saturation lines b Determine the change in volume c Find the change in total internal energy Answers b 0187 m3 c 1984 kJ 3122 A 4L rigid tank contains 2 kg of saturated liquid vapor mixture of water at 50C The water is now slowly heated until it exists in a single phase At the final state will the water be in the liquid phase or the vapor phase What would your answer be if the volume of the tank were 400 L instead of 4 L FIGURE P3123 Steam 02 kg 200 kPa 300C Q FIGURE P3122 3116 A rigid tank contains nitrogen gas at 227C and 100 kPa gage The gas is heated until the gage pressure reads 250 kPa If the atmospheric pressure is 100 kPa determine the final temperature of the gas in C 3117 One kilogram of R134a fills a 0090m3 rigid con tainer at an initial temperature of 40C The container is then heated until the pressure is 280 kPa Determine the initial pres sure and final temperature Answers 5125 kPa 50C 3118 A rigid tank with a volume of 0117 m3 contains 1 kg of refrigerant134a vapor at 240 kPa The refrigerant is now allowed to cool Determine the pressure when the refrigerant first starts condensing Also show the process on a Pv dia gram with respect to saturation lines 3119 Water initially at 300 kPa and 250C is contained in a constantvolume tank The water is allowed to cool until its pressure is 150 kPa On the Pv and Tv diagrams sketch with respect to the saturation lines the process curve passing through both the initial and final states of the water Label the end states on the process curve Also on both the Pv and Tv diagrams sketch the isotherms passing through both states and show their values in C on the isotherms 3120 A 9m3 tank contains nitrogen at 17C and 600 kPa Some nitrogen is allowed to escape until the pressure in the tank drops to 400 kPa If the temperature at this point is 15C deter mine the amount of nitrogen that has escaped Answer 206 kg 3123 A pistoncylinder device initially contains 02 kg of steam at 200 kPa and 300C Now the steam is cooled at constant pressure until it is at 150C Determine the volume change of the cylinder during this process using the compress ibility factor and compare the result to the actual value FIGURE P3115 Combustion chamber 12 MPa 450C FIGURE P3116 227C 100 kPa gage Nitrogen gas Patm 100 kPa Q 3124 A tank whose volume is unknown is divided into two parts by a partition One side of the tank contains 003 m3 of refrigerant134a that is a saturated liquid at 09 MPa while the other side is evacuated The partition is now removed and the refrigerant fills the entire tank If the final state of the refriger ant is 20C and 280 kPa determine the volume of the tank FIGURE P3124 Evacuated P 09 MPa V 003 m3 R134a Final PDF to printer 158 PROPERTIES OF PURE SUBSTANCES cen22672ch03109160indd 158 092217 1145 AM 3125 Reconsider Prob 3124 Using appropriate software investigate the effect of the initial pressure of refrigerant134a on the volume of the tank Let the initial pressure vary from 05 to 15 MPa Plot the volume of the tank versus the initial pressure and discuss the results 3126 A tank contains helium at 37C and 140 kPa gage The helium is heated in a process by heat transfer from the surroundings such that the helium reaches a final equilib rium state at 200C Determine the final gage pressure of the helium Assume atmospheric pressure is 100 kPa 3127 On the property diagrams indicated below sketch not to scale with respect to the saturated liquid and saturated vapor lines and label the following processes and states for steam Use arrows to indicate the direction of the process and label the initial and final states a On the Pv diagram sketch the constanttemperature pro cess through the state P 300 kPa v 0525 m3kg as pressure changes from P1 200 kPa to P2 400 kPa Place the value of the temperature on the process curve on the Pv diagram b On the Tv diagram sketch the constantspecificvolume process through the state T 120C v 07163 m3kg from P1 100 kPa to P2 300 kPa For this data set place the temperature values at states 1 and 2 on its axis Place the value of the specific volume on its axis 3128 On the property diagrams indicated below sketch not to scale with respect to the saturated liquid and saturated vapor lines and label the following processes and states for refrigerant134a Use arrows to indicate the direction of the process and label the initial and final states a On the Pv diagram sketch the constanttemperature pro cess through the state P 280 kPa v 006 m3kg as pres sure changes from P1 400 kPa to P2 200 kPa Place the value of the temperature on the process curve on the Pv diagram b On the Tv diagram sketch the constantspecificvolume process through the state T 20C v 002 m3kg from P1 1200 kPa to P2 300 kPa For this data set place the temperature values at states 1 and 2 on its axis Place the value of the specific volume on its axis 3129 Water initially at 300 kPa and 05 m3kg is contained in a pistoncylinder device fitted with stops so that the water supports the weight of the piston and the force of the atmo sphere The water is heated until it reaches the saturated vapor state and the piston rests against the stops With the piston against the stops the water is further heated until the pressure is 600 kPa On the Pv and Tv diagrams sketch with respect to the saturation lines the process curves passing through both the initial and final states of the water Label the states on the process as 1 2 and 3 On both the Pv and Tv diagrams sketch the isotherms passing through the states and show their values in C on the isotherms 3130 Ethane at 10 MPa and 100C is heated at constant pressure until its volume has increased by 60 percent Deter mine the final temperature using a the idealgas equation of state and b the compressibility factor Which of these two results is the more accurate 3131 Steam at 400C has a specific volume of 002 m3kg Determine the pressure of the steam based on a the idealgas equation b the generalized compressibility chart and c the steam tables Answers a 15529 kPa b 12574 kPa c 12515 kPa 3132E Nitrogen is maintained at 400 psia and 100F Compare the specific volume of this nitrogen as predicted by a the idealgas equation of state b the BenedictWebb Rubin equation of state and c the compressibility factor 3133 Consider an 18mdiameter hotair balloon that together with its cage has a mass of 120 kg when empty The air in the balloon which is now carrying two 85kg people is heated by propane burners at a location where the atmospheric pressure and temperature are 93 kPa and 12C respectively Determine the average temperature of the air in the balloon when the balloon first starts rising What would your response be if the atmospheric air temperature were 25C 3134 Oxygen is maintained at 4 MPa and 20C Compare the specific volume of the oxygen under this condition as predicted by a the idealgas equation of state b the BeattieBridgeman equation of state and c the compressibility factor Fundamentals of Engineering FE Exam Problems 3135 A 1m3 rigid tank contains 10 kg of water in any phase or phases at 160C The pressure in the tank is a 738 kPa b 618 kPa c 370 kPa d 2000 kPa e 1618 kPa 3136 A 3m3 rigid vessel contains steam at 2 MPa and 500C The mass of the steam is a 13 kg b 17 kg c 22 kg d 28 kg e 35 kg 3137 A 240m3 rigid tank is filled with a saturated liquid vapor mixture of water at 200 kPa If 25 percent of the mass FIGURE P3129 Water 300 kPa 05 m3kg Q Final PDF to printer 159 CHAPTER 3 cen22672ch03109160indd 159 092217 1145 AM is liquid and 75 percent of the mass is vapor the total mass in the tank is a 240 kg b 265 kg c 307 kg d 361 kg e 450 kg 3138 Water is boiled at 1 atm pressure in a coffeemaker equipped with an immersiontype electric heating element The coffeemaker initially contains 1 kg of water Once boiling has begun it is observed that half of the water in the coffeemaker evaporates in 10 min If the heat loss from the coffeemaker is negligible the power rating of the heating element is a 38 kW b 22 kW c 19 kW d 16 kW e 08 kW 3139 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range It is observed that 125 kg of liquid water evaporates in 30 min The rate of heat transfer to the water is a 157 kW b 186 kW c 209 kW d 243 kW e 251 kW 3140 Water is boiled in a pan on a stove at sea level During 10 min of boiling it is observed that 200 g of water has evapo rated Then the rate of heat transfer to the water is a 084 kJmin b 451 kJmin c 418 kJmin d 535 kJmin e 2257 kJmin 3141 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40C Now a valve is opened and half of mass of the gas is allowed to escape If the final pressure in the tank is 22 atm the final temperature in the tank is a 71C b 44C c 100C d 20C e 172C 3142 The pressure of an automobile tire is measured to be 190 kPa gage before a trip and 215 kPa gage after the trip at a location where the atmospheric pressure is 95 kPa If the temperature of air in the tire before the trip is 25C the air temperature after the trip is a 511C b 642C c 272C d 283C e 250C 3143 Consider a sealed can that is filled with refrigerant 134a The contents of the can are at the room temperature of 25C Now a leak develops and the pressure in the can drops to the local atmospheric pressure of 90 kPa The temperature of the refrigerant in the can is expected to drop to rounded to the nearest integer a 0C b 29C c 16C d 5C e 25C Design and Essay Problems 3144 A solid normally absorbs heat as it melts but there is a known exception at temperatures close to absolute zero Find out which solid it is and give a physical explanation for it 3145 In an article on tire maintenance it is stated that tires lose air over time and pressure losses as high as 90 kPa 13 psi per year are measured The article recommends checking tire pressure at least once a month to avoid low tire pressure that hurts fuel efficiency and causes uneven thread wear on tires Taking the beginning tire pressure to be 220 kPa gage and the atmospheric pressure to be 100 kPa determine the fraction of air that can be lost from a tire per year 3146 It is well known that water freezes at 0C at atmo spheric pressure The mixture of liquid water and ice at 0C is said to be at stable equilibrium since it cannot undergo any changes when it is isolated from its surroundings However when water is free of impurities and the inner surfaces of the container are smooth the temperature of water can be lowered to 2C or even lower without any formation of ice at atmo spheric pressure But at that state even a small disturbance can initiate the formation of ice abruptly and the water tempera ture stabilizes at 0C following this sudden change The water at 2C is said to be in a metastable state Write an essay on metastable states and discuss how they differ from stable equi librium states Final PDF to printer cen22672ch03109160indd 160 092217 1145 AM Final PDF to printer cen22672ch04161210indd 161 100217 0123 PM 161 4 OBJECTIVES The objectives of Chapter 4 are to Examine the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors Identify the first law of thermodynamics as simply a statement of the conservation of energy principle for closed fixedmass systems Develop the general energy balance applied to closed systems Define the specific heat at constant volume and the specific heat at constant pressure Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases Describe incompressible substances and determine the changes in their internal energy and enthalpy Solve energy balance problems for closed fixedmass systems that involve heat and work interactions for general pure substances ideal gases and incompressible substances EN E R GY A N A LYSI S O F C LOS E D SYSTE M S I n Chap 2 we considered various forms of energy and energy transfer and we developed a general relation for the conservation of energy principle or energy balance Then in Chap 3 we learned how to determine the ther modynamics properties of substances In this chapter we apply the energy balance relation to systems that do not involve any mass flow across their boundariesthat is closed systems We start this chapter with a discussion of the moving boundary work or P dV work commonly encountered in reciprocating devices such as automotive engines and compressors We continue by applying the general energy balance relation which is simply expressed as Ein Eout ΔEsystem to systems that involve pure substance Then we define specific heats obtain relations for the internal energy and enthalpy of ideal gases in terms of specific heats and temperature changes and perform energy balances on various systems that involve ideal gases We repeat this for systems that involve solids and liquids which are approximated as incompressible substances CHAPTER Final PDF to printer 162 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 162 100217 0123 PM 41 MOVING BOUNDARY WORK One form of mechanical work often encountered in practice is associated with the expansion or compression of a gas in a pistoncylinder device During this process part of the boundary the inner face of the piston moves back and forth Therefore the expansion and compression work is often called moving boundary work or simply boundary work Fig 41 Some call it the P dV work for reasons explained later Moving boundary work is the primary form of work involved in automobile engines During their expan sion the combustion gases force the piston to move which in turn forces the crankshaft to rotate The moving boundary work associated with real engines or compressors cannot be determined exactly from a thermodynamic analysis alone because the piston usually moves at very high speeds making it difficult for the gas inside to maintain equilibrium Then the states through which the system passes during the process cannot be specified and no process path can be drawn Work being a path function cannot be determined analytically with out a knowledge of the path Therefore the boundary work in real engines or compressors is determined by direct measurements In this section we analyze the moving boundary work for a quasi equilibrium process a process during which the system remains nearly in equilibrium at all times A quasiequilibrium process also called a quasi static process is closely approximated by real engines especially when the piston moves at low velocities Under identical conditions the work output of the engines is found to be a maximum and the work input to the compres sors to be a minimum when quasiequilibrium processes are used in place of nonquasiequilibrium processes In the following example the work associ ated with a moving boundary is evaluated for a quasiequilibrium process Consider the gas enclosed in the pistoncylinder device shown in Fig 42 The initial pressure of the gas is P the total volume is V and the cross sectional area of the piston is A If the piston is allowed to move a distance ds in a quasi equilibrium manner the differential work done during this process is δ W b F ds PA ds P dV 41 That is the boundary work in the differential form is equal to the product of the absolute pressure P and the differential change in the volume dV of the system This expression also explains why the moving boundary work is sometimes called the P dV work Note in Eq 41 that P is the absolute pressure which is always positive However the volume change dV is positive during an expansion process vol ume increasing and negative during a compression process volume decreas ing Thus the boundary work is positive during an expansion process and negative during a compression process Therefore Eq 41 can be viewed as an expression for boundary work output Wbout A negative result indicates boundary work input compression The total boundary work done during the entire process as the piston moves is obtained by adding all the differential works from the initial state to the final state W b 1 2 P dV kJ 42 FIGURE 41 The work associated with a moving boundary is called boundary work The moving boundary Gas FIGURE 42 A gas does a differential amount of work δWb as it forces the piston to move by a differential amount ds Gas P A F ds Final PDF to printer 163 CHAPTER 4 cen22672ch04161210indd 163 100217 0123 PM This integral can be evaluated only if we know the functional relationship between P and V during the process That is P f V should be available Note that P f V is simply the equation of the process path on a PV diagram The quasiequilibrium expansion process described is shown on a PV dia gram in Fig 43 On this diagram the differential area dA is equal to P dV which is the differential work The total area A under the process curve 12 is obtained by adding these differential areas Area A 1 2 dA 1 2 P dV 43 A comparison of this equation with Eq 42 reveals that the area under the process curve on a PV diagram is equal in magnitude to the work done dur ing a quasiequilibrium expansion or compression process of a closed system On the Pv diagram it represents the boundary work done per unit mass A gas can follow several different paths as it expands from state 1 to state 2 In general each path will have a different area underneath it and since this area represents the magnitude of the work the work done will be different for each process Fig 44 This is expected since work is a path function ie it depends on the path followed as well as the end states If work were not a path function no cyclic devices car engines power plants could operate as workproducing devices The work produced by these devices during one part of the cycle would have to be consumed during another part and there would be no net work output The cycle shown in Fig 45 produces a net work output because the work done by the system during the expansion process area under path A is greater than the work done on the system during the compression part of the cycle area under path B and the difference between these two is the net work done during the cycle the colored area If the relationship between P and V during an expansion or a compression process is given in terms of experimental data instead of in a functional form obviously we cannot perform the integration analytically We can however plot the PV diagram of the process using these data points and calculate the area underneath graphically to determine the work done Strictly speaking the pressure P in Eq 42 is the pressure at the inner sur face of the piston It becomes equal to the pressure of the gas in the cylinder only if the process is quasiequilibrium and thus the entire gas in the cylin der is at the same pressure at any given time Equation 42 can also be used for nonquasiequilibrium processes provided that the pressure at the inner face of the piston is used for P Besides we cannot speak of the pressure of a system during a nonquasiequilibrium process since properties are defined for equilibrium states only Therefore we can generalize the boundary work relation by expressing it as W b 1 2 P i dV 44 where Pi is the pressure at the inner face of the piston Note that work is a mechanism for energy interaction between a system and its surroundings and Wb represents the amount of energy transferred from the system during an expansion process or to the system during a compression process Therefore it has to appear somewhere else and we must be able FIGURE 43 The area under the process curve on a PV diagram represents the boundary work Process path 2 1 P dV V dA PdV P V1 V2 FIGURE 44 The boundary work done during a process depends on the path followed as well as the end states V2 WA 10 kJ 1 2 P V V1 A B C WB 8 kJ WC 5 kJ FIGURE 45 The net work done during a cycle is the difference between the work done by the system and the work done on the system Wnet 2 1 P V V2 V1 A B Final PDF to printer 164 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 164 100217 0123 PM to account for it since energy is conserved In a car engine for example the boundary work done by the expanding hot gases is used to overcome friction between the piston and the cylinder to push atmospheric air out of the way and to rotate the crankshaft Therefore W b W friction W atm W crank 1 2 F friction P atm A F crank dx 45 Of course the work used to overcome friction appears as frictional heat and the energy transmitted through the crankshaft is transmitted to other com ponents such as the wheels to perform certain functions But note that the energy transferred by the system as work must equal the energy received by the crankshaft the atmosphere and the energy used to overcome friction The use of the boundary work relation is not limited to the quasiequilibrium pro cesses of gases only It can also be used for solids and liquids EXAMPLE 41 Boundary Work for a ConstantVolume Process A rigid tank contains air at 500 kPa and 150C As a result of heat transfer to the sur roundings the temperature and pressure inside the tank drop to 65C and 400 kPa respectively Determine the boundary work done during this process SOLUTION Air in a rigid tank is cooled and both the pressure and temperature drop The boundary work done is to be determined Analysis A sketch of the system and the PV diagram of the process are shown in Fig 46 The boundary work can be determined from Eq 42 to be W b 1 2 P dV 0 0 Discussion This is expected since a rigid tank has a constant volume and dV 0 in this equation Therefore there is no boundary work done during this process That is the boundary work done during a constantvolume process is always zero This is also evident from the PV diagram of the process the area under the process curve is zero FIGURE 46 Schematic and PV diagram for Example 41 P1 500 kPa Heat Air T1 150C P2 400 kPa T2 65C 2 1 P kPa V 400 500 EXAMPLE 42 Boundary Work for a ConstantPressure Process A frictionless pistoncylinder device contains 10 lbm of steam at 60 psia and 320F Heat is now transferred to the steam until the temperature reaches 400F If the piston is not attached to a shaft and its mass is constant determine the work done by the steam during this process SOLUTION Steam in a pistoncylinder device is heated and the temperature rises at constant pressure The boundary work done is to be determined Assumptions The expansion process is quasiequilibrium Analysis A sketch of the system and the Pv diagram of the process are shown in Fig 47 Even though it is not explicitly stated the pressure of the steam within the cylinder remains constant during this process since both the atmospheric pressure FIGURE 47 Schematic and Pv diagram for Example 42 Final PDF to printer 165 CHAPTER 4 cen22672ch04161210indd 165 100217 0123 PM and the weight of the piston remain constant Therefore this is a constantpressure process and from Eq 42 W b 1 2 P dV P 0 1 2 dV P 0 V 2 V 1 46 or W b m P 0 v 2 v 1 since V mv From the superheated vapor table Table A6E the specific vol umes are determined to be v1 74863 ft3lbm at state 1 60 psia 320F and v2 83548 ft3lbm at state 2 60 psia 400F Substituting these values yields W b 10 lbm60 psia 83548 74863 ft 3 lbm 1 Btu 5404 psiaft 3 964 Btu Discussion The positive sign indicates that the work is done by the system That is the steam used 964 Btu of its energy to do this work The magnitude of this work could also be determined by calculating the area under the process curve on the PV diagram which is simply P0 ΔV for this case EXAMPLE 43 Isothermal Compression of an Ideal Gas A pistoncylinder device initially contains 04 m3 of air at 100 kPa and 80C The air is now compressed to 01 m3 in such a way that the temperature inside the cylinder remains constant Determine the work done during this process SOLUTION Air in a pistoncylinder device is compressed isothermally The boundary work done is to be determined Assumptions 1 The compression process is quasiequilibrium 2 At specified con ditions air can be considered to be an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values Analysis A sketch of the system and the PV diagram of the process are shown in Fig 48 For an ideal gas at constant temperature T0 PV mR T 0 C or P C V where C is a constant Substituting this into Eq 42 we have W b 1 2 P dV 1 2 C V dV C 1 2 dV V C ln V 2 V 1 P 1 V 1 ln V 2 V 1 47 In Eq 47 P1V1 can be replaced by P2V2 or mRT0 Also V2V1 can be replaced by P1P2 for this case since P1V1 P2V2 Substituting the numerical values into Eq 47 yields W b 100 kPa04 m 3 ln 01 04 1 kJ 1 kPam 3 555 kJ Discussion The negative sign indicates that this work is done on the system a work input which is always the case for compression processes FIGURE 48 Schematic and PV diagram for Example 43 Final PDF to printer 166 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 166 100217 0123 PM Polytropic Process During actual expansion and compression processes of gases pressure and volume are often related by PVn C where n and C are constants A process of this kind is called a polytropic process Fig 49 Next we develop a gen eral expression for the work done during a polytropic process The pressure for a polytropic process can be expressed as P C V n 48 Substituting this relation into Eq 42 we obtain W b 1 2 P dV 1 2 C V n dV C V 2 n 1 V 1 n 1 n 1 P 2 V 2 P 1 V 1 1 n 49 since C P 1 V 1 n P 2 V 2 n For an ideal gas PV mRT this equation can also be written as W b mR T 2 T 1 1 n n 1 kJ 410 For the special case of n 1 the boundary work becomes W b 1 2 P dV 1 2 C V 1 dV PV ln V 2 V 1 For an ideal gas this result is equivalent to the isothermal process discussed in the previous example FIGURE 49 Schematic and PV diagram for a polytropic process Gas PVn C const PV n const 2 1 P V P1 P2 V1 V2 P1V1 P2V2 n n EXAMPLE 44 Expansion of a Gas Against a Spring A pistoncylinder device contains 005 m3 of a gas initially at 200 kPa At this state a linear spring that has a spring constant of 150 kNm is touching the piston but exerting no force on it Now heat is transferred to the gas causing the piston to rise and to compress the spring until the volume inside the cylinder doubles If the cross sectional area of the piston is 025 m2 determine a the final pressure inside the cyl inder b the total work done by the gas and c the fraction of this work done against the spring to compress it SOLUTION A gas in a pistoncylinder device equipped with a linear spring expands as a result of heating The final gas pressure the total work done and the fraction of the work done to compress the spring are to be determined Assumptions 1 The expansion process is quasiequilibrium 2 The spring is linear in the range of interest Analysis A sketch of the system and the PV diagram of the process are shown in Fig 410 a The enclosed volume at the final state is V 2 2 V 1 2005 m 3 01 m 3 Then the displacement of the piston and of the spring becomes x ΔV A 01 005 m 3 025 m 3 02 m Final PDF to printer 167 CHAPTER 4 cen22672ch04161210indd 167 100217 0123 PM The force applied by the linear spring at the final state is F kx 150 kN m 02 m 30 kN The additional pressure applied by the spring on the gas at this state is P F A 30 kN 025 m 2 120 kPa Without the spring the pressure of the gas would remain constant at 200 kPa while the piston is rising But under the effect of the spring the pressure rises linearly from 200 kPa to 200 120 320 kPa at the final state b An easy way of finding the work done is to plot the process on a PV diagram and find the area under the process curve From Fig 410 the area under the process curve a trapezoid is determined to be W area 200 320 kPa 2 01 005 m 3 1 kJ 1 kPa m 3 13 kJ Note that the work is done by the system c The work represented by the rectangular area region I is done against the piston and the atmosphere and the work represented by the triangular area region II is done against the spring Thus W spring 1 2 320 200 kPa005 m 3 1 kJ 1 kPam 3 3 kJ Discussion This result could also be obtained from W spring 1 2 k x 2 2 x 1 2 1 2 150 kN m 02 m 2 0 2 1 kJ 1 kNm 3 kJ FIGURE 410 Schematic and PV diagram for Example 44 P kPa V m3 II 01 005 I 320 200 P1 200 kPa V1 005 m3 Heat A 025 m2 k 150 kNm 42 ENERGY BALANCE FOR CLOSED SYSTEMS Energy balance for any system undergoing any kind of process was expressed as see Chap 2 E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies kJ 411 or in the rate form as E in E out Rate of net energy transfer by heat work and mass dE system dt Rate of change in internal kinetic potential etc energies kW 412 For constant rates the total quantities during a time interval Δt are related to the quantities per unit time as Q Q Δt W W Δt and ΔE dE dt Δt kJ 413 Final PDF to printer 168 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 168 100217 0123 PM The energy balance can be expressed on a per unit mass basis as e in e out Δ e system kJ kg 414 which is obtained by dividing all the quantities in Eq 411 by the mass m of the system Energy balance can also be expressed in the differential form as δ E in δ E out d E system or δ e in δ e out d e system 415 For a closed system undergoing a cycle the initial and final states are identi cal and thus ΔEsystem E2 E1 0 Then the energy balance for a cycle simplifies to Ein Eout 0 or Ein Eout Noting that a closed system does not involve any mass flow across its boundaries the energy balance for a cycle can be expressed in terms of heat and work interactions as W netout Q netin or W netout Q netin for a cycle 416 That is the net work output during a cycle is equal to net heat input Fig 411 The energy balance or the firstlaw relations already given are intuitive in nature and are easy to use when the magnitudes and directions of heat and work transfers are known However when performing a general analytical study or solving a problem that involves an unknown heat or work interaction we need to assume a direction for the heat or work interactions In such cases it is common practice to use the classical thermodynamics sign convention and to assume heat to be transferred into the system heat input in the amount of Q and work to be done by the system work output in the amount of W and then to solve the problem The energy balance relation in that case for a closed system becomes Q netin W netout Δ E system or Q W ΔE 417 where Q Qnetin Qin Qout is the net heat input and W Wnetout Wout Win is the net work output Obtaining a negative quantity for Q or W simply means that the assumed direction for that quantity is wrong and should be reversed Various forms of this traditional firstlaw relation for closed systems are given in Fig 412 The first law cannot be proven mathematically but no process in nature is known to have violated the first law and this should be taken as sufficient proof Note that if it were possible to prove the first law on the basis of other physical principles the first law then would be a consequence of those prin ciples instead of being a fundamental physical law itself As energy quantities heat and work are not that different and you prob ably wonder why we keep distinguishing them After all the change in the energy content of a system is equal to the amount of energy that crosses the system boundaries and it makes no difference whether the energy crosses the boundary as heat or work It seems as if the firstlaw rela tions would be much simpler if we had just one quantity that we could call energy interaction to represent both heat and work Well from the firstlaw point of view heat and work are not different at all From the secondlaw point of view however heat and work are very different as is discussed in later chapters FIGURE 411 For a cycle ΔE 0 thus Q W P V Qnet Wnet FIGURE 412 Various forms of the firstlaw relation for closed systems General Q W ΔE Stationary systems Q W ΔU Per unit mass q w Δe Differential form δq δw de Final PDF to printer 169 CHAPTER 4 cen22672ch04161210indd 169 100217 0123 PM EXAMPLE 45 Electric Heating of a Gas at Constant Pressure A pistoncylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa A resistance heater within the cylinder is turned on and passes a current of 02 A for 5 min from a 120V source At the same time a heat loss of 37 kJ occurs a Show that for a closed system the boundary work Wb and the change in internal energy ΔU in the firstlaw relation can be combined into one term ΔH for a constantpressure process b Determine the final temperature of the steam SOLUTION Saturated water vapor in a pistoncylinder device expands at constant pressure as a result of heating It is to be shown that ΔU Wb ΔH and the final temperature is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 Therefore ΔE ΔU and internal energy is the only form of energy of the system that may change during this process 2 Electrical wires constitute a very small part of the system and thus the energy change of the wires can be neglected Analysis We take the contents of the cylinder including the resistance wires as the system Fig 413 This is a closed system since no mass crosses the system bound ary during the process We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work Wb The pressure remains constant during the process and thus P2 P1 Also heat is lost from the system and electrical work We is done on the system a This part of the solution involves a general analysis for a closed system under going a quasiequilibrium constantpressure process and thus we consider a general closed system We take the direction of heat transfer Q to be to the system and the work W to be done by the system We also express the work as the sum of boundary and other forms of work such as electrical and shaft Then the energy balance can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q W ΔU ΔKE 0 ΔPE 0 Q W other W b U 2 U 1 For a constantpressure process the boundary work is given as Wb P0V2 V1 Substituting this into the preceding relation gives Q W other P 0 V 2 V 1 U 2 U 1 However P 0 P 2 P 1 Q W other U 2 P 2 V 2 U 1 P 1 V 1 Also H U PV and thus Q W other H 2 H 1 kJ 418 which is the desired relation Fig 414 This equation is very convenient to use in the analysis of closed systems undergoing a constantpressure quasiequilibrium FIGURE 414 For a closed system undergoing a quasiequilibrium P constant process ΔU Wb ΔH Note that this relation is NOT valid for closed systems processes during which pressure DOES NOT remain constant ΔH Q Wother Wb ΔU P const Q Wother ΔH FIGURE 413 Schematic and Pv diagram for Example 45 Final PDF to printer 170 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 170 100217 0123 PM process since the boundary work is automatically taken care of by the enthalpy terms and one no longer needs to determine it separately b The only other form of work in this case is the electrical work which can be determined from W e VIΔt 120 V02 A300 s 1 kJ s 1000 VA 72 kJ State 1 P 1 300 kPa Sat vapor h 1 h g 300 kPa 27249 kJ kg Table A5 The enthalpy at the final state can be determined directly from Eq 418 by express ing heat transfer from the system and work done on the system as negative quantities since their directions are opposite to the assumed directions Alternately we can use the general energy balance relation with the simplification that the boundary work is considered automatically by replacing ΔU with ΔH for a constantpressure expansion or compression process E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W ein Q out W b ΔU W ein Q out ΔH m h 2 h 1 since P constant 72 kJ 37 kJ 0025 kg h 2 27249 kJ kg h 2 28649 kJ kg Now the final state is completely specified since we know both the pressure and the enthalpy The temperature at this state is State 2 P 2 300 kPa h 2 28649 kJ kg T 2 200C Table A6 Therefore the steam will be at 200C at the end of this process Discussion Strictly speaking the potential energy change of the steam is not zero for this process since the center of gravity of the steam rose somewhat Assuming an elevation change of 1 m which is rather unlikely the change in the potential energy of the steam would be 00002 kJ which is very small compared to the other terms in the firstlaw relation Therefore in problems of this kind the potential energy term is always neglected EXAMPLE 46 Unrestrained Expansion of Water A rigid tank is divided into two equal parts by a partition Initially one side of the tank contains 5 kg of water at 200 kPa and 25C and the other side is evacuated The partition is then removed and the water expands into the entire tank The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25C Determine a the volume of the tank b the final pressure and c the heat transfer for this process SOLUTION Onehalf of a rigid tank is filled with liquid water while the other side is evacuated The partition between the two parts is removed and water is allowed Final PDF to printer 171 CHAPTER 4 cen22672ch04161210indd 171 100217 0123 PM to expand and fill the entire tank while the temperature is maintained constant The volume of the tank the final pressure and the heat transfer are to be to determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 and ΔE ΔU 2 The direction of heat transfer is to the system heat gain Qin A negative result for Qin indicates the assumed direction is wrong and thus it is a heat loss 3 The volume of the rigid tank is constant and thus there is no energy transfer as boundary work 4 There is no electrical shaft or any other kind of work involved Analysis We take the contents of the tank including the evacuated space as the system Fig 415 This is a closed system since no mass crosses the system boundary during the process We observe that the water fills the entire tank when the partition is removed possibly as a liquidvapor mixture a Initially the water in the tank exists as a compressed liquid since its pressure 200 kPa is greater than the saturation pressure at 25C 31698 kPa Approximating the compressed liquid as a saturated liquid at the given temperature we find v 1 v f 25C 0001003 m 3 kg 0001 m 3 kg Table A4 Then the initial volume of the water is V 1 m v 1 5 kg0001 m 3 kg 0005 m 3 The total volume of the tank is twice this amount V tank 20005 m 3 001 m 3 b At the final state the specific volume of the water is v 2 V 2 m 001 m 3 5 kg 0002 m 3 kg which is twice the initial value of the specific volume This result is expected since the volume doubles while the amount of mass remains constant At 25 C v f 0001003 m 3 kg and v g 43340 m 3 kg Table A4 Since vf v2 vg the water is a saturated liquidvapor mixture at the final state and thus the pressure is the saturation pressure at 25C P 2 P sat 25C 31698 kPa Table A4 c Under stated assumptions and observations the energy balance on the system can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q in ΔU m u 2 u 1 Notice that even though the water is expanding during this process the system chosen involves fixed boundaries only the dashed lines and therefore the moving bound ary work is zero Fig 416 Then W 0 since the system does not involve any other forms of work Can you reach the same conclusion by choosing the water as our system Initially u 1 u f 25C 10483 kJ kg FIGURE 415 Schematic and Pv diagram for Example 46 Evacuated space P1 200 kPa T1 25C m 5 kg H2O 2 P kPa 1 317 200 v Partition System boundary Qin FIGURE 416 Expansion against a vacuum involves no work and thus no energy transfer Vacuum P 0 H2O Heat W 0 Final PDF to printer 172 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 172 100217 0123 PM The quality at the final state is determined from the specific volume information x 2 v 2 v f v fg 0002 0001 4334 0001 23 10 5 Then u 2 u f x 2 u fg 10483 kJ kg 10488 kJ kg 23 10 5 23043 kJ kg Substituting yields Q in 5 kg 10488 10483 kJ kg 025 kJ Discussion The positive sign indicates that the assumed direction is correct and heat is transferred to the water 43 SPECIFIC HEATS We know from experience that it takes different amounts of energy to raise the temperature of identical masses of different substances by one degree For example we need about 45 kJ of energy to raise the temperature of 1 kg of iron from 20 to 30C whereas it takes about nine times this much energy 418 kJ to be exact to raise the temperature of 1 kg of liquid water by the same amount Fig 417 Therefore it is useful to have a property that will enable us to compare the energy storage capabilities of various substances This property is the specific heat The specific heat is defined as the energy required to raise the tempera ture of a unit mass of a substance by one degree Fig 418 In general this energy depends on how the process is executed In thermodynamics we are interested in two kinds of specific heats specific heat at constant volume cv and specific heat at constant pressure cp Physically the specific heat at constant volume cv can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant The energy required to do the same as the pressure is maintained constant is the specific heat at constant pressure cp This is illustrated in Fig 419 The specific heat at constant pres sure cp is always greater than cv because at constant pressure the system is allowed to expand and the energy for this expansion work must also be sup plied to the system Now we try to express the specific heats in terms of other thermodynamic properties First consider a fixed mass in a stationary closed system undergo ing a constantvolume process and thus no expansion or compression work is involved The conservation of energy principle ein eout Δesystem for this process can be expressed in the differential form as δ e in δ e out du FIGURE 417 It takes different amounts of energy to raise the temperature of different substances by the same amount 20 30C Iron 1 kg 45 kJ 20 30C Water 1 kg 418 kJ FIGURE 418 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way Specific heat 5 kJkgC ΔT 1C m 1 kg 5 kJ FIGURE 419 Constantvolume and constant pressure specific heats cv and cp values given are for helium gas ΔT 1C cv 312 kJ m 1 kg 312 kJ V constant kgC ΔT 1C cp 519 kJ m 1 kg 519 kJ P constant kgC 1 2 Final PDF to printer 173 CHAPTER 4 cen22672ch04161210indd 173 100217 0123 PM The lefthand side of this equation represents the net amount of energy trans ferred to the system From the definition of cv this energy must be equal to cv dT where dT is the differential change in temperature Thus c v dT du at constant volume or c v u T v 419 Similarly an expression for the specific heat at constant pressure cp can be obtained by considering a constantpressure expansion or compression pro cess It yields c p u T p 420 Equations 419 and 420 are the defining equations for cv and cp and their interpretation is given in Fig 420 Note that cv and cp are expressed in terms of other properties thus they must be properties themselves Like any other property the specific heats of a substance depend on the state that in general is specified by two indepen dent intensive properties That is the energy required to raise the temperature of a substance by one degree is different at different temperatures and pres sures Fig 421 But this difference is usually not very large A few observations can be made from Eqs 419 and 420 First these equations are property relations and as such are independent of the type of processes They are valid for any substance undergoing any process The only relevance cv has to a constantvolume process is that cv happens to be the energy transferred to a system during a constantvolume process per unit mass per unit degree rise in temperature This is how the values of cv are determined This is also how the name specific heat at constant volume origi nated Likewise the energy transferred to a system per unit mass per unit temperature rise during a constantpressure process happens to be equal to cp This is how the values of cp can be determined and it also explains the origin of the name specific heat at constant pressure Another observation that can be made from Eqs 419 and 420 is that cv is related to the changes in internal energy and cp to the changes in enthalpy In fact it would be more proper to define cv as the change in the internal energy of a substance per unit change in temperature at constant volume Likewise cp can be defined as the change in the enthalpy of a substance per unit change in temperature at constant pressure In other words cv is a measure of the variation of internal energy of a substance with temperature and cp is a mea sure of the variation of enthalpy of a substance with temperature Both the internal energy and enthalpy of a substance can be changed by the transfer of energy in any form with heat being only one of them Therefore the term specific energy is probably more appropriate than the term specific heat which implies that energy is transferred and stored in the form of heat A common unit for specific heats is kJkgC or kJkgK Notice that these two units are identical since ΔTC ΔTK and a 1C change in FIGURE 420 Formal definitions of cv and cp T the change in internal energy with temperature at constant volume the change in enthalpy with temperature at constant pressure cv v cp p u T h FIGURE 421 The specific heat of a substance changes with temperature 300 301 K Air m 1 kg Air m 1 kg 1000 1001 K 0718 kJ 0855 kJ Final PDF to printer 174 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 174 100217 0123 PM temperature is equivalent to a change of 1 K The specific heats are some times given on a molar basis They are then denoted by c v and c p and have the unit kJkmolC or kJkmolK 44 INTERNAL ENERGY ENTHALPY AND SPECIFIC HEATS OF IDEAL GASES We defined an ideal gas as a gas whose temperature pressure and specific volume are related by Pv RT It has been demonstrated mathematically Chap 12 and experimentally Joule 1843 that for an ideal gas the internal energy is a function of the temperature only That is u u T 421 In his classical experiment Joule submerged two tanks connected with a pipe and a valve in a water bath as shown in Fig 422 Initially one tank contained air at a high pressure and the other tank was evacuated When ther mal equilibrium was attained he opened the valve to let air pass from one tank to the other until the pressures equalized Joule observed no change in the temperature of the water bath and assumed that no heat was transferred to or from the air Since there was also no work done he concluded that the internal energy of the air did not change even though the volume and the pressure changed Therefore he reasoned the internal energy is a function of temperature only and not a function of pressure or specific volume Joule later showed that for gases that deviate significantly from idealgas behavior the internal energy is not a function of temperature alone Using the definition of enthalpy and the equation of state of an ideal gas we have h u Pv Pv RT h u RT Since R is constant and u uT it follows that the enthalpy of an ideal gas is also a function of temperature only h h T 422 Since u and h depend only on temperature for an ideal gas the specific heats cv and cp also depend at most on temperature only Therefore at a given tempera ture u h cv and cp of an ideal gas have fixed values regardless of the specific volume or pressure Fig 423 Thus for ideal gases the partial derivatives in Eqs 419 and 420 can be replaced by ordinary derivatives Then the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as du c v T dT 423 and dh c p T dT 424 FIGURE 422 Schematic of the experimental apparatus used by Joule Thermometer Air high pressure Evacuated Water FIGURE 423 For ideal gases u h cv and cp vary with temperature only u uT h hT cv cvT cp cpT Final PDF to printer 175 CHAPTER 4 cen22672ch04161210indd 175 100217 0123 PM The change in internal energy or enthalpy for an ideal gas during a process from state 1 to state 2 is determined by integrating these equations Δu u 2 u 1 1 2 c v T dT kJ kg 425 and Δh h 2 h 1 1 2 c p T dT kJ kg 426 To carry out these integrations we need to have relations for cv and cp as func tions of temperature At low pressures all real gases approach idealgas behavior and therefore their specific heats depend on temperature only The specific heats of real gases at low pressures are called idealgas specific heats or zeropressure specific heats and they are often denoted cp0 and cv0 Accurate analytical expressions for idealgas specific heats based on direct measurements or cal culations from statistical behavior of molecules are available and are given as thirddegree polynomials in the appendix Table A2c for several gases A plot of c p0 T data for some common gases is given in Fig 424 The use of idealgas specific heat data is limited to low pressures but these data can also be used at moderately high pressures with reasonable accuracy as long as the gas does not deviate from idealgas behavior significantly The integrations in Eqs 425 and 426 are straightforward but rather time consuming and thus impractical To avoid these laborious calculations u and h data for a number of gases have been tabulated over small temperature inter vals These tables are obtained by choosing an arbitrary reference point and performing the integrations in Eqs 425 and 426 by treating state 1 as the reference state In the idealgas tables given in the appendix zero kelvin is chosen as the reference state and both the enthalpy and the internal energy are assigned zero values at that state Fig 425 The choice of the reference state has no effect on Δu or Δh calculations The u and h data are given in kJkg for air Table A17 and usually in kJkmol for other gases The unit kJkmol is very convenient in the thermodynamic analysis of chemical reactions Some observations can be made from Fig 424 First the specific heats of gases with complex molecules molecules with two or more atoms are higher and increase with temperature Also the variation of specific heats with temperature is smooth and may be approximated as linear over small temperature intervals a few hundred degrees or less Therefore the specific heat functions in Eqs 425 and 426 can be replaced by the constant average specific heat values Then the integrations in these equations can be per formed yielding u 2 u 1 c vavg T 2 T 1 kJ kg 427 and h 2 h 1 c pavg T 2 T 1 kJ kg 428 The specific heat values for some common gases are listed as a function of temperature in Table A2b The average specific heats cpavg and cvavg are FIGURE 424 Idealgas constantpressure specific heats for some gases see Table A2c for cp equations 1000 20 2000 3000 Temperature K Ar He Ne Kr Xe Rn 30 40 50 60 CO2 H2O O2 H2 Air cp0 kJkmolK FIGURE 425 In the preparation of idealgas tables 0 K is chosen as the reference temperature 0 0 0 T K Air u kJkg h kJkg 300 21407 30019 310 22125 31024 Final PDF to printer 176 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 176 100217 0123 PM evaluated from this table at the average temperature T1 T22 as shown in Fig 426 If the final temperature T2 is not known the specific heats may be evaluated at T1 or at the anticipated average temperature Then T2 can be determined by using these specific heat values The value of T2 can be refined if necessary by evaluating the specific heats at the new average temperature Another way of determining the average specific heats is to evaluate them at T1 and T2 and then take their average Usually both methods give reason ably good results and one is not necessarily better than the other Another observation that can be made from Fig 424 is that the idealgas specific heats of monatomic gases such as argon neon and helium remain constant over the entire temperature range Thus Δu and Δh of monatomic gases can easily be evaluated from Eqs 427 and 428 Note that the Δu and Δh relations given previously are not restricted to any kind of process They are valid for all processes The presence of the constantvolume specific heat cv in an equation should not lead one to believe that this equation is valid for a constantvolume process only On the contrary the relation Δu cvavg ΔT is valid for any ideal gas undergoing any process Fig 427 A similar argument can be given for cp and Δh To summarize there are three ways to determine the internal energy and enthalpy changes of ideal gases Fig 428 1 By using the tabulated u and h data This is the easiest and most accurate way when tables are readily available 2 By using the cv or cp relations as a function of temperature and performing the integrations This is very inconvenient for hand calculations but quite desirable for computerized calculations The results obtained are very accurate 3 By using average specific heats This is very simple and certainly very convenient when property tables are not available The results obtained are reasonably accurate if the temperature interval is not very large Specific Heat Relations of Ideal Gases A special relationship between cp and cv for ideal gases can be obtained by differentiating the relation h u RT which yields dh du R dT Replacing dh with cpdT and du with cvdT and dividing the resulting expres sion by dT we obtain c p c v R kJ kgK 429 This is an important relationship for ideal gases since it enables us to deter mine cv from a knowledge of cp and the gas constant R When the specific heats are given on a molar basis R in Eq 429 should be replaced by the universal gas constant Ru Fig 429 c p c v R u kJ kmolK 430 FIGURE 426 For small temperature intervals the specific heats may be assumed to vary linearly with temperature Actual 1 T1 Tavg T2 T 2 Approximation cpavg cp FIGURE 427 The relation Δu cv ΔT is valid for any kind of process constantvolume or not Δu cv ΔT Q2 T1 20C T2 30C V constant Air T1 20C T2 30C P constant Air 718 kJkg Δu cv ΔT 718 kJkg Q1 FIGURE 428 Three ways of calculating Δu Δu u2 u1 table Δu 2 1 cv T dT Δu cvavg ΔT Final PDF to printer 177 CHAPTER 4 cen22672ch04161210indd 177 100217 0123 PM At this point we introduce another idealgas property called the specific heat ratio k defined as k c p c v 431 The specific ratio also varies with temperature but this variation is very mild For monatomic gases its value is essentially constant at 1667 Many diatomic gases including air have a specific heat ratio of about 14 at room temperature EXAMPLE 47 Evaluation of the Δu of an Ideal Gas Air at 300 K and 200 kPa is heated at constant pressure to 600 K Determine the change in internal energy of air per unit mass using a data from the air table Table A17 b the functional form of the specific heat Table A2c and c the average specific heat value Table A2b SOLUTION The internal energy change of air is to be determined in three differ ent ways Assumptions At specified conditions air can be considered to be an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values Analysis The internal energy change Δu of ideal gases depends on the initial and final temperatures only and not on the type of process Thus the following solution is valid for any kind of process a One way of determining the change in internal energy of air is to read the u values at T1 and T2 from Table A17 and take the difference u 1 u 300 K 21407 kJ kg u 2 u 600 K 43478 kJ kg Thus Δu u 2 u 1 43478 21407 kJ kg 22071 kJ kg b The c p T of air is given in Table A2c in the form of a thirddegree polynomial expressed as c p T a bT c T 2 d T 3 where a 2811 b 01967 102 c 04802 105 and d 1966 109 From Eq 430 c v T c p R u a R u bT c T 2 d T 3 From Eq 425 Δ u dT 1 2 c v T dT T 1 T 2 a R u bT c T 2 d T 3 dT Performing the integration and substituting the values we obtain Δ u 6447 kJ kmol FIGURE 429 The cp of an ideal gas can be determined from a knowledge of cv and R Air at 300 K cv 0718 kJkgK cv 2080 kJkmolK R 0287 kJkgK Ru 8314 kJkmolK cp 1005 kJkgK cp 29114 kJkmolK or Final PDF to printer 178 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 178 100217 0123 PM The change in the internal energy on a unitmass basis is determined by dividing this value by the molar mass of air Table A1 Δu Δ u M 6447 kJ kmol 2897 kg kmol 2225 kJ kg which differs from the tabulated value by 08 percent c The average value of the constantvolume specific heat cvavg is determined from Table A2b at the average temperature of T1 T22 450 K to be c vavg c v 450 K 0733 kJ kgK Thus Δu c vavg T 2 T 1 0733 kJ kgK 600 300 K 220 kJ kg Discussion This answer differs from the tabulated value 22071 kJkg by only 04 percent This close agreement is not surprising since the assumption that cv varies linearly with temperature is a reasonable one at temperature intervals of only a few hundred degrees If we had used the cv value at T1 300 K instead of at Tavg the result would be 2154 kJkg which is in error by about 2 percent Errors of this magnitude are acceptable for most engineering purposes FIGURE 430 Schematic and PV diagram for Example 48 1 P psia P2 2 50 He m 15 lbm T1 80F P1 50 psia V2 V1 V Wsh EXAMPLE 48 Heating of a Gas in a Tank by Stirring An insulated rigid tank initially contains 15 lbm of helium at 80F and 50 psia A paddle wheel with a power rating of 002 hp is operated within the tank for 30 min Determine a the final temperature and b the final pressure of the helium gas SOLUTION Helium gas in an insulated rigid tank is stirred by a paddle wheel The final temperature and pressure of helium are to be determined Assumptions 1 Helium is an ideal gas since it is at a very high temperature relative to its criticalpoint value of 451F 2 Constant specific heats can be used for helium 3 The system is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 and ΔE ΔU 4 The volume of the tank is constant and thus there is no boundary work 5 The system is adiabatic and thus there is no heat transfer Analysis We take the contents of the tank as the system Fig 430 This is a closed system since no mass crosses the system boundary during the process We observe that there is shaft work done on the system a The amount of paddlewheel work done on the system is W sh W sh Δt 002 hp 05 h 2545 Btu h 1 hp 2545 Btu Under the stated assumptions and observations the energy balance on the system can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W shin ΔU m u 2 u 1 m c vavg T 2 T 1 Final PDF to printer 179 CHAPTER 4 cen22672ch04161210indd 179 100217 0123 PM As we pointed out earlier the idealgas specific heats of monatomic gases helium being one of them are constant The cv value of helium is determined from Table A2Ea to be cv 0753 BtulbmF Substituting this and other known quanti ties into the preceding equation we obtain 2545 Btu 15 lbm 0753 Btu lbmF T 2 80 F T 2 1025F b The final pressure is determined from the idealgas relation P 1 V 1 T 1 P 2 V 2 T 2 where V1 and V2 are identical and cancel out Then the final pressure becomes 50 psia 80 460 R P 2 1025 460 R P 2 521 psia Discussion Note that the pressure in the idealgas relation is always the absolute pressure FIGURE 431 Schematic and PV diagram for Example 49 1 P kPa V m3 2 400 05 2800 J 120 V 2A N2 P1 400 kPa V1 05 m3 P const T1 27C EXAMPLE 49 Heating of a Gas by a Resistance Heater A pistoncylinder device initially contains 05 m3 of nitrogen gas at 400 kPa and 27C An electric heater within the device is turned on and is allowed to pass a cur rent of 2 A for 5 min from a 120V source Nitrogen expands at constant pressure and a heat loss of 2800 J occurs during the process Determine the final temperature of nitrogen SOLUTION Nitrogen gas in a pistoncylinder device is heated by an electric resistance heater Nitrogen expands at constant pressure while some heat is lost The final temperature of nitrogen is to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values of 147C and 339 MPa 2 The system is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 and ΔE ΔU 3 The pressure remains constant during the process and thus P2 P1 4 Nitrogen has constant specific heats at room temperature Analysis We take the contents of the cylinder as the system Fig 431 This is a closed system since no mass crosses the system boundary during the process We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work Wb Also heat is lost from the system and electrical work We is done on the system First let us determine the electrical work done on the nitrogen W e VI Δt 120 V2 A5 60 s 1 kJ s 1000 VA 72 kJ The mass of nitrogen is determined from the idealgas relation m P 1 V 1 R T 1 400 kPa05 m 3 0297 kPa m 3 kgK300 K 2245 kg Final PDF to printer 180 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 180 100217 0123 PM Under the stated assumptions and observations the energy balance on the system can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W ein Q out W bout ΔU W ein Q out ΔH m h 2 h 1 m c p T 2 T 1 since ΔU Wb ΔH for a closed system undergoing a quasiequilibrium expansion or compression process at constant pressure From Table A2a cp 1039 kJkgK for nitrogen at room temperature The only unknown quantity in the previous equation is T2 and it is found to be 72 kJ 28 kJ 2245 kg 1039 kJ kgK T 2 27 C T 2 567C Discussion Note that we could also solve this problem by determining the bound ary work and the internal energy change rather than the enthalpy change FIGURE 432 Schematic and PV diagram for Example 410 3 2 350 04 150 1 A 08 P kPa V m3 Q Air V1 400 L P1 150 kPa T1 27C EXAMPLE 410 Heating of a Gas at Constant Pressure A pistoncylinder device initially contains air at 150 kPa and 27C At this state the piston is resting on a pair of stops as shown in Fig 432 and the enclosed volume is 400 L The mass of the piston is such that a 350kPa pressure is required to move it The air is now heated until its volume has doubled Determine a the final tempera ture b the work done by the air and c the total heat transferred to the air SOLUTION Air in a pistoncylinder device with a set of stops is heated until its volume is doubled The final temperature work done and the total heat transfer are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pres sure relative to its criticalpoint values 2 The system is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 and ΔE ΔU 3 The volume remains constant until the piston starts moving and the pressure remains constant afterwards 4 There are no electrical shaft or other forms of work involved Analysis We take the contents of the cylinder as the system Fig 432 This is a closed system since no mass crosses the system boundary during the process We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work Wb Also the boundary work is done by the system and heat is trans ferred to the system a The final temperature can be determined easily by using the idealgas relation between states 1 and 3 in the following form P 1 V 1 T 1 P 3 V 3 T 3 150 kPa V 1 300 K 350 kPa 2 V 1 T 3 T 3 1400 K b The work done could be determined by integration but for this case it is much easier to find it from the area under the process curve on a PV diagram shown in Fig 432 A V 2 V 1 P 2 04 m 3 350 kPa 140 m 3 kPa Final PDF to printer 181 CHAPTER 4 cen22672ch04161210indd 181 100217 0123 PM 45 INTERNAL ENERGY ENTHALPY AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS A substance whose specific volume or density is constant is called an incompressible substance The specific volumes of solids and liquids essen tially remain constant during a process Fig 433 Therefore liquids and solids can be approximated as incompressible substances without sacrificing much in accuracy The constantvolume assumption should be taken to imply that the energy associated with the volume change is negligible compared with other forms of energy Otherwise this assumption would be ridiculous for studying the thermal stresses in solids caused by volume change with temperature or analyzing liquidinglass thermometers It can be mathematically shown see Chap 12 that the constantvolume and constantpressure specific heats are identical for incompressible substances Fig 434 Therefore for solids and liquids the subscripts on cp and cv can be dropped and both specific heats can be represented by a single symbol c That is c p c v c 432 Therefore W 13 140 kJ The work is done by the system to raise the piston and to push the atmospheric air out of the way and thus it is work output c Under the stated assumptions and observations the energy balance on the system between the initial and final states process 13 can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q in W bout ΔU m u 3 u 1 The mass of the system can be determined from the idealgas relation m P 1 V 1 R T 1 150 kPa04 m 3 0287 kPa m 3 kgK300 K 0697 kg The internal energies are determined from the air table Table A17 to be u 1 u 300 K 21407 kJ kg u 3 u 1400 K 111352 kJ kg Thus Q in 140 kJ 0697 kg 111352 21407 kJ kg Q in 767 kJ Discussion The positive sign verifies that heat is transferred to the system FIGURE 433 The specific volumes of incompressible substances remain constant during a process vs constant Solid Liquid vl constant FIGURE 434 The cv and cp values of incompressible substances are identical and are denoted by c Iron 25C c cv cp 045 kJkgK Final PDF to printer 182 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 182 100217 0123 PM This result could also be deduced from the physical definitions of constant volume and constantpressure specific heats Specific heat values for several common liquids and solids are given in Table A3 Internal Energy Changes Like those of ideal gases the specific heats of incompressible substances depend on temperature only Thus the partial differentials in the defining equation of cv can be replaced by ordinary differentials which yield du c v dT cT dT 433 The change in internal energy between states 1 and 2 is then obtained by integration Δu u 2 u 1 1 2 c T dT kJ kg 434 The variation of specific heat c with temperature should be known before this integration can be carried out For small temperature intervals a c value at the average temperature can be used and treated as a constant yielding Δu c avg T 2 T 1 kJ kg 435 Enthalpy Changes Using the definition of enthalpy h u Pv and noting that v constant the differential form of the enthalpy change of incompressible substances can be determined by differentiation to be dh du v dP P dv 0 du v dP 436 Integrating Δh Δu v ΔP c avg ΔT v ΔP kJ kg 437 For solids the term v ΔP is insignificant and thus Δh Δu cavg ΔT For liquids two special cases are commonly encountered 1 Constantpressure processes as in heaters ΔP 0 Δh Δu cavg ΔT 2 Constanttemperature processes as in pumps ΔT 0 Δh v ΔP For a process between states 1 and 2 the last relation can be expressed as h2 h1 vP2 P1 By taking state 2 to be the compressed liquid state at a given T and P and state 1 to be the saturated liquid state at the same tempera ture the enthalpy of the compressed liquid can be expressed as h PT h f T v f T P P sat T 438 as discussed in Chap 3 This is an improvement over the assumption that the enthalpy of the compressed liquid could be taken as hf at the given tempera ture that is h PT hf T However the contribution of the last term is often very small and is neglected Note that at high temperatures and pressures Eq 438 may overcorrect the enthalpy and result in a larger error than the approximation h hf T Final PDF to printer 183 CHAPTER 4 cen22672ch04161210indd 183 100217 0123 PM EXAMPLE 411 Enthalpy of Compressed Liquid Determine the enthalpy of liquid water at 100C and 15 MPa a by using compressed liquid tables b by approximating it as a saturated liquid and c by using the correc tion given by Eq 438 SOLUTION The enthalpy of liquid water is to be determined exactly and approximately Analysis At 100C the saturation pressure of water is 10142 kPa and since P Psat the water exists as a compressed liquid at the specified state a From compressed liquid tables we read P 15 MPa T 100C h 43039 kJ kg Table A7 This is the exact value b Approximating the compressed liquid as a saturated liquid at 100C as is commonly done we obtain h h f 100C 41917 kJ kg This value is in error by about 26 percent c From Eq 438 h PT h f T v f T P P sat T 41917 kJ kg 0001 m 3 kg15000 10142 kPa 1 kJ 1 kPa m 3 43407 kJ kg Discussion Note that the correction term reduced the error from 26 to about 1 percent in this case However this improvement in accuracy is often not worth the extra effort involved EXAMPLE 412 Cooling of an Iron Block by Water A 50kg iron block at 80C is dropped into an insulated tank that contains 05 m3 of liquid water at 25C Determine the temperature when thermal equilibrium is reached SOLUTION An iron block is dropped into water in an insulated tank The final temperature when thermal equilibrium is reached is to be determined Assumptions 1 Both water and the iron block are incompressible substances 2 Constant specific heats at room temperature can be used for water and the iron 3 The system is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 and ΔE ΔU 4 There are no electrical shaft or other forms of work involved 5 The system is well insulated and thus there is no heat transfer Analysis We take the entire contents of the tank as the system Fig 435 This is a closed system since no mass crosses the system boundary during the process We FIGURE 435 Schematic for Example 412 Final PDF to printer 184 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 184 100217 0123 PM FIGURE 436 Schematic for Example 413 Furnace 900C 100C Steel ball Air 35C observe that the volume of a rigid tank is constant and thus there is no boundary work The energy balance on the system can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies 0 ΔU The total internal energy U is an extensive property and therefore it can be expressed as the sum of the internal energies of the parts of the system Then the total internal energy change of the system becomes Δ U sys Δ U iron Δ U water 0 mc T 2 T 1 iron mc T 2 T 1 water 0 The specific volume of liquid water at or about room temperature can be taken to be 0001 m3kg Then the mass of the water is m water V v 05 m 3 0001 m 3 kg 500 kg The specific heats of iron and liquid water are determined from Table A3 to be ciron 045 kJkgC and cwater 418 kJkgC Substituting these values into the energy equation we obtain 50 kg045 kJ kgC T 2 80C 500 kg418 kJ kgC T 2 25 C 0 T 2 256 C Therefore when thermal equilibrium is established both the water and iron will be at 256C Discussion The small rise in water temperature is due to its large mass and large specific heat EXAMPLE 413 Cooling of Carbon Steel Balls in Air Carbon steel balls ρ 7833 kgm3 and cp 0465 kJkgC 8 mm in diameter are annealed by heating them first to 900C in a furnace and then allowing them to cool slowly to 100C in ambient air at 35C as shown in Fig 436 If 2500 balls are to be annealed per hour determine the total rate of heat transfer from the balls to the ambient air SOLUTION Carbon steel balls are to be annealed at a rate of 2500h by heating them first and then allowing them to cool slowly in ambient air at a specified rate The total rate of heat transfer from the balls to the ambient air is to be determined Assumptions 1 The thermal properties of the balls are constant 2 There are no changes in kinetic and potential energies 3 The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the balls are given to be ρ 7833 kg m3 and cp 0465 kJkgC Final PDF to printer 185 CHAPTER 4 cen22672ch04161210indd 185 100217 0123 PM Analysis We take a single ball as the system The energy balance for this closed system can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q out Δ U ball m u 2 u 1 Q out mc T 1 T 2 The amount of heat transfer from a single ball is m ρV ρ π D 3 6 7833 kg m 3 π 0008 m 3 6 000210 kg Q out mc T 1 T 2 000210 kg 0465 kJ kgC 900 100C 0781 kJ per ball Then the total rate of heat transfer from the balls to the ambient air becomes Q out n ball Q out 2500 ballh0781 kJ ball 1953 kJh 542 W Discussion For solids and liquids constantpressure and constantvolume specific heats are identical and can be represented by a single symbol c However it is custom ary to use the symbol cp for the specific heat of incompressible substances An important and exciting application area of thermodynamics is biological sys tems which are the sites of rather complex and intriguing energy transfer and trans formation processes Biological systems are not in thermodynamic equilibrium and thus they are not easy to analyze Despite their complexity biological systems are primarily made up of four simple elements hydrogen oxygen carbon and nitrogen In the human body hydrogen accounts for 63 percent oxygen 255 per cent carbon 95 percent and nitrogen 14 percent of all the atoms The remaining 06 percent of the atoms comes from 20 other elements essential for life By mass about 72 percent of the human body is water The building blocks of living organisms are cells which resemble miniature factories performing functions that are vital for the survival of organisms A biological system can be as simple as a single cell The human body contains about 100 trillion cells with an average diameter of 001 mm The membrane of the cell is a semipermeable wall that allows some substances to pass through it while excluding others In a typical cell thousands of chemical reactions occur every second dur ing which some molecules are broken down and energy is released and some new molecules are formed This high level of chemical activity in the cells which maintains the human body at a temperature of 37C while performing the necessary bodily tasks is called metabolism In simple terms metabolism TOPIC OF SPECIAL INTEREST Thermodynamic Aspects of Biological Systems This section can be skipped without a loss in continuity Final PDF to printer 186 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 186 100217 0123 PM FIGURE 437 An average person dissipates energy to the surroundings at a rate of 84 W when resting Janis ChristieGetty Images RF refers to the burning of foods such as carbohydrates fat and protein The rate of metabolism in the resting state is called the basal metabolic rate which is the rate of metabolism required to keep a body performing the necessary func tions such as breathing and blood circulation at zero external activity level The metabolic rate can also be interpreted as the energy consumption rate for a body For an average male 30 years old 70 kg 18m2 body surface area the basal metabolic rate is 84 W That is the body dissipates energy to the environ ment at a rate of 84 W which means that the body is converting the chemical energy of food or of body fat if the person has not eaten into thermal energy at a rate of 84 W Fig 437 The metabolic rate increases with the level of activ ity and it may exceed 10 times the basal metabolic rate when a body is doing strenuous exercise That is two people doing heavy exercising in a room may be supplying more energy to the room than a 1kW electrical resistance heater Fig 438 The fraction of sensible heat varies from about 40 percent in the case of heavy work to about 70 percent in the case of light work The rest of the energy is rejected from the body by perspiration in the form of latent heat The basal metabolic rate varies with sex body size general health condi tions and so forth and decreases considerably with age This is one of the reasons people tend to put on weight in their late twenties and thirties even though they do not increase their food intake The brain and the liver are the major sites of metabolic activity These two organs are responsible for almost 50 percent of the basal metabolic rate of an adult human body although they constitute only about 4 percent of the body mass In small children it is remark able that about half of the basal metabolic activity occurs in the brain alone The biological reactions in cells occur essentially at constant temperature pressure and volume The temperature of the cell tends to rise when some chemical energy is converted to heat but this energy is quickly transferred to the circulatory system which transports it to outer parts of the body and even tually to the environment through the skin The muscle cells function very much like an engine converting the chemical energy into mechanical energy work with a conversion efficiency of close to 20 percent When the body does no net work on the environment such as mov ing some furniture upstairs the entire work is also converted to heat In that case the entire chemical energy in the food released during metabolism in the body is eventually transferred to the environment A TV set that consumes elec tricity at a rate of 300 W must reject heat to its environment at a rate of 300 W in steady operation regardless of what goes on in the set That is turning on a 300W TV set or three 100W lightbulbs will produce the same heating effect in a room as a 300W resistance heater Fig 439 This is a consequence of the conservation of energy principle which requires that the energy input into a system must equal the energy output when the total energy content of a sys tem remains constant during a process Food and Exercise The energy requirements of a body are met by the food we eat The nutrients in the food are considered in three major groups carbohydrates proteins and fats Carbohydrates are characterized by having hydrogen and oxygen atoms in a 21 ratio in their molecules The molecules of carbohydrates range from very sim ple as in plain sugar to very complex or large as in starch Bread and plain FIGURE 438 Two fastdancing people supply more energy to a room than a 1kW electric resistance heater 12 kJs 1 kJs Final PDF to printer 187 CHAPTER 4 cen22672ch04161210indd 187 100217 0123 PM sugar are the major sources of carbohydrates Proteins are very large molecules that contain carbon hydrogen oxygen and nitrogen and they are essential for the building and repairing of the body tissues Proteins are made up of smaller building blocks called amino acids Complete proteins such as meat milk and eggs have all the amino acids needed to build body tissues Plant source pro teins such as those in fruits vegetables and grains lack one or more amino acids and are called incomplete proteins Fats are relatively small molecules that consist of carbon hydrogen and oxygen Vegetable oils and animal fats are major sources of fats Most foods we eat contain all three nutrition groups at varying amounts The typical average American diet consists of 45 percent car bohydrate 40 percent fat and 15 percent protein although it is recommended that in a healthy diet less than 30 percent of the calories should come from fat The energy content of a given food is determined by burning a small sample of the food in a device called a bomb calorimeter which is basically a well insulated rigid tank Fig 440 The tank contains a small combustion cham ber surrounded by water The food is ignited and burned in the combustion chamber in the presence of excess oxygen and the energy released is trans ferred to the surrounding water The energy content of the food is calculated on the basis of the conservation of energy principle by measuring the temperature rise of the water The carbon in the food is converted into CO2 and hydrogen into H2O as the food burns The same chemical reactions occur in the body and thus the same amount of energy is released Using dry free of water samples the average energy contents of the three basic food groups are determined by bomb calorimeter measurements to be 180 MJkg for carbohydrates 222 MJkg for proteins and 398 MJkg for fats These food groups are not entirely metabolized in the human body however The fraction of metabolizable energy contents are 955 percent for carbohydrates 775 percent for proteins and 977 percent for fats That is the fats we eat are almost entirely metabolized in the body but close to onequarter of the protein we eat is discarded from the body unburned This corresponds to 41 Caloriesg for proteins and carbohydrates and 93 Caloriesg for fats Fig 441 com monly seen in nutrition books and on food labels The energy contents of the foods we normally eat are much lower than the preceding values because of the large water content water adds bulk to the food but it cannot be metabolized or burned and thus it has no energy value Most vegetables fruits and meats for example are mostly water The average metabolizable energy contents of the three basic food groups are 42 MJkg for carbohydrates 84 MJkg for proteins and 331 MJkg for fats Note that 1 kg of natural fat contains almost 8 times the metabolizable energy of 1 kg of natural carbohydrates Thus a person who fills his stomach with fatty foods is consuming much more energy than a person who fills his stomach with carbohydrates such as bread or rice The metabolizable energy content of foods is usually expressed by nutri tionists in terms of the capitalized Calories One Calorie is equivalent to one kilocalorie 1000 calories which is equivalent to 41868 kJ That is 1 Cal Calorie 1000 calories 1 kcal kilocalorie 41868 kJ The calorie notation often causes confusion since it is not always followed in the tables or articles on nutrition When the topic is food or fitness a calorie normally means a kilocalorie whether it is capitalized or not FIGURE 439 Some arrangements that supply a room with the same amount of energy as a 300W electric resistance heater A 300W refrigerator Two people each dissipating 150 W A 100W computer with a 200W monitor Solar energy 300 W Three light bulbs 100 W each A 300W fan A 300W TV A 300W resistance heater FIGURE 440 Schematic of a bomb calorimeter used to determine the energy content of food samples Bomb combustion chamber Food sample Mixer and motor Electrical switch Thermometer Insulation Water Final PDF to printer 188 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 188 100217 0123 PM FIGURE 441 Evaluating the calorie content of one serving of chocolate chip cookies values are for Chips Ahoy cookies made by Nabisco ComstockGetty Images RF 3 cookies 32 g Fat 8 g93 Calg 744 Cal Protein 2 g41 Calg 82 Cal Carbohydrates 21 g41 Calg 861 Cal Other 1 g0 Calg 0 TOTAL for 32 g 169 Cal The daily calorie needs of people vary greatly with age gender the state of health the activity level the body weight and the composition of the body as well as other factors A small person needs fewer calories than a larger person of the same sex and age An average man needs about 2400 to 2700 Calories a day The daily need of an average woman varies from 1800 to 2200 Calories The daily calorie needs are about 1600 for sedentary women and some older adults 2000 for sedentary men and most older adults 2200 for most children teenage girls and active women 2800 for teenage boys active men and some very active women and above 3000 for very active men The average value of calorie intake is usually taken to be 2000 Calories per day The daily calorie needs of a person can be determined by multiplying the body weight in pounds which is 2205 times the body weight in kg by 11 for a sedentary person 13 for a moderately active person 15 for a moderate exer ciser or physical laborer and 18 for an extremely active exerciser or physical laborer The extra calories a body consumes are usually stored as fat which serves as a spare energy supply for use when the energy intake of the body is less than the needed amount Like other natural fat 1 kg of human body fat contains about 331 MJ of metabolizable energy Therefore a starving person zero energy intake who uses up 2200 Calories 9211 kJ a day can meet his daily energy intake require ments by burning only 921133100 028 kg of body fat So it is no surprise that people are known to survive over 100 days without eating They still need to drink water however to replenish the water lost through the lungs and the skin to avoid the dehydration that may occur in just a few days Although the desire to get rid of the excess fat in a thin world may be overwhelming at times starvation diets are not recommended because the body soon starts to consume its own muscle tissue in addition to fat A healthy diet should involve regular exercise while allowing for a reasonable calorie intake The average metabolizable energy contents of various foods and the energy consumption during various activities are given in Tables 41 and 42 Con sidering that no two hamburgers are alike and that no two people walk exactly the same way there is some uncertainty in these values as you would expect Therefore you may encounter somewhat different values in other books or magazines for the same items TABLE 41 Approximate metabolizable energy content of some common foods 1 Calorie 41868 kJ 3968 Btu Food Calories Food Calories Food Calories Apple one medium Baked potato plain Baked potato with cheese Bread white one slice Butter one teaspoon Cheeseburger Chocolate candy bar 20 g Cola 200 ml Egg one 70 250 550 70 35 325 105 87 80 Fish sandwich French fries regular Hamburger Hot dog Ice cream 100 ml 10 fat Lettuce salad with French dressing 450 250 275 300 110 150 Milk skim 200 ml Milk whole 200 ml Peach one medium Pie one 1 8 slice 23 cm diameter Pizza large cheese one 1 8 slice 76 136 65 300 350 Final PDF to printer 189 CHAPTER 4 cen22672ch04161210indd 189 100217 0123 PM The rates of energy consumption listed in Table 42 during some activities are for a 68kg adult The energy consumed for smaller or larger adults can be determined using the proportionality of the metabolism rate and the body size For example the rate of energy consumption by a 68kg bicyclist is listed in Table 42 to be 639 Caloriesh Then the rate of energy consumption by a 50kg bicyclist is 50 kg 639 Cal h 68 kg 470 Cal h For a 100kg person it would be 940 Calh The thermodynamic analysis of the human body is rather complicated since it involves mass transfer during breathing perspiring etc as well as energy transfer Thus it should be treated as an open system However the energy transfer with mass is difficult to quantify Therefore the human body is often modeled as a closed system for simplicity by treating energy transported with mass as just energy transfer For example eating is modeled as the transfer of energy into the human body in the amount of the metabolizable energy content of the food Dieting Most diets are based on calorie counting that is the conservation of energy principle a person who consumes more calories than his or her body burns will gain weight whereas a person who consumes fewer calories than his or her body burns will lose weight Yet people who eat whatever they want whenever they want without gaining any weight are living proof that the caloriecounting technique alone does not work in dieting Obviously there is more to dieting than keeping track of calories It should be noted that the phrases weight gain and weight loss are misnomers The correct phrases should be mass gain and mass loss A man who goes to space loses practically all of his weight but none of his mass When the topic is food and fitness weight is understood to mean mass and weight is expressed in mass units Researchers on nutrition proposed several theories on dieting One theory suggests that some people have very food efficient bodies These people need fewer calories than other people do for the same activity just like a fuel efficient car needing less fuel for traveling a given distance It is interesting that we want our cars to be fuel efficient but we do not want the same high efficiency for our bodies One thing that frustrates the dieters is that the body interprets dieting as starvation and starts using the energy reserves of the body more stringently Shifting from a normal 2000Calorie daily diet to an 800Calorie diet without exercise is observed to lower the basal metabolic rate by 10 to 20 percent Although the metabolic rate returns to normal once the dieting stops extended periods of lowcalorie dieting without adequate exer cise may result in the loss of considerable muscle tissue together with fat With less muscle tissue to burn calories the metabolic rate of the body declines and stays below normal even after a person starts eating normally As a result the person regains the weight he or she has lost in the form of fat plus more The basal metabolic rate remains about the same in people who exercise while dieting TABLE 42 Approximate energy consumption of a 68kg adult during some activities 1 Calorie 41868 kJ 3968 Btu Activity Caloriesh Basal metabolism 72 Basketball 550 Bicycling 21 kmh 639 Crosscountry skiing 13 kmh 936 Driving a car 180 Eating 99 Fast dancing 600 Fast running 13 kmh 936 Jogging 8 kmh 540 Swimming fast 860 Swimming slow 288 Tennis advanced 480 Tennis beginner 288 Walking 72 kmh 432 Watching TV 72 Final PDF to printer 190 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 190 100217 0123 PM FIGURE 442 The body tends to keep the body fat level at a set point by speeding up metabolism when a person splurges and by slowing it down when the person starves Set point New set point Body fat level Regular moderate exercise is part of any healthy dieting program for good reason It builds or preserves muscle tissue that burns calories much faster than the fat tissue does It is interesting that aerobic exercise continues to burn calories for several hours after the workout raising the overall metabolic rate considerably Another theory suggests that people with too many fat cells developed dur ing childhood or adolescence are much more likely to gain weight Some peo ple believe that the fat content of our bodies is controlled by the setting of a fat control mechanism much like the temperature of a house is controlled by the thermostat setting Some people put the blame for weight problems simply on the genes Considering that 80 percent of the children of overweight parents are also overweight heredity may indeed play an important role in the way a body stores fat Researchers from the University of Washington and the Rocke feller University have identified a gene called the RIIbeta that seems to control the rate of metabolism The body tries to keep the body fat content at a particular level called the set point that differs from person to person Fig 442 This is done by speeding up the metabolism and thus burning calories faster when a person tends to gain weight and by slowing down the metabolism and thus burning calories more slowly when a person tends to lose weight Therefore a person who just became slim burns fewer calories than does a person of the same size who has always been slim Even exercise does not seem to change that Then to keep the weight off the newly slim person should consume no more calories than he or she can burn Note that in people with high metabolic rates the body dissipates the extra calories as body heat instead of storing them as fat and thus there is no violation of the conservation of energy principle In some people a genetic flaw is believed to be responsible for the extremely low rates of metabolism Several studies concluded that losing weight for such people is nearly impossible That is obesity is a biological phenomenon How ever even such people will not gain weight unless they eat more than their body can burn They just must learn to be content with little food to remain slim and forget about ever having a normal eating life For most people genetics determine the range of normal weights A person may end up at the high or low end of that range depending on eating and exercise habits This also explains why some genetically identical twins are not so identical when it comes to body weight Hormone imbalance is also believed to cause excessive weight gain or loss Based on his experience the first author of this book has also developed a diet called the sensible diet It consists of two simple rules eat whatever you want whenever you want as much as you want provided that 1 you do not eat unless you are hungry and 2 you stop eating before you get stuffed In other words listen to your body and dont impose on it Dont expect to see this unscientific diet advertised anywhere since there is nothing to be sold and thus no money to be made Also it is not as easy as it sounds since food is at the center stage of most leisure activities in social life and eating and drinking have become synonymous with having a good time However it is comforting to know that the human body is quite forgiving of occasional impositions Final PDF to printer 191 CHAPTER 4 cen22672ch04161210indd 191 100217 0123 PM Being overweight is associated with a long list of health risks from high blood pressure to some forms of cancer especially for people who have a weight related medical condition such as diabetes hypertension and heart disease Therefore people often wonder if their weight is in the proper range Well the answer to this question is not written in stone but if you cannot see your toes or you can pinch your love handles more than an inch you dont need an expert to tell you that you went over your range On the other hand some people who are obsessed with the weight issue try to lose more weight even though they are actually underweight Therefore it is useful to have a scientific criterion to determine physical fitness The range of healthy weight for adults is usually expressed in terms of the body mass index BMI defined in SI units as BMI 19 underweight BMI W kg H 2 m 2 with 19 BMI 25 healthy weight BMI 25 overweight 439 where W is the weight actually the mass of the person in kg and H is the height in m Therefore a BMI of 25 is the upper limit for the healthy weight and a person with a BMI of 27 is 8 percent overweight It can be shown that the formula is equivalent in English units to BMI 705 WH2 where W is in pounds and H is in inches The proper range of weight for adults of various heights is given in Table 43 in both SI and English units TABLE 43 The range of healthy weight for adults of various heights Source National Institute of Health English units SI units Height in Healthy weight lbm Height m Healthy weight kg 58 91119 145 4053 60 97127 150 4356 62 103136 155 4660 64 111146 160 4964 66 118156 165 5268 68 125165 170 5572 70 133175 175 5877 72 140185 180 6281 74 148195 185 6586 76 156205 190 6990 The upper and lower limits of healthy range corre spond to body mass indexes of 25 and 19 respectively EXAMPLE 414 Burning Off Lunch Calories A 90kg man had two hamburgers a regular serving of french fries and a 200ml Coke for lunch Fig 443 Determine how long it will take for him to burn the lunch calories off a by watching TV and b by fast swimming What would your answers be for a 45kg man SOLUTION A man had lunch at a restaurant The time it will take for him to burn the lunch calories by watching TV and by fast swimming are to be determined Assumptions The values in Tables 41 and 42 are applicable for food and exercise Analysis a We take the human body as our system and treat it as a closed system whose energy content remains unchanged during the process Then the conservation of energy principle requires that the energy input into the body must be equal to the energy output The net energy input in this case is the metabolizable energy content of the food eaten It is determined from Table 41 to be E in 2 E hamburger E fries E cola 2 275 250 87 887 Cal The rate of energy output for a 68kg man watching TV is given in Table 42 to be 72 Caloriesh For a 90kg man it becomes E out 90 kg 72 Cal h 68 kg 953 Cal h FIGURE 443 A typical lunch discussed in Example 414 Copyright Food Collection RF Final PDF to printer 192 ENERGY ANALYSIS OF CLOSED SYSTEMS cen22672ch04161210indd 192 100217 0123 PM EXAMPLE 415 Losing Weight by Switching to FatFree Chips The fake fat olestra passes through the body undigested and thus adds zero calorie to the diet Although foods cooked with olestra taste pretty good they may cause abdominal discomfort and the longterm effects are unknown A 1oz 283g serv ing of regular potato chips has 10 g of fat and 150 Calories whereas 1 oz of the so called fatfree chips fried in olestra has only 75 Calories Consider a person who eats 1 oz of regular potato chips every day at lunch without gaining or losing any weight Determine how much weight this person will lose in one year if he or she switches to fatfree chips Fig 444 SOLUTION A person switches from regular potato chips to fatfree ones The weight the person loses in one year is to be determined Assumptions Exercising and other eating habits remain the same Analysis The person who switches to the fatfree chips consumes 75 fewer Calories a day Then the annual reduction in calories consumed becomes E reduced 75 Cal day365 day year 27375 Cal year The metabolizable energy content of 1 kg of body fat is 33100 kJ Therefore assum ing the deficit in the calorie intake is made up by burning body fat the person who switches to fatfree chips will lose m fat E reduced Energy content of fat 27375 Cal 33100 kJ kg 41868 kJ 1 Cal 346 kg about 76 pounds of body fat that year FIGURE 444 Schematic for Example 415 Therefore it will take Δt 887 Cal 953 Cal h 93 h to burn the lunch calories off by watching TV b It can be shown in a similar manner that it takes only 47 min to burn the lunch calories off by fast swimming Discussion The 45kg man is half as large as the 90kg man Therefore expending the same amount of energy takes twice as long in each case 186 h by watching TV and 94 min by fast swimming SUMMARY Work is the energy transferred as a force acts on a system through a distance The most common form of mechani cal work is the boundary work which is the work associ ated with the expansion and compression of substances On a PV diagram the area under the process curve represents the boundary work for a quasiequilibrium process Various forms of boundary work are expressed as follows 1 General W b 1 2 P dV Final PDF to printer 193 CHAPTER 4 cen22672ch04161210indd 193 100217 0123 PM REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Fundamentals SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 2 ASHRAE Handbook of Refrigeration SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1994 2 Isobaric process W b P 0 V 2 V 1 P 1 P 2 P 0 constant 3 Polytropic process W b P 2 V 2 P 1 V 1 1 n n 1 P V n constant 4 Isothermal process of an ideal gas W b P 1 V 1 ln V 2 V 1 mR T 0 ln V 2 V 1 PV mR T 0 constant The first law of thermodynamics is essentially an expression of the conservation of energy principle also called the energy balance The general energy balance for any system undergo ing any process can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies It can also be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass dE system dt Rate of change in internal kinetic potential etc energies Taking heat transfer to the system and work done by the system to be positive quantities the energy balance for a closed system can also be expressed as Q W ΔU ΔKE ΔPE where W W other W b ΔU m u 2 u 1 ΔKE 1 2 m V 2 2 V 1 2 ΔPE mg z 2 z 1 For a constantpressure process Wb ΔU ΔH Thus Q W other ΔH ΔKE ΔPE Note that the relation above is limited to constantpressure processes of closed systems and is NOT valid for processes during which pressure varies The amount of energy needed to raise the temperature of a unit mass of a substance by one degree is called the spe cific heat at constant volume cv for a constantvolume process and the specific heat at constant pressure cp for a constant pressure process They are defined as c v u T v and c p h T p For ideal gases u h cv and cp are functions of temperature alone The Δu and Δh of ideal gases are expressed as Δu u 2 u 1 1 2 c v T dT c vavg T 2 T 1 Δh h 2 h 1 1 2 c p T dT c pavg T 2 T 1 For ideal gases cv and cp are related by c p c v R where R is the gas constant The specific heat ratio k is defined as k c p c v For incompressible substances liquids and solids both the constantpressure and constantvolume specific heats are iden tical and denoted by c c p c v c The Δu and Δh of incompressible substances are given by Δu 1 2 c T dT c avg T 2 T 1 Δh Δu v ΔP Final PDF to printer cen22672ch04161210indd 194 100217 0123 PM 194 ENERGY ANALYSIS OF CLOSED SYSTEMS PROBLEMS Moving Boundary Work 41C Is the boundary work associated with constantvolume systems always zero 42C On a Pv diagram what does the area under the pro cess curve represent 43C An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant tempera ture For which case is the work done greater 44 Calculate the total work in kJ for process 13 shown in Fig P44 when the system consists of 2 kg of nitrogen 46 Nitrogen at an initial state of 300 K 150 kPa and 02 m3 is compressed slowly in an isothermal process to a final pres sure of 800 kPa Determine the work done during this process 47 The volume of 1 kg of helium in a pistoncylinder device is initially 5 m3 Now helium is compressed to 2 m3 while its pressure is maintained constant at 130 kPa Deter mine the initial and final temperatures of helium as well as the work required to compress it in kJ 48 A pistoncylinder device with a set of stops initially contains 06 kg of steam at 10 MPa and 400C The loca tion of the stops corresponds to 40 percent of the initial volume Now the steam is cooled Determine the compres sion work if the final state is a 10 MPa and 250C and b 500 kPa c Also determine the temperature at the final state in part b FIGURE P44 05 1 500 400 100 P kPa 1 2 3 v m3kg 45E Calculate the total work in Btu produced by the pro cess of Fig P45E FIGURE P45E P psia 2 4 500 100 1 2 V ft3 Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software FIGURE P48 49 A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200C Calculate the work done by the steam during this pro cess Answer 214 kJ 410E A frictionless pistoncylinder device contains 16 lbm of superheated water vapor at 40 psia and 600F Steam is now cooled at constant pressure until 70 percent of it by mass con denses Determine the work done during this process 411 1 m3 of saturated liquid water at 200C is expanded isothermally in a closed system until its quality is 80 percent Determine the total work produced by this expansion in kJ 412 Argon is compressed in a polytropic process with n 12 from 120 kPa and 30C to 1200 kPa in a piston cylinder device Determine the final temperature of the argon 413 A gas is compressed from an initial volume of 042 m3 to a final volume of 012 m3 During the quasi equilibrium process the pressure changes with volume according to the relation P aV b where a 1200 kPam3 and b 600 kPa Calculate the work done during this process a by plotting the process on a PV diagram and finding the Final PDF to printer cen22672ch04161210indd 195 100217 0123 PM 195 CHAPTER 4 414 A mass of 15 kg of air at 120 kPa and 24C is con tained in a gastight frictionless pistoncylinder device The air is now compressed to a final pressure of 600 kPa During the process heat is transferred from the air such that the tem perature inside the cylinder remains constant Calculate the work input during this process Answer 206 kJ 415 During some actual expansion and compression processes in pistoncylinder devices the gases have been observed to satisfy the relationship PVn C where n and C are constants Calculate the work done when a gas expands from 350 kPa and 003 m3 to a final volume of 02 m3 for the case of n 15 416 Reconsider Prob 415 Using appropriate soft ware plot the process described in the problem on a PV diagram and investigate the effect of the polytropic exponent n on the boundary work Let the polytropic exponent vary from 11 to 16 Plot the boundary work versus the poly tropic exponent and discuss the results 417 A frictionless pistoncylinder device contains 5 kg of nitrogen at 100 kPa and 250 K Nitrogen is now compressed slowly according to the relation PV14 constant until it reaches a final temperature of 450 K Calculate the work input during this process Answer 742 kJ 419 A pistoncylinder device initially contains 04 kg of nitrogen gas at 160 kPa and 140C The nitrogen is now expanded isothermally to a pressure of 100 kPa Determine the boundary work done during this process Answer 230 kJ FIGURE P413 Gas P aV b FIGURE P417 N2 PV14 const 418E During an expansion process the pressure of a gas changes from 15 to 100 psia according to the relation P aV b where a 5 psiaft3 and b is a constant If the ini tial volume of the gas is 7 ft3 calculate the work done during the process Answer 181 Btu FIGURE P419 N2 160 kPa 140C 420 A pistoncylinder device contains 015 kg of air ini tially at 2 MPa and 350C The air is first expanded isother mally to 500 kPa then compressed polytropically with a polytropic exponent of 12 to the initial pressure and finally compressed at the constant pressure to the initial state Deter mine the boundary work for each process and the net work of the cycle 421 Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa 1 L 290 kPa 11 L 270 kPa 12 L 250 kPa 14 L 220 kPa 17 L and 200 kPa 2 L 422 1 kg of water that is initially at 90C with a quality of 10 percent occupies a springloaded pistoncylinder device such as that in Fig P422 This device is now heated until the pressure rises to 800 kPa and the temperature is 250C Determine the total work produced during this process in kJ Answer 245 kJ FIGURE P422 Water 90C x 010 Q 423 An ideal gas undergoes two processes in a piston cylinder device as follows 12 Polytropic compression from T1 and P1 with a polytropic exponent n and a compression ratio of r V1V2 23 Constant pressure expansion at P3 P2 until V3 V1 a Sketch the processes on a single PV diagram b Obtain an expression for the ratio of the compression toexpansion work as a function of n and r area under the process curve and b by performing the nec essary integrations Final PDF to printer cen22672ch04161210indd 196 100217 0123 PM 196 ENERGY ANALYSIS OF CLOSED SYSTEMS c Find the value of this ratio for values of n 14 and r 6 Answers b 1 n 1 1 r 1n r 1 c 0256 424 A pistoncylinder device contains 50 kg of water at 250 kPa and 25C The crosssectional area of the piston is 01 m2 Heat is now transferred to the water causing part of it to evaporate and expand When the volume reaches 02 m3 the piston reaches a linear spring whose spring constant is 100 kNm More heat is transferred to the water until the piston rises 20 cm more Deter mine a the final pressure and temperature and b the work done during this process Also show the process on a PV diagram Answers a 450 kPa 1479C b 445 kJ 428 A rigid container equipped with a stirring device con tains 25 kg of motor oil Determine the rate of specific energy increase when heat is transferred to the oil at a rate of 1 W and 15 W of power is applied to the stirring device 429 Complete each line of the following table on the basis of the conservation of energy principle for a closed system FIGURE P424 A 01 m2 H2O m 50 kg Q 425 Reconsider Prob 424 Using appropriate soft ware investigate the effect of the spring constant on the final pressure in the cylinder and the boundary work done Let the spring constant vary from 50 kNm to 500 kNm Plot the final pressure and the boundary work against the spring constant and discuss the results 426 Carbon dioxide contained in a pistoncylinder device is compressed from 03 to 01 m3 During the process the pres sure and volume are related by P aV2 where a 8 kPam6 Calculate the work done on the carbon dioxide during this pro cess Answer 533 kJ Closed System Energy Analysis 427E A closed system like that shown in Fig P427E is operated in an adiabatic manner First 15000 lbfft of work are done by this system Then work is applied to the stirring device to raise the internal energy of the fluid by 1028 Btu Calculate the net increase in the internal energy of this system FIGURE P427E 430 A substance is contained in a wellinsulated rigid container that is equipped with a stirring device as shown in Fig P430 Determine the change in the internal energy of this substance when 15 kJ of work is applied to the stirring device FIGURE P430 431 A 05m3 rigid tank contains refrigerant134a initially at 160 kPa and 40 percent quality Heat is now transferred to the refrigerant until the pressure reaches 700 kPa Determine a the mass of the refrigerant in the tank and b the amount of heat transferred Also show the process on a Pv diagram with respect to saturation lines 432E A 20ft3 rigid tank initially contains saturated refrigerant 134a vapor at 160 psia As a result of heat transfer from the refrigerant the pressure drops to 50 psia Show the process on a Pv diagram with respect to saturation lines and determine a the final temperature b the amount of refriger ant that has condensed and c the heat transfer 433 A rigid 10L vessel initially contains a mixture of liq uid water and vapor at 100C with 123 percent quality The Qin kJ Wout kJ E1 kJ E2 kJ m kg e2 e1 kJkg 280 1020 860 3 350 130 550 5 260 300 2 150 300 750 500 1 200 300 2 100 FIGURE P429 Final PDF to printer cen22672ch04161210indd 197 100217 0123 PM 197 CHAPTER 4 434E A rigid 1ft3 vessel contains R134a originally at 20F and 277 percent quality The refrigerant is then heated until its temperature is 100F Calculate the heat transfer required to do this Answer 847 Btu the liquid is evaporated during this constantpressure process and the paddlewheel work amounts to 400 kJ determine the voltage of the source Also show the process on a Pv diagram with respect to saturation lines Answer 224 V 439 A 40L electrical radiator containing heating oil is placed in a 50m3 room Both the room and the oil in the radia tor are initially at 10C The radiator with a rating of 24 kW is now turned on At the same time heat is lost from the room at an average rate of 035 kJs After some time the average temperature is measured to be 20C for the air in the room and 50C for the oil in the radiator Taking the density and the specific heat of the oil to be 950 kgm3 and 22 kJkgC respectively determine how long the heater is kept on Assume the room is well sealed so that there are no air leaks FIGURE P438 FIGURE P433 Water 10 L 100C x 0123 Q FIGURE P434E R134a 1 ft3 20F x 0277 Heat 435 A pistoncylinder device contains 5 kg of refrigerant 134a at 800 kPa and 70C The refrigerant is now cooled at constant pressure until it exists as a liquid at 15C Determine the amount of heat loss and show the process on a Tv diagram with respect to saturation lines Answer 1173 kJ 436E A pistoncylinder device contains 05 lbm of water initially at 120 psia and 2 ft3 Now 200 Btu of heat is trans ferred to the water while its pressure is held constant Deter mine the final temperature of the water Also show the process on a Tv diagram with respect to saturation lines 437 2 kg of saturated liquid water at 150C is heated at con stant pressure in a pistoncylinder device until it is saturated vapor Determine the heat transfer required for this process 438 An insulated pistoncylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water If onehalf of FIGURE P439 Room 10C Q Radiator 440 Steam at 75 kPa and 8 percent quality is contained in a springloaded pistoncylinder device as shown in Fig P440 with an initial volume of 2 m3 Steam is now heated until its vol ume is 5 m3 and its pressure is 225 kPa Determine the heat trans ferred to and the work produced by the steam during this process FIGURE P440 441 A pistoncylinder device initially contains 06 m3 of satu rated water vapor at 250 kPa At this state the piston is resting on a set of stops and the mass of the piston is such that a pressure of 300 kPa is required to move it Heat is now slowly transferred to the steam until the volume doubles Show the process on a Pv diagram with respect to saturation lines and determine a the final temperature b the work done during this process and c the total heat transfer Answers a 662C b 180 kJ c 910 kJ 442 An insulated tank is divided into two parts by a parti tion One part of the tank contains 25 kg of compressed liquid water at 60C and 600 kPa while the other part is evacuated The partition is now removed and the water expands to fill the mixture is then heated until its temperature is 180C Calculate the heat transfer required for this process Answer 925 kJ Final PDF to printer cen22672ch04161210indd 198 100217 0123 PM 198 ENERGY ANALYSIS OF CLOSED SYSTEMS entire tank Determine the final temperature of the water and the volume of the tank for a final pressure of 10 kPa 448C A fixed mass of an ideal gas is heated from 50 to 80C a at constant volume and b at constant pressure For which case do you think the energy required will be greater Why 449C Is the relation Δu mcvavgΔT restricted to constant volume processes only or can it be used for any kind of pro cess of an ideal gas 450C Is the relation Δh mcpavgΔT restricted to constant pressure processes only or can it be used for any kind of pro cess of an ideal gas 451E What is the change in the internal energy in Btulbm of air as its temperature changes from 100 to 200F Is there any difference if the temperature were to change from 0 to 100F 452 Neon is compressed from 100 kPa and 20C to 500 kPa in an isothermal compressor Determine the change in the specific volume and specific enthalpy of neon caused by this compression 453 What is the change in the enthalpy in kJkg of oxygen as its temperature changes from 150 to 250C Is there any dif ference if the temperature change were from 0 to 100C Does the pressure at the beginning and end of this process have any effect on the enthalpy change 454 A mass of 10 g of nitrogen is contained in the spring loaded pistoncylinder device shown in Fig P454 The spring constant is 1 kNm and the piston diameter is 10 cm When the spring exerts no force against the piston the nitrogen is at 120 kPa and 27C The device is now heated until its vol ume is 10 percent greater than the original volume Determine the change in the specific internal energy and enthalpy of the nitrogen Answers 468 kJkg 655 kJkg FIGURE P442 Evacuated H2O Partition 443 Reconsider Prob 442 Using appropriate soft ware investigate the effect of the initial pressure of water on the final temperature in the tank Let the initial pressure vary from 100 to 600 kPa Plot the final temperature against the initial pressure and discuss the results 444 Two tanks Tank A and Tank B are separated by a par tition Initially Tank A contains 2 kg of steam at 1 MPa and 300C while Tank B contains 3 kg of saturated liquidvapor mixture at 150C with a vapor mass fraction of 50 percent The partition is now removed and the two sides are allowed to mix until mechanical and thermal equilibrium are established If the pressure at the final state is 300 kPa determine a the temperature and quality of the steam if mixture at the final state and b the amount of heat lost from the tanks FIGURE P444 Tank A 2 kg 1 MPa 300C Tank B 3 kg 150C x 05 Q Specific Heats Δu and Δh of Ideal Gases 445C Is the energy required to heat air from 295 to 305 K the same as the energy required to heat it from 345 to 355 K Assume the pressure remains constant in both cases 446C A fixed mass of an ideal gas is heated from 50 to 80C at a constant pressure of a 1 atm and b 3 atm For which case do you think the energy required will be greater Why 447C A fixed mass of an ideal gas is heated from 50 to 80C at a constant volume of a 1 m3 and b 3 m3 For which case do you think the energy required will be greater Why FIGURE P454 Nitrogen D Spring 455 Determine the internal energy change Δu of hydrogen in kJkg as it is heated from 200 to 800 K using a the empirical specific heat equation as a function of temperature Table A2c b the cv value at the average temperature Table A2b and c the cv value at room temperature Table A2a Final PDF to printer cen22672ch04161210indd 199 100217 0123 PM 199 CHAPTER 4 FIGURE P462 Oxygen 1 kg T1 20C T2 120C Q Oxygen 1 kg T1 20C T2 120C Q 456E Determine the enthalpy change Δh of oxygen in Btu lbm as it is heated from 800 to 1500 R using a the empirical specific heat equation as a function of temperature Table A2Ec b the cp value at the average temperature Table A2Eb and c the cp value at room temperature Table A2Ea Answers a 170 Btulbm b 169 Btulbm c 153 Btulbm ClosedSystem Energy Analysis Ideal Gases 457C Is it possible to compress an ideal gas isothermally in an adiabatic pistoncylinder device Explain 458 Nitrogen in a rigid vessel is cooled by rejecting 100 kJkg of heat Determine the internal energy change of the nitrogen in kJkg 459E Nitrogen at 100 psia and 300F in a rigid container is cooled until its pressure is 50 psia Determine the work done and the heat transferred during this process in Btulbm Answers 0 Btulbm 673 Btulbm 460E A pistoncylinder device containing carbondioxide gas undergoes an isobaric process from 15 psia and 80F to 200F Determine the work and the heat transfer associated with this process in Btulbm Answers 542 Btulbm 244 Btulbm 461 A 3m3 rigid tank contains hydrogen at 250 kPa and 550 K The gas is now cooled until its temperature drops to 350 K Determine a the final pressure in the tank and b the amount of heat transfer 462 1 kg of oxygen is heated from 20 to 120C Determine the amount of heat transfer required when this is done during a a constantvolume process and b isobaric process 465 An insulated rigid tank is divided into two equal parts by a partition Initially one part contains 4 kg of an ideal gas at 800 kPa and 50C and the other part is evacuated The parti tion is now removed and the gas expands into the entire tank Determine the final temperature and pressure in the tank FIGURE P465 Evacuated Ideal gas 800 kPa 50C 463E A 10ft3 tank contains oxygen initially at 147 psia and 80F A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia During the process 20 Btu of heat is lost to the surroundings Determine the paddlewheel work done Neglect the energy stored in the paddle wheel 464 A 4m 5m 7m room is heated by the radiator of a steamheating system The steam radiator transfers heat at a rate of 10000 kJh and a 100W fan is used to distribute the warm air in the room The rate of heat loss from the room is estimated to be about 5000 kJh If the initial temperature of the room air is 10C determine how long it will take for the air FIGURE P464 4 m 5 m 7 m 5000 kJh Room Steam Wpw 10000 kJh 466 An ideal gas contained in a pistoncylinder device undergoes an isothermal compression process which begins with an initial pressure and volume of 100 kPa and 06 m3 respectively During the process there is a heat transfer of 60 kJ from the ideal gas to the surroundings Determine the volume and pressure at the end of the process Answers 0221 m3 272 kPa 467 A 4m 5m 6m room is to be heated by a base board resistance heater It is desired that the resistance heater be able to raise the air temperature in the room from 5 to 25C within 17 min Assuming no heat losses from the room and an atmospheric pressure of 100 kPa determine the required power of the resistance heater Assume constant specific heats at room temperature Answer 212 kW 468 An insulated pistoncylinder device initially contains 03 m3 of carbon dioxide at 200 kPa and 27C An electric switch is turned on and a 110V source supplies current to a resistance heater inside the cylinder for a period of 10 min The pressure is held constant during the process while the volume is doubled Determine the current that passes through the resistance heater temperature to rise to 20C Assume constant specific heats at room temperature Final PDF to printer cen22672ch04161210indd 200 100217 0123 PM 200 ENERGY ANALYSIS OF CLOSED SYSTEMS 469 Argon is compressed in a polytropic process with n 12 from 120 kPa and 10C to 800 kPa in a pistoncylinder device Determine the work produced and heat transferred dur ing this compression process in kJkg 473E A 3ft3 adiabatic rigid container is divided into two equal volumes by a thin membrane as shown in Fig P473E Initially one of these chambers is filled with air at 100 psia and 100F while the other chamber is evacuated Determine the internal energy change of the air when the membrane is ruptured Also determine the final air pressure in the container FIGURE P472 FIGURE P469 Argon 120 kPa 10C Pvn constant Q 470 An insulated pistoncylinder device contains 100 L of air at 400 kPa and 25C A paddle wheel within the cylinder is rotated until 15 kJ of work is done on the air while the pressure is held constant Determine the final temperature of the air Neglect the energy stored in the paddle wheel 471 Air is contained in a variableload pistoncylinder device equipped with a paddle wheel Initially air is at 400 kPa and 17C The paddle wheel is now turned by an external elec tric motor until 75 kJkg of work has been transferred to air During this process heat is transferred to maintain a constant air temperature while allowing the gas volume to triple Calculate the required amount of heat transfer in kJkg Answer 164 kJkg FIGURE P471 472 A mass of 15 kg of air in a pistoncylinder device is heated from 25 to 95C by passing current through a resistance heater inside the cylinder The pressure inside the cylinder is held constant at 300 kPa during the process and a heat loss of 60 kJ occurs Determine the electric energy supplied in kWh Answer 0310 kWh FIGURE P473E Air 15 ft3 100 psia 100F Vacuum 15 ft3 474 A pistoncylinder device contains 22 kg of nitrogen initially at 100 kPa and 25C The nitrogen is now compressed slowly in a polytropic process during which PV13 constant until the volume is reduced by onehalf Determine the work done and the heat transfer for this process 475 Reconsider Prob 474 Using appropriate software plot the process described in the problem on a PV diagram and investigate the effect of the polytropic exponent n on the boundary work and heat transfer Let the polytropic expo nent vary from 10 to 14 Plot the boundary work and the heat transfer versus the polytropic exponent and discuss the results 476 A pistoncylinder device contains 4 kg of argon at 250 kPa and 35C During a quasiequilibrium isothermal expansion process 15 kJ of boundary work is done by the sys tem and 3 kJ of paddlewheel work is done on the system Determine the heat transfer for this process 477 A springloaded pistoncylinder device contains 5 kg of helium as the system as shown in Fig P477 This system is heated from 100 kPa and 20C to 800 kPa and 160C Determine the heat transferred to and the work produced by this system Helium Spring FIGURE P477 478 A pistoncylinder device whose piston is resting on top of a set of stops initially contains 05 kg of helium gas at 100 kPa and 25C The mass of the piston is such that 500 kPa of pressure Final PDF to printer cen22672ch04161210indd 201 100217 0123 PM 201 CHAPTER 4 is required to raise it How much heat must be transferred to the helium before the piston starts rising Answer 1857 kJ 479 A pistoncylinder device whose piston is resting on a set of stops initially contains 3 kg of air at 200 kPa and 27C The mass of the piston is such that a pressure of 400 kPa is required to move it Heat is now transferred to the air until its volume doubles Determine the work done by the air and the total heat transferred to the air during this process Also show the process on a Pv diagram Answers 516 kJ 2674 kJ ClosedSystem Energy Analysis Solids and Liquids 480 A 1kg block of iron is heated from 25 to 75C What is the change in the irons total internal energy and enthalpy 481E The state of liquid water is changed from 50 psia and 50F to 2000 psia and 100F Determine the change in the internal energy and enthalpy of water on the basis of the a compressed liquid tables b incompressible substance approximation and property tables and c specificheat model 482E During a picnic on a hot summer day all the cold drinks disappear quickly and the only available drinks are those at the ambient temperature of 85F In an effort to cool a 12fluidoz drink in a can a person grabs the can and starts shaking it in the iced water of the chest at 32F Using the properties of water for the drink determine the mass of ice that will melt by the time the canned drink cools to 37F 483 An ordinary egg can be approximated as a 55cmdiameter sphere The egg is initially at a uniform temper ature of 8C and is dropped into boiling water at 97C Taking the properties of the egg to be ρ 1020 kgm3 and cp 332 kJkgC determine how much heat is transferred to the egg by the time the average temperature of the egg rises to 80C 484 Consider a 1000W iron whose base plate is made of 05cmthick aluminum alloy 2024T6 ρ 2770 kgm3 and cp 875 JkgC The base plate has a surface area of 003 m2 Initially the iron is in thermal equilibrium with the ambient air at 22C Assuming 90 percent of the heat generated in the resistance wires is transferred to the plate determine the mini mum time needed for the plate temperature to reach 200C 485 Stainless steel ball bearings ρ 8085 kgm3 and cp 0480 kJkgC having a diameter of 12 cm are to be quenched in water at a rate of 800 per minute The balls leave the oven at a uniform temperature of 900C and are exposed to air at 25C for a while before they are dropped into the water If the temperature of the balls drops to 850C prior to quench ing determine the rate of heat transfer from the balls to the air 486E In a production facility 16inthick 2ft 2ft square brass plates ρ 5325 lbmft3 and cp 0091 BtulbmF that are initially at a uniform temperature of 75F are heated by passing them through an oven at 1500F at a rate of 300 per minute If the plates remain in the oven until their average tem perature rises to 900F determine the rate of heat transfer to the plates in the furnace FIGURE P484 FIGURE P486E 16 in Furnace 1500F Brass plate 75F 487 Long cylindrical steel rods ρ 7833 kgm3 and cp 0465 kJkgC of 8 cm diameter are heattreated by drawing them at a velocity of 2 mmin through an oven main tained at 900C If the rods enter the oven at 30C and leave at a mean temperature of 500C determine the rate of heat transfer to the rods in the oven 488 An electronic device dissipating 25 W has a mass of 20 g and a specific heat of 850 JkgC The device is lightly used and it is on for 5 min and then off for several hours dur ing which it cools to the ambient temperature of 25C Deter mine the highest possible temperature of the device at the end of the 5min operating period What would your answer be if the device were attached to a 05kg aluminum heat sink Assume the device and the heat sink to be nearly isothermal 489 Reconsider Prob 488 Using appropriate soft ware investigate the effect of the mass of the heat sink on the maximum device temperature Let the mass of the heat sink vary from 0 to 1 kg Plot the maximum temperature against the mass of the heat sink and discuss the results 490 If you ever slapped someone or got slapped yourself you probably remember the burning sensation Imagine you had the unfortunate occasion of being slapped by an angry per son which caused the temperature of the affected area of your face to rise by 24C ouch Assuming the slapping hand has a mass of 09 kg and about 0150 kg of the tissue on the face and the hand is affected by the incident estimate the velocity Final PDF to printer cen22672ch04161210indd 202 100217 0123 PM 202 ENERGY ANALYSIS OF CLOSED SYSTEMS of the hand just before impact Take the specific heat of the tissue to be 38 kJkgK Special Topic Biological Systems 491C For what is the energy released during metabolism in humans used 492C Is the metabolizable energy content of a food the same as the energy released when it is burned in a bomb calo rimeter If not how does it differ 493C Is the number of prospective occupants an important consideration in the design of heating and cooling systems of classrooms Explain 494C What do you think of a diet program that allows for generous amounts of bread and rice provided that no butter or margarine is added 495 Consider two identical rooms one with a 2kW electric resistance heater and the other with three couples fast dancing In which room will the air temperature rise more quickly 496 The average specific heat of the human body is 36 kJkgC If the body temperature of an 80kg man rises from 37C to 39C during strenuous exercise determine the increase in the thermal energy of the body as a result of this rise in body temperature 497 Consider two identical 80kg men who are eating identical meals and doing identical things except that one of them jogs for 30 min every day while the other watches TV Determine the weight difference between the two in a month Answer 104 kg 498 A 68kg woman is planning to bicycle for an hour If she is to meet her entire energy needs while bicycling by eating 30g chocolate candy bars determine how many candy bars she needs to take with her 499 A 90kg man gives in to temptation and eats an entire 1L box of ice cream How long does this man need to jog to burn off the calories he consumed from the ice cream Answer 154 h 4100 A 60kg man used to have an apple every day after dinner without losing or gaining any weight He now eats a 200ml serving of ice cream instead of an apple and walks 20 min every day On this new diet how much weight will he lose or gain per month Answer 0087kg gain 4101 Consider a man who has 20 kg of body fat when he goes on a hunger strike Determine how long he can survive on his body fat alone 4102 Consider two identical 50kg women Candy and Wendy who are doing identical things and eating identical food except that Candy eats her baked potato with four tea spoons of butter while Wendy eats hers plain every evening Determine the difference in the weights of Candy and Wendy after one year Answer 65 kg 4103E A 190pound man and a 130pound woman went to Burger King for lunch The man had a BK Big Fish sand wich 720 Cal medium french fries 400 Cal and a large Coke 225 Cal The woman had a basic hamburger 330 Cal medium french fries 400 Cal and a diet Coke 0 Cal After lunch they start shoveling snow and burn calories at a rate of 420 Calh for the woman and 610 Calh for the man Deter mine how long each one of them needs to shovel snow to burn off the lunch calories 4104 Consider two friends who go to Burger King every day for lunch One of them orders a Double Whopper sand wich large fries and a large Coke total Calories 1600 while the other orders a Whopper Junior small fries and a small Coke total Calories 800 every day If these two friends are very much alike otherwise and they have the same metabolic rate determine the weight difference between these two friends in a year 4105 A person eats a McDonalds Big Mac sandwich 530 Cal a second person eats a Burger King Whopper sand wich 640 Cal and a third person eats 50 olives with regular french fries 350 Cal for lunch Determine who consumes the most calories An olive contains about 5 Calories 4106 A 5oz serving of a Bloody Mary contains 14 g of alcohol and 5 g of carbohydrates and thus 116 Calories A 25oz serving of a martini contains 22 g of alcohol and a negligible amount of carbohydrates and thus 156 Calories An average person burns 600 Calories per hour while exer cising on a crosscountry ski machine Determine how long it will take to burn the calories from one serving of a a Bloody Mary and b a martini on this crosscountry ski machine 4107E The range of healthy weight for adults is usually expressed in terms of the body mass index BMI defined in SI units as BMI W kg H 2 m 2 where W is the weight actually the mass of the person in kg and H is the height in m and the range of healthy weight is 19 BMI 25 Convert the previous formula to English units such that the weight is in pounds and the height in inches Also calculate your own BMI and if it is not in the healthy range determine how many pounds or kg you need to gain or lose to be fit 4108 The body mass index BMI of a 16mtall woman who normally has 3 large slices of cheese pizza and a 400ml Coke for lunch is 30 She now decides to change her lunch to 2 slices of pizza and a 200ml Coke Assuming that the deficit in the calorie intake is made up by burning body fat determine how long it will take for the BMI of this person to drop to 20 Use the data in the text for calories and take the metabolizable energy content of 1 kg of body fat to be 33100 kJ Answer 463 days Final PDF to printer cen22672ch04161210indd 203 100217 0123 PM 203 CHAPTER 4 Review Problems 4109 Which of two gasesneon or airrequires less work when compressed in a closed system from P1 to P2 using a polytropic process with n 15 4110 Which of two gasesneon or airproduces more work when expanded from P1 to P2 in a closedsystem poly tropic process with n 12 4111 Consider a classroom that is losing heat to the out doors at a rate of 12000 kJh If there are 40 students in class each dissipating sensible heat at a rate of 84 W determine if it is necessary to turn the heater in the classroom on to prevent the room temperature from dropping 4112 The temperature of air changes from 0 to 10C while its velocity changes from zero to a final velocity and its eleva tion changes from zero to a final elevation At which values of final air velocity and final elevation will the internal kinetic and potential energy changes be equal Answers 120 ms 732 m 4113 A rigid tank contains a gas mixture with a specific heat of cv 0748 kJkgK The mixture is cooled from 200 kPa and 200C until its pressure is 100 kPa Determine the heat transfer during this process in kJkg 4114 Consider a pistoncylinder device that contains 05 kg air Now heat is transferred to the air at constant pressure and the air temperature increases by 5C Determine the expansion work done during this process 4115 A mass of 02 kg of saturated refrigerant134a is contained in a pistoncylinder device at 200 kPa Initially 75 percent of the mass is in the liquid phase Now heat is trans ferred to the refrigerant at constant pressure until the cylinder contains vapor only Show the process on a Pv diagram with respect to saturation lines Determine a the volume occupied by the refrigerant initially b the work done and c the total heat transfer 4116E Air in the amount of 2 lbm is contained in a well insulated rigid vessel equipped with a stirring paddle wheel The initial state of this air is 30 psia and 60F How much work in Btu must be transferred to the air with the paddle wheel to raise the air pressure to 40 psia Also what is the final temperature of the air 4117 Air is expanded in a polytropic process with n 12 from 1 MPa and 400C to 110 kPa in a pistoncylinder device Determine the final temperature of the air 4118 Nitrogen at 100 kPa and 25C in a rigid vessel is heated until its pressure is 300 kPa Calculate the work done and the heat transferred during this process in kJkg 4119 A wellinsulated rigid vessel contains 3 kg of satu rated liquid water at 40C The vessel also contains an electri cal resistor that draws 10 A when 50 V are applied Determine the final temperature in the vessel after the resistor has been operating for 30 min 4120 A mass of 3 kg of saturated liquidvapor mixture of water is contained in a pistoncylinder device at 160 kPa Initially 1 kg of the water is in the liquid phase and the rest is in the vapor phase Heat is now transferred to the water and the piston which is resting on a set of stops starts moving when the pressure inside reaches 500 kPa Heat transfer continues until the total volume increases by 20 percent Determine a the initial and final temperatures b the mass of liquid water when the piston first starts moving and c the work done during this process Also show the process on a Pv diagram FIGURE P4116E Wsh Air 2 lbm 30 psia 60F FIGURE P4120 4121 A mass of 12 kg of saturated refrigerant134a vapor is contained in a pistoncylinder device at 240 kPa Now 300 kJ of heat is transferred to the refrigerant at constant pressure while a 110V source supplies current to a resistor within the cylinder for 6 min Determine the current supplied if the final temperature is 70C Also show the process on a Tv diagram with respect to the saturation lines Answer 128 A FIGURE P4121 Final PDF to printer cen22672ch04161210indd 204 100217 0123 PM 204 ENERGY ANALYSIS OF CLOSED SYSTEMS 4122 Saturated water vapor at 200C is condensed to a satu rated liquid at 50C in a springloaded pistoncylinder device Determine the heat transfer for this process in kJkg 4123 A pistoncylinder device contains helium gas initially at 100 kPa 10C and 02 m3 The helium is now compressed in a polytropic process PVn constant to 700 kPa and 290C Determine the heat loss or gain during this process Answer 651 kJ loss If the cylinder contains saturated liquid at 120C when thermal equilibrium is established determine the amount of ice added The melting temperature and the heat of fusion of ice at atmo spheric pressure are 0C and 3337 kJkg respectively 4127 A passive solar house that is losing heat to the out doors at an average rate of 50000 kJh is maintained at 22C at all times during a winter night for 10 h The house is to be heated by 50 glass containers each containing 20 L of water that is heated to 80C during the day by absorbing solar energy A thermostatcontrolled 15kW backup electric resis tance heater turns on whenever necessary to keep the house at 22C a How long did the electric heating system run that night b How long would the electric heater run that night if the house incorporated no solar heating Answers a 477 h b 926 h FIGURE P4127 Pump 22C Water 80C FIGURE P4124 Air P1 100 kPa V1 015 m3 FIGURE P4123 He PVn constant 4124 A frictionless pistoncylinder device initially con tains air at 100 kPa and 015 m3 At this state a linear spring F x is touching the piston but exerts no force on it The air is now heated to a final state of 045 m3 and 800 kPa Deter mine a the total work done by the air and b the work done against the spring Also show the process on a PV diagram Answers a 135 kJ b 105 kJ 4125 A frictionless pistoncylinder device and a rigid tank initially contain 12 kg of an ideal gas each at the same tem perature pressure and volume We wish to raise the tempera tures of both systems by 15C Determine the amount of extra heat that must be supplied to the gas in the cylinder which is maintained at constant pressure to achieve this result Assume the molar mass of the gas is 25 4126 An insulated pistoncylinder device initially contains 001 m3 of saturated liquidvapor mixture with a quality of 02 at 120C Now some ice at 0C is added to the cylinder 4128 Water is boiled at sea level in a coffeemaker equipped with an immersiontype electric heating element The coffee maker contains 1 L of water when full Once boiling starts it is observed that half of the water in the coffeemaker evaporates in 13 min Determine the power rating of the electric heating element immersed in water Also determine how long it will take for this heater to raise the temperature of 1 L of cold water from 18C to the boiling temperature FIGURE P4128 Coffee maker 1 atm 1 L Final PDF to printer cen22672ch04161210indd 205 100217 0123 PM 205 CHAPTER 4 4129 The energy content of a certain food is to be deter mined in a bomb calorimeter that contains 3 kg of water by burning a 2g sample of it in the presence of 100 g of air in the reaction chamber If the water temperature rises by 32C when equilibrium is established determine the energy content of the food in kJkg by neglecting the thermal energy stored in the reaction chamber and the energy supplied by the mixer What is a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber Answer 20060 kJkg 4133 In order to cool 1 ton of water at 20C in an insulated tank a person pours 130 kg of ice at 5C into the water Deter mine the final equilibrium temperature in the tank The melt ing temperature and the heat of fusion of ice at atmospheric pressure are 0C and 3337 kJkg respectively Answer 82C 4134 A 03L glass of water at 20C is to be cooled with ice to 5C Determine how much ice needs to be added to the water in grams if the ice is at a 0C and b 20C Also determine how much water would be needed if the cooling is to be done with cold water at 0C The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 3337 kJkg respectively and the density of water is 1 kgL 4135 Reconsider Prob 4134 Using appropriate software investigate the effect of the initial tem perature of the ice on the final mass required Let the ice tem perature vary from 26 to 0C Plot the mass of ice against the initial temperature of ice and discuss the results 4136 A wellinsulated 3m 4m 6m room initially at 7C is heated by the radiator of a steam heating system The radiator has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200C At this moment both the inlet and the exit valves to the radiator are closed A 120W fan is used to distribute the air in the room The pressure of the steam is observed to drop to 100 kPa after 45 min as a result of heat transfer to the room Assuming constant specific heats for air at room temperature determine the average temperature of air in 45 min Assume the air pressure in the room remains constant at 100 kPa 4130 A 68kg man whose average body temperature is 39C drinks 1 L of cold water at 3C in an effort to cool down Taking the average specific heat of the human body to be 36 kJkgC determine the drop in the average body tempera ture of this person under the influence of this cold water 4131 An insulated pistoncylinder device initially contains 18 kg of saturated liquid water at 120C Now an electric resistor placed in the cylinder is turned on for 10 min until the volume quadruples Determine a the volume of the cylinder b the final temperature and c the electrical power rating of the resistor Answers a 000763 m3 b 120C c 00236 kW 4132 An insulated rigid tank initially contains 14 kg of saturated liquid water at 200C and air At this state 25 percent of the volume is occupied by liquid water and the rest by air Now an electric resistor placed in the tank is turned on and the tank is observed to contain saturated water vapor after 20 min Determine a the volume of the tank b the final temperature FIGURE P4132 Water 14 kg 200C Air We FIGURE P4131 Water 18 kg 120C sat liquid We FIGURE P4129 Reaction chamber Food ΔT 32C FIGURE P4136 7C 3 m 4 m 6 m Fan Steam radiator and c the electric power rating of the resistor Neglect energy added to the air Answers a 000648 m3 b 371C c 158 kW Final PDF to printer cen22672ch04161210indd 206 100217 0123 PM 206 ENERGY ANALYSIS OF CLOSED SYSTEMS 4137 Two adiabatic chambers 2 m3 each are intercon nected by a valve as shown in Fig P4137 with one chamber containing oxygen at 1000 kPa and 127C and the other cham ber evacuated The valve is now opened until the oxygen fills both chambers and both tanks have the same pressure Deter mine the total internal energy change and the final pressure in the tanks 4141 A pistoncylinder device initially contains 035 kg of steam at 35 MPa superheated by 74C Now the steam loses heat to the surroundings and the piston moves down hitting a set of stops at which point the cylinder contains saturated liquid water The cooling continues until the cylinder contains water at 200C Determine a the final pressure and the qual ity if mixture b the boundary work c the amount of heat transfer when the piston first hits the stops and d the total heat transfer FIGURE P4137 Valve Chamb re B hamb re A C FIGURE P4138 200 kPa H2O B 400 kPa H2O A Q 4138 Two rigid tanks are connected by a valve Tank A contains 02 m3 of water at 400 kPa and 80 percent quality Tank B contains 05 m3 of water at 200 kPa and 250C The valve is now opened and the two tanks eventually come to the same state Determine the pressure and the amount of heat transfer when the system reaches thermal equilibrium with the surroundings at 25C Answers 317 kPa 2170 kJ 4139 Reconsider Prob 4138 Using appropriate software investigate the effect of the environ ment temperature on the final pressure and the heat transfer Let the environment temperature vary from 0 to 50C Plot the final results against the environment temperature and discuss the results 4140 A vertical 10cmdiameter pistoncylinder device contains an ideal gas at the ambient conditions of 1 bar and 24C Initially the inner face of the piston is 20 cm from the base of the cylinder Now an external shaft connected to the piston exerts a force corresponding to a boundary work input of 01 kJ The temperature of the gas remains constant during the process Determine a the amount of heat transfer b the final pressure in the cylinder and c the distance that the pis ton is displaced FIGURE P4141 FIGURE P4142E Water 10 ft3 450 psia x 010 Water 10 ft3 15 psia x 075 4142E Two 10ft3 adiabatic tanks are connected by a valve Initially one tank contains water at 450 psia with 10 percent quality while the second contains water at 15 psia with 75 percent quality The valve is now opened allowing the water vapor from the highpressure tank to move to the low pressure tank until the pressure in the two becomes equal Determine the final pressure and the final mass in each tank Answers 313 psia 416 Ibm 4143 An insulated rigid tank is divided into two compart ments of different volumes Initially each compartment con tains the same ideal gas at identical pressure but at different temperatures and masses The wall separating the two com partments is removed and the two gases are allowed to mix Assuming constant specific heats find the simplest expression for the mixture temperature written in the form T 3 f m 1 m 2 m 2 m 3 T 1 T 2 Final PDF to printer cen22672ch04161210indd 207 100217 0123 PM 207 CHAPTER 4 4144 In solarheated buildings energy is often stored as sensible heat in rocks concrete or water during the day for use at night To minimize the storage space it is desirable to use a material that can store a large amount of heat while experiencing a small temperature change A large amount of heat can be stored essentially at constant temperature during a phasechange process and thus materials that change phase at about room temperature such as glaubers salt sodium sulfate decahydrate which has a melting point of 32C and a heat of fusion of 329 kJL are very suitable for this purpose Deter mine how much heat can be stored in a 5m3 storage space using a glaubers salt undergoing a phase change b granite rocks with a heat capacity of 232 kJkgC and a tempera ture change of 20C and c water with a heat capacity of 400 kJkgC and a temperature change of 20C Fundamentals of Engineering FE Exam Problems 4145 A 3m3 rigid tank contains nitrogen gas at 500 kPa and 300 K Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa The work done during this process is a 500 kJ b 1500 kJ c 0 kJ d 900 kJ e 2400 kJ 4146 A 05m3 rigid tank contains nitrogen gas at 600 kPa and 300 K Now the gas is compressed isothermally to a volume of 02 m3 The work done on the gas during this compression process is a 82 kJ b 180 kJ c 240 kJ d 275 kJ e 315 kJ 4147 A wellsealed room contains 60 kg of air at 200 kPa and 25C Now solar energy enters the room at an average rate of 08 kJs while a 120W fan is turned on to circulate the air in the room If heat transfer through the walls is negligible the air temperature in the room in 30 min will be a 256C b 498C c 534C d 525C e 634C 4148 A room contains 75 kg of air at 100 kPa and 15C The room has a 250W refrigerator the refrigerator consumes 250 W of electricity when running a 120W TV a 18kW electric resistance heater and a 50W fan During a cold winter day it is observed that the refrigerator the TV the fan and the electric resistance heater are running continuously but the air temperature in the room remains constant The rate of heat loss from the room that day is a 5832 kJh b 6192 kJh c 7560 kJh d 7632 kJh e 7992 kJh 4149 A frictionless pistoncylinder device and a rigid tank contain 3 kmol of an ideal gas at the same temperature pres sure and volume Now heat is transferred and the temperature of both systems is raised by 10C The amount of extra heat that must be supplied to the gas in the cylinder that is main tained at constant pressure is a 0 kJ b 27 kJ c 83 kJ d 249 kJ e 300 kJ 4150 A pistoncylinder device contains 5 kg of air at 400 kPa and 30C During a quasiequilibium isothermal expansion process 15 kJ of boundary work is done by the sys tem and 3 kJ of paddlewheel work is done on the system The heat transfer during this process is a 12 kJ b 18 kJ c 24 kJ d 35 kJ e 60 kJ 4151 A glass of water with a mass of 032 kg at 20C is to be cooled to 0C by dropping ice cubes at 0C into it The latent heat of fusion of ice is 334 kJkg and the specific heat of water is 418 kJkgC The amount of ice that needs to be added is a 32 g b 40 g c 80 g d 93 g e 110 g 4152 A 2kW electric resistance heater submerged in 5 kg of water is turned on and kept on for 10 min During the pro cess 300 kJ of heat is lost from the water The temperature rise of the water is a 04C b 431C c 574C d 718C e 180C 4153 A 2kW baseboard electric resistance heater in a vacant room is turned on and kept on for 15 min The mass of the air in the room is 75 kg and the room is tightly sealed so that no air can leak in or out The temperature rise of air at the end of 15 min is a 85C b 124C c 240C d 334C e 548C 4154 15 kg of liquid water initially at 12C is to be heated to 95C in a teapot equipped with an 800W electric heating element inside The specific heat of water can be taken to be 418 kJkgC and the heat loss from the water during heating can be neglected The time it takes to heat water to the desired temperature is a 59 min b 73 min c 108 min d 140 min e 170 min 4155 A container equipped with a resistance heater and a mixer is initially filled with 36 kg of saturated water vapor at 120C Now the heater and the mixer are turned on the steam is compressed and there is heat loss to the FIGURE P4143 Side 1 Mass m1 Temperature T1 Side 2 Mass m2 Temperature T2 where m3 and T3 are the mass and temperature of the final mix ture respectively Final PDF to printer cen22672ch04161210indd 208 100217 0123 PM 208 ENERGY ANALYSIS OF CLOSED SYSTEMS surrounding air At the end of the process the temperature and pressure of steam in the container are measured to be 300C and 05 MPa The net energy transfer to the steam during this process is a 274 kJ b 914 kJ c 1213 kJ d 988 kJ e 1291 kJ 4156 An ordinary egg with a mass of 01 kg and a specific heat of 332 kJkgC is dropped into boiling water at 95C If the initial temperature of the egg is 5C the maximum amount of heat transfer to the egg is a 12 kJ b 30 kJ c 24 kJ d 18 kJ e infinity 4157 An apple with an average mass of 018 kg and average specific heat of 365 kJkgC is cooled from 17C to 5C The amount of heat transferred from the apple is a 79 kJ b 112 kJ c 145 kJ d 176 kJ e 191 kJ 4158 A 6pack of canned drinks is to be cooled from 18C to 3C The mass of each canned drink is 0355 kg The drinks can be treated as water and the energy stored in the aluminum can itself is negligible The amount of heat transfer from the six canned drinks is a 22 kJ b 32 kJ c 134 kJ d 187 kJ e 223 kJ 4159 A room is filled with saturated steam at 100C Now a 5kg bowling ball at 25C is brought to the room Heat is transferred to the ball from the steam and the temperature of the ball rises to 100C while some steam condenses on the ball as it loses heat but it still remains at 100C The specific heat of the ball can be taken to be 18 kJkgC The mass of steam that condensed during this process is a 80 g b 128 g c 299 g d 351 g e 405 g 4160 An ideal gas has a gas constant R 03 kJkgK and a constantvolume specific heat cv 07 kJkgK If the gas has a temperature change of 100C choose the correct answer for each of the following 1 The change in enthalpy is in kJkg a 30 b 70 c 100 d insufficient information to determine 2 The change in internal energy is in kJkg a 30 b 70 c 100 d insufficient information to determine 3 The work done is in kJkg a 30 b 70 c 100 d insufficient information to determine 4 The heat transfer is in kJkg a 30 b 70 c 100 d insufficient information to determine 5 The change in the pressurevolume product is in kJkg a 30 b 70 c 100 d insufficient information to determine 4161 Saturated steam vapor is contained in a piston cylinder device While heat is added to the steam the piston is held stationary and the pressure and temperature become 12 MPa and 700C respectively Additional heat is added to the steam until the temperature rises to 1200C and the piston moves to maintain a constant pressure 1 The initial pressure of the steam is most nearly a 250 kPa b 500 kPa c 750 kPa d 1000 kPa e 1250 kPa 2 The work done by the steam on the piston is most nearly a 230 kJkg b 1100 kJkg c 2140 kJkg d 2340 kJkg e 840 kJkg 3 The total heat transferred to the steam is most nearly a 230 kJkg b 1100 kJkg c 2140 kJkg d 2340 kJkg e 840 kJkg 4162 A pistoncylinder device contains an ideal gas The gas undergoes two successive cooling processes by rejecting heat to the surroundings First the gas is cooled at constant pressure until T2 3 4 T1 Then the piston is held stationary while the gas is further cooled to T3 1 2 T1 where all tempera tures are in K 1 The ratio of the final volume to the initial volume of the gas is a 025 b 050 c 067 d 075 e 10 2 The work done on the gas by the piston is a RT14 b cvT12 c cpT12 d cv cpT14 e cvT1 T22 3 The total heat transferred from the gas is a RT14 b cvT12 c cpT12 d cv cpT14 e cvT1 T32 Design and Essay Problems 4163 Find out how the specific heats of gases liquids and solids are determined in national laboratories Describe the experimental apparatus and the procedures used 4164 You are asked to design a heating system for a swim ming pool that is 2 m deep 25 m long and 25 m wide Your client wants the heating system be large enough to raise the water temperature from 20 to 30C in 3 h The rate of heat loss from the water to the air at the outdoor design conditions is determined to be 960 Wm2 and the heater must also be able to maintain the pool at 30C at those conditions Heat losses to the ground are expected to be small and can be disre garded The heater considered is a natural gas furnace whose efficiency is 80 percent What heater size in kW input would you recommend to your client Final PDF to printer cen22672ch04161210indd 209 100217 0123 PM 209 CHAPTER 4 4165 Using a thermometer measure the boiling tempera ture of water and calculate the corresponding saturation pres sure From this information estimate the altitude of your town and compare it with the actual altitude value 4166 Compressed gases and phasechanging liquids are used to store energy in rigid containers What are the advantages and disadvantages of each substance as a means of storing energy 4167 A 1982 US Department of Energy article FS204 states that a leak of one drip of hot water per second can cost 100 per month Making reasonable assumptions about the drop size and the unit cost of energy determine if this claim is reasonable 4168 Design an experiment complete with instrumentation to determine the specific heats of a liquid using a resistance heater Discuss how the experiment will be conducted what measurements need to be taken and how the specific heats will be determined What are the sources of error in your system How can you minimize the experimental error How would you modify this system to determine the specific heat of a solid Final PDF to printer cen22672ch04161210indd 210 100217 0123 PM Final PDF to printer cen22672ch05211270indd 211 110917 1149 AM 211 OBJECTIVES The objectives of Chapter 5 are to Develop the conservation of mass principle Apply the conservation of mass principle to various systems including steady and unsteady flow control volumes Apply the first law of thermo dynamics as the statement of the conservation of energy principle to control volumes Identify the energy carried by a fluid stream crossing a control surface as the sum of internal energy flow work kinetic energy and potential energy of the fluid and to relate the combination of the internal energy and the flow work to the property enthalpy Solve energy balance problems for common steadyflow devices such as nozzles compressors turbines throttling valves mixing chambers and heat exchangers Apply the energy balance to general unsteadyflow processes with particular emphasis on the uniform flow process as the model for commonly encountered charging and discharging processes M AS S AN D E N E RGY AN ALYS I S O F C O N TRO L VOLUM E S I n Chap 4 we applied the general energy balance relation expressed as Ein Eout ΔEsystem to closed systems In this chapter we extend the energy analysis to systems that involve mass flow across their boundaries that is control volumes with particular emphasis on steadyflow systems We start this chapter with the development of the general conservation of mass relation for control volumes and we continue with a discussion of flow work and the energy of fluid streams We then apply the energy bal ance to systems that involve steadyflow processes and analyze the common steadyflow devices such as nozzles diffusers compressors turbines throt tling valves mixing chambers and heat exchangers Finally we apply the energy balance to general unsteadyflow processes such as the charging and discharging of vessels 5 CHAPTER Final PDF to printer 212 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 212 110917 1149 AM 51 CONSERVATION OF MASS The conservation of mass principle is one of the most fundamental prin ciples in nature We are all familiar with this principle and it is not diffi cult to understand A person does not have to be a rocket scientist to figure out how much vinegarandoil dressing will be obtained by mixing 100 g of oil with 25 g of vinegar Even chemical equations are balanced on the basis of the conservation of mass principle When 16 kg of oxygen reacts with 2 kg of hydrogen 18 kg of water is formed Fig 51 In an elec trolysis process the water separates back to 2 kg of hydrogen and 16 kg of oxygen Technically mass is not exactly conserved It turns out that mass m and energy E can be converted to each other according to the wellknown formula proposed by Albert Einstein 18791955 E m c 2 51 where c is the speed of light in a vacuum which is c 29979 108 ms This equation suggests that there is equivalence between mass and energy All physical and chemical systems exhibit energy interactions with their sur roundings but the amount of energy involved is equivalent to an extremely small mass compared to the systems total mass For example when 1 kg of liquid water is formed from oxygen and hydrogen at normal atmospheric conditions the amount of energy released is 158 MJ which corresponds to a mass of only 176 1010 kg However even in nuclear reactions the mass equivalence of the amount of energy interacted is a very small fraction of the total mass involved Therefore in most engineering analyses we consider both mass and energy as conserved quantities For closed systems the conservation of mass principle is implicitly used by requiring that the mass of the system remain constant during a process For control volumes however mass can cross the boundaries and so we must keep track of the amount of mass entering and leaving the control volume Mass and Volume Flow Rates The amount of mass flowing through a cross section per unit time is called the mass flow rate and is denoted by m The dot over a symbol is used to indicate time rate of change A fluid flows into or out of a control volume usually through pipes or ducts The differential mass flow rate of fluid flowing across a small area element dAc in a cross section of a pipe is proportional to dAc itself the fluid density ρ and the component of the flow velocity normal to dAc which we denote as Vn and is expressed as Fig 52 δ m ρ V n d A c 52 Note that both δ and d are used to indicate differential quantities but δ is typically used for quantities such as heat work and mass transfer that are path functions and have inexact differentials while d is used for quantities such as properties that are point functions and have exact differentials For flow through an annulus of inner radius r1 and outer radius r2 for example FIGURE 51 Mass is conserved even during chemical reactions 2 kg H2 16 kg O2 18 kg H2O FIGURE 52 The normal velocity Vn for a surface is the component of velocity perpendicular to the surface dAc Vn V n Control surface Final PDF to printer 213 CHAPTER 5 cen22672ch05211270indd 213 110917 1149 AM 1 2 d A c A c2 A c1 π r 2 2 r 1 2 but 1 2 δ m m total total mass flow rate through the annulus not m 2 m 1 For specified values of r1 and r2 the value of the integral of dAc is fixed thus the names point function and exact differential but this is not the case for the integral of δ m thus the names path function and inexact differential The mass flow rate through the entire crosssectional area of a pipe or duct is obtained by integration m A c δ m A c ρ V n d A c kg s 53 While Eq 53 is always valid in fact it is exact it is not always practical for engineering analyses because of the integral We would like instead to express mass flow rate in terms of average values over a cross section of the pipe In a general compressible flow both ρ and Vn vary across the pipe In many practical applications however the density is essentially uniform over the pipe cross section and we can take ρ outside the integral of Eq 53 Velocity however is never uniform over a cross section of a pipe because of the no slip condition at the walls Rather the velocity varies from zero at the walls to some maximum value at or near the centerline of the pipe We define the average velocity Vavg as the average value of Vn across the entire cross section of the pipe Fig 53 Average velocity V avg 1 A c A c V n d A c 54 where Ac is the area of the cross section normal to the flow direction Note that if the speed were Vavg all through the cross section the mass flow rate would be identical to that obtained by integrating the actual velocity pro file Thus for incompressible flow or even for compressible flow where ρ is approximated as uniform across Ac Eq 53 becomes m ρ V avg A c kg s 55 For compressible flow we can think of ρ as the bulk average density over the cross section and then Eq 55 can be used as a reasonable approximation For simplicity we drop the subscript on the average velocity Unless other wise stated V denotes the average velocity in the flow direction Also Ac denotes the crosssectional area normal to the flow direction The volume of the fluid flowing through a cross section per unit time is called the volume flow rate V Fig 54 and is given by V A c V n d A c V avg A c V A c m 3 s 56 An early form of Eq 56 was published in 1628 by the Italian monk Bene detto Castelli circa 15771644 Note that many fluid mechanics textbooks use Q instead of V for volume flow rate We use V to avoid confusion with heat transfer FIGURE 53 The average velocity Vavg is defined as the average speed through a cross section Vavg FIGURE 54 The volume flow rate is the volume of fluid flowing through a cross section per unit time Vavg Cross section Ac V VavgAc Final PDF to printer 214 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 214 110917 1149 AM The mass and volume flow rates are related by m ρ V V v 57 where v is the specific volume This relation is analogous to m ρV Vv which is the relation between the mass and the volume of a fluid in a container Conservation of Mass Principle The conservation of mass principle for a control volume can be expressed as The net mass transfer to or from a control volume during a time interval Δt is equal to the net change increase or decrease of the total mass within the control volume during Δt That is Total mass entering the CV during Δt Total mass leaving the CV during Δt Net change of mass within the CV during Δt or m in m out Δ m CV kg 58 where ΔmCV mfinal minitial is the change in the mass of the control volume during the process Fig 55 It can also be expressed in rate form as m in m out d m CV dt kg s 59 where m in and m out are the total rates of mass flow into and out of the control volume and dmCVdt is the rate of change of mass within the control vol ume boundaries Equations 58 and 59 are often referred to as the mass balance and are applicable to any control volume undergoing any kind of process Consider a control volume of arbitrary shape as shown in Fig 56 The mass of a differential volume dV within the control volume is dm ρ dV The total mass within the control volume at any instant in time t is determined by integration to be Total mass within the CV m cv CV ρ dV 510 Then the time rate of change of the amount of mass within the control volume is expressed as Rate of change of mass within the CV d m CV dt d dt CV ρ dV 511 For the special case of no mass crossing the control surface ie the control volume is a closed system the conservation of mass principle reduces to dmCVdt 0 This relation is valid whether the control volume is fixed mov ing or deforming Now consider mass flow into or out of the control volume through a dif ferential area dA on the control surface of a fixed control volume Let n be the outward unit vector of dA normal to dA and V be the flow velocity at dA relative to a fixed coordinate system as shown in Fig 56 In general FIGURE 55 Conservation of mass principle for an ordinary bathtub Water mbathtub minmout 20 kg mout 30 kg min 50 kg FIGURE 56 The differential control volume dV and the differential control surface dA used in the derivation of the conservation of mass relation Control volume CV Control surface CS dV dm dA n V θ Final PDF to printer 215 CHAPTER 5 cen22672ch05211270indd 215 110917 1149 AM the velocity may cross dA at an angle θ off the normal of dA and the mass flow rate is proportional to the normal component of velocity V n V cos θ ranging from a maximum outflow of V for θ 0 flow is normal to dA to a minimum of zero for θ 90 flow is tangent to dA to a maximum inflow of V for θ 180 flow is normal to dA but in the opposite direction Making use of the concept of dot product of two vectors the magnitude of the normal component of velocity is Normal component of velocity V n V cos θ V n 512 The mass flow rate through dA is proportional to the fluid density ρ normal velocity Vn and the flow area dA and is expressed as Differential mass flow rate δ m ρ V n dA ρ V cos θ dA ρ V n dA 513 The net flow rate into or out of the control volume through the entire control surface is obtained by integrating δ m over the entire control surface Net mass flow rate m net CS δ m CS ρ V n dA CS ρ V n dA 514 Note that V n V n V cos θ is positive for θ 90 outflow and negative for θ 90 inflow Therefore the direction of flow is automatically accounted for and the surface integral in Eq 514 directly gives the net mass flow rate A positive value for m net indicates a net outflow of mass and a negative value indicates a net inflow of mass Rearranging Eq 59 as dmCVdt m out m in 0 the conservation of mass relation for a fixed control volume is then expressed as General conservation of mass d dt CV ρ dV CS ρ V n dA 0 515 It states that the time rate of change of mass within the control volume plus the net mass flow rate through the control surface is equal to zero Splitting the surface integral in Eq 515 into two partsone for the out going flow streams positive and one for the incoming flow streams negative the general conservation of mass relation can also be expressed as d dt CV ρ dV out ρ V n A in ρ V n A 0 516 where A represents the area for an inlet or outlet and the summation signs are used to emphasize that all the inlets and outlets are to be considered Using the definition of mass flow rate Eq 516 can also be expressed as d dt CV ρ dV in m out m or d m CV dt in m out m 517 There is considerable flexibility in the selection of a control volume when solving a problem Many control volume choices are available but some are Final PDF to printer 216 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 216 110917 1149 AM more convenient to work with A control volume should not introduce any unnecessary complications A wise choice of a control volume can make the solution of a seemingly complicated problem rather easy A simple rule in selecting a control volume is to make the control surface normal to the flow at all locations where it crosses the fluid flow whenever possible This way the dot product V n simply becomes the magnitude of the velocity and the integral A ρ V n dA becomes simply ρVA Fig 57 Equations 515 and 516 are also valid for moving or deforming control volumes provided that the absolute velocity V is replaced by the relative veloc ity V r which is the fluid velocity relative to the control surface Mass Balance for SteadyFlow Processes During a steadyflow process the total amount of mass contained within a control volume does not change with time mCV constant Then the con servation of mass principle requires that the total amount of mass entering a control volume equal the total amount of mass leaving it For a garden hose nozzle in steady operation for example the amount of water entering the nozzle per unit time is equal to the amount of water leaving it per unit time When dealing with steadyflow processes we are not interested in the amount of mass that flows in or out of a device over time instead we are interested in the amount of mass flowing per unit time that is the mass flow rate m The conservation of mass principle for a general steadyflow system with multiple inlets and outlets is expressed in rate form as Fig 58 Steady flow in m out m kg s 518 It states that the total rate of mass entering a control volume is equal to the total rate of mass leaving it Many engineering devices such as nozzles diffusers turbines compres sors and pumps involve a single stream only one inlet and one outlet For these cases we typically denote the inlet state by the subscript 1 and the out let state by the subscript 2 and drop the summation signs Then Eq 518 reduces for singlestream steadyflow systems to Steady flow single stream m 1 m 2 ρ 1 V 1 A 1 ρ 2 V 2 A 2 519 Special Case Incompressible Flow The conservation of mass relations can be simplified even further when the fluid is incompressible which is usually the case for liquids Canceling the density from both sides of the general steadyflow relation gives Steady incompressible flow in V out V m 3 s 520 For singlestream steadyflow systems Eq 520 becomes Steady incompressible flow single stream V 1 V 2 V 1 A 1 V 2 A 2 521 FIGURE 58 Conservation of mass principle for a twoinletoneoutlet steadyflow system m CV 1 2 kgs m 2 3 kgs m3 m1 m2 5 kgs a Control surface at an angle to the flow V θ n Vn V cos θ Acos θ A m ρV cos θAcos θ ρVA b Control surface normal to the flow FIGURE 57 A control surface should always be selected normal to the flow at all locations where it crosses the fluid flow to avoid complications even though the result is the same n A m ρVA V Final PDF to printer 217 CHAPTER 5 cen22672ch05211270indd 217 110917 1149 AM It should always be kept in mind that there is no such thing as a conserva tion of volume principle Therefore the volume flow rates into and out of a steadyflow device may be different The volume flow rate at the outlet of an air compressor is much less than that at the inlet even though the mass flow rate of air through the compressor is constant Fig 59 This is due to the higher density of air at the compressor exit For steady flow of liquids how ever the volume flow rates remain nearly constant since liquids are essen tially incompressible constantdensity substances Water flow through the nozzle of a garden hose is an example of the latter case The conservation of mass principle requires every bit of mass to be accounted for during a process If you can balance your checkbook by keep ing track of deposits and withdrawals or by simply observing the conser vation of money principle you should have no difficulty applying the conservation of mass principle to engineering systems FIGURE 59 During a steadyflow process volume flow rates are not necessarily conserved although mass flow rates are Air compressor m1 2 kgs V1 14 m3s m2 2 kgs V2 08 m3s FIGURE 510 Schematic for Example 51 John M Cimbala EXAMPLE 51 Water Flow Through a Garden Hose Nozzle A garden hose attached with a nozzle is used to fill a 10gal bucket The inner diam eter of the hose is 2 cm and it reduces to 08 cm at the nozzle exit Fig 510 If it takes 50 s to fill the bucket with water determine a the volume and mass flow rates of water through the hose and b the average velocity of water at the nozzle exit SOLUTION A garden hose is used to fill a water bucket The volume and mass flow rates of water and the exit velocity are to be determined Assumptions 1 Water is a nearly incompressible substance 2 Flow through the hose is steady 3 There is no waste of water by splashing Properties We take the density of water to be 1000 kgm3 1 kgL Analysis a Noting that 10 gal of water are discharged in 50 s the volume and mass flow rates of water are V V Δt 10 gal 50 s 37854 L 1 gal 0757 L s m ρ V 1 kg L 0757 L s 0757 L s b The crosssectional area of the nozzle exit is A e π r e 2 π 04 cm 2 05027 cm 2 05027 10 4 m 2 The volume flow rate through the hose and the nozzle is constant Then the average velocity of water at the nozzle exit becomes V e V A e 0757 L s 05027 10 4 m 2 1 m 3 1000 L 151 m s Discussion It can be shown that the average velocity in the hose is 24 ms Therefore the nozzle increases the water velocity by over six times Final PDF to printer 218 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 218 110917 1149 AM EXAMPLE 52 Discharge of Water from a Tank A 4fthigh 3ftdiameter cylindrical water tank whose top is open to the atmosphere is initially filled with water Now the discharge plug near the bottom of the tank is pulled out and a water jet whose diameter is 05 in streams out Fig 511 The aver age velocity of the jet is approximated as V 2gh where h is the height of water in the tank measured from the center of the hole a variable and g is the gravitational acceleration Determine how long it takes for the water level in the tank to drop to 2 ft from the bottom SOLUTION The plug near the bottom of a water tank is pulled out The time it takes for half of the water in the tank to empty is to be determined Assumptions 1 Water is a nearly incompressible substance 2 The distance between the bottom of the tank and the center of the hole is negligible compared to the total water height 3 The gravitational acceleration is 322 fts2 Analysis We take the volume occupied by water as the control volume The size of the control volume decreases in this case as the water level drops and thus this is a variable control volume We could also treat this as a fixed control volume that consists of the interior volume of the tank by disregarding the air that replaces the space vacated by the water This is obviously an unsteadyflow problem since the properties such as the amount of mass within the control volume change with time The conservation of mass relation for a control volume undergoing any process is given in rate form as m in m out d m CV dt 1 During this process no mass enters the control volume m in 0 and the mass flow rate of discharged water is m out ρVA out ρ 2gh A jet 2 where A jet 𝜋 D jet 2 4 is the crosssectional area of the jet which is constant Noting that the density of water is constant the mass of water in the tank at any time is m CV ρV ρ A tank h 3 where A tank D tank 2 4 is the base area of the cylindrical tank Substituting Eqs 2 and 3 into the mass balance relation Eq 1 gives ρ 2gh A jet d ρ A tank h dt ρ 2gh π D jet 2 4 ρ π D tank 2 4 dh dt Canceling the densities and other common terms and separating the variables give dt D tank 2 D jet 2 dh 2gh Integrating from t 0 at which h h0 to t t at which h h2 gives 0 t dt D tank 2 D jet 2 2g h 0 h 2 dh h t h 0 h 2 g 2 D tank D jet 2 FIGURE 511 Schematic for Example 52 Air Dtank Djet 0 h2 h0 h Water Final PDF to printer 219 CHAPTER 5 cen22672ch05211270indd 219 110917 1149 AM 52 FLOW WORK AND THE ENERGY OF A FLOWING FLUID Unlike closed systems control volumes involve mass flow across their bound aries and some work is required to push the mass into or out of the control volume This work is known as the flow work or flow energy and is neces sary for maintaining a continuous flow through a control volume To obtain a relation for flow work consider a fluid element of volume V as shown in Fig 512 The fluid immediately upstream forces this fluid element to enter the control volume thus it can be regarded as an imaginary piston The fluid element can be chosen to be sufficiently small so that it has uniform properties throughout If the fluid pressure is P and the crosssectional area of the fluid element is A Fig 513 the force applied on the fluid element by the imaginary piston is F PA 522 To push the entire fluid element into the control volume this force must act through a distance L Thus the work done in pushing the fluid element across the boundary ie the flow work is W flow FL PAL PV kJ 523 The flow work per unit mass is obtained by dividing both sides of this equa tion by the mass of the fluid element w flow Pv kJ kg 524 The flow work relation is the same whether the fluid is pushed into or out of the control volume Fig 514 It is interesting that unlike other work quantities flow work is expressed in terms of properties In fact it is the product of two properties of the fluid For that reason some people view it as a combination property like enthalpy and refer to it as flow energy convected energy or transport energy instead of flow work Others however argue rightfully that the product PV Substituting the time of discharge is determined to be t 4 ft 2 ft 322 2 ft s 2 3 12 in 05 in 2 757 s 126 min Therefore it takes 126 min after the discharge hole is unplugged for half of the tank to be emptied Discussion Using the same relation with h2 0 gives t 431 min for the dis charge of the entire amount of water in the tank Therefore emptying the bottom half of the tank takes much longer than emptying the top half This is due to the decrease in the average discharge velocity of water with decreasing h FIGURE 512 Schematic for flow work Imaginary piston CV A V P m L F FIGURE 513 In the absence of acceleration the force applied on a fluid by a piston is equal to the force applied on the piston by the fluid P F A FIGURE 514 Flow work is the energy needed to push a fluid into or out of a control volume and it is equal to Pv a Before entering P v wflow b After entering CV CV P v wflow Final PDF to printer 220 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 220 110917 1149 AM represents energy for flowing fluids only and does not represent any form of energy for nonflow closed systems Therefore it should be treated as work This controversy is not likely to end but it is comforting to know that both arguments yield the same result for the energy balance equation In the discussions that follow we consider the flow energy to be part of the energy of a flowing fluid since this greatly simplifies the energy analysis of control volumes Total Energy of a Flowing Fluid As we discussed in Chap 2 the total energy of a simple compressible system consists of three parts internal kinetic and potential energies Fig 515 On a unitmass basis it is expressed as e u ke pe u V 2 2 gz kJ kg 525 where V is the velocity and z is the elevation of the system relative to some external reference point The fluid entering or leaving a control volume possesses an additional form of energythe flow energy Pv as already discussed Then the total energy of a flowing fluid on a unitmass basis denoted by θ becomes θ Pv e Pv u ke pe 526 But the combination Pv u has been previously defined as the enthalpy h So the relation in Eq 526 reduces to θ h ke pe h V 2 2 gz kJ kg 527 By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid one does not need to be concerned about the flow work The energy associated with pushing the fluid into or out of the control volume is automatically taken care of by enthalpy In fact this is the main reason for defining the property enthalpy From now on the energy of a fluid stream flowing into or out of a control volume is represented by Eq 527 and no reference will be made to flow work or flow energy FIGURE 515 The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid Nonflowing fluid e u gz V2 2 Flowing fluid Pv u gz 2 Internal energy Potential energy Kinetic energy Internal energy Potential energy Kinetic energy Flow energy V2 θ Final PDF to printer 221 CHAPTER 5 cen22672ch05211270indd 221 110917 1149 AM Energy Transport by Mass Noting that θ is total energy per unit mass the total energy of a flowing fluid of mass m is simply mθ provided that the properties of the mass m are uniform Also when a fluid stream with uniform properties is flowing at a mass flow rate of m the rate of energy flow with that stream is m θ Fig 516 That is Amount of energy transport by mass E mass mθ m h V 2 2 gz kJ 528 Rate of energy transport by mass E mass m θ m h V 2 2 gz kW 529 When the kinetic and potential energies of a fluid stream are negligible as is often the case these relations simplify to E mass mh and E mass m h In general the total energy transported by mass into or out of the control volume is not easy to determine since the properties of the mass at each inlet or exit may be changing with time as well as over the cross section Thus the only way to determine the energy transport through an open ing as a result of mass flow is to consider sufficiently small differential masses δm that have uniform properties and to add their total energies during flow Again noting that θ is total energy per unit mass the total energy of a flowing fluid of mass δm is θ δm Then the total energy transported by mass through an inlet or exit miθi and meθe is obtained by integration At an inlet for example it becomes E inmass m i θ i δ m i m i h i V i 2 2 g z i δ m i 530 Most flows encountered in practice can be approximated as being steady and onedimensional and thus the simple relations in Eqs 528 and 529 can be used to represent the energy transported by a fluid stream FIGURE 516 The product m iθi is the energy transported into control volume by mass per unit time mi kgs θi kJkg CV miθi kW EXAMPLE 53 Energy Transport by Flowing Air Air flows steadily in a pipe at 300 kPa 77C and 25 ms at a rate of 18 kgmin Fig 517 Determine a the diameter of the pipe b the rate of flow energy c the rate of energy transport by mass and d the error involved in part c if the kinetic energy is neglected SOLUTION Air flows steadily in a pipe at a specified state The diameter of the pipe the rate of flow energy and the rate of energy transport by mass are to be deter mined Also the error involved in the determination of energy transport by mass is to be determined Assumptions 1 The flow is steady 2 The potential energy is negligible Properties The properties of air are R 0287 kJkgK and cp 1008 kJkgK at 350 K from Table A2b FIGURE 517 Schematic for Example 53 300 kPa 77C Air 25 ms 18 kgmin Final PDF to printer 222 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 222 110917 1149 AM Analysis a The diameter is determined as follows v RT P 0287 kJ kgK 77 273 K 300 kPa 03349 m 3 kg A m v V 18 60 kg s 03349 m 3 kg 25 m s 0004018 m 2 D 4A π 4 0004018 m 2 π 00715 m b The rate of flow energy is determined from W flow m Pv 1860 kg s300 kPa03349 m 3 kg 3014 kW c The rate of energy transport by mass is E mass m h ke m c p T 1 2 V 2 18 60 kgs 1008 kJ kgK77 273 K 1 2 25 ms 2 1 kJ kg 1000 m 2 s 2 10594 kW d If we neglect kinetic energy in the calculation of energy transport by mass E mass m h m c p T 1860 kg s1005 kJ kgK77 273 K 10584 kW Therefore the error involved if we neglect the kinetic energy is only 009 percent Discussion The numerical value of the energy transport with air alone does not mean much since this value depends on the reference point selected for enthalpy it could even be negative The significant quantity is the difference between the enthalpies of the air in the pipe and the ambient air since it relates directly to the amount of energy supplied to heat air from ambient temperature to 77C 53 ENERGY ANALYSIS OF STEADYFLOW SYSTEMS A large number of engineering devices such as turbines compressors and nozzles operate for long periods of time under the same conditions once the transient startup period is completed and steady operation is established and they are classified as steadyflow devices Fig 518 Processes involving such devices can be represented reasonably well by a somewhat idealized process called the steadyflow process which was defined in Chap 1 as a process during which a fluid flows through a control volume steadily That is the fluid properties can change from point to point within the control volume but at any point they remain constant during the entire process Remember steady means no change with time During a steadyflow process no intensive or extensive properties within the control volume change with time Thus the volume V the mass m and the total energy content E of the control volume remain constant FIGURE 518 Many engineering systems such as power plants operate under steady conditions Malcolm FifeGetty Images RF Final PDF to printer 223 CHAPTER 5 cen22672ch05211270indd 223 110917 1149 AM Fig 519 As a result the boundary work is zero for steadyflow systems since VCV constant and the total mass or energy entering the control vol ume must be equal to the total mass or energy leaving it since mCV constant and ECV constant These observations greatly simplify the analysis The fluid properties at an inlet or exit remain constant during a steadyflow process The properties may however be different at different inlets and exits They may even vary over the cross section of an inlet or an exit However all properties including the velocity and elevation must remain constant with time at a fixed point at an inlet or exit It follows that the mass flow rate of the fluid at an opening must remain constant during a steadyflow process Fig 520 As an added simplification the fluid properties at an opening are usually considered to be uniform at some average value over the cross section Thus the fluid properties at an inlet or exit may be specified by the average single values Also the heat and work interactions between a steady flow system and its surroundings do not change with time Thus the power delivered by a system and the rate of heat transfer to or from a system remain constant during a steadyflow process The mass balance for a general steadyflow system was given in Sec 51 as in m out m kg s 531 The mass balance for a singlestream oneinlet and oneoutlet steadyflow system was given as m 1 m 2 ρ 1 V 1 A 1 ρ 2 V 2 A 2 532 where the subscripts 1 and 2 denote the inlet and the exit states respectively ρ is density V is the average flow velocity in the flow direction and A is the crosssectional area normal to flow direction During a steadyflow process the total energy content of a control volume remains constant ECV constant and thus the change in the total energy of the control volume is zero ΔECV 0 Therefore the amount of energy enter ing a control volume in all forms by heat work and mass must be equal to the amount of energy leaving it Then the rate form of the general energy bal ance reduces for a steadyflow process to E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 533 or Energy balance E in Rate of net energy transfer in by heat work and mass E out Rate of net energy transfer out by heat work and mass kW 534 Noting that energy can be transferred by heat work and mass only the energy balance in Eq 534 for a general steadyflow system can also be written more explicitly as Q in W in in m θ Q out W out out m θ 535 FIGURE 519 Under steadyflow conditions the mass and energy contents of a control volume remain constant Control volume mCV constant ECV constant Mass in Mass out FIGURE 520 Under steadyflow conditions the fluid properties at an inlet or exit remain constant do not change with time m1 h1 m2 h2 m3 h3 Control volume Final PDF to printer 224 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 224 110917 1149 AM or Q in W in in m h V 2 2 gz for each inlet Q out W out out m h V 2 2 gz for each exit 536 since the energy of a flowing fluid per unit mass is θ h ke pe h V 22 gz The energy balance relation for steadyflow systems first appeared in 1859 in a German thermodynamics book written by Gustav Zeuner Consider for example an ordinary electric hotwater heater under steady operation as shown in Fig 521 A coldwater stream with a mass flow rate m is continuously flowing into the water heater and a hotwater stream of the same mass flow rate is continuously flowing out of it The water heater the control volume is losing heat to the surrounding air at a rate of Q out and the electric heating element is supplying electrical work heating to the water at a rate of W in On the basis of the conservation of energy principle we can say that the water stream experiences an increase in its total energy as it flows through the water heater that is equal to the electric energy supplied to the water minus the heat losses The energy balance relation just given is intuitive in nature and is easy to use when the magnitudes and directions of heat and work transfers are known When performing a general analytical study or solving a problem that involves an unknown heat or work interaction however we need to assume a direction for the heat or work interactions In such cases it is common prac tice to assume heat to be transferred into the system heat input at a rate of Q and work produced by the system work output at a rate of W and then solve the problem The firstlaw or energy balance relation in that case for a general steadyflow system becomes Q W out m h V 2 2 gz for each exit in m h V 2 2 gz for each inlet 537 Obtaining a negative quantity for Q or W simply means that the assumed direc tion is wrong and should be reversed For singlestream devices the steady flow energy balance equation becomes Q W m h 2 h 1 V 2 2 V 1 2 2 g z 2 z 1 538 Dividing Eq 538 by m gives the energy balance on a unitmass basis as q w h 2 h 1 V 2 2 V 1 2 2 g z 2 z 1 539 where q Q m and w W m are the heat transfer and work done per unit mass of the working fluid respectively When the fluid experiences negli gible changes in its kinetic and potential energies that is Δke 0 Δpe 0 the energy balance equation is reduced further to q w h 2 h 1 540 FIGURE 521 A water heater in steady operation Qout m1 m2 m1 Electric heating element Win CV Hotwater tank Cold water in Hot water out Final PDF to printer 225 CHAPTER 5 cen22672ch05211270indd 225 110917 1149 AM The various terms appearing in the above equations are as follows Q rate of heat transfer between the control volume and its surroundings When the control volume is losing heat as in the case of the water heater Q is negative If the control volume is well insulated ie adiabatic then Q 0 W power For steadyflow devices the control volume is constant thus there is no boundary work involved The work required to push mass into and out of the control volume is also taken care of by using enthalpies for the energy of fluid streams instead of internal energies Then W represents the remaining forms of work done per unit time Fig 522 Many steadyflow devices such as turbines compressors and pumps transmit power through a shaft and W simply becomes the shaft power for those devices If the control surface is crossed by electric wires as in the case of an electric water heater W represents the electrical work done per unit time If neither is present then W 0 Δh h2 h1 The enthalpy change of a fluid can easily be determined by reading the enthalpy values at the exit and inlet states from the tables For ideal gases it can be approximated by Δh cpavgT2 T1 Note that kgskJkg kW Δke V 2 2 V 1 2 2 The unit of kinetic energy is m2s2 which is equivalent to Jkg Fig 523 The enthalpy is usually given in kJkg To add these two quantities the kinetic energy should be expressed in kJkg This is easily accomplished by dividing it by 1000 A velocity of 45 ms corresponds to a kinetic energy of only 1 kJkg which is a very small value compared with the enthalpy values encountered in practice Thus the kinetic energy term at low velocities can be neglected When a fluid stream enters and leaves a steadyflow device at about the same velocity V1 V2 the change in the kinetic energy is close to zero regardless of the velocity Caution should be exercised at high velocities however since small changes in velocities may cause significant changes in kinetic energy Fig 524 Δpe gz2 z1 A similar argument can be given for the potential energy term A potential energy change of 1 kJkg corresponds to an elevation difference of 102 m The elevation difference between the inlet and exit of most industrial devices such as turbines and compressors is well below this value and the potential energy term is always neglected for these devices The only time the potential energy term is significant is when a process involves pumping a fluid to high elevations and we are interested in the required pumping power 54 SOME STEADYFLOW ENGINEERING DEVICES Many engineering devices operate essentially under the same conditions for long periods of time The components of a steam power plant turbines compressors heat exchangers and pumps for example operate nonstop for months before the system is shut down for maintenance Fig 525 There fore these devices can be conveniently analyzed as steadyflow devices FIGURE 522 Under steady operation shaft work and electrical work are the only forms of work a simple compressible system may involve CV We Wsh FIGURE 523 The units m2s2 and Jkg are equivalent lbm s2 kg kg s2 kg s2 Also Btu J Nm kg m m m2 25037 ft2 FIGURE 524 At very high velocities even small changes in velocities can cause significant changes in the kinetic energy of the fluid ms kJkg 200 205 1 500 502 1 0 45 1 50 67 1 100 110 1 ms V2 V1 ke Final PDF to printer 226 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 226 110917 1149 AM In this section some common steadyflow devices are described and the thermodynamic aspects of the flow through them are analyzed The conserva tion of mass and the conservation of energy principles for these devices are illustrated with examples 1 Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines rockets space craft and even garden hoses A nozzle is a device that increases the velocity of a fluid at the expense of pressure A diffuser is a device that increases the pressure of a fluid by slowing it down That is nozzles and diffusers perform opposite tasks The crosssectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows The reverse is true for diffusers The rate of heat transfer between the fluid flowing through a nozzle or a diffuser and the surroundings is usually very small Q 0 since the fluid has high velocities and thus it does not spend enough time in the device for any significant heat transfer to take place Nozzles and diffusers typically involve no work W 0 and any change in potential energy is negligible Δpe 0 But nozzles and diffusers usually involve very high velocities and as a fluid passes through a nozzle or diffuser it experiences large changes in its velocity Fig 526 Therefore the kinetic energy changes must be accounted for in analyzing the flow through these devices Δke 0 FIGURE 525 A modern landbased gas turbine used for electric power production This is a General Electric LM5000 turbine It has a length of 62 m weighs 125 tons and produces 552 MW at 3600 rpm with steam injection Courtesy of GE Power Systems LPC bleed air collector 14Stage high pressure compressor Fuel system manifolds Combustor 2Stage high pressure turbine 5Stage low pressure turbine Hot end drive flange 5Stage low pressure compressor LPC Cold end drive flange FIGURE 526 Nozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies Nozzle V2 V1 V2 V1 V1 V1 Diffuser Final PDF to printer 227 CHAPTER 5 cen22672ch05211270indd 227 110917 1149 AM FIGURE 527 The diffuser of a jet engine discussed in Example 54 Yunus Çengel EXAMPLE 54 Deceleration of Air in a Diffuser Air at 10C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 ms The inlet area of the diffuser is 04 m2 The air leaves the diffuser with a velocity that is very small compared with the inlet velocity Determine a the mass flow rate of the air and b the temperature of the air leaving the diffuser SOLUTION Air enters the diffuser of a jet engine steadily at a specified velocity The mass flow rate of air and the temperature at the diffuser exit are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values 3 The potential energy change is zero Δpe 0 4 Heat transfer is negligible 5 Kinetic energy at the diffuser exit is negligible 6 There are no work interactions Analysis We take the diffuser as the system Fig 527 This is a control volume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus m 1 m 2 m a To determine the mass flow rate we need to find the specific volume of the air first This is determined from the idealgas relation at the inlet conditions v 1 R T 1 P 1 0287 kPa m 3 kgK283 K 80 kPa 1015 m 3 kg Then m 1 v 1 V 1 A 1 1 1015 m 3 kg 200 m s04 m 2 788 kg s Since the flow is steady the mass flow rate through the entire diffuser remains con stant at this value b Under stated assumptions and observations the energy balance for this steady flow system can be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m h 1 V 1 2 2 m h 2 V 2 2 2 since Q 0 W 0 and Δpe 0 h 2 h 1 V 2 2 V 1 2 2 The exit velocity of a diffuser is usually small compared with the inlet velocity V2 V1 thus the kinetic energy at the exit can be neglected The enthalpy of air at the diffuser inlet is determined from the air table Table A17 to be h 1 h 283 K 28314 kJ kg Substituting we get h 2 28314 kJ kg 0 200 m s 2 2 1 kJ kg 1000 m 2 s 2 30314 kJ kg Final PDF to printer 228 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 228 110917 1149 AM From Table A17 the temperature corresponding to this enthalpy value is T 2 303 K Discussion This result shows that the temperature of the air increases by about 20C as it is slowed down in the diffuser The temperature rise of the air is mainly due to the conversion of kinetic energy to internal energy EXAMPLE 55 Acceleration of Steam in a Nozzle Steam at 250 psia and 700F steadily enters a nozzle whose inlet area is 02 ft2 The mass flow rate of steam through the nozzle is 10 lbms Steam leaves the nozzle at 200 psia with a velocity of 900 fts Heat losses from the nozzle per unit mass of the steam are estimated to be 12 Btulbm Determine a the inlet velocity and b the exit temperature of the steam SOLUTION Steam enters a nozzle steadily at a specified flow rate and velocity The inlet velocity of steam and the exit temperature are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 There are no work interactions 3 The potential energy change is zero Δpe 0 Analysis We take the nozzle as the system Fig 528 This is a control volume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus m 1 m 2 m a The specific volume and enthalpy of steam at the nozzle inlet are P 1 250 psia T 1 700F v 1 26883 ft 3 lbm h 1 13714 Btu lbm Table A6E Then m 1 v 1 V 1 A 1 10 lbm s 1 26883 ft 3 lbm V 1 02 ft 2 V 1 1344 ft s b Under stated assumptions and observations the energy balance for this steadyflow system can be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m h 1 V 1 2 2 m h 2 V 2 2 2 since Q 0 W 0 and Δpe 0 FIGURE 528 Schematic for Example 55 Steam m 10 lbms P2 200 psia V2 900 fts P1 250 psia T1 700F A1 02 ft2 qout 12 Btulbm Final PDF to printer 229 CHAPTER 5 cen22672ch05211270indd 229 110917 1149 AM 2 Turbines and Compressors In steam gas or hydroelectric power plants the device that drives the elec tric generator is the turbine As the fluid passes through the turbine work is done against the blades which are attached to the shaft As a result the shaft rotates and the turbine produces work Fig 529 Compressors as well as pumps and fans are devices used to increase the pressure of a fluid Work is supplied to these devices from an external source through a rotating shaft Therefore compressors involve work inputs Even though these three devices function similarly they do differ in the tasks they perform A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas A compressor is capable of compressing the gas to very high pressures Pumps work very much like compressors except that they handle liquids instead of gases Note that turbines produce power output whereas compressors pumps and fans require power input Heat transfer from turbines is usually negligible Q 0 since they are typically well insulated Heat transfer is also negligible for compressors unless there is intentional cooling Potential energy changes are negligible for all of these devices Δpe 0 The velocities involved in these devices with the exception of turbines and fans are usually too low to cause any significant change in the kinetic energy Δke 0 The fluid veloc ities encountered in most turbines are very high and the fluid experiences a significant change in its kinetic energy However this change is usually very small relative to the change in enthalpy and thus it is often disregarded Dividing by the mass flow rate m and substituting h2 is determined to be h 2 h 1 q out V 2 2 V 1 2 2 13714 12 Btu lbm 900 ft s 2 1344 ft s 2 1 Btu lbm 25037 ft 2 s 2 13544 Btu lbm Then P 2 200 psia h 2 13544 Btu lbm T 2 6620F Table A6E Discussion Note that the temperature of steam drops by 380F as it flows through the nozzle This drop in temperature is mainly due to the conversion of internal energy to kinetic energy The heat loss is too small to cause any significant effect in this case EXAMPLE 56 Compressing Air with a Compressor Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K The mass flow rate of the air is 002 kgs and a heat loss of 16 kJkg occurs during the process Assuming the changes in kinetic and potential energies are negligible determine the necessary power input to the compressor FIGURE 529 Turbine blades attached to the turbine shaft Miss Kanithar AiumlaOrShutterstock RF Final PDF to printer 230 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 230 110917 1149 AM FIGURE 530 Schematic for Example 56 Win qout 16 kJkg P1 100 kPa T1 280 K P2 600 kPa T2 400 K Air m 002 kgs SOLUTION Air is compressed steadily by a compressor to a specified tempera ture and pressure The power input to the compressor is to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values 3 The kinetic and potential energy changes are zero Δke Δpe 0 Analysis We take the compressor as the system Fig 530 This is a control vol ume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus m 1 m 2 m Also heat is lost from the system and work is supplied to the system Under stated assumptions and observations the energy balance for this steadyflow system can be expressed in the rate form as E in E out Rate of net energy tranfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out W in m h 1 Q out m h 2 since Δke Δpe 0 W in m q out m h 2 h 1 The enthalpy of an ideal gas depends on temperature only and the enthalpies of the air at the specified temperatures are determined from the air table Table A17 to be h 1 h 280 K 28013 kJ kg h 2 h 400 K 40098 kJ kg Substituting the power input to the compressor is determined to be W in 002 kg s 16 kJ kg 002 kg s 40098 28013 kJ kg 274 kW Discussion Note that the mechanical energy input to the compressor manifests itself as a rise in enthalpy of air and heat loss from the compressor EXAMPLE 57 Power Generation by a Steam Turbine The power output of an adiabatic steam turbine is 5 MW and the inlet and the exit conditions of the steam are as indicated in Fig 531 a Compare the magnitudes of Δh Δke and Δpe b Determine the work done per unit mass of the steam flowing through the turbine c Calculate the mass flow rate of the steam SOLUTION The inlet and exit conditions of a steam turbine and its power output are given The changes in kinetic energy potential energy and enthalpy of steam as well as the work done per unit mass and the mass flow rate of steam are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 The system is adiabatic and thus there is no heat transfer FIGURE 531 Schematic for Example 57 Steam turbine P1 2 MPa T1 400C V1 50 ms z1 10 m P2 15 kPa x2 090 V2 180 ms z2 6 m Wout 5 MW Final PDF to printer 231 CHAPTER 5 cen22672ch05211270indd 231 110917 1149 AM Analysis We take the turbine as the system This is a control volume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus m 1 m 2 m Also work is done by the system The inlet and exit velocities and elevations are given and thus the kinetic and potential energies are to be considered a At the inlet steam is in a superheated vapor state and its enthalpy is P 1 2 MPa T 1 400C h 1 32484 kJ kg Table A6 At the turbine exit we obviously have a saturated liquidvapor mixture at 15kPa pressure The enthalpy at this state is h 2 h f x 2 h fg 22594 0923723 kJ kg 236101 kJ kg Then Δh h 2 h 1 236101 32484 kJ kg 88739 kJ kg Δke V 2 2 V 1 2 2 180 m s 2 50 m s 2 2 1 kJ kg 1000 m 2 s 2 1495 kJ kg Δpe g z 2 z 1 981 m s 2 6 10 m 1 kJ kg 1000 m 2 s 2 004 kJ kg b The energy balance for this steadyflow system can be expressed in the rate form as E in E out Rate of net energy transfer in by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m h 1 V 1 2 2 g z 1 W out m h 2 V 2 2 2 g z 2 since Q 0 Dividing by the mass flow rate m and substituting the work done by the turbine per unit mass of the steam is determined to be w out h 2 h 1 V 2 2 V 1 2 2 g z 2 z 1 Δh Δke Δpe 88739 1495 004 kJ kg 87248 kJ kg c The required mass flow rate for a 5MW power output is m W out w out 5000 kJ s 87248 kJ kg 573 kg s Discussion Two observations can be made from these results First the change in potential energy is insignificant in comparison to the changes in enthalpy and kinetic energy This is typical for most engineering devices Second as a result of low pres sure and thus high specific volume the steam velocity at the turbine exit can be very high Yet the change in kinetic energy is a small fraction of the change in enthalpy less than 2 percent in our case and is therefore often neglected Final PDF to printer 232 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 232 110917 1149 AM 3 Throttling Valves Throttling valves are any kind of flowrestricting devices that cause a sig nificant pressure drop in the fluid Some familiar examples are ordinary adjustable valves capillary tubes and porous plugs Fig 532 Unlike turbines they produce a pressure drop without involving any work The pressure drop in the fluid is often accompanied by a large drop in tempera ture and for that reason throttling devices are commonly used in refrigera tion and airconditioning applications The magnitude of the temperature drop or sometimes the temperature rise during a throttling process is governed by a property called the JouleThomson coefficient discussed in Chap 12 Throttling valves are usually small devices and the flow through them may be assumed to be adiabatic q 0 since there is neither sufficient time nor large enough area for any effective heat transfer to take place Also there is no work done w 0 and the change in potential energy if any is very small Δpe 0 Even though the exit velocity is often considerably higher than the inlet velocity in many cases the increase in kinetic energy is insignificant Δke 0 Then the conservation of energy equation for this singlestream steadyflow device reduces to h 2 h 1 kJ kg 541 That is enthalpy values at the inlet and exit of a throttling valve are the same For this reason a throttling valve is sometimes called an isenthalpic device Note however that for throttling devices with large exposed surface areas such as capillary tubes heat transfer may be significant To gain some insight into how throttling affects fluid properties let us express Eq 541 as follows u 1 P 1 v 1 u 2 P 2 v 2 or Internal energy Flow energy Constant Thus the final outcome of a throttling process depends on which of the two quantities increases during the process If the flow energy increases during the process P2v2 P1v1 it can do so at the expense of the internal energy As a result internal energy decreases which is usually accompanied by a drop in temperature If the product Pv decreases the internal energy and the temperature of a fluid will increase during a throttling process In the case of an ideal gas h hT and thus the temperature has to remain constant during a throttling process Fig 533 FIGURE 532 Throttling valves are devices that cause large pressure drops in the fluid a An adjustable valve b A porous plug c A capillary tube EXAMPLE 58 Expansion of Refrigerant134a in a Refrigerator Refrigerant134a enters the capillary tube of a refrigerator as saturated liquid at 08 MPa and is throttled to a pressure of 012 MPa Determine the quality of the refrigerant at the final state and the temperature drop during this process FIGURE 533 The temperature of an ideal gas does not change during a throttling h constant process since h hT Throttling valve Ideal gas T1 T2 T1 h2 h1 h1 Final PDF to printer 233 CHAPTER 5 cen22672ch05211270indd 233 110917 1149 AM FIGURE 534 During a throttling process the enthalpy flow energy internal energy of a fluid remains constant But internal and flow energies may be converted to each other Throttling valve u1 9480 kJkg P1v1 068 kJkg h1 9548 kJkg u2 8880 kJkg P2v2 668 kJkg h2 9548 kJkg SOLUTION Refrigerant134a that enters a capillary tube as saturated liquid is throttled to a specified pressure The exit quality of the refrigerant and the tempera ture drop are to be determined Assumptions 1 Heat transfer from the tube is negligible 2 Kinetic energy change of the refrigerant is negligible Analysis A capillary tube is a simple flowrestricting device that is commonly used in refrigeration applications to cause a large pressure drop in the refrigerant Flow through a capillary tube is a throttling process thus the enthalpy of the refrigerant remains constant Fig 534 At inlet P 1 08 MPa sat liquid T 1 T sat 08 MPa 3131C h 1 h f 08 MPa 9548 kJ kg Table A12 At exit P 2 012 MPa h 2 h 1 h f 2247 kJ kg T sat 2232C h g 23699 kJ kg Obviously hf h2 hg thus the refrigerant exists as a saturated mixture at the exit state The quality at this state is x 2 h 2 h f h fg 9548 2247 23699 2247 0340 Since the exit state is a saturated mixture at 012 MPa the exit temperature must be the saturation temperature at this pressure which is 2232C Then the temperature change for this process becomes ΔT T 2 T 1 2232 3131 C 5363C Discussion Note that the temperature of the refrigerant drops by 5363C during this throttling process Also note that 340 percent of the refrigerant vaporizes during this throttling process and the energy needed to vaporize this refrigerant is absorbed from the refrigerant itself 4a Mixing Chambers In engineering applications mixing two streams of fluids is not a rare occurrence The section where the mixing process takes place is commonly referred to as a mixing chamber The mixing chamber does not have to be a distinct chamber An ordinary Telbow or a Yelbow in a shower for example serves as the mixing chamber for the cold and hotwater streams Fig 535 The conservation of mass principle for a mixing chamber requires that the sum of the incoming mass flow rates equal the mass flow rate of the outgoing mixture Mixing chambers are usually well insulated q 0 and usually do not involve any kind of work w 0 Also the kinetic and potential ener gies of the fluid streams are usually negligible ke 0 pe 0 Then all there is left in the energy equation is the total energies of the incoming streams and the outgoing mixture The conservation of energy principle requires that these two equal each other Therefore the conservation of energy equation becomes analogous to the conservation of mass equation for this case FIGURE 535 The Telbow of an ordinary shower serves as the mixing chamber for the hot and the coldwater streams Hot water Cold water Telbow Final PDF to printer 234 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 234 110917 1149 AM EXAMPLE 59 Mixing of Hot and Cold Waters in a Shower Consider an ordinary shower where hot water at 140F is mixed with cold water at 50F If it is desired that a steady stream of warm water at 110F be supplied determine the ratio of the mass flow rates of the hot to cold water Assume the heat losses from the mixing chamber to be negligible and the mixing to take place at a pressure of 20 psia SOLUTION In a shower cold water is mixed with hot water at a specified tem perature For a specified mixture temperature the ratio of the mass flow rates of the hot to cold water is to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 The kinetic and potential energies are negligible ke pe 0 3 Heat losses from the system are negligible and thus Q 0 4 There is no work interaction involved Analysis We take the mixing chamber as the system Fig 536 This is a control volume since mass crosses the system boundary during the process We observe that there are two inlets and one exit Under the stated assumptions and observations the mass and energy balances for this steadyflow system can be expressed in the rate form as follows Mass balance m in m out d m system dt 0 steady 0 m in m out m 1 m 2 m 3 Energy balance E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m 1 h 1 m 1 h 2 m 3 h 3 since Q 0 W 0 ke pe 0 Combining the mass and energy balances m 1 h 1 m 2 h 2 m 1 m 2 h 3 Dividing this equation by m 2 yields y h 1 h 2 y 1 1 h 3 where y m 1 m 2 is the desired mass flow rate ratio The saturation temperature of water at 20 psia is 22792F Since the temperatures of all three streams are below this value T Tsat the water in all three streams exists as a compressed liquid Fig 537 A compressed liquid can be approximated as a saturated liquid at the given temperature Thus h 1 h f 140F 10799 Btu lbm h 2 h f 50F 1807 Btu lbm h 3 h f 110F 7802 Btu lbm Solving for y and substituting yields y h 3 h 2 h 1 h 3 7802 1807 10799 7802 20 Discussion Note that the mass flow rate of the hot water must be twice the mass flow rate of the cold water for the mixture to leave at 110F FIGURE 537 A substance exists as a compressed liquid at temperatures below the saturation temperatures at the given pressure Compressed liquid states P const T Tsat v FIGURE 536 Schematic for Example 59 T1 140F T2 50F T3 110F m2 m1 m3 P 20 psia Mixing chamber Final PDF to printer 235 CHAPTER 5 cen22672ch05211270indd 235 110917 1149 AM 4b Heat Exchangers As the name implies heat exchangers are devices where two moving fluid streams exchange heat without mixing Heat exchangers are widely used in various industries and they come in various designs The simplest form of a heat exchanger is a doubletube heat exchanger shown in Fig 538 It is composed of two concentric pipes of different diam eters One fluid flows in the inner pipe and the other in the annular space between the two pipes Heat is transferred from the hot fluid to the cold one through the wall separating them Sometimes the inner tube makes a couple of turns inside the shell to increase the heat transfer area and thus the rate of heat transfer The mixing chambers discussed earlier are sometimes classified as directcontact heat exchangers The conservation of mass principle for a heat exchanger in steady operation requires that the sum of the inbound mass flow rates equal the sum of the out bound mass flow rates This principle can also be expressed as follows Under steady operation the mass flow rate of each fluid stream flowing through a heat exchanger remains constant Heat exchangers typically involve no work interactions w 0 and negligible kinetic and potential energy changes Δke 0 Δpe 0 for each fluid stream The heat transfer rate associated with heat exchang ers depends on how the control volume is selected Heat exchangers are intended for heat transfer between two fluids within the device and the outer shell is usually well insulated to prevent any heat loss to the sur rounding medium When the entire heat exchanger is selected as the control volume Q becomes zero since the boundary for this case lies just beneath the insu lation and little or no heat crosses the boundary Fig 539 If however only one of the fluids is selected as the control volume then heat will cross this boundary as it flows from one fluid to the other and Q will not be zero In fact Q in this case will be the rate of heat transfer between the two fluids FIGURE 538 A heat exchanger can be as simple as two concentric pipes Hot fluid Cold fluid FIGURE 539 The heat transfer associated with a heat exchanger may be zero or nonzero depending on how the control volume is selected a System Entire heat exchanger QCV 0 b System Fluid A QCV 0 Fluid B Fluid A Fluid B Fluid A CV boundary CV boundary Heat Heat Final PDF to printer 236 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 236 110917 1149 AM EXAMPLE 510 Cooling of Refrigerant134a by Water Refrigerant134a is to be cooled by water in a condenser The refrigerant enters the condenser with a mass flow rate of 6 kgmin at 1 MPa and 70C and leaves at 35C The cooling water enters at 300 kPa and 15C and leaves at 25C Neglecting any pressure drops determine a the mass flow rate of the cooling water required and b the heat transfer rate from the refrigerant to water SOLUTION Refrigerant134a is cooled by water in a condenser The mass flow rate of the cooling water and the rate of heat transfer from the refrigerant to the water are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 The kinetic and potential energies are negligible ke pe 0 3 Heat losses from the system are negligible and thus Q 0 4 There is no work interaction Analysis We take the entire heat exchanger as the system Fig 540 This is a control volume since mass crosses the system boundary during the process In gen eral there are several possibilities for selecting the control volume for multiple stream steadyflow devices and the proper choice depends on the situation at hand We observe that there are two fluid streams and thus two inlets and two exits but no mixing a Under the stated assumptions and observations the mass and energy balances for this steadyflow system can be expressed in the rate form as follows Mass balance m in m out for each fluid stream since there is no mixing Thus m 1 m 2 m w m 3 m 4 m R Energy balance E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m 1 h 1 m 3 h 3 m 2 h 2 m 4 h 4 since Q 0 W 0 ke pe 0 Combining the mass and energy balances and rearranging give m w h 1 h 2 m R h 4 h 3 Now we need to determine the enthalpies at all four states Water exists as a compressed liquid at both the inlet and the exit since the temperatures at both locations are below the saturation temperature of water at 300 kPa 13352C Approximating the compressed liquid as a saturated liquid at the given tempera tures we have h 1 h f 15C 62982 kJ kg h 2 h f 25C 10483 kJ kg Table A4 FIGURE 540 Schematic for Example 510 2 25C 3 70C 1 MPa R134a 1 Water 15C 300 kPa 4 35C Final PDF to printer 237 CHAPTER 5 cen22672ch05211270indd 237 110917 1149 AM 5 Pipe and Duct Flow The transport of liquids or gases in pipes and ducts is of great importance in many engineering applications Flow through a pipe or a duct usually satisfies the steadyflow conditions and thus can be analyzed as a steadyflow process This of course excludes the transient startup and shutdown periods The control volume can be selected to coincide with the interior surfaces of the portion of the pipe or the duct that we are interested in analyzing Under normal operating conditions the amount of heat gained or lost by the fluid may be very significant particularly if the pipe or duct is long Fig 542 Sometimes heat transfer is desirable and is the sole purpose of the flow Water flow through the pipes in the furnace of a power plant the flow of refrigerant in a freezer and the flow in heat exchangers are some exam ples of this case At other times heat transfer is undesirable and the pipes or ducts are insulated to prevent any heat loss or gain particularly when the FIGURE 541 In a heat exchanger the heat transfer depends on the choice of the control volume Qwin QRout R134a Control volume boundary The refrigerant enters the condenser as a superheated vapor and leaves as a com pressed liquid at 35C From refrigerant134a tables P 3 1 MPa T 3 70C h 3 30387 kJ kg Table A13 P 4 1 MPa T 4 70C h 4 h f 35C 10088 kJ kg Table A11 Substituting we find m w 62982 10483 kJ kg 6 kg min 10088 30387 kJ kg m w 291 kg min b To determine the heat transfer from the refrigerant to the water we have to choose a control volume whose boundary lies on the path of heat transfer We can choose the volume occupied by either fluid as our control volume For no particular reason we choose the volume occupied by the water All the assumptions stated earlier apply except that the heat transfer is no longer zero Then assuming heat to be transferred to water the energy balance for this singlestream steadyflow system reduces to E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out Q win m w h 1 m w h 2 Rearranging and substituting Q win m w h 2 h 1 291 kg min 10483 62982 kJ kg 1218 kJ min Discussion Had we chosen the volume occupied by the refrigerant as the control volume Fig 541 we would have obtained the same result for Q Rout since the heat gained by the water is equal to the heat lost by the refrigerant FIGURE 542 Heat losses from a hot fluid flowing through an uninsulated pipe or duct to the cooler environment may be very significant Surroundings 20C 70C Hot fluid Qout Final PDF to printer 238 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 238 110917 1149 AM temperature difference between the flowing fluid and the surroundings is large Heat transfer in this case is negligible If the control volume involves a heating section electric wires a fan or a pump shaft the work interactions should be considered Fig 543 Of these fan work is usually small and often neglected in energy analysis The velocities involved in pipe and duct flow are relatively low and the kinetic energy changes are usually insignificant This is particularly true when the pipe or duct diameter is constant and the heating effects are neg ligible Kinetic energy changes may be significant however for gas flow in ducts with variable crosssectional areas especially when the compressibil ity effects are significant The potential energy term may also be significant when the fluid undergoes a considerable elevation change as it flows in a pipe or duct FIGURE 543 Pipe or duct flow may involve more than one form of work at the same time Control volume We Wsh EXAMPLE 511 Electric Heating of Air in a House The electric heating systems used in many houses consist of a simple duct with resis tance heaters Air is heated as it flows over resistance wires Consider a 15kW elec tric heating system Air enters the heating section at 100 kPa and 17C with a volume flow rate of 150 m3min If heat is lost from the air in the duct to the surroundings at a rate of 200 W determine the exit temperature of air SOLUTION The electric heating system of a house is considered For speci fied electric power consumption and air flow rate the air exit temperature is to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 Air is an ideal gas since it is at a high temperature and low pressure relative to its criticalpoint values 3 The kinetic and potential energy changes are negligible Δke Δpe 0 4 Constant specific heats at room temperature can be used for air Analysis We take the heating section portion of the duct as the system Fig 544 This is a control volume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus m 1 m 2 m Also heat is lost from the system and electrical work is supplied to the system At temperatures encountered in heating and airconditioning applications Δh can be replaced by cp ΔT where cp 1005 kJkgCthe value at room temperature with negligible error Fig 545 Then the energy balance for this steadyflow system can be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out W ein m h 1 Q out m h 2 since Δke Δpe 0 W ein Q out m c p T 2 T 1 From the idealgas relation the specific volume of air at the inlet of the duct is v 1 R T 1 P 1 0287 kPa m 3 kgK 290 K 100 kPa 0832 m 3 kg FIGURE 544 Schematic for Example 511 T2 T1 17C P1 100 kPa V1 150 m3min Qout 200 W Wein 15 kW FIGURE 545 The error involved in Δh cp ΔT where cp 1005 kJkgC is less than 05 percent for air in the temperature range 20 to 70C h 1005 T kJkg Air 20 to 70C Final PDF to printer 239 CHAPTER 5 cen22672ch05211270indd 239 110917 1149 AM The mass flow rate of the air through the duct is determined from m V 1 v 1 150 m 3 min 0832 m 3 kg 1 min 60 s 30 kg s Substituting the known quantities the exit temperature of the air is determined to be 15 kJ s 02 kJ s 3 kg s1005 kJ kg C T 2 17C T 2 219C Discussion Note that heat loss from the duct reduces the exit temperature of air 55 ENERGY ANALYSIS OF UNSTEADYFLOW PROCESSES During a steadyflow process no changes occur within the control volume thus one does not need to be concerned about what is going on within the boundaries Not having to worry about any changes within the control volume with time greatly simplifies the analysis Many processes of interest however involve changes within the control volume with time Such processes are called unsteadyflow or transient flow processes The steadyflow relations developed earlier are obviously not applicable to these processes When an unsteadyflow process is analyzed it is important to keep track of the mass and energy contents of the control vol ume as well as the energy interactions across the boundary Some familiar unsteadyflow processes are the charging of rigid vessels from supply lines Fig 546 discharging a fluid from a pressurized vessel driving a gas turbine with pressurized air stored in a large container inflating tires or balloons and even cooking with an ordinary pressure cooker Unlike steadyflow processes unsteadyflow processes start and end over some finite time period instead of continuing indefinitely Therefore in this section we deal with changes that occur over some time interval Δt instead of with the rate of changes changes per unit time An unsteadyflow system in some respects is similar to a closed system except that the mass within the system boundaries does not remain constant during a process Another difference between steady and unsteadyflow systems is that steady flow systems are fixed in space size and shape Unsteadyflow systems how ever are not Fig 547 They are usually stationary that is they are fixed in space but they may involve moving boundaries and thus boundary work The mass balance for any system undergoing any process can be expressed as see Sec 51 m in m out Δ m system kg 542 where Δmsystem mfinal minitial is the change in the mass of the system For control volumes it can also be expressed more explicitly as m i m e m 2 m 1 CV 543 where i inlet e exit 1 initial state and 2 final state of the control volume Often one or more terms in the equation above are zero For example FIGURE 546 Charging of a rigid tank from a supply line is an unsteadyflow process since it involves changes within the control volume Control volume CV boundary Supply line FIGURE 547 The shape and size of a control volume may change during an unsteadyflow process Control volume CV boundary Final PDF to printer 240 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 240 110917 1149 AM mi 0 if no mass enters the control volume during the process me 0 if no mass leaves and m1 0 if the control volume is initially evacuated The energy content of a control volume changes with time during an unsteadyflow process The magnitude of change depends on the amount of energy transfer across the system boundaries as heat and work as well as on the amount of energy transported into and out of the control volume by mass during the process When analyzing an unsteadyflow process we must keep track of the energy content of the control volume as well as the energies of the incoming and outgoing flow streams The general energy balance was given earlier as Energy balance E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies kJ 544 The general unsteadyflow process in general is difficult to analyze because the properties of the mass at the inlets and exits may change during a process Most unsteadyflow processes however can be represented reasonably well by the uniformflow process which involves the following idealization The fluid flow at any inlet or exit is uniform and steady and thus the fluid properties do not change with time or position over the cross section of an inlet or exit If they do they are averaged and treated as constants for the entire process Note that unlike the steadyflow systems the state of an unsteadyflow system may change with time and that the state of the mass leaving the control volume at any instant is the same as the state of the mass in the control volume at that instant The initial and final properties of the control volume can be determined from the knowledge of the initial and final states which are completely speci fied by two independent intensive properties for simple compressible systems Then the energy balance for a uniformflow system can be expressed explicitly as Q in W in in mθ Q out W out out mθ m 2 e 2 m 1 e 1 system 545 where θ h ke pe is the energy of a fluid stream at any inlet or exit per unit mass and e u ke pe is the energy of the nonflowing fluid within the control volume per unit mass When the kinetic and potential energy changes associated with the control volume and fluid streams are negligible as is usually the case the energy balance above simplifies to Q W out mh in mh m 2 u 2 m 1 u 1 system 546 where Q Qnetin Qin Qout is the net heat input and W Wnetout Wout Win is the net work output Note that if no mass enters or leaves the control vol ume during a process mi me 0 and m1 m2 m this equation reduces to the energy balance relation for closed systems Fig 548 Also note that an unsteadyflow system may involve boundary work as well as electrical and shaft work Fig 549 Although both the steadyflow and uniformflow processes are somewhat idealized many actual processes can be approximated reasonably well by one of these with satisfactory results The degree of satisfaction depends on the desired accuracy and the degree of validity of the assumptions made FIGURE 548 The energy equation of a uniform flow system reduces to that of a closed system when all the inlets and exits are closed Closed Closed system Closed Q W U Q W FIGURE 549 A uniformflow system may involve electrical shaft and boundary work all at once We Moving boundary Wsh Final PDF to printer 241 CHAPTER 5 cen22672ch05211270indd 241 110917 1149 AM FIGURE 550 Schematic for Example 512 Steam m1 0 a Flow of steam into an evacuated tank P2 1 MPa T2 Pi 1 MPa Ti 300C Imaginary piston Pi 1 MPa constant mi m2 b The closedsystem equivalence EXAMPLE 512 Charging of a Rigid Tank by Steam A rigid insulated tank that is initially evacuated is connected through a valve to a sup ply line that carries steam at 1 MPa and 300C Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa at which point the valve is closed Determine the final temperature of the steam in the tank SOLUTION A valve connecting an initially evacuated tank to a steam line is opened and steam flows in until the pressure inside rises to the line level The final temperature in the tank is to be determined Assumptions 1 This process can be analyzed as a uniformflow process since the properties of the steam entering the control volume remain constant during the entire process 2 The kinetic and potential energies of the streams are negligible ke pe 0 3 The tank is stationary and thus its kinetic and potential energy changes are zero that is ΔKE ΔPE 0 and ΔEsystem ΔUsystem 4 There are no boundary electrical or shaft work interactions involved 5 The tank is well insulated and thus there is no heat transfer Analysis We take the tank as the system Fig 550 This is a control volume since mass crosses the system boundary during the process We observe that this is an unsteadyflow process since changes occur within the control volume The control volume is initially evacuated and thus m1 0 and m1u1 0 Also there is one inlet and no exits for mass flow Noting that microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance m in m out Δ m system m i m 2 m 1 0 m 2 Energy balance E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies m i h i m 2 u 2 since W Q 0 ke pe 0 m 1 0 Final PDF to printer 242 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 242 110917 1149 AM Combining the mass and energy balances gives u 2 h i That is the final internal energy of the steam in the tank is equal to the enthalpy of the steam entering the tank The enthalpy of the steam at the inlet state is P i 1 MPa T i 300C h i 30516 kJ kg Table A6 which is equal to u2 Since we now know two properties at the final state it is fixed and the temperature at this state is determined from the same table to be P 2 1 MPa u 2 30516 kJ kg T 2 4561C Discussion Note that the temperature of the steam in the tank has increased by 1561C This result may be surprising at first and you may be wondering where the energy to raise the temperature of the steam came from The answer lies in the enthalpy term h u Pv Part of the energy represented by enthalpy is the flow energy Pv and this flow energy is converted to sensible internal energy once the flow ceases to exist in the control volume and it shows up as an increase in tempera ture Fig 551 Alternative solution This problem can also be solved by considering the region within the tank and the mass that is destined to enter the tank as a closed system as shown in Fig 550b Since no mass crosses the boundaries viewing this as a closed system is appropriate During the process the steam upstream the imaginary piston will push the enclosed steam in the supply line into the tank at a constant pressure of 1 MPa Then the boundary work done during this process is W bin 1 2 P i dV P i V 2 V 1 P i V tank V tank V i P i V i where Vi is the volume occupied by the steam before it enters the tank and Pi is the pressure at the moving boundary the imaginary piston face The energy balance for the closed system gives E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W bin ΔU m i P i v i m 2 u 2 m i u i u 2 u i P i v i h i since the initial state of the system is simply the line conditions of the steam This result is identical to the one obtained with the uniformflow analysis Once again the temperature rise is caused by the socalled flow energy or flow work which is the energy required to move the fluid during flow FIGURE 551 The temperature of steam rises from 300 to 4561C as it enters a tank as a result of flow energy being converted to internal energy Steam Ti 300C T2 4561C Final PDF to printer 243 CHAPTER 5 cen22672ch05211270indd 243 110917 1149 AM FIGURE 552 Schematic for Example 513 Wein V 8 m3 P 600 kPa T 400 K Air EXAMPLE 513 Discharge of Heated Air at Constant Temperature An insulated 8m3 rigid tank contains air at 600 kPa and 400 K A valve connected to the tank is now opened and air is allowed to escape until the pressure inside drops to 200 kPa The air temperature during the process is maintained constant by an electric resistance heater placed in the tank Determine the electrical energy supplied to air during this process SOLUTION Pressurized air in an insulated rigid tank equipped with an electric heater is allowed to escape at constant temperature until the pressure inside drops to a specified value The amount of electrical energy supplied to air is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniformflow process since the exit conditions remain constant 2 Kinetic and potential energies are negli gible 3 The tank is insulated and thus heat transfer is negligible 4 Air is an ideal gas with variable specific heats Analysis We take the contents of the tank as the system which is a control volume since mass crosses the boundary Fig 552 Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u respectively the mass and energy balances for this uniformflow system can be expressed as Mass balance m in m out Δ m system m e m 1 m 2 Energy balance E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W ein m e h e m 2 u 2 m 1 u 1 since Q ke pe 0 The gas constant of air is R 0287 kPam3kgK Table A1 The initial and final masses of air in the tank and the discharged amount are determined from the ideal gas relation to be m 1 P 1 V 1 R T 1 600 kPa8 m 3 0287 kPa m 3 kgK400 K 4181 kg m 2 P 2 V 2 R T 2 200 kPa8 m 3 0287 kPa m 3 kgK400 K 1394 kg m e m 1 m 2 4181 1394 2787 kg The enthalpy and internal energy of air at 400 K are he 40098 kJkg and u1 u2 28616 kJkg Table A17 The electrical energy supplied to air is determined from the energy balance to be W ein m e h e m 2 u 2 m 1 u 1 2787 kg40098 kJ kg 1394 kg28616 kJ kg 4181 kg28616 kJ kg 3200 kJ 0889 kWh since 1 kWh 3600 kJ Discussion If the temperature of discharged air changes during the process the problem can be solved with reasonable accuracy by evaluating he at the average dis charge temperature Te T2 T12 and treating it as constant Final PDF to printer 244 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 244 110917 1149 AM One of the most fundamental laws in nature is the first law of thermodynamics also known as the conservation of energy principle which provides a sound basis for studying the relationships among the various forms of energy and energy interactions It states that energy can be neither created nor destroyed during a process it can only change forms The energy content of a fixed quantity of mass a closed system can be changed by two mechanisms heat transfer Q and work transfer W Then the con servation of energy for a fixed quantity of mass can be expressed in rate form as Q W d E sys dt or Q W d dt sys ρe dV 547 where Q Q netin Q in Q out is the net rate of heat transfer to the system negative if from the system W W netout W out W in is the net power output from the system in all forms negative if power input and dEsysdt is the rate of change of the total energy content of the system The overdot stands for time rate For simple compressible systems total energy consists of internal kinetic and potential energies and it is expressed on a unitmass basis as e u ke pe u V 2 2 gz 548 Note that total energy is a property and its value does not change unless the state of the system changes An energy interaction is heat if its driving force is a temperature differ ence and it is work if it is associated with a force acting through a distance as explained in Chap 2 A system may involve numerous forms of work and the total work can be expressed as W total W shift W pressure W viscous W other 549 where Wshaft is the work transmitted by a rotating shaft Wpressure is the work done by the pressure forces on the control surface Wviscous is the work done by the normal and shear components of viscous forces on the control surface and Wother is the work done by other forces such as electric magnetic and surface tension which are insignificant for simple compressible systems and are not considered in this text We do not consider Wviscous either since it is usually small relative to other terms in control volume analysis But it should be kept in mind that the work done by shear forces as the blades shear through the fluid may need to be considered in a refined analysis of turbomachinery Work Done by Pressure Forces Consider a gas being compressed in the pistoncylinder device shown in Fig 553a When the piston moves down a differential distance ds under the influence of the pressure force PA where A is the crosssectional area of the TOPIC OF SPECIAL INTEREST General Energy Equation This section can be skipped without a loss in continuity FIGURE 553 The pressure force acting on a the moving boundary of a system in a pistoncylinder device and b the differential surface area of a system of arbitrary shape System System boundary A dV dm dA P n u V b a ds P A Vpiston System gas in cylinder Final PDF to printer 245 CHAPTER 5 cen22672ch05211270indd 245 110917 1149 AM piston the boundary work done on the system is δWboundary PA ds Dividing both sides of this relation by the differential time interval dt gives the time rate of boundary work ie power W pressure W boundary PA V piston where Vpiston dsdt is the piston velocity which is the velocity of the moving boundary at the piston face Now consider a material chunk of fluid a system of arbitrary shape which moves with the flow and is free to deform under the influence of pressure as shown in Fig 553b Pressure always acts inward and normal to the surface and the pressure force acting on a differential area dA is P dA Again noting that work is force times distance and distance traveled per unit time is velocity the time rate at which work is done by pressure forces on this differential part of the system is δ W pressure P dA V n P dA V n 550 since the normal component of velocity through the differential area dA is Vn V cos θ V n Note that n is the outer normal of dA and thus the quan tity V n is positive for expansion and negative for compression The total rate of work done by pressure forces is obtained by integrating δ W pressure over the entire surface A W pressurenet out A P V n dA A P ρ ρ V n dA 551 In light of these discussions the net power transfer can be expressed as W netout W shaftnet out W pressurenet out W shaftnet out A V n dA 552 Then the rate form of the conservation of energy relation for a closed system becomes Q netin W shaftnet out W pressurenet out d E sys dt 553 To obtain a relation for the conservation of energy for a control volume we apply the Reynolds transport theorem by replacing the extensive property B with total energy E and its associated intensive property b with total energy per unit mass e which is e u ke pe u V 22 gz Fig 554 This yields d E sys dt d dt CV eρ dV CS eρ V n A 554 Substituting the lefthand side of Eq 553 into Eq 554 the general form of the energy equation that applies to fixed moving or deforming control vol umes becomes Q netin W shaftnet out W pressurenet out d dt CV eρ dV CS eρ V r n dA 555 FIGURE 554 The conservation of energy equation is obtained by replacing an extensive property B in the Reynolds transport theorem with energy E and its associated intensive property b with e Ref 3 CV CS dt B E b e b e dBsys bρbV dt d d dt bρVr n dA CV CS dEsys eρ dV dt eρVr n dA Final PDF to printer 246 MASS AND ENERGY ANALYSIS cen22672ch05211270indd 246 110917 1149 AM which can be stated as Here V r V V CS is the fluid velocity relative to the control surface and the product ρ V n dA represents the mass flow rate through area element dA into or out of the control volume Again noting that n is the outer normal of dA the quantity V r n and thus mass flow is positive for outflow and negative for inflow Substituting the surface integral for the rate of pressure work from Eq 551 into Eq 555 and combining it with the surface integral on the right give Q netin W shaftnet out d dt CV eρ dV CS P ρ e ρ V r n dA 556 This is a very convenient form for the energy equation since pressure work is now combined with the energy of the fluid crossing the control surface and we no longer have to deal with pressure work The term Pρ Pv wflow is the flow work which is the work associ ated with pushing a fluid into or out of a control volume per unit mass Note that the fluid velocity at a solid surface is equal to the velocity of the solid surface because of the noslip condition and is zero for nonmoving surfaces As a result the pressure work along the portions of the control surface that coincide with nonmoving solid surfaces is zero Therefore pressure work for fixed control volumes can exist only along the imaginary part of the control surface where the fluid enters and leaves the control volume ie inlets and outlets This equation is not in a convenient form for solving practical engineering problems because of the integrals and thus it is desirable to rewrite it in terms of average velocities and mass flow rates through inlets and outlets If Pρ e is nearly uniform across an inlet or outlet we can simply take it outside the integral Noting that m A c ρ V r n d A c is the mass flow rate across an inlet or outlet the rate of inflow or outflow of energy through the inlet or outlet can be approximated as m Pρ e Then the energy equation becomes Fig 555 Q netin W shaftnet out d dt CV eρ dV out m P ρ e in m P ρ e 557 where e u V 22 gz is the total energy per unit mass for both the control volume and flow streams Then Q netin W shaftnet out d dt CV eρ dV out m P ρ u V 2 2 gz in m P ρ u V 2 2 gz 558 The net rate of energy transfer into a CV by heat and work transfer The time rate of change of the energy content of the CV The net flow rate of energy out of the control surface by mass flow FIGURE 555 In a typical engineering problem the control volume may contain many inlets and outlets energy flows in at each inlet and energy flows out at each outlet Energy also enters the control volume through net heat trans fer and net shaft work min energyin In mout Out mout Out Wshaft net in mout energyout energyout min energyin energyout Qnet in In Out Fixed control volume Final PDF to printer 247 CHAPTER 5 cen22672ch05211270indd 247 110917 1149 AM or Q netin W shaftnet out d dt CV eρ dV out m h V 2 2 gz in m h V 2 2 gz 559 where we used the definition of enthalpy h u Pv u Pρ The last two equations are fairly general expressions of conservation of energy but their use is still limited to uniform flow at inlets and outlets and negligible work due to viscous forces and other effects Also the subscript net in stands for net input and thus any heat or work transfer is positive if to the system and nega tive if from the system SUMMARY The conservation of mass principle states that the net mass transfer to or from a system during a process is equal to the net change increase or decrease in the total mass of the sys tem during that process and it is expressed as m in m out Δ m system and m in m out d m system dt where Δmsystem mfinal minitial is the change in the mass of the system during the process m in and m out are the total rates of mass flow into and out of the system and dmsystemdt is the rate of change of mass within the system boundaries These relations are also referred to as the mass balance and are applicable to any system undergoing any kind of process The amount of mass flowing through a cross section per unit time is called the mass flow rate and is expressed as m ρVA where ρ density of fluid V average fluid velocity normal to A and A crosssectional area normal to flow direction The volume of the fluid flowing through a cross section per unit time is called the volume flow rate and is expressed as V VA m ρ The work required to push a unit mass of fluid into or out of a control volume is called flow work or flow energy and is expressed as wflow Pv In the analysis of control volumes it is convenient to combine the flow energy and internal energy into enthalpy Then the total energy of a flowing fluid is expressed as θ h ke pe h V 2 2 gz The total energy transported by a flowing fluid of mass m with uniform properties is mθ The rate of energy transport by a fluid with a mass flow rate of m is m θ When the kinetic and potential energies of a fluid stream are negligible the amount and rate of energy transport become Emass mh and E mass m h respectively The first law of thermodynamics is essentially an expres sion of the conservation of energy principle also called the energy balance The general mass and energy balances for any system undergoing any process can be expressed as E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies It can also be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies Thermodynamic processes involving control volumes can be considered in two groups steadyflow processes and unsteady flow processes During a steadyflow process the fluid flows through the control volume steadily experiencing no change with time at a fixed position The mass and energy content of the control volume remain constant during a steadyflow pro cess Taking heat transfer to the system and work done by the system to be positive quantities the conservation of mass and energy equations for steadyflow processes are expressed as in m out m Q W out m h V 2 2 gz for each exit in m h V 2 2 gz for each inlet These are the most general forms of the equations for steady flow processes For singlestream oneinletoneexit systems Final PDF to printer cen22672ch05211270indd 248 110917 1149 AM 248 MASS AND ENERGY ANALYSIS such as nozzles diffusers turbines compressors and pumps they simplify to m 1 m 2 1 v 1 V 1 A 1 1 v 2 V 2 A 2 Q W m h 2 h 1 V 2 2 V 1 2 2 g z 2 z 1 In these relations subscripts 1 and 2 denote the inlet and exit states respectively Most unsteadyflow processes can be modeled as a uniform flow process which requires that the fluid flow at any inlet or exit is uniform and steady and thus the fluid properties do not change with time or position over the cross section of an inlet or exit If they do they are averaged and treated as constants for the entire process When kinetic and potential energy changes associated with the control volume and the fluid streams are negligible the mass and energy balance rela tions for a uniformflow system are expressed as m in m out Δ m system Q W out mh in mh m 2 u 2 m 1 u 1 system where Q Qnetin Qin Qout is the net heat input and W Wnetout Wout Win is the net work output When solving thermodynamic problems it is recommended that the general form of the energy balance Ein Eout ΔEsystem be used for all problems and that we simplify it for the particular problem instead of using the specific relations given here for different processes REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Fundamentals SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 2 ASHRAE Handbook of Refrigeration SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1994 3 Y A Çengel and J M Cimbala Fluid Mechanics Fundamentals and Applications 4th ed New York McGrawHill Education 2018 Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software PROBLEMS Conservation of Mass 51C Name four physical quantities that are conserved and two quantities that are not conserved during a process 52C Define mass and volume flow rates How are they related to each other 53C Does the amount of mass entering a control volume have to be equal to the amount of mass leaving during an unsteadyflow process 54C Consider a device with one inlet and one outlet If the volume flow rates at the inlet and at the outlet are the same is the flow through this device necessarily steady Why 55 The ventilating fan of the bathroom of a building has a volume flow rate of 30 Ls and runs continuously If the density of air inside is 120 kgm3 determine the mass of air vented out in one day 56 Air enters a 16cmdiameter pipe steadily at 200 kPa and 20C with a velocity of 5 ms Air is heated as it flows and it leaves the pipe at 180 kPa and 40C Determine a the volume flow rate of air at the inlet b the mass flow rate of air and c the velocity and volume flow rate at the exit FIGURE P56 Air 200 kPa 20C 5 ms 180 kPa 40C Q 57E A steam pipe is to transport 200 lbms of steam at 200 psia and 600F Calculate the minimum diameter this pipe can have so that the steam velocity does not exceed 59 fts Answer 363 ft 58E A garden hose attached with a nozzle is used to fill a 20gal bucket The inner diameter of the hose is 1 in and it reduces to 05 in at the nozzle exit If the average velocity in the hose is 8 fts determine a the volume and mass flow rates of water through the hose b how long it will take to fill the bucket with water and c the average velocity of water at the nozzle exit Final PDF to printer cen22672ch05211270indd 249 110917 1149 AM 249 CHAPTER 5 59E A steadyflow compressor is used to compress helium from 15 psia and 70F at the inlet to 200 psia and 600F at the outlet The outlet area and velocity are 001 ft2 and 100 fts respectively and the inlet velocity is 50 fts Deter mine the mass flow rate and the inlet area Answers 00704 lbms 0133 ft2 510 Air enters the 1m2 inlet of an aircraft engine at 100 kPa and 20C with a velocity of 180 ms Determine the volume flow rate in m3s at the engines inlet and the mass flow rate in kgs at the engines exit 511 A 2m3 rigid tank initially contains air whose den sity is 118 kgm3 The tank is connected to a highpressure supply line through a valve The valve is opened and air is allowed to enter the tank until the density in the tank rises to 530 kgm3 Determine the mass of air that has entered the tank Answer 824 kg 512 Air enters a nozzle steadily at 221 kgm3 and 40 ms and leaves at 0762 kgm3 and 180 ms If the inlet area of the nozzle is 90 cm2 determine a the mass flow rate through the nozzle and b the exit area of the nozzle Answers a 0796 kgs b 580 cm2 513 A spherical hotair balloon is initially filled with air at 120 kPa and 20C with an initial diameter of 5 m Air enters this balloon at 120 kPa and 20C with a velocity of 3 ms through a 1mdiameter opening How many minutes will it take to inflate this balloon to a 17m diameter when the pres sure and temperature of the air in the balloon remain the same as the air entering the balloon Answer 177 min FIGURE P513 Getty Images RF 514 Water enters the constant 130mm insidediameter tubes of a boiler at 7 MPa and 65C and leaves the tubes at 6 MPa and 450C with a velocity of 80 ms Calculate the velocity of the water at the tube inlet and the inlet volume flow rate 515 A desktop computer is to be cooled by a fan whose flow rate is 034 m3min Determine the mass flow rate of air through the fan at an elevation of 3400 m where the air density is 07 kgm3 Also if the average velocity of air is not to exceed 110 mmin determine the diameter of the casing of the fan Answers 0238 kgmin 63 cm FIGURE P515 Air outlet Exhaust fan Air inlet 516 A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed A small fan pulls the air in and forces it through the resistors where it is heated If the density of air is 120 kgm3 at the inlet and 095 kgm3 at the exit determine the percent increase in the velocity of air as it flows through the dryer FIGURE P516 095 kgm3 120 kgm3 517 Refrigerant134a enters a 28cmdiameter pipe steadily at 200 kPa and 20C with a velocity of 5 ms The refriger ant gains heat as it flows and leaves the pipe at 180 kPa and 40C Determine a the volume flow rate of the refrigerant at the inlet b the mass flow rate of the refrigerant and c the velocity and volume flow rate at the exit Final PDF to printer cen22672ch05211270indd 250 110917 1149 AM 250 MASS AND ENERGY ANALYSIS Flow Work and Energy Transfer by Mass 518C What are the different mechanisms for transferring energy to or from a control volume 519C How do the energies of a flowing fluid and a fluid at rest compare Name the specific forms of energy associated with each case 520 An air compressor compresses 6 L of air at 120 kPa and 20C to 1000 kPa and 400C Determine the flow work in kJkg required by the compressor Answer 109 kJkg 521 A house is maintained at 1 atm and 24C and warm air inside a house is forced to leave the house at a rate of 90 m3h as a result of outdoor air at 5C infiltrating into the house through the cracks Determine the rate of net energy loss of the house due to mass transfer Answer 0567 kW 522 Refrigerant134a enters the compressor of a refrig eration system as saturated vapor at 014 MPa and leaves as superheated vapor at 08 MPa and 60C at a rate of 006 kgs Determine the rates of energy transfers by mass into and out of the compressor Assume the kinetic and potential energies to be negligible 523E Steam is leaving a pressure cooker whose operat ing pressure is 20 psia It is observed that the amount of liquid in the cooker has decreased by 06 gal in 45 min utes after the steady operating conditions are established and the crosssectional area of the exit opening is 015 in2 Determine a the mass flow rate of the steam and the exit velocity b the total and flow energies of the steam per unit mass and c the rate at which energy is leaving the cooker by steam SteadyFlow Energy Balance Nozzles and Diffusers 524C How is a steadyflow system characterized 525C Can a steadyflow system involve boundary work 526C A diffuser is an adiabatic device that decreases the kinetic energy of the fluid by slowing it down What happens to this lost kinetic energy 527C The kinetic energy of a fluid increases as it is accel erated in an adiabatic nozzle Where does this energy come from 528E The stators in a gas turbine are designed to increase the kinetic energy of the gas passing through them adiabati cally Air enters a set of these nozzles at 300 psia and 700F with a velocity of 80 fts and exits at 250 psia and 645F Cal culate the velocity at the exit of the nozzles 529 The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30C enters FIGURE P529 StockbyteGetty Images RF 530E Air enters a nozzle steadily at 50 psia 140F and 150 fts and leaves at 147 psia and 900 fts The heat loss from the nozzle is estimated to be 65 Btulbm of air flowing The inlet area of the nozzle is 01 ft2 Determine a the exit temperature of air and b the exit area of the nozzle Answers a 507 R b 00480 ft2 531 Air at 600 kPa and 500 K enters an adiabatic nozzle that has an inlettoexit area ratio of 21 with a velocity of 120 ms and leaves with a velocity of 380 ms Determine a the exit temperature and b the exit pressure of the air Answers a 437 K b 331 kPa 532 Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa and 500C with a mass flow rate of 6000 kgh and leaves at 100 kPa and 450 ms The inlet area of the nozzle is 40 cm2 Determine a the inlet velocity and b the exit temperature 533 Steam enters a nozzle at 400C and 800 kPa with a velocity of 10 ms and leaves at 375C and 400 kPa while losing heat at a rate of 25 kW For an inlet area of 800 cm2 determine the velocity and the volume flow rate of the steam at the nozzle exit Answers 260 ms 155 m3s FIGURE P533 400C 800 kPa 10 ms Steam 375C 400 kPa Q 534 Air at 80 kPa and 127C enters an adiabatic diffuser steadily at a rate of 6000 kgh and leaves at 100 kPa The velocity of the airstream is decreased from 230 to 30 ms as it passes through the diffuser Find a the exit temperature of the air and b the exit area of the diffuser 535E Air at 13 psia and 65F enters an adiabatic diffuser steadily with a velocity of 750 fts and leaves with a low veloc ity at a pressure of 145 psia The exit area of the diffuser is it with a velocity of 350 ms and the exit state is 200 kPa and 90C Final PDF to printer cen22672ch05211270indd 251 110917 1149 AM 251 CHAPTER 5 536 Refrigerant134a at 700 kPa and 120C enters an adia batic nozzle steadily with a velocity of 20 ms and leaves at 400 kPa and 30C Determine a the exit velocity and b the ratio of the inlet to exit area A1A2 537 Refrigerant134a enters a diffuser steadily as satu rated vapor at 600 kPa with a velocity of 160 ms and it leaves at 700 kPa and 40C The refrigerant is gaining heat at a rate of 2 kJs as it passes through the diffuser If the exit area is 80 percent greater than the inlet area determine a the exit velocity and b the mass flow rate of the refrigerant Answers a 821 ms b 0298 kgs 538 Air at 80 kPa 27C and 220 ms enters a diffuser at a rate of 25 kgs and leaves at 42C The exit area of the diffuser is 400 cm2 The air is estimated to lose heat at a rate of 18 kJs during this process Determine a the exit velocity and b the exit pressure of the air Answer a 620 ms b 911 kPa 539 Air enters an adiabatic nozzle steadily at 300 kPa 200C and 45 ms and leaves at 100 kPa and 180 ms The inlet area of the nozzle is 110 cm2 Determine a the mass flow rate through the nozzle b the exit temperature of the air and c the exit area of the nozzle 543C Somebody proposes the following system to cool a house in the summer Compress the regular outdoor air let it cool back to the outdoor temperature pass it through a turbine and discharge the cold air leaving the turbine into the house From a thermodynamic point of view is the proposed system sound 544 Air is expanded from 1000 kPa and 600C at the inlet of a steadyflow turbine to 100 kPa and 200C at the outlet The inlet area and velocity are 01 m2 and 30 ms respectively and the outlet velocity is 10 ms Determine the mass flow rate and outlet area 545E Air enters a gas turbine at 150 psia and 700F and leaves at 15 psia and 100F Determine the inlet and outlet volume flow rates when the mass flow rate through this turbine is 5 lbms 546 Refrigerant134a enters a compressor at 100 kPa and 24C with a flow rate of 135 m3min and leaves at 800 kPa and 60C Determine the mass flow rate of R134a and the power input to the compressor 547 Refrigerant134a enters a compressor at 180 kPa as a saturated vapor with a flow rate of 035 m3min and leaves at 900 kPa The power supplied to the refrigerant during the compression process is 235 kW What is the temperature of R134a at the exit of the compressor Answer 525C 548 Steam flows steadily through an adiabatic turbine The inlet conditions of the steam are 4 MPa 500C and 80 ms and the exit conditions are 30 kPa 92 percent quality and 50 ms The mass flow rate of the steam is 12 kgs Determine a the change in kinetic energy b the power output and c the tur bine inlet area Answers a 195 kJkg b 121 MW c 00130 m2 FIGURE P535E P1 13 psia T1 65F V1 750 fts P2 145 psia V2 V1 A2 3A1 Air FIGURE P539 Air P1 300 kPa T1 200C V1 45 ms A1 110 cm2 P2 100 kPa V2 180 ms 540 Reconsider Prob 539 Using appropriate soft ware investigate the effect of the inlet area on the mass flow rate exit temperature and the exit area Let the inlet area vary from 50 cm2 to 150 cm2 Plot the final results against the inlet area and discuss the results Turbines and Compressors 541C Consider an adiabatic turbine operating steadily Does the work output of the turbine have to be equal to the decrease in the energy of the steam flowing through it 542C Will the temperature of air rise as it is compressed by an adiabatic compressor Why FIGURE P548 Wout P1 4 MPa T1 500C V1 80 ms P2 30 kPa x2 092 V2 50 ms Steam m 12 kgs 549 Reconsider Prob 548 Using appropriate soft ware investigate the effect of the turbine exit pressure on the power output of the turbine Let the exit pres sure vary from 10 to 200 kPa Plot the power output against the exit pressure and discuss the results 3 times the inlet area Determine a the exit temperature and b the exit velocity of the air Final PDF to printer cen22672ch05211270indd 252 110917 1149 AM 252 MASS AND ENERGY ANALYSIS 550E Steam flows steadily through a turbine at a rate of 45000 lbmh entering at 1000 psia and 900F and leaving at 5 psia as saturated vapor If the power generated by the turbine is 4 MW determine the rate of heat loss from the steam 551 Steam enters an adiabatic turbine at 8 MPa and 500C at a rate of 3 kgs and leaves at 20 kPa If the power output of the turbine is 25 MW determine the temperature of the steam at the turbine exit Neglect kinetic energy changes Answer 601C 552 An adiabatic air compressor compresses 10 Ls of air at 120 kPa and 20C to 1000 kPa and 300C Determine a the work required by the compressor in kJkg and b the power required to drive the air compressor in kW 556 Air enters the compressor of a gasturbine plant at ambient conditions of 100 kPa and 25C with a low velocity and exits at 1 MPa and 347C with a velocity of 90 ms The compressor is cooled at a rate of 1500 kJmin and the power input to the compressor is 250 kW Determine the mass flow rate of air through the compressor 557 A portion of the steam passing through a steam turbine is sometimes removed for the purposes of feedwater heating as shown in Fig P557 Consider an adiabatic steam turbine with 125 MPa and 550C steam entering at a rate of 20 kgs Steam is bled from this turbine at 1000 kPa and 200C with a mass flow rate of 1 kgs The remaining steam leaves the turbine at 100 kPa and 100C Determine the power produced by this turbine Answer 15860 kW FIGURE P552 120 kPa 20C 10 Ls 1 MPa 300C Compressor 553 Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 05 kgs and leaves at 600 kPa and 450 K Neglecting kinetic energy changes determine a the volume flow rate of the carbon dioxide at the compres sor inlet and b the power input to the compressor Answers a 0283 m3s b 688 kW 554 Steam flows steadily into a turbine with a mass flow rate of 26 kgs and a negligible velocity at 6 MPa and 600C The steam leaves the turbine at 05 MPa and 200C with a velocity of 180 ms The rate of work done by the steam in the turbine is measured to be 20350 kW If the elevation change between the turbine inlet and exit is negligible deter mine the rate of heat transfer associated with this process Answer 105 kW 555 Air is compressed by an adiabatic compressor from 100 kPa and 20C to 18 MPa and 400C Air enters the com pressor through a 015m2 opening with a velocity of 30 ms It exits through a 008m2 opening Calculate the mass flow rate of air and the required power input FIGURE P557 125 MPa 550C 20 kgs Steam turbine 1st stage 2nd stage 1 MPa 200C 1 kgs 100 kPa 100C Throttling Valves 558C Why are throttling devices commonly used in refrig eration and airconditioning applications 559C Would you expect the temperature of air to drop as it undergoes a steadyflow throttling process Explain 560C During a throttling process the temperature of a fluid drops from 30 to 20C Can this process occur adiabatically 561C Someone claims based on temperature measure ments that the temperature of a fluid rises during a throttling process in a wellinsulated valve with negligible friction How do you evaluate this claim Does this process violate any ther modynamic laws 562 Refrigerant134a is throttled from the saturated liq uid state at 700 kPa to a pressure of 160 kPa Determine the Final PDF to printer cen22672ch05211270indd 253 110917 1149 AM 253 CHAPTER 5 563 A saturated liquidvapor mixture of water called wet steam in a steam line at 1500 kPa is throttled to 50 kPa and 100C What is the quality in the steam line Answer 0944 570C Consider a steadyflow heat exchanger involving two different fluid streams Under what conditions will the amount of heat lost by one fluid be equal to the amount of heat gained by the other 571C When two fluid streams are mixed in a mixing cham ber can the mixture temperature be lower than the temperature of both streams Explain 572 Refrigerant134a at 700 kPa 70C and 8 kgmin is cooled by water in a condenser until it exists as a saturated liquid at the same pressure The cooling water enters the con denser at 300 kPa and 15C and leaves at 25C at the same pressure Determine the mass flow rate of the cooling water required to cool the refrigerant Answer 420 kgmin 573 Hot and cold streams of a fluid are mixed in a rigid mixing chamber The hot fluid flows into the chamber at a mass flow rate of 5 kgs with an energy in the amount of 150 kJkg The cold fluid flows into the chamber with a mass flow rate of 15 kgs and carries energy in the amount of 50 kJkg There is heat transfer to the surroundings from the mixing chamber in the amount of 55 kW The mixing cham ber operates in a steadyflow manner and does not gain or lose energy or mass with time Determine the energy carried from the mixing chamber by the fluid mixture per unit mass of fluid in kJkg 574 A hotwater stream at 80C enters a mixing chamber with a mass flow rate of 05 kgs where it is mixed with a stream of cold water at 20C If it is desired that the mixture leave the chamber at 42C determine the mass flow rate of the coldwater stream Assume all the streams are at a pressure of 250 kPa Answer 0865 kgs FIGURE P562 P1 700 kPa sat liquid P2 160 kPa R134a FIGURE P563 Throttling valve Steam 15 MPa 50 kPa 100C 564 An adiabatic capillary tube is used in some refrigeration systems to drop the pressure of the refrigerant from the condenser level to the evaporator level The R134a enters the capillary tube as a saturated liquid at 50C and leaves at 20C Determine the quality of the refrigerant at the inlet of the evaporator 565 A wellinsulated valve is used to throttle steam from 8 MPa and 350C to 2 MPa Determine the final temperature of the steam Answer 285C 566 Reconsider Prob 565 Using appropriate soft ware investigate the effect of the exit pressure of steam on the exit temperature after throttling Let the exit pres sure vary from 6 to 1 MPa Plot the exit temperature of steam against the exit pressure and discuss the results 567E Refrigerant134a enters the expansion valve of a refrigeration system at 120 psia as a saturated liquid and leaves at 20 psia Determine the temperature and internal energy changes across the valve 568E Air at 200 psia and 90F is throttled to the atmospheric pressure of 147 psia Determine the final temperature of the air Mixing Chambers and Heat Exchangers 569C Consider a steadyflow mixing process Under what conditions will the energy transported into the control volume by the incoming streams be equal to the energy transported out of it by the outgoing stream FIGURE P574 H2O T1 80C P 250 kPa m1 05 kgs T3 42C T2 20C m2 575E Water at 80F and 20 psia is heated in a chamber by mixing it with saturated water vapor at 20 psia If both streams enter the mixing chamber at the same mass flow rate deter mine the temperature and the quality of the exiting stream Answers 228F 0423 576 An adiabatic open feedwater heater in an electric power plant mixes 02 kgs of steam at 100 kPa and 160C with 10 kgs of feedwater at 100 kPa and 50C to produce feedwater at 100 kPa temperature drop during this process and the final specific vol ume of the refrigerant Answers 423C 00345 m3kg Final PDF to printer cen22672ch05211270indd 254 110917 1149 AM 254 MASS AND ENERGY ANALYSIS open feedwater heater that mixes 01 lbms of steam at 10 psia and 200F with 20 lbms of feedwater at 10 psia and 100F to produce 10 psia and 120F feedwater at the outlet The diam eter of the outlet pipe is 05 ft Determine the mass flow rate and feedwater velocity at the outlet Would the outlet flow rate and velocity be significantly different if the temperature at the outlet were 180F 581 Refrigerant134a at 1 MPa and 90C is to be cooled to 1 MPa and 30C in a condenser by air The air enters at 100 kPa and 27C with a volume flow rate of 600 m3min and leaves at 95 kPa and 60C Determine the mass flow rate of the refrigerant Answer 100 kgmin FIGURE P583 24C Cold air 7C Warm air 34C Room FIGURE P576 Warm feedwater Steam Cool feedwater 577 Cold water cp 418 kJkgC leading to a shower enters a thinwalled doublepipe counterflow heat exchanger at 15C at a rate of 060 kgs and is heated to 45C by hot water cp 419 kJkgC that enters at 100C at a rate of 3 kgs Determine the rate of heat transfer in the heat exchanger and the exit temperature of the hot water 578E Steam is to be condensed on the shell side of a heat exchanger at 75F Cooling water enters the tubes at 50F at a rate of 45 lbms and leaves at 65F Assuming the heat exchanger to be wellinsulated determine the rate of heat transfer in the heat exchanger and the rate of condensation of the steam 579 Air cp 1005 kJkgC is to be preheated by hot exhaust gases in a crossflow heat exchanger before it enters the furnace Air enters the heat exchanger at 95 kPa and 20C at a rate of 06 m3s The combustion gases cp 110 kJkgC enter at 160C at a rate of 095 kgs and leave at 95C Deter mine the rate of heat transfer to the air and its outlet temperature FIGURE P579 Air 95 kPa 20C 06 m3s Exhaust gases 095 kgs 95C FIGURE P581 R134a V3 600 m3min P3 100 kPa T3 27C Air P4 95 kPa T4 60C P2 1 MPa T2 30C P1 1 MPa T1 90C 582 The evaporator of a refrigeration cycle is basically a heat exchanger in which a refrigerant is evaporated by absorb ing heat from a fluid Refrigerant22 enters an evaporator at 200 kPa with a quality of 22 percent and a flow rate of 265 Lh R22 leaves the evaporator at the same pressure superheated by 5C The refrigerant is evaporated by absorbing heat from air whose flow rate is 075 kgs Determine a the rate of heat absorbed from the air and b the temperature change of air The properties of R22 at the inlet and exit of the condenser are h1 2202 kJkg v1 00253 m3kg and h2 3980 kJkg 583 An airconditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the condi tioned room in steady operation Cold air enters the mixing cham ber at 7C and 105 kPa at a rate of 055 m3s while warm air enters at 34C and 105 kPa The air leaves the room at 24C The ratio of the mass flow rates of the hot to cold airstreams is 16 Using vari able specific heats determine a the mixture temperature at the inlet of the room and b the rate of heat gain of the room 580E An open feedwater heater heats the feedwater by mix ing it with hot steam Consider an electric power plant with an and 60C at the outlet Determine the outlet mass flow rate and the outlet velocity when the outlet pipe diameter is 003 m Final PDF to printer cen22672ch05211270indd 255 110917 1149 AM 255 CHAPTER 5 FIGURE P584 Water 15C 2 MPa sat vap Exhaust gases 400C Q Heat exchanger 584 Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at 2 MPa pressure The exhaust gases enter the heat exchanger at 400C at a rate of 32 kgmin while water enters at 15C The heat exchanger is not well insulated and it is estimated that 10 percent of heat given up by the exhaust gases is lost to the surroundings If the mass flow rate of the exhaust gases is 15 times that of the water determine the temperature of the exhaust gases at the heat exchanger exit and the rate of heat transfer to the water Use the constant specific heat properties of air for the exhaust gases at a rate of 101 kgs and leaves at 27C Determine the rate of condensation of the steam in the condenser Answer 160 kgs 587 Reconsider Prob 586 Using appropriate soft ware investigate the effect of the inlet tempera ture of cooling water on the rate of condensation of steam Let the inlet temperature vary from 10 to 20C and assume the exit temperature to remain constant Plot the rate of condensa tion of steam against the inlet temperature of the cooling water and discuss the results 588 Two streams of water are mixed in an insulated container to form a third stream leaving the container The first stream has a flow rate of 30 kgs and a temperature of 90C The flow rate of the second stream is 200 kgs and its temperature is 50C What is the temperature of the third stream 589 Two mass streams of the same ideal gas are mixed in a steadyflow chamber while receiving energy by heat transfer from the surroundings The mixing process takes place at con stant pressure with no work and negligible changes in kinetic and potential energies Assume the gas has constant specific heats a Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the inlet and exit mass flow rates b Obtain an expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber c For the special case of adiabatic mixing show that the exit volume flow rate is the sum of the two inlet volume flow rates Pipe and Duct Flow 590 Water is heated in an insulated constantdiameter tube by a 7kW electric resistance heater If the water enters the heater steadily at 20C and leaves at 75C determine the mass flow rate of water 591 A 110volt electrical heater is used to warm 03 m3s of air at 100 kPa and 15C to 100 kPa and 30C How much cur rent in amperes must be supplied to this heater 592 The ducts of an air heating system pass through an unheated area As a result of heat losses the temperature of the air in the duct drops by 4C If the mass flow rate of air is 120 kgmin determine the rate of heat loss from the air to the cold environment 593E The fan on a personal computer draws 03 ft3s of air at 147 psia and 70F through the box containing the CPU and other components Air leaves at 147 psia and 83F Calculate 585 A wellinsulated shellandtube heat exchanger is used to heat water cp 418 kJkgC in the tubes from 20 to 70C at a rate of 45 kgs Heat is supplied by hot oil cp 230 kJkgC that enters the shell side at 170C at a rate of 10 kgs Determine the rate of heat transfer in the heat exchanger and the exit temperature of oil 586 Steam is to be condensed in the condenser of a steam power plant at a temperature of 50C with cooling water from a nearby lake which enters the tubes of the condenser at 18C FIGURE P586 50C Steam 50C Cooling water 18C 27C Final PDF to printer cen22672ch05211270indd 256 110917 1149 AM 256 MASS AND ENERGY ANALYSIS 594 Saturated liquid water is heated in a steadyflow steam boiler at a constant pressure of 2 MPa at a rate of 4 kgs to an outlet temperature of 250C Determine the rate of heat trans fer in the boiler 595E Water enters the tubes of a cold plate at 70F with an average velocity of 40 ftmin and leaves at 105F The diameter of the tubes is 025 in Assuming 15 percent of the heat generated is dissipated from the components to the surroundings by con vection and radiation and the remaining 85 percent is removed by the cooling water determine the amount of heat generated by the electronic devices mounted on the cold plate Answer 614 W 596 Consider a hollowcore printed circuit board 9 cm high and 18 cm long dissipating a total of 15 W The width of the air gap in the middle of the PCB is 025 cm If the cooling air enters the 12cmwide core at 25C and 1 atm at a rate of 08 Ls determine the average temperature at which the air leaves the hollow core Answer 460C 597 A computer cooled by a fan contains eight PCBs each dissipating 10 W power The height of the PCBs is 12 cm and the length is 18 cm The cooling air is supplied by a 25W fan mounted at the inlet If the temperature rise of air as it flows through the case of the computer is not to exceed 10C determine a the flow rate of the air that the fan needs to deliver and b the fraction of the temperature rise of air that is due to the heat gener ated by the fan and its motor Answers a 00104 kgs b 24 percent 598 A desktop computer is to be cooled by a fan The elec tronic components of the computer consume 60 W of power under fullload conditions The computer is to operate in envi ronments at temperatures up to 45C and at elevations up to 3400 m where the average atmospheric pressure is 6663 kPa The exit temperature of air is not to exceed 60C to meet the reliability requirements Also the average velocity of air is not to exceed 110 mmin at the exit of the computer case where the fan is installed to keep the noise level down Determine the flow rate of the fan that needs to be installed and the diameter of the casing of the fan 599 Repeat Prob 598 for a computer that consumes 100 W of power 5100 A 4m 5m 6m room is to be heated by an electric resistance heater placed in a short duct in the room Initially the room is at 15C and the local atmospheric pressure is 98 kPa The room is losing heat steadily to the outside at a rate of 150 kJmin A 200W fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of 40 kgmin The duct can be assumed to be adiabatic and there is no air leaking in or out of the room If it takes 25 min for the room air to reach an average temperature of 25C find a the power rating of the electric heater and b the temperature rise that the air experi ences each time it passes through the heater 5101 A house has an electric heating system that consists of a 300W fan and an electric resistance heating element placed in a duct Air flows steadily through the duct at a rate of 06 kgs and experiences a temperature rise of 7C The rate of heat loss from the air in the duct is estimated to be 300 W Determine the power rating of the electric resistance heating element Answer 422 kW 5102 A long roll of 2mwide and 05cmthick 1Mn man ganese steel plate ρ 7854 kgm3 and cp 0434 kJkgC coming off a furnace at 820C is to be quenched in an oil bath at 45C to a temperature of 511C If the metal sheet is mov ing at a steady velocity of 10 mmin determine the required rate of heat removal from the oil to keep its temperature con stant at 45C Answer 4368 kW FIGURE P593E PhotoDiscGetty Images RF FIGURE P597 Air outlet PCB 10 W Air inlet 10 mmin Steel plate Oil bath 45C Furnace FIGURE P5102 5103 Reconsider Prob 5102 Using appropriate soft ware investigate the effect of the moving veloc ity of the steel plate on the rate of heat transfer from the oil the electrical power in kW dissipated by the PC components Answer 00740 kW Final PDF to printer cen22672ch05211270indd 257 110917 1149 AM 257 CHAPTER 5 bath Let the velocity vary from 5 to 50 mmin Plot the rate of heat transfer against the plate velocity and discuss the results 5104E The hotwater needs of a household are to be met by heating water at 55F to 180F with a parabolic solar collector at a rate of 4 lbms Water flows through a 125indiameter thin aluminum tube whose outer surface is blackanodized in order to maximize its solar absorption ability The centerline of the tube coincides with the focal line of the collector and a glass sleeve is placed outside the tube to minimize the heat losses If solar energy is transferred to water at a net rate of 400 Btuh per ft length of the tube determine the required length of the para bolic collector to meet the hotwater requirements of this house 5105 Argon steadily flows into a constantpressure heater at 300 K and 100 kPa with a mass flow rate of 624 kgs Heat transfer in the rate of 150 kW is supplied to the argon as it flows through the heater a Determine the argon temperature at the heater exit in C b Determine the argon volume flow rate at the heater exit in m3s 5106 Steam enters a long horizontal pipe with an inlet diameter of D1 16 cm at 2 MPa and 300C with a velocity of 25 ms Farther downstream the conditions are 18 MPa and 250C and the diameter is D2 14 cm Determine a the mass flow rate of the steam and b the rate of heat transfer Answers a 0401 kgs b 451 kJs 5107 Refrigerant134a enters the condenser of a refrigera tor at 900 kPa and 60C and leaves as a saturated liquid at the same pressure Determine the heat transfer from the refrigerant per unit mass 5109 Reconsider Prob 5108 Using appropriate software investigate the effect of the exit cross sectional area of the hair dryer on the exit velocity Let the exit area vary from 25 to 75 cm2 Plot the exit velocity against the exit crosssectional area and discuss the results Include the effect of the flow kinetic energy in the analysis 5110E Air enters the duct of an airconditioning system at 15 psia and 50F at a volume flow rate of 450 ft3min The diameter of the duct is 10 in and heat is transferred to the air in the duct from the surroundings at a rate of 2 Btus Determine a the velocity of the air at the duct inlet and b the tempera ture of the air at the exit 5111 Steam enters an insulated pipe at 200 kPa and 200C and leaves at 150 kPa and 150C The inlettooutlet diameter ratio for the pipe is D1D2 180 Determine the inlet and exit velocities of the steam FIGURE P5107 900 kPa 60C 900 kPa sat liq R134a qout 5108 A hair dryer is basically a duct in which a few layers of electric resistors are placed A small fan pulls the air in and forces it through the resistors where it is heated Air enters a 1200W hair dryer at 100 kPa and 22C and leaves at 47C The crosssectional area of the hair dryer at the exit is 60 cm2 Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer determine a the volume flow rate of air at the inlet and b the velocity of the air at the exit Answers a 00404 m3s b 731 ms FIGURE P5108 T1 22C P1 100 kPa T2 47C A2 60 cm2 We 1200 W FIGURE P5111 D1 200 kPa 200C Steam D2 150 kPa 150C Charging and Discharging Processes 5112 An insulated rigid tank is initially evacuated A valve is opened and atmospheric air at 95 kPa and 17C enters the tank until the pressure in the tank reaches 95 kPa at which point the valve is closed Determine the final temperature of the air in the tank Assume constant specific heats 5113 A rigid insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 4 MPa Now the valve is opened and steam is allowed to flow into the tank until the pressure reaches 4 MPa at which point the valve is closed If the final temperature of the steam in the tank is 550C determine the temperature of the steam in the supply line and the flow work per unit mass of the steam 5114 A 2m3 rigid insulated tank initially containing satu rated water vapor at 1 MPa is connected through a valve to a supply line that carries steam at 400C Now the valve is opened and steam is allowed to flow slowly into the tank until the pres sure in the tank rises to 2 MPa At this instant the tank tempera ture is measured to be 300C Determine the mass of the steam that has entered and the pressure of the steam in the supply line FIGURE P5114 Steam Sat vapor 2 m3 1 MPa 400C Final PDF to printer cen22672ch05211270indd 258 110917 1149 AM 258 MASS AND ENERGY ANALYSIS 5115 Consider a 35L evacuated rigid bottle that is sur rounded by the atmosphere at 100 kPa and 22C A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle The air trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere Determine the net heat transfer through the wall of the bottle during this filling process Answer 350 kJ 5118E A 3ft3 rigid tank initially contains saturated water vapor at 300F The tank is connected by a valve to a supply line that carries steam at 200 psia and 400F Now the valve is opened and steam is allowed to enter the tank Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at 300F at all times The valve is closed when it is observed that onehalf of the volume of the tank is occupied by liquid water Find a the final pres sure in the tank b the amount of steam that has entered the tank and c the amount of heat transfer Answers a 6703 psia b 8574 lbm c 80900 Btu 5119E An insulated 40ft3 rigid tank contains air at 50 psia and 120F A valve connected to the tank is now opened and air is allowed to escape until the pressure inside drops to 25 psia The air temperature during this process is kept constant by an electric resistance heater placed in the tank Determine the electrical work done during this process 5116 A 2m3 rigid tank initially contains air at 100 kPa and 22C The tank is connected to a supply line through a valve Air is flowing in the supply line at 600 kPa and 22C The valve is opened and air is allowed to enter the tank until the pressure in the tank reaches the line pressure at which point the valve is closed A thermometer placed in the tank indicates that the air temperature at the final state is 77C Determine a the mass of air that has entered the tank and b the amount of heat transfer Answers a 958 kg b 339 kJ FIGURE P5115 35 L evacuated Air 100 kPa 22C FIGURE P5116 Pi 600 kPa Ti 22C T1 22C V 2 m3 P1 100 kPa Qout 5117 A 02m3 rigid tank equipped with a pressure regula tor contains steam at 2 MPa and 300C The steam in the tank is now heated The regulator keeps the steam pressure constant by letting out some steam but the temperature inside rises Determine the amount of heat transferred when the steam tem perature reaches 500C FIGURE P5119E Air V 40 ft3 P 50 psia T 120F Wein 5120 A 4L pressure cooker has an operating pressure of 175 kPa Initially onehalf of the volume is filled with liquid and the other half with vapor If it is desired that the pressure cooker not run out of liquid water for 75 min determine the highest rate of heat transfer allowed FIGURE P5120 Qin V 4 L P 175 kPa Final PDF to printer cen22672ch05211270indd 259 110917 1149 AM 259 CHAPTER 5 5121 An airconditioning system is to be filled from a rigid container that initially contains 5 kg of liquid R134a at 24C The valve connecting this container to the airconditioning sys tem is now opened until the mass in the container is 025 kg at which time the valve is closed During this time only liquid R134a flows from the container Presuming that the process is isothermal while the valve is open determine the final qual ity of the R134a in the container and the total heat transfer Answers 0506 226 kJ 5126 The airrelease flap on a hotair balloon is used to release hot air from the balloon when appropriate On one hotair balloon the air release opening has an area of 05 m2 and the filling opening has an area of 1 m2 During a two minute adiabatic flight maneuver hot air enters the balloon at 100 kPa and 35C with a velocity of 2 ms the air in the bal loon remains at 100 kPa and 35C and air leaves the balloon through the airrelease flap at velocity 1 ms At the start of this maneuver the volume of the balloon is 75 m3 Determine the final volume of the balloon and the work produced by the air inside the balloon as it expands the balloon skin FIGURE P5121 Liquid R134a 5 kg 24C AC line 5122E Oxygen is supplied to a medical facility from ten 15ft3 compressed oxygen tanks Initially these tanks are at 1500 psia and 80F The oxygen is removed from these tanks slowly enough that the temperature in the tanks remains at 80F After two weeks the pressure in the tanks is 300 psia Determine the mass of oxygen used and the total heat transfer to the tanks 5123 A 005m3 rigid tank initially contains refrigerant 134a at 08 MPa and 100 percent quality The tank is con nected by a valve to a supply line that carries refrigerant134a at 12 MPa and 40C Now the valve is opened and the refrig erant is allowed to enter the tank The valve is closed when it is observed that the tank contains saturated liquid at 12 MPa Determine a the mass of the refrigerant that has entered the tank and b the amount of heat transfer Answers a 540 kg b 202 kJ 5124 A 012m3 rigid tank contains saturated refrigerant 134a at 800 kPa Initially 25 percent of the volume is occupied by liquid and the rest by vapor A valve at the bottom of the tank is now opened and liquid is withdrawn from the tank Heat is transferred to the refrigerant such that the pressure inside the tank remains constant The valve is closed when no liquid is left in the tank and vapor starts to come out Deter mine the total heat transfer for this process Answer 201 kJ 5125 A 03m3 rigid tank is filled with saturated liquid water at 200C A valve at the bottom of the tank is opened and liquid is withdrawn from the tank Heat is transferred to the water such that the temperature in the tank remains constant FIGURE P5125 me m1 12 H2O V 03 m3 T 200C sat liquid Qin FIGURE P5126 Getty Images RF 5127 A balloon initially contains 40 m3 of helium gas at atmospheric conditions of 100 kPa and 17C The balloon is connected by a valve to a large reservoir that supplies helium gas at 125 kPa and 25C Now the valve is opened and helium Determine the amount of heat that must be transferred by the time onehalf of the total mass has been withdrawn Final PDF to printer cen22672ch05211270indd 260 110917 1149 AM 260 MASS AND ENERGY ANALYSIS is allowed to enter the balloon until pressure equilibrium with the helium at the supply line is reached The material of the balloon is such that its volume increases linearly with pressure If no heat transfer takes place during this process determine the final temperature in the balloon Answer 315 K 5131 A vertical pistoncylinder device initially contains 001 m3 of steam at 200C The mass of the frictionless pis ton is such that it maintains a constant pressure of 500 kPa inside Now steam at 1 MPa and 350C is allowed to enter the cylinder from a supply line until the volume inside doubles Neglecting any heat transfer that may have taken place during the process determine a the final temperature of the steam in the cylinder and b the amount of mass that has entered Answers a 2617C b 00176 kg 5132 A pistoncylinder device initially contains 06 kg of steam with a volume of 01 m3 The mass of the piston is such that it maintains a constant pressure of 800 kPa The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500C Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylin der doubles and the temperature in the cylinder reaches 250C at which point the valve is closed Determine a the mass of steam that has entered and b the amount of heat transfer FIGURE P5127 Ti 25C Pi 125 kPa He P1 100 kPa T1 17C 5128 An insulated 015m3 tank contains helium at 3 MPa and 130C A valve is now opened allowing some helium to escape The valve is closed when onehalf of the initial mass has escaped Determine the final temperature and pressure in the tank Answers 257 K 956 kPa 5129 A vertical pistoncylinder device initially contains 02 m3 of air at 20C The mass of the piston is such that it maintains a constant pressure of 300 kPa inside Now a valve connected to the cylinder is opened and air is allowed to escape until the volume inside the cylinder is decreased by onehalf Heat transfer takes place during the process so that the temperature of the air in the cylinder remains constant Determine a the amount of air that has left the cylinder and b the amount of heat transfer Answers a 0357 kg b 0 5130 A vertical pistoncylinder device initially contains 025 m3 of air at 600 kPa and 300C A valve connected to the cylinder is now opened and air is allowed to escape until threequarters of the mass leaves the cylinder at which point the volume is 005 m3 Determine the final temperature in the cylinder and the boundary work during this process FIGURE P5130 Air 025 m3 600 kPa 300C Air FIGURE P5132 Steam 06 kg 01 m3 800 kPa Steam 5 MPa 500C Q 5133 The air in an insulated rigid compressedair tank whose volume is 05 m3 is initially at 2400 kPa and 20C Enough air is now released from the tank to reduce the pres sure to 2000 kPa Following this release what is the tempera ture of the remaining air in the tank FIGURE P5133 C Squared StudiosGetty Images RF 5134E The weighted piston of the device shown in Fig P5134E maintains the pressure of the pistoncylinder contents at 200 psia Initially this system contains no mass Final PDF to printer cen22672ch05211270indd 261 110917 1149 AM 261 CHAPTER 5 The valve is now opened and steam from the line flows into the cylinder until the volume is 10 ft3 This process is adia batic and the steam in the line remains at 300 psia and 450F Determine the final temperature and quality if appropriate of the steam in the cylinder and the total work produced as the device is filled 5139 Air at 418 kgm3 enters a nozzle that has an inletto exit area ratio of 21 with a velocity of 120 ms and leaves with a velocity of 380 ms Determine the density of air at the exit Answer 264 kgm3 5140 Water is boiled at 100C electrically by a 3kW resis tance wire Determine the rate of evaporation of water FIGURE P5142 16 MPa 350C 22 kgs 30C sat vapor Heat Turbine FIGURE P5134E Valve Supply line Cylinder Weighted piston 5135E Repeat Prob 5134E when the supply line carries oxygen at 300 psia and 450F Answers 450F 370 Btu Review Problems 5136 Underground water is being pumped into a pool whose cross section is 6 m 9 m while water is discharged through a 7cmdiameter orifice at a constant average velocity of 4 ms If the water level in the pool rises at a rate of 25 cm min determine the rate at which water is supplied to the pool in m3s 5137 A long roll of 1mwide and 05cmthick 1Mn man ganese steel plate ρ 7854 kgm3 coming off a furnace is to be quenched in an oil bath to a specified temperature If the metal sheet is moving at a steady velocity of 10 mmin deter mine the mass flow rate of the steel plate through the oil bath FIGURE P5137 10 mmin Steel plate Oil bath Furnace 5138 Helium steadily enters a pipe with a mass flow rate of 8 kgs at 427C and 100 kPa and leaves the pipe at 27C The pressure during the process is constant at 100 kPa a Deter mine the heat transfer for the process in kW b Determine the volume flow rate of the helium at the pipe exit in m3s FIGURE P5140 Steam Water 100C 5141 An air compressor compresses 15 Ls of air at 120 kPa and 20C to 800 kPa and 300C while consuming 62 kW of power How much of this power is being used to increase the pressure of the air versus the power needed to move the fluid through the compressor Answers 448 kW 172 kW 5142 A steam turbine operates with 16 MPa and 350C steam at its inlet and saturated vapor at 30C at its exit The mass flow rate of the steam is 22 kgs and the turbine pro duces 12350 kW of power Determine the rate at which heat is lost through the casing of this turbine Final PDF to printer cen22672ch05211270indd 262 110917 1149 AM 262 MASS AND ENERGY ANALYSIS 5143E Refrigerant134a enters an adiabatic compressor at 15 psia and 20F with a volume flow rate of 10 ft3s and leaves at a pressure of 100 psia The power input to the compressor is 45 hp Find a the mass flow rate of the refrigerant and b the exit temperature 5150 Saturated steam at 1 atm condenses on a vertical plate that is maintained at 90C by circulating cooling water through the other side If the rate of heat transfer by condensation to the plate is 180 kJs determine the rate at which the condensate drips off the plate at the bottom FIGURE P5149E Water 20 psia 50F 05 lbms Air 20 psia 200F 100 ft3min FIGURE P5143E R134a 45 hp P1 15 psia T V 1 20F 1 10 ft3s P2 100 psia 5144E Nitrogen gas flows through a long constant diameter adiabatic pipe It enters at 100 psia and 120F and leaves at 50 psia and 70F Calculate the velocity of the nitro gen at the pipes inlet and outlet 5145 A 110V electric water heater warms 01 Ls of water from 18 to 30C Calculate the current in amperes that must be supplied to this heater Answer 456 A 5146 A fan is powered by a 05hp motor and delivers air at a rate of 85 m3min Determine the highest value for the aver age velocity of air mobilized by the fan Take the density of air to be 118 kgm3 5147 Steam enters a long insulated pipe at 1200 kPa 250C and 4 ms and exits at 1000 kPa The diameter of the pipe is 015 m at the inlet and 01 m at the exit Calculate the mass flow rate of the steam and its speed at the pipe outlet 5148 Steam enters a nozzle with a low velocity at 150C and 200 kPa and leaves as a saturated vapor at 75 kPa There is a heat transfer from the nozzle to the surroundings in the amount of 26 kJ for every kilogram of steam flowing through the nozzle Determine a the exit velocity of the steam and b the mass flow rate of the steam at the nozzle entrance if the nozzle exit area is 0001 m2 5149E Consider a heat exchanger that uses hot air to heat cold water Air enters this heat exchanger at 20 psia and 200F at a rate of 100 ft3min and leaves at 17 psia and 100F Water enters this unit at 20 psia and 50F at a rate of 05 lbms and exits at 17 psia and 90F Determine the total flow power in 5151 Steam at 40C condenses on the outside of a 5mlong 3cmdiameter thin horizontal copper tube by cooling water that enters the tube at 25C at an average velocity of 2 ms and FIGURE P5150 1 atm Steam 90C m FIGURE P5151 35C Steam 40C Cooling water 25C hp required for this unit and the flow work in Btulbm for both the air and water streams Final PDF to printer cen22672ch05211270indd 263 110917 1149 AM 263 CHAPTER 5 leaves at 35C Determine the rate of condensation of steam Answer 00245 kgs 5152 In large steam power plants the feedwater is fre quently heated in a closed feedwater heater by using steam extracted from the turbine at some stage Steam enters the feedwater heater at 1 MPa and 200C and leaves as saturated liquid at the same pressure Feedwater enters the heater at 25 MPa and 50C and leaves at 10C below the exit tempera ture of the steam Determine the ratio of the mass flow rates of the extracted steam and the feedwater 5153 In large gasturbine power plants air is preheated by the exhaust gases in a heat exchanger called the regenerator before it enters the combustion chamber Air enters the regen erator at 1 MPa and 550 K at a mass flow rate of 800 kgmin Heat is transferred to the air at a rate of 2700 kJs Exhaust gases enter the regenerator at 140 kPa and 800 K and leave at 130 kPa and 600 K Treating the exhaust gases as air deter mine a the exit temperature of the air and b the mass flow rate of exhaust gases Answers a 741 K b 126 kgs 5154 Cold water enters a steam generator at 20C and leaves as saturated vapor at 200C Determine the fraction of heat used in the steam generator to preheat the liquid water from 20C to the saturation temperature of 200C 5155 An ideal gas expands in an adiabatic turbine from 1200 K and 900 kPa to 800 K Determine the turbine inlet volume flow rate of the gas in m3s required to produce turbine work output at the rate of 650 kW The average values of the specific heats for this gas over the temperature range and the gas constant are cp 113 kJkgK cv 083 kJkgK and R 030 kJkgK 5156 Determine the power input for a compressor that compresses helium from 110 kPa and 20C to 400 kPa and 200C Helium enters this compressor through a 01m2 pipe at a velocity of 7 ms 5157 Chickens with an average mass of 22 kg and average specific heat of 354 kJkgC are to be cooled by chilled water that enters a continuousflowtype immersion chiller at 05C Chickens are dropped into the chiller at a uniform temperature of 15C at a rate of 500 chickens per hour and are cooled to an average temperature of 3C before they are taken out The chiller gains heat from the surroundings at a rate of 200 kJh Determine a the rate of heat removal from the chickens in kW and b the mass flow rate of water in kgs if the tem perature rise of water is not to exceed 2C 5158 Repeat Prob 5157 assuming heat gain of the chiller is negligible 5159E A refrigeration system is being designed to cool eggs ρ 674 lbmft3 and cp 080 BtulbmF with an average mass of 014 lbm from an initial temperature of 90F to a final average temperature of 50F by air at 34F at a rate of 3000 eggs per hour Determine a the rate of heat removal from the eggs in Btuh and b the required volume flow rate of air in ft3h if the temperature rise of air is not to exceed 10F 5160 A glass bottle washing facility uses a wellagitated hotwater bath at 50C that is placed on the ground The bot tles enter at a rate of 450 per minute at an ambient temperature of 20C and leave at the water temperature Each bottle has a mass of 150 g and removes 02 g of water as it leaves the bath wet Makeup water is supplied at 15C Disregarding any heat losses from the outer surfaces of the bath determine the rate at which a water and b heat must be supplied to maintain steady operation 5161 In a dairy plant milk at 4C is pasteurized continuously at 72C at a rate of 20 Ls for 24 h a day and 365 days a year The milk is heated to the pasteurizing temperature by hot water heated in a naturalgasfired boiler that has an efficiency of 90 percent The pasteurized milk is then cooled by cold water at 18C before it is finally refrigerated back to 4C To save energy and money the plant installs a regenerator that has an effec tiveness of 82 percent If the cost of natural gas is 110therm 1 therm 105500 kJ determine how much energy and money the regenerator will save this company per year FIGURE P5161 5162 Long aluminum wires of diameter 5 mm ρ 2702 kgm3 and cp 0896 kJkgC are extruded at a tem perature of 350C and are cooled to 50C in atmospheric air at 25C If the wire is extruded at a velocity of 8 mmin determine the rate of heat transfer from the wire to the extrusion room FIGURE P5162 Aluminum wire Tair 25C 350C 8 mmin 5163 Repeat Prob 5162 for a copper wire ρ 8950 kgm3 and cp 0383 kJkgC Final PDF to printer cen22672ch05211270indd 264 110917 1149 AM 264 MASS AND ENERGY ANALYSIS 5164E Steam at 80 psia and 400F is mixed with water at 60F and 80 psia steadily in an adiabatic device Steam enters the device at a rate of 005 lbms while the water enters at 1 lbms Determine the temperature of the mixture leaving this device when the outlet pressure is 80 psia Answer 117F 5165 A constantpressure R134a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams Determine the flow power needed to pass 6 Ls of R134a at 320 kPa and 55 percent qual ity through this unit What is the mass flow rate in kgs of the two outlet streams 5168 The ventilating fan of the bathroom of a building has a volume flow rate of 30 Ls and runs continuously The build ing is located in San Francisco California where the average winter temperature is 122C and it is maintained at 22C at all times The building is heated by electricity whose unit cost is 012kWh Determine the amount and cost of the heat vented out per month in winter FIGURE P5168 122C Bathroom 22C Fan 30 Ls FIGURE P5165 3 Saturated liquid 1 Liquidvapor mixture Vapor separation unit Saturated vapor 2 5166E It is well established that indoor air quality IAQ has a significant effect on general health and productivity of employees at a workplace A study showed that enhanc ing IAQ by increasing the building ventilation from 5 cfm cubic feet per minute to 20 cfm increased the productivity by 025 percent valued at 90 per person per year and decreased the respiratory illnesses by 10 percent for an average annual savings of 39 per person while increasing the annual energy consumption by 6 and the equipment cost by about 4 per person per year ASHRAE Journal December 1998 For a workplace with 120 employees determine the net monetary benefit of installing an enhanced IAQ system to the employer per year Answer 14280yr 5167E The average atmospheric pressure in Spokane Washington elevation 2350 ft is 135 psia and the aver age winter temperature is 365F The pressurization test of a 9fthigh 4500ft2 older home revealed that the seasonal aver age infiltration rate of the house is 22 air changes per hour ACH That is the entire air in the house is replaced com pletely 22 times per hour by the outdoor air It is suggested that the infiltration rate of the house can be reduced by half to 11 ACH by winterizing the doors and the windows If the house is heated by natural gas whose unit cost is 124therm and the heating season can be taken to be six months deter mine how much the homeowner will save from the heating costs per year by this winterization project Assume the house is maintained at 72F at all times and the efficiency of the fur nace is 092 Also assume the latent heat load during the heat ing season to be negligible 5169E A small positioning control rocket in a satellite is driven by a 2ft3 container filled with R134a at 10F Upon launch the container is completely filled with saturated liquid R134a The rocket is designed for short bursts of 5s dura tion During each burst the mass flow rate leaving the rocket is 005 lbms How many such bursts can this rocket experience before the quality in the container is 90 percent or more pre suming that the temperature of the container contents is main tained at 10F Answer 680 5170 Determine the rate of sensible heat loss from a build ing due to infiltration if the outdoor air at 5C and 95 kPa enters the building at a rate of 60 Ls when the indoors is main tained at 25C 5171 Consider a large classroom on a hot summer day with 150 students each dissipating 60 W of sensible heat All the lights with 60 kW of rated power are kept on The room has no external walls and thus heat gain through the walls and the roof is negligible Chilled air is available at 15C and the temperature of the return air is not to exceed 25C Determine the required flow rate of air in kgs that needs to be supplied to the room to keep the average temperature of the room constant Answer 149 kgs Final PDF to printer cen22672ch05211270indd 265 110917 1149 AM 265 CHAPTER 5 5172 An airconditioning system requires airflow at the main supply duct at a rate of 130 m3min The average velocity of air in the circular duct is not to exceed 8 ms to avoid exces sive vibration and pressure drops Assuming the fan converts 80 percent of the electrical energy it consumes into kinetic energy of air determine the size of the electric motor needed to drive the fan and the diameter of the main duct Take the density of air to be 120 kgm3 5174 The maximum flow rate of standard shower heads is about 35 gpm 133 Lmin and can be reduced to 275 gpm 105 Lmin by switching to lowflow shower heads that are equipped with flow controllers Consider a family of four with each person taking a 5min shower every morning City water at 15C is heated to 55C in an electric water heater and tempered to 42C by cold water at the Telbow of the shower before being routed to the shower heads Assuming a constant specific heat of 418 kJkgC for water determine a the ratio of the flow rates of the hot and cold water as they enter the Telbow and b the amount of electricity that will be saved per year in kWh by replacing the standard shower heads with the lowflow ones 5175 Reconsider Prob 5174 Using appropriate software investigate the effect of the inlet tem perature of cold water on the energy saved by using the low flow shower head Let the inlet temperature vary from 10C to 20C Plot the electric energy savings against the water inlet temperature and discuss the results 5176 Submarines change their depth by adding or removing air from rigid ballast tanks thereby displacing seawater in the tanks Consider a submarine that has a 700m3 airballast tank originally partially filled with 100 m3 of air at 1500 kPa and 15C For the submarine to surface air at 1500 kPa and 20C is pumped into the ballast tank until it is entirely filled with air The tank is filled so quickly that the process is adiabatic and the seawater leaves the tank at 15C Determine the final temperature and mass of the air in the ballast tank 5177 In Prob 5176 presume that air is added to the tank in such a way that the temperature and pressure of the air in the tank remain constant Determine the final mass of the air in the ballast tank under this condition Also determine the total heat transfer while the tank is being filled in this manner 5178 Steam enters a turbine steadily at 7 MPa and 600C with a velocity of 60 ms and leaves at 25 kPa with a quality of 95 percent A heat loss of 20 kJkg occurs during the pro cess The inlet area of the turbine is 150 cm2 and the exit area is 1400 cm2 Determine a the mass flow rate of the steam b the exit velocity and c the power output 5179 Reconsider Prob 5178 Using appropriate software investigate the effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine Let the exit pressure vary from 10 to 50 kPa with the same quality and let the exit area vary from 1000 to 3000 cm2 Plot the exit velocity and the power outlet against the exit pressure for the exit areas of 1000 2000 and 3000 cm2 and discuss the results 5180 It is proposed to have a water heater that consists of an insulated pipe of 75cm diameter and an electric resistor inside Cold water at 20C enters the heating section steadily at a rate of 24 Lmin If water is to be heated to 48C determine a the power rating of the resistance heater and b the average velocity of the water in the pipe FIGURE P5172 130 m3min 8 ms FIGURE P5173 P 95 kPa V 400 m3 T1 450 kJmin m Wein 30 kW 250 W 14C 24C T2 T1 5C 5173 A building with an internal volume of 400 m3 is to be heated by a 30kW electric resistance heater placed in the duct inside the building Initially the air in the building is at 14C and the local atmospheric pressure is 95 kPa The building is losing heat to the surroundings at a steady rate of 450 kJmin Air is forced to flow through the duct and the heater steadily by a 250W fan and it experiences a temperature rise of 5C each time it passes through the duct which may be assumed to be adiabatic a How long will it take for the air inside the building to reach an average temperature of 24C b Determine the average mass flow rate of air through the duct Answers a 146 s b 602 kgs Final PDF to printer cen22672ch05211270indd 266 110917 1149 AM 266 MASS AND ENERGY ANALYSIS 5181 A liquid R134a bottle has an internal volume of 00015 m3 Initially it contains 055 kg of R134a saturated mixture at 26C A valve is opened and R134a vapor only no liquid is allowed to escape slowly such that tempera ture remains constant until the mass of R134a remaining is 015 kg Find the heat transfer with the surroundings that is needed to maintain the temperature and pressure of the R134a constant 5182 A pistoncylinder device initially contains 2 kg of refrigerant134a at 800 kPa and 80C At this state the piston is touching on a pair of stops at the top The mass of the piston is such that a 500kPa pressure is required to move it A valve at the bottom of the cylinder is opened and R134a is with drawn from the cylinder After a while the piston is observed to move and the valve is closed when half of the refrigerant is withdrawn from the cylinder and the temperature in the tank drops to 20C Determine a the work done and b the heat transfer Answers a 116 kJ b 607 kJ 5183 A pistoncylinder device initially contains 12 kg of air at 700 kPa and 200C At this state the piston is touching on a pair of stops The mass of the piston is such that 600kPa pressure is required to move it A valve at the bottom of the tank is opened and air is withdrawn from the cylinder The valve is closed when the volume of the cylinder decreases to 80 percent of the initial volume If it is estimated that 40 kJ of heat is lost from the cylinder determine a the final tempera ture of the air in the cylinder b the amount of mass that has escaped from the cylinder and c the work done Use constant specific heats at the average temperature The newer pressure cookers use a spring valve with several pressure settings rather than a weight on the cover A certain pressure cooker has a volume of 6 L and an operat ing pressure of 75 kPa gage Initially it contains 1 kg of water Heat is supplied to the pressure cooker at a rate of 500 W for 30 min after the operating pressure is reached Assuming an atmospheric pressure of 100 kPa determine a the tem perature at which cooking takes place and b the amount of water left in the pressure cooker at the end of the process Answers a 11604C b 06 kg 5185 A tank with an internal volume of 1 m3 contains air at 800 kPa and 25C A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to 150 kPa at which point the valve is closed Assume there is negligible heat transfer from the tank to the air left in the tank a Using the approximation he constant heavg 05 h1 h2 calculate the mass withdrawn during the process b Consider the same process but broken into two parts That is consider an intermediate state at P2 400 kPa calculate the mass removed during the process from P1 800 kPa to P2 and then the mass removed during the process from P2 to P3 150 kPa using the type of approximation used in part a and add the two to get the total mass removed c Calculate the mass removed if the variation of he is accounted for FIGURE P5183 Q Air 12 kg 700 kPa 200C 5184 A pressure cooker is a pot that cooks food much faster than ordinary pots by maintaining a higher pressure and temper ature during cooking The pressure inside the pot is controlled by a pressure regulator the petcock that keeps the pressure at a constant level by periodically allowing some steam to escape thus preventing any excess pressure buildup Pressure cookers in general maintain a gage pressure of 2 atm or 3 atm abso lute inside Therefore pressure cookers cook at a temperature of about 133C instead of 100C cutting the cooking time by as much as 70 percent while minimizing the loss of nutrients FIGURE P5185 Air 800 kPa 25C 1 m3 5186 In a singleflash geothermal power plant geothermal water enters the flash chamber a throttling valve at 230C as a saturated liquid at a rate of 50 kgs The steam result ing from the flashing process enters a turbine and leaves at 20 kPa with a moisture content of 5 percent Determine the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam Final PDF to printer cen22672ch05211270indd 267 110917 1149 AM 267 CHAPTER 5 5187 An adiabatic air compressor is to be powered by a directcoupled adiabatic steam turbine that is also driving a generator Steam enters the turbine at 125 MPa and 500C at a rate of 25 kgs and exits at 10 kPa and a quality of 092 Air enters the compressor at 98 kPa and 295 K at a rate of 10 kgs and exits at 1 MPa and 620 K Determine the net power deliv ered to the generator by the turbine the compressor at 50C and 100 kPa and leaves at 130 kPa at a rate of 0018 kgs The compressor increases the air pressure with a side effect It also increases the air temperature which increases the possibility that a gasoline engine will experience an engine knock To avoid this an aftercooler is placed after the compressor to cool the warm air with cold ambient air before it enters the engine cylinders It is estimated that the aftercooler must decrease the air temperature below 80C if knock is to be avoided The cold ambient air enters the aftercooler at 30C and leaves at 40C Disregarding any frictional losses in the turbine and the compressor and treating the exhaust gases as air determine a the temperature of the air at the compressor outlet and b the minimum volume flow rate of ambient air required to avoid knock FIGURE P5188 Aftercooler Cold air 30C 40C Exhaust gases 120 kPa 400C Compressor 130 kPa Air 100 kPa 50C Turbine 350C FIGURE P5186 Flash chamber Separator Steam turbine 20 kPa x 095 Liquid 230C sat liq 1 2 3 4 FIGURE P5187 Air compressor Steam turbine 98 kPa 295 K 1 MPa 620 K 125 MPa 500C 10 kPa 5188 The turbocharger of an internal combustion engine consists of a turbine and a compressor Hot exhaust gases flow through the turbine to produce work and the work output from the turbine is used as the work input to the compressor The pressure of ambient air is increased as it flows through the compressor before it enters the engine cylinders Thus the pur pose of a turbocharger is to increase the pressure of air so that more air gets into the cylinder Consequently more fuel can be burned and more power can be produced by the engine In a turbocharger exhaust gases enter the turbine at 400C and 120 kPa at a rate of 002 kgs and leave at 350C Air enters 5189 A D0 10mdiameter tank is initially filled with water 2 m above the center of a D 10cmdiameter valve near the bot tom The tank surface is open to the atmosphere and the tank drains through a L 100mlong pipe connected to the valve The friction factor of the pipe is given to be f 0015 and the discharge velocity is expressed as V 2gz 15 f L D where z is the water height above the center of the valve Determine a the initial discharge velocity from the tank and b the time required to empty the tank The tank can be considered to be empty when the water level drops to the center of the valve 5190 Consider an evacuated rigid bottle of volume V that is surrounded by the atmosphere at pressure P0 and tempera ture T0 A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle The air at the exit of the flash chamber is a 1 MPa b 500 kPa c 100 kPa d 50 kPa Final PDF to printer cen22672ch05211270indd 268 110917 1149 AM 268 MASS AND ENERGY ANALYSIS trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere Determine the net heat transfer through the wall of the bottle during this filling process in terms of the properties of the system and the surrounding atmosphere Fundamentals of Engineering FE Exam Problems 5191 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kgs with hot air at 90C enter ing also at a rate of 5 kgs If the exit temperature of hot air is 20C the exit temperature of cold water is a 27C b 32C c 52C d 85C e 90C 5192 A heat exchanger is used to heat cold water at 15C entering at a rate of 2 kgs with hot air at 85C entering at a rate of 3 kgs The heat exchanger is not insulated and is los ing heat at a rate of 25 kJs If the exit temperature of hot air is 20C the exit temperature of cold water is a 28C b 35C c 38C d 41C e 80C 5193 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kgs with hot water at 90C entering at a rate of 4 kgs If the exit temperature of hot water is 50C the exit temperature of cold water is a 42C b 47C c 55C d 78C e 90C 5194 In a shower cold water at 10C flowing at a rate of 5 kgmin is mixed with hot water at 60C flowing at a rate of 2 kgmin The exit temperature of the mixture is a 243C b 350C c 400C d 443C e 552C 5195 In a heating system cold outdoor air at 7C flowing at a rate of 4 kgmin is mixed adiabatically with heated air at 70C flowing at a rate of 5 kgmin The exit temperature of the mixture is a 34C b 39C c 42C d 57C e 70C 5196 Refrigerant134a expands in an adiabatic turbine from 12 MPa and 100C to 018 MPa and 50C at a rate of 125 kgs The power output of the turbine is a 447 kW b 664 kW c 727 kW d 892 kW e 1120 kW 5197 Hot combustion gases assumed to have the properties of air at room temperature enter a gas turbine at 1 MPa and 1500 K at a rate of 01 kgs and exit at 02 MPa and 900 K If heat is lost from the turbine to the surroundings at a rate of 15 kJs the power output of the gas turbine is a 15 kW b 30 kW c 45 kW d 60 kW e 75 kW 5198 Steam expands in a turbine from 4 MPa and 500C to 05 MPa and 250C at a rate of 1350 kgh Heat is lost from the turbine at a rate of 25 kJs during the process The power output of the turbine is a 157 kW b 207 kW c 182 kW d 287 kW e 246 kW 5199 Steam is compressed by an adiabatic compressor from 02 MPa and 150C to 08 MPa and 350C at a rate of 130 kgs The power input to the compressor is a 511 kW b 393 kW c 302 kW d 717 kW e 901 kW 5200 Refrigerant134a is compressed by a compressor from the saturated vapor state at 014 MPa to 09 MPa and 60C at a rate of 0108 kgs The refrigerant is cooled at a rate of 110 kJs during compression The power input to the compressor is a 494 kW b 604 kW c 714 kW d 750 kW e 813 kW 5201 Refrigerant134a at 14 MPa and 70C is throttled to a pressure of 06 MPa The temperature of the refrigerant after throttling is a 70C b 66C c 57C d 49C e 22C 5202 Steam enters a diffuser steadily at 05 MPa 300C and 90 ms at a rate of 35 kgs The inlet area of the diffuser is a 22 cm2 b 53 cm2 c 126 cm2 d 175 cm2 e 203 cm2 5203 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 280 ms at a rate of 25 kgs If the temperature and pressure of the steam at the nozzle exit are 400C and 2 MPa the exit area of the nozzle is a 84 cm2 b 107 cm2 c 135 cm2 d 196 cm2 e 230 cm2 5204 Air at 27C and 5 atm is throttled by a valve to 1 atm If the valve is adiabatic and the change in kinetic energy is negligible the exit temperature of air will be a 10C b 15C c 20C d 23C e 27C 5205 Steam at 1 MPa and 300C is throttled adiabati cally to a pressure of 04 MPa If the change in kinetic energy is negligible the specific volume of the steam after throttling is a 0358 m3kg b 0233 m3kg c 0375 m3kg d 0646 m3kg e 0655 m3kg 5206 Air is to be heated steadily by an 8kW electric resistance heater as it flows through an insulated duct If the air enters at 50C at a rate of 2 kgs the exit temperature of air is a 460C b 500C c 540C d 554C e 580C Final PDF to printer cen22672ch05211270indd 269 110917 1149 AM 269 CHAPTER 5 5207 Saturated water vapor at 40C is to be condensed as it flows through a tube at a rate of 020 kgs The condensate leaves the tube as a saturated liquid at 40C The rate of heat transfer from the tube is a 34 kJs b 481 kJs c 2406 kJs d 514 kJs e 548 kJs Design and Essay Problems 5208 Pneumatic nail drivers used in construction require 002 ft3 of air at 100 psia and 1 Btu of energy to drive a single nail You have been assigned the task of designing a compressedair storage tank with enough capacity to drive 500 nails The pressure in this tank cannot exceed 500 psia and the temperature cannot exceed that normally found at a construction site What is the maximum pressure to be used in the tank and what is the tanks volume 5209 You have been given the responsibility of picking a steam turbine for an electricalgeneration station that is to pro duce 300 MW of electrical power that will sell for 008 per kilowatthour The boiler will produce steam at 700 psia and 700F and the condenser is planned to operate at 80F The cost of generating and condensing the steam is 0015 per kilowatthour of electricity produced You have narrowed your selection to the three turbines in the following table Your cri terion for selection is to pay for the equipment as quickly as possible Which turbine should you choose 5210 You are to design a small directional control rocket to operate in space by providing as many as 100 bursts of 5 seconds each with a mass flow rate of 05 lbms at a veloc ity of 400 fts Storage tanks that will contain up to 3000 psia are available and the tanks will be located in an environ ment whose temperature is 40F Your design criterion is to minimize the volume of the storage tank Should you use a compressed air or an R134a system 5211 An air cannon uses compressed air to propel a pro jectile from rest to a final velocity Consider an air cannon that is to accelerate a 10gram projectile to a speed of 300 ms using compressed air whose temperature cannot exceed 20C The volume of the storage tank is not to exceed 01 m3 Select the storage volume size and maximum storage pressure that require the minimum amount of energy to fill the tank 5212 Design a 1200W electric hair dryer such that the air temperature and velocity in the dryer will not exceed 50C and 3 ms respectively 5213 To maintain altitude the temperature of the air inside a hotair balloon must remain within a 1C band while the volume cannot vary by more than 1 percent At a 300m alti tude the air in a 1000 m3 hotair balloon needs to maintain a 35C average temperature This balloon loses heat at a rate of 3 kW through the fabric When the burner is activated it adds 30 kgs of air at 200C and 100 kPa to the balloon When the flap that allows air to escape is opened air leaves the balloon at a rate of 20 kgs Design the burner and exhaustflap control cycles on time and off time necessary to maintain the balloon at a 300m altitude Turbine Capacity MW η Cost Million Operating Cost kWh A 50 09 5 001 B 100 092 11 001 C 100 093 105 0015 Final PDF to printer cen22672ch05211270indd 270 110917 1149 AM Final PDF to printer cen22672ch06271322indd 271 110317 0917 AM 271 T H E S E C O N D L AW O F T H E R M O DY N AMICS T o this point we have focused our attention on the first law of thermo dynamics which requires that energy be conserved during a process In this chapter we introduce the second law of thermodynamics which asserts that processes occur in a certain direction and that energy has quality as well as quantity A process cannot take place unless it satisfies both the first and second laws of thermodynamics In this chapter the thermal energy reservoirs reversible and irreversible processes heat engines refrigerators and heat pumps are introduced first Various statements of the second law are followed by a discussion of perpetualmotion machines and the thermody namic temperature scale The Carnot cycle is introduced next and the Carnot principles are discussed Finally the idealized Carnot heat engines refrigera tors and heat pumps are examined CHAPTER6 OBJECTIVES The objectives of Chapter 6 are to Introduce the second law of thermodynamics Identify valid processes as those that satisfy both the first and second laws of thermodynamics Discuss thermal energy reservoirs reversible and irreversible processes heat engines refrigerators and heat pumps Describe the KelvinPlanck and Clausius statements of the second law of thermodynamics Discuss the concepts of perpetualmotion machines Apply the second law of thermodynamics to cycles and cyclic devices Apply the second law to develop the absolute thermodynamic temperature scale Describe the Carnot cycle Examine the Carnot principles idealized Carnot heat engines refrigerators and heat pumps Determine the expressions for the thermal efficiencies and coefficients of performance for reversible heat engines heat pumps and refrigerators Final PDF to printer 272 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 272 110317 0917 AM 61 INTRODUCTION TO THE SECOND LAW In Chaps 4 and 5 we applied the first law of thermodynamics or the conser vation of energy principle to processes involving closed and open systems As pointed out repeatedly in those chapters energy is a conserved property and no process is known to have taken place in violation of the first law of thermodynamics Therefore it is reasonable to conclude that a process must satisfy the first law to occur However as explained here satisfying the first law alone does not ensure that the process will actually take place It is common experience that a cup of hot coffee left in a cooler room even tually cools off Fig 61 This process satisfies the first law of thermody namics since the amount of energy lost by the coffee is equal to the amount gained by the surrounding air Now let us consider the reverse processthe hot coffee getting even hotter in a cooler room as a result of heat transfer from the room air We all know that this process never takes place Yet doing so would not violate the first law as long as the amount of energy lost by the air is equal to the amount gained by the coffee As another familiar example consider the heating of a room by the passage of electric current through a resistor Fig 62 Again the first law dictates that the amount of electric energy supplied to the resistance wires be equal to the amount of energy transferred to the room air as heat Now let us try to reverse this process It will come as no surprise that transferring some heat to the wires does not cause an equivalent amount of electric energy to be gener ated in the wires Finally consider a paddlewheel mechanism that is operated by the fall of a mass Fig 63 The paddle wheel rotates as the mass falls and stirs a fluid within an insulated container As a result the potential energy of the mass decreases and the internal energy of the fluid increases in accordance with the conservation of energy principle However the reverse process raising the mass by transferring heat from the fluid to the paddle wheel does not occur in nature although doing so would not violate the first law of thermodynamics It is clear from these arguments that processes proceed in a certain direction and not in the reverse direction Fig 64 The first law places no restriction on the direction of a process but satisfying the first law does not ensure that the process can actually occur This inadequacy of the first law to identify whether a process can take place is remedied by introducing another general principle the second law of thermodynamics We show later in this chapter that the reverse processes we just mentioned violate the second law of ther modynamics This violation is easily detected with the help of a property called entropy defined in Chap 7 A process cannot occur unless it satisfies both the first and the second laws of thermodynamics Fig 65 There are many valid statements of the second law of thermodynamics Two such statements are presented and discussed later in this chapter in relation to some engineering devices that operate on cycles The use of the second law of thermodynamics is not limited to identifying the direction of processes The second law also asserts that energy has quality as well as quantity The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality Preserving the quality of energy is a major concern to engineers and the second law provides the necessary means to determine the quality as well FIGURE 61 A cup of hot coffee does not get hotter in a cooler room Heat Hot coffee FIGURE 62 Transferring heat to a wire will not generate electricity Heat I 0 FIGURE 63 Transferring heat to a paddle wheel will not cause it to rotate Heat FIGURE 64 Processes occur in a certain direction and not in the reverse direction ONE WAY FIGURE 65 A process must satisfy both the first and second laws of thermodynamics to proceed Process 2nd law 1st law Final PDF to printer 273 CHAPTER 6 cen22672ch06271322indd 273 110317 0917 AM as the degree of degradation of energy during a process As discussed later in this chapter more of hightemperature energy can be converted to work and thus it has a higher quality than the same amount of energy at a lower temperature The second law of thermodynamics is also used in determining the theoretical limits for the performance of commonly used engineering systems such as heat engines and refrigerators as well as predicting the degree of completion of chemical reactions The second law is also closely associated with the concept of perfection In fact the second law defines perfection for thermodynamic processes It can be used to quantify the level of perfection of a process and to point in the direction to eliminate imperfections effectively 62 THERMAL ENERGY RESERVOIRS In the development of the second law of thermodynamics it is very convenient to have a hypothetical body with a relatively large thermal energy capac ity mass specific heat that can supply or absorb finite amounts of heat without undergoing any change in temperature Such a body is called a thermal energy reservoir or just a reservoir In practice large bodies of water such as oceans lakes and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or thermal masses Fig 66 The atmosphere for example does not warm up as a result of heat losses from residential buildings in winter Likewise megajoules of waste energy dumped into large rivers by power plants do not cause any significant change in water temperature A twophase system can also be modeled as a reservoir since it can absorb and release large quantities of heat while remaining at constant temperature Another familiar example of a thermal energy reservoir is the industrial fur nace The temperatures of most furnaces are carefully controlled and they can supply large quantities of thermal energy as heat in an essentially isother mal manner Therefore they can be modeled as reservoirs A body does not actually have to be very large to be considered a reser voir Any physical body whose thermal energy capacity is large relative to the amount of energy it supplies or absorbs can be modeled as one The air in a room for example can be treated as a reservoir in the analysis of the heat dissipation from a TV set in the room since the amount of heat transfer from the TV set to the room air is not large enough to have a noticeable effect on the room air temperature A reservoir that supplies energy in the form of heat is called a source and one that absorbs energy in the form of heat is called a sink Fig 67 Thermal energy reservoirs are often referred to as heat reservoirs since they supply or absorb energy in the form of heat Heat transfer from industrial sources to the environment is of major con cern to environmentalists as well as to engineers Irresponsible management of waste energy can significantly increase the temperature of portions of the environment causing what is called thermal pollution If it is not carefully controlled thermal pollution can seriously disrupt marine life in lakes and riv ers However by careful design and management the waste energy dumped into large bodies of water can be used to improve the quality of marine life by keeping the local temperature increases within safe and desirable levels FIGURE 66 Bodies with relatively large thermal masses can be modeled as thermal energy reservoirs Atmosphere River Ocean Lake FIGURE 67 A source supplies energy in the form of heat and a sink absorbs it Thermal energy Sink Thermal energy Source Heat Heat Final PDF to printer 274 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 274 110317 0917 AM 63 HEAT ENGINES As pointed out earlier work can easily be converted to other forms of energy but converting other forms of energy to work is not that easy The mechanical work done by the shaft shown in Fig 68 for example is first converted to the internal energy of the water This energy may then leave the water as heat We know from experience that any attempt to reverse this process will fail That is transferring heat to the water does not cause the shaft to rotate From this and other observations we conclude that work can be converted to heat directly and completely but converting heat to work requires the use of some special devices These devices are called heat engines Heat engines differ considerably from one another but all can be character ized by the following Fig 69 1 They receive heat from a hightemperature source solar energy oil furnace nuclear reactor etc 2 They convert part of this heat to work usually in the form of a rotating shaft 3 They reject the remaining waste heat to a lowtemperature sink the atmosphere rivers etc 4 They operate on a cycle Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle This fluid is called the working fluid The term heat engine is often used in a broader sense to include work producing devices that do not operate in a thermodynamic cycle Engines that involve internal combustion such as gas turbines and car engines fall into this category These devices operate in a mechanical cycle but not in a thermodynamic cycle since the working fluid the combustion gases does not undergo a complete cycle Instead of being cooled to the initial temperature the exhaust gases are purged and replaced by fresh airandfuel mixture at the end of the cycle The workproducing device that best fits into the definition of a heat engine is the steam power plant which is an externalcombustion engine That is combustion takes place outside the engine and the thermal energy released during this process is transferred to the steam as heat The schematic of a basic steam power plant is shown in Fig 610 This is a rather simplified diagram and the discussion of actual steam power plants is given in later chapters The various quantities shown on this figure are as follows Q in amount of heat supplied to steam in boiler from a hightemperature source furnace Q out amount of heat rejected from steam in condenser to a low temperature sink the atmosphere a river etc W out amount of work delivered by steam as it expands in turbine W in amount of work required to compress water to boiler pressure Notice that the directions of the heat and work interactions are indicated by the subscripts in and out Therefore all four of the described quantities are always positive FIGURE 68 Work can always be converted to heat directly and completely but the reverse is not true Heat Work Water Water Heat No work FIGURE 69 Part of the heat received by a heat engine is converted to work while the rest is rejected to a sink Lowtemperature Sink Wnetout Qout Qin Hightemperature Source Heat engine Final PDF to printer 275 CHAPTER 6 cen22672ch06271322indd 275 110317 0917 AM The net work output of this power plant is simply the difference between the total work output of the plant and the total work input Fig 611 W netout W out W in kJ 61 The net work can also be determined from the heat transfer data alone The four components of the steam power plant involve mass flow in and out and therefore should be treated as open systems These components together with the connecting pipes however always contain the same fluid not counting the steam that may leak out of course No mass enters or leaves this combination system which is indicated by the shaded area on Fig 610 thus it can be analyzed as a closed system Recall that for a closed system undergoing a cycle the change in internal energy ΔU is zero and therefore the net work output of the system is also equal to the net heat transfer to the system W netout Q in Q out kJ 62 Thermal Efficiency In Eq 62 Qout represents the magnitude of the energy wasted in order to complete the cycle But Qout is never zero thus the net work output of a heat engine is always less than the amount of heat input That is only part of the heat transferred to the heat engine is converted to work The fraction of the heat input that is converted to net work output is a measure of the performance of a heat engine and is called the thermal efficiency ηth Fig 612 FIGURE 610 Schematic of a steam power plant System boundary Qout Qin Energy source such as a furnace Energy sink such as the atmosphere Boiler Turbine Win Wout Condenser Pump FIGURE 611 A portion of the work output of a heat engine is consumed internally to main tain continuous operation Heat engine Wout Wnetout Win Final PDF to printer 276 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 276 110317 0917 AM For heat engines the desired output is the net work output and the required input is the amount of heat supplied to the working fluid Then the thermal efficiency of a heat engine can be expressed as Thermal efficiency Net work output Total heat input 63 or η th W netout Q in 64 It can also be expressed as η th 1 Q out Q in 65 since Wnetout Qin Qout Cyclic devices of practical interest such as heat engines refrigerators and heat pumps operate between a hightemperature medium or reservoir at temperature TH and a lowtemperature medium or reservoir at temperature TL To bring uniformity to the treatment of heat engines refrigerators and heat pumps we define these two quantities Q H magnitude of heat transfer between the cyclic device and the high temperature medium at temperature TH Q L magnitude of heat transfer between the cyclic device and the low temperature medium at temperature TL Notice that both QL and QH are defined as magnitudes and therefore are positive quantities The direction of QH and QL is easily determined by inspection Then the net work output and thermal efficiency relations for any heat engine shown in Fig 613 can also be expressed as W netout Q H Q L and η th W netout Q H or η th 1 Q L Q H 66 The thermal efficiency of a heat engine is always less than unity since both QL and QH are defined as positive quantities Thermal efficiency is a measure of how efficiently a heat engine converts the heat that it receives to work Heat engines are built for the purpose of converting heat to work and engineers are constantly trying to improve the efficiencies of these devices since increased efficiency means less fuel con sumption and thus lower fuel bills and less pollution The thermal efficiencies of workproducing devices are relatively low Ordinary sparkignition automobile engines have a thermal efficiency of about 25 percent That is an automobile engine converts about 25 percent of the chemical energy of the gasoline to mechanical work This number is as high as 40 percent for diesel engines and large gasturbine plants and as high FIGURE 612 Some heat engines perform better than others convert more of the heat they receive to work Sink Waste heat 80 kJ Waste heat 70 kJ ηth1 20 ηth2 30 Heat input 100 kJ 100 kJ Net work output 20 kJ Net work output 30 kJ 1 2 Source FIGURE 613 Schematic of a heat engine Lowtemperature reservoir at TL QH Wnetout QL Hightemperature reservoir at TH HE Final PDF to printer 277 CHAPTER 6 cen22672ch06271322indd 277 110317 0917 AM as 60 percent for large combined gassteam power plants Thus even with the most efficient heat engines available today almost onehalf of the energy sup plied ends up in the rivers lakes or the atmosphere as waste or useless energy Fig 614 Can We Save Qout In a steam power plant the condenser is the device where large quantities of waste heat are rejected to rivers lakes or the atmosphere Then one may ask can we not just take the condenser out of the plant and save all that waste energy The answer to this question is unfortunately a firm no for the simple reason that without a heat rejection process in a condenser the cycle cannot be completed Cyclic devices such as steam power plants cannot run continu ously unless the cycle is completed This is demonstrated next with the help of a simple heat engine Consider the simple heat engine shown in Fig 615 that is used to lift weights It consists of a pistoncylinder device with two sets of stops The working fluid is the gas contained within the cylinder Initially the gas tem perature is 30C The piston which is loaded with the weights is resting on top of the lower stops Now 100 kJ of heat is transferred to the gas in the cylinder from a source at 100C causing it to expand and to raise the loaded piston until the piston reaches the upper stops as shown in the figure At this point the load is removed and the gas temperature is observed to be 90C The work done on the load during this expansion process is equal to the increase in its potential energy say 15 kJ Even under ideal conditions weight less piston no friction no heat losses and quasiequilibrium expansion the amount of heat supplied to the gas is greater than the work done since part of the heat supplied is used to raise the temperature of the gas Now let us try to answer this question Is it possible to transfer the 85 kJ of excess heat at 90C back to the reservoir at 100C for later use If it is then we will have a heat engine that can have a thermal efficiency of 100 percent under ideal conditions The answer to this question is again no for the very simple reason that heat is always transferred from a hightemperature FIGURE 614 Even the most efficient heat engines reject almost onehalf of the energy they receive as waste heat The atmosphere QH 100 MJ Wnetout 55 MJ QL 45 MJ Furnace HE FIGURE 615 A heatengine cycle cannot be completed without rejecting some heat to a lowtemperature sink Gas 30C Heat in 100 kJ Reservoir at 100C Load Gas 90C Load Gas 30C Heat out 85 kJ Reservoir at 20C 15 kJ Final PDF to printer 278 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 278 110317 0917 AM medium to a lowtemperature one and never the other way around Therefore we cannot cool this gas from 90 to 30C by transferring heat to a reservoir at 100C Instead we have to bring the system into contact with a low temperature reservoir say at 20C so that the gas can return to its initial state by rejecting its 85 kJ of excess energy as heat to this reservoir This energy cannot be recycled and it is properly called waste energy We conclude from this discussion that every heat engine must waste some energy by transferring it to a lowtemperature reservoir in order to complete the cycle even under idealized conditions The requirement that a heat engine exchange heat with at least two reservoirs for continuous operation forms the basis for the KelvinPlanck expression of the second law of thermody namics discussed later in this section EXAMPLE 62 Fuel Consumption Rate of a Car A car engine with a power output of 65 hp has a thermal efficiency of 24 percent Determine the fuel consumption rate of this car if the fuel has a heating value of 19000 Btulbm that is 19000 Btu of energy is released for each lbm of fuel burned SOLUTION The power output and the efficiency of a car engine are given The rate of fuel consumption of the car is to be determined Assumptions The power output of the car is constant FIGURE 616 Schematic for Example 61 River QH 80 MW Wnetout QL 50 MW Furnace HE EXAMPLE 61 Net Power Production of a Heat Engine Heat is transferred to a heat engine from a furnace at a rate of 80 MW If the rate of waste heat rejection to a nearby river is 50 MW determine the net power output and the thermal efficiency for this heat engine SOLUTION The rates of heat transfer to and from a heat engine are given The net power output and the thermal efficiency are to be determined Assumptions Heat losses through the pipes and other components are negligible Analysis A schematic of the heat engine is given in Fig 616 The furnace serves as the hightemperature reservoir for this heat engine and the river as the lowtemperature reservoir The given quantities can be expressed as Q H 80 MW and Q L 50 MW The net power output of this heat engine is W netout Q H Q L 80 50 MW 30 MW Then the thermal efficiency is easily determined to be η th W netout Q H 30 MW 80 MW 0375 or 375 percent Discussion Note that the heat engine converts 375 percent of the heat it receives to work Final PDF to printer 279 CHAPTER 6 cen22672ch06271322indd 279 110317 0917 AM The Second Law of Thermodynamics KelvinPlanck Statement We demonstrated earlier with reference to the heat engine shown in Fig 615 that even under ideal conditions a heat engine must reject some heat to a low temperature reservoir in order to complete the cycle That is no heat engine can convert all the heat it receives to useful work This limitation on the ther mal efficiency of heat engines forms the basis for the KelvinPlanck state ment of the second law of thermodynamics which is expressed as follows It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work That is a heat engine must exchange heat with a lowtemperature sink as well as a hightemperature source to keep operating The KelvinPlanck statement can also be expressed as no heat engine can have a thermal efficiency of 100 percent Fig 618 or as for a power plant to operate the working fluid must exchange heat with the environment as well as the furnace Note that the impossibility of having a 100 percent efficient heat engine is not due to friction or other dissipative effects It is a limitation that applies to both the idealized and the actual heat engines Later in this chapter we develop a relation for the maximum thermal efficiency of a heat engine We also demonstrate that this maximum value depends on the reservoir tem peratures only 64 REFRIGERATORS AND HEAT PUMPS We all know from experience that heat is transferred in the direction of decreasing temperature that is from hightemperature mediums to low temperature ones This heat transfer process occurs in nature without requir ing any devices The reverse process however cannot occur by itself The transfer of heat from a lowtemperature medium to a hightemperature one requires special devices called refrigerators FIGURE 617 Schematic for Example 62 Atmosphere QH Wnetout 65 hp QL Combustion chamber Car engine idealized mfuel FIGURE 618 A heat engine that violates the KelvinPlanck statement of the second law QH 100 kW Wnetout 100 kW QL 0 Thermal energy reservoir Heat engine Analysis A schematic of the car engine is given in Fig 617 The car engine is powered by converting 24 percent of the chemical energy released during the combustion process to work The amount of energy input required to produce a power output of 65 hp is determined from the definition of thermal efficiency to be Q H W netout η th 65 hp 024 2545 Btu h 1 hp 689270 Btu h To supply energy at this rate the engine must burn fuel at a rate of m fuel 689270 Btu h 19000 Btu lbm 363 lbm h since 19000 Btu of thermal energy is released for each lbm of fuel burned Discussion Note that if the thermal efficiency of the car could be doubled the rate of fuel consumption would be reduced by half Final PDF to printer 280 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 280 110317 0917 AM Refrigerators like heat engines are cyclic devices The working fluid used in the refrigeration cycle is called a refrigerant The most frequently used refrigeration cycle is the vaporcompression refrigeration cycle which involves four main components a compressor a condenser an expansion valve and an evaporator as shown in Fig 619 The refrigerant enters the compressor as a vapor and is compressed to the condenser pressure It leaves the compressor at a relatively high temperature and cools down and condenses as it flows through the coils of the condenser by rejecting heat to the surrounding medium It then enters a capillary tube where its pressure and temperature drop drastically due to the throttling effect The lowtemperature refrigerant then enters the evaporator where it evapo rates by absorbing heat from the refrigerated space The cycle is completed as the refrigerant leaves the evaporator and reenters the compressor In a household refrigerator the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator and the coils usually behind the refrigerator where heat is dissipated to the kitchen air serve as the condenser A refrigerator is shown schematically in Fig 620 Here QL is the magni tude of the heat removed from the refrigerated space at temperature TL QH is the magnitude of the heat rejected to the warm environment at temperature TH and Wnetin is the net work input to the refrigerator As discussed before QL and QH represent magnitudes and thus are positive quantities Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance COP denoted by COPR The objective of a refrigerator is to remove heat QL from the refrigerated space To accomplish this FIGURE 619 Basic components of a refrigeration system and typical operating conditions Surrounding medium such as the kitchen air QL QH Expansion valve 120 kPa 25C 800 kPa 30C 800 kPa 60C 120 kPa 25C Compressor Refrigerated space Evaporator Wnetin Condenser FIGURE 620 The objective of a refrigerator is to remove QL from the cooled space Cold refrigerated space at TL Warm environment at TH TL R Wnetin QH QL Required input Desired output Final PDF to printer 281 CHAPTER 6 cen22672ch06271322indd 281 110317 0917 AM objective it requires a work input of Wnetin Then the COP of a refrigerator can be expressed as COP R Desired output Required input Q L W netin 67 This relation can also be expressed in rate form by replacing QL with Q L and Wnetin with W netin The conservation of energy principle for a cyclic device requires that W netin Q H Q L kJ 68 Then the COP relation becomes COP R Q L Q H Q L 1 Q H Q L 1 69 Notice that the value of COPR can be greater than unity That is the amount of heat removed from the refrigerated space can be greater than the amount of work input This is in contrast to the thermal efficiency which can never be greater than 1 In fact one reason for expressing the efficiency of a refrigera tor with another termthe coefficient of performanceis the desire to avoid the oddity of having efficiencies greater than unity Heat Pumps Another device that transfers heat from a lowtemperature medium to a high temperature one is the heat pump shown schematically in Fig 621 Refrig erators and heat pumps operate on the same cycle but differ in their objectives The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it Discharging this heat to a higher temperature medium is merely a necessary part of the operation not the purpose The objective of a heat pump however is to maintain a heated space at a high temperature This is accomplished by absorbing heat from a low temperature source such as well water or cold outside air in winter and supplying this heat to the hightemperature medium such as a house Fig 622 An ordinary refrigerator that is placed in the window of a house with its door open to the cold outside air in winter will function as a heat pump since it will try to cool the outside by absorbing heat from it and rejecting this heat into the house through the coils behind it The measure of performance of a heat pump is also expressed in terms of the coefficient of performance COPHP defined as COP HP Desired output Required input Q H W netin 610 which can also be expressed as COP HP Q H Q H Q L 1 1 Q L Q H 611 A comparison of Eqs 67 and 610 reveals that COP HP COP R 1 612 FIGURE 621 The objective of a heat pump is to sup ply heat QH into the warmer space Warm heated space at TH TL HP Wnetin QH QL Required input Desired output Cold environment at TL FIGURE 622 The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors Warm indoors at 20C HP Wnetin 2 kJ QH 7 kJ QL 5 kJ COP 35 Cold outdoors at 4C Final PDF to printer 282 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 282 110317 0917 AM for fixed values of QL and QH This relation implies that the coefficient of performance of a heat pump is always greater than unity since COPR is a posi tive quantity That is a heat pump will function at worst as a resistance heater supplying as much energy to the house as it consumes In reality however part of QH is lost to the outside air through piping and other devices and COPHP may drop below unity when the outside air temperature is too low When this happens the system usually switches to a resistance heating mode Most heat pumps in operation today have a seasonally averaged COP of 2 to 3 Most existing heat pumps use the cold outside air as the heat source in winter and they are referred to as airsource heat pumps The COP of such heat pumps is about 30 at design conditions Airsource heat pumps are not appropriate for cold climates since their efficiency drops considerably when temperatures are below the freezing point In such cases geothermal also called groundsource heat pumps that use the ground as the heat source can be used Geothermal heat pumps require the burial of pipes in the ground 1 to 2 m deep Such heat pumps are more expensive to install but they are also more efficient up to 45 percent more efficient than airsource heat pumps The COP of groundsource heat pumps can be as high as 6 or more Air conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment A window air conditioning unit cools a room by absorbing heat from the room air and discharging it to the outside The same airconditioning unit can be used as a heat pump in winter by installing it backwards In this mode the unit absorbs heat from the cold outside and delivers it to the room Airconditioning systems that are equipped with proper controls and a reversing valve operate as air conditioners in summer and as heat pumps in winter Performance of Refrigerators Air Conditioners and Heat Pumps The performance of air conditioners and heat pumps is often expressed in terms of the energy efficiency ratio EER or seasonal energy efficiency ratio SEER determined by following certain testing standards SEER is the ratio of the total amount of heat removed by an air conditioner or heat pump during a normal cooling season in Btu to the total amount of electricity consumed in watthours Wh and it is a measure of seasonal performance of cooling equipment EER on the other hand is a measure of the instantaneous energy efficiency and is defined as the ratio of the rate of heat removal from the cooled space by the cooling equipment to the rate of electricity consumption in steady operation Therefore both EER and SEER have the unit BtuWh Considering that 1 kWh 3412 Btu and thus 1 Wh 3412 Btu a device that removes 1 kWh of heat from the cooled space for each kWh of electric ity it consumes COP 1 will have an EER of 3412 Therefore the relation between EER or SEER and COP is EER 3 412 COP R To promote the efficient use of energy governments worldwide have man dated minimum standards for the performance of energyconsuming equip ment Most air conditioners or heat pumps in the market have SEER values Final PDF to printer 283 CHAPTER 6 cen22672ch06271322indd 283 110317 0917 AM from 13 to 21 which correspond to COP values of 38 to 62 Best perfor mance is achieved using units equipped with variablespeed drives also called inverters Variablespeed compressors and fans allow the unit to oper ate at maximum efficiency for varying heatingcooling needs and weather conditions as determined by a microprocessor In the airconditioning mode for example they operate at higher speeds on hot days and at lower speeds on cooler days enhancing both efficiency and comfort The EER or COP of a refrigerator decreases with decreasing refrigeration temperature Therefore it is not economical to refrigerate to a lower tem perature than needed The COPs of refrigerators are in the range of 2630 for cutting and preparation rooms 2326 for meat deli dairy and produce 1215 for frozen foods and 1012 for ice cream units Note that the COP of freezers is about half of the COP of meat refrigerators and thus it costs twice as much to cool the meat products with refrigerated air that is cold enough to cool frozen foods It is good energy conservation practice to use separate refrigeration systems to meet different refrigeration needs EXAMPLE 63 Analysis of a Household Refrigerator A household refrigerator with a COP of 12 removes heat from the refrigerated space at a rate of 60 kJmin Fig 623 Determine a the electric power consumed by the refrigerator and b the rate of heat transfer to the kitchen air SOLUTION The COP and the refrigeration rate of a refrigerator are given The power consumption and the rate of heat rejection are to be determined Assumptions The refrigerator operates steadily Analysis a Using the definition of the coefficient of performance the power input to the refrigerator is determined to be W netin Q L COP R 60 kJmin 12 50 kJmin 0833 kW b The heat transfer rate to the kitchen air is determined from the energy balance Q H Q L W netin 60 50 110 kJ min Discussion Notice that both the energy removed from the refrigerated space as heat and the energy supplied to the refrigerator as electrical work eventually show up in the room air and become part of the internal energy of the air This demonstrates that energy can change from one form to another and can move from one place to another but it is never destroyed during a process FIGURE 623 Schematic for Example 63 Wnetin QH QL 60 kJmin COP 12 Refrigerator Kitchen R EXAMPLE 64 Heating a House with a Heat Pump A heat pump is used to meet the heating requirements of a house and maintain it at 20C On a day when the outdoor air temperature drops to 2C the house is estimated to lose heat at a rate of 80000 kJh If the heat pump under these conditions has a COP of 25 determine a the power consumed by the heat pump and b the rate at which heat is absorbed from the cold outdoor air Final PDF to printer 284 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 284 110317 0917 AM The Second Law of Thermodynamics Clausius Statement There are two classic statements of the second lawthe KelvinPlanck state ment which is related to heat engines and discussed in the preceding section and the Clausius statement which is related to refrigerators or heat pumps The Clausius statement is expressed as follows It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lowertemperature body to a highertemperature body It is common knowledge that heat does not of its own volition transfer from a cold medium to a warmer one The Clausius statement does not imply that a cyclic device that transfers heat from a cold medium to a warmer one is impossible to construct In fact this is precisely what a common household refrigerator does It simply states that a refrigerator cannot operate unless its compressor is driven by an external power source such as an electric motor Fig 625 This way the net effect on the surroundings involves the con sumption of some energy in the form of work in addition to the transfer of heat from a colder body to a warmer one That is it leaves a trace in the surroundings Therefore a household refrigerator is in complete compliance with the Clausius statement of the second law Both the KelvinPlanck and the Clausius statements of the second law are negative statements and a negative statement cannot be proved Like any FIGURE 624 Schematic for Example 64 House 20C Heat loss 80000 kJh Wnetin COP 25 QH QL Outdoor air at 2C HP SOLUTION The COP of a heat pump is given The power consumption and the rate of heat absorption are to be determined Assumptions Steady operating conditions exist Analysis a The power consumed by this heat pump shown in Fig 624 is deter mined from the definition of the coefficient of performance to be W netin Q H COP HP 80000 kJ h 25 32000 kJ h or 89 kW b The house is losing heat at a rate of 80000 kJh If the house is to be maintained at a constant temperature of 20C the heat pump must deliver heat to the house at the same rate that is at a rate of 80000 kJh Then the rate of heat transfer from the outdoor becomes Q L Q H W netin 80000 32000 kJ h 48000 kJ h Discussion Note that 48000 of the 80000 kJh heat delivered to the house is actually extracted from the cold outdoor air Therefore we are paying only for the 32000 kJh energy that is supplied as electrical work to the heat pump If we were to use an electric resistance heater instead we would have to supply the entire 80000 kJh to the resistance heater as electric energy This would mean a heating bill that is 25 times higher This explains the popularity of heat pumps as heating systems and why they are preferred to simple electric resistance heaters despite their considerably higher initial cost FIGURE 625 A refrigerator that violates the Clausius statement of the second law Warm environment R Wnetin 0 QH 5 kJ QL 5 kJ Cold refrigerated space Final PDF to printer 285 CHAPTER 6 cen22672ch06271322indd 285 110317 0917 AM other physical law the second law of thermodynamics is based on experimen tal observations To date no experiment has been conducted that contradicts the second law and this should be taken as sufficient proof of its validity Equivalence of the Two Statements The KelvinPlanck and the Clausius statements are equivalent in their conse quences and either statement can be used as the expression of the second law of thermodynamics Any device that violates the KelvinPlanck statement also violates the Clausius statement and vice versa This can be demonstrated as follows Consider the heat enginerefrigerator combination shown in Fig 626a operating between the same two reservoirs The heat engine is assumed to have in violation of the KelvinPlanck statement a thermal efficiency of 100 percent and therefore it converts all the heat QH it receives to work W This work is now supplied to a refrigerator that removes heat in the amount of QL from the lowtemperature reservoir and rejects heat in the amount of QL QH to the hightemperature reservoir During this process the high temperature reservoir receives a net amount of heat QL the difference between QL QH and QH Thus the combination of these two devices can be viewed as a refrigerator as shown in Fig 626b that transfers heat in an amount of QL from a cooler body to a warmer one without requiring any input from outside This is clearly a violation of the Clausius statement Therefore a violation of the KelvinPlanck statement results in the violation of the Clausius statement FIGURE 626 Proof that the violation of the KelvinPlanck statement leads to the violation of the Clausius statement Hightemperature reservoir at TH Heat engine ηth 100 Wnet QH QL QH QL QH Refrigerator QL QL Lowtemperature reservoir at TL a A refrigerator that is powered by a 100 percent efficient heat engine Lowtemperature reservoir at TL b The equivalent refrigerator Hightemperature reservoir at TH Refrigerator Final PDF to printer 286 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 286 110317 0917 AM It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the KelvinPlanck statement Therefore the Clausius and the KelvinPlanck statements are two equivalent expressions of the second law of thermodynamics 65 PERPETUALMOTION MACHINES We have repeatedly stated that a process cannot take place unless it satis fies both the first and second laws of thermodynamics Any device that vio lates either law is called a perpetualmotion machine and despite numerous attempts no perpetualmotion machine is known to have worked But this has not stopped inventors from trying to create new ones A device that violates the first law of thermodynamics by creating energy is called a perpetualmotion machine of the first kind PMM1 and a device that violates the second law of thermodynamics is called a perpetualmotion machine of the second kind PMM2 Consider the steam power plant shown in Fig 627 It is proposed to heat the steam by resistance heaters placed inside the boiler instead of by the energy supplied from fossil or nuclear fuels Part of the electricity generated by the plant is to be used to power the resistors as well as the pump The rest of the electric energy is to be supplied to the electric network as the net work output The inventor claims that once the system is started this power plant will produce electricity indefinitely without requiring any energy input from the outside Well here is an invention that could solve the worlds energy problemif it works of course A careful examination of this invention reveals that the system enclosed by the shaded area is continuously supplying energy to the outside at a rate of Q out W netout without receiving any energy That is this FIGURE 627 A perpetualmotion machine that violates the first law of thermodynamics PMM1 System boundary Boiler Turbine Condenser Pump Generator Resistance heater Wnetout Qout Final PDF to printer 287 CHAPTER 6 cen22672ch06271322indd 287 110317 0917 AM system is creating energy at a rate of Q out W netout which is clearly a viola tion of the first law Therefore this wonderful device is nothing more than a PMM1 and does not warrant any further consideration Now let us consider another novel idea by the same inventor Convinced that energy cannot be created the inventor suggests the following modifi cation that will greatly improve the thermal efficiency of that power plant without violating the first law Aware that more than onehalf of the heat transferred to the steam in the furnace is discarded in the condenser to the environment the inventor suggests getting rid of this wasteful component and sending the steam to the pump as soon as it leaves the turbine as shown in Fig 628 This way all the heat transferred to the steam in the boiler will be converted to work and thus the power plant will have a theoretical effi ciency of 100 percent The inventor realizes that some heat losses and friction between the moving components are unavoidable and that these effects will hurt the efficiency somewhat but still expects the efficiency to be no less than 80 percent as opposed to 40 percent in most actual power plants for a carefully designed system Well the possibility of doubling the efficiency would certainly be very tempting to plant managers and if not properly trained they would probably give this idea a chance since intuitively they see nothing wrong with it A student of thermodynamics however will immediately label this device as a PMM2 since it works on a cycle and does a net amount of work while exchanging heat with a single reservoir the furnace only It satisfies the first law but violates the second law and therefore it will not work Countless perpetualmotion machines have been proposed throughout his tory with many more still being proposed Some proposers have even gone so far as to patent their inventions only to find out that what they actually have in their hands is a worthless piece of paper Some perpetualmotion machine inventors were very successful in fund raising For example a Philadelphia carpenter named J W Kelly col lected millions of dollars between 1874 and 1898 from investors in his hydropneumatic pulsatingvacuengine which supposedly could push a railroad train 3000 miles on 1 L of water Of course it never did After his death in 1898 the investigators discovered that the demonstration machine was powered by a hidden motor In another case a group of investors was set to invest 25 million in a mysterious energy augmentor which multiplied whatever power it took in but their lawyer wanted an expert opinion first Confronted by the scientists the inventor fled the scene without even trying to run his demo machine Tired of applications for perpetualmotion machines the US Patent Office decreed in 1918 that it would no longer consider any perpetualmotion machine applications However several such patent applications were still filed and some made it through the patent office undetected Some applicants whose patent applications were denied sought legal action For example in 1982 the US Patent Office dismissed as just another perpetualmotion machine a huge device that involves several hundred kilograms of rotating magnets and kilometers of copper wire that is supposed to be generating more electricity than it is consuming from a battery pack However the inventor challenged the decision and in 1985 the National Bureau of Standards finally tested the machine just to certify that it is batteryoperated However it did not convince the inventor that his machine will not work FIGURE 628 A perpetualmotion machine that violates the second law of thermodynamics PMM2 System boundary Boiler Turbine Pump Wnetout Qin Final PDF to printer 288 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 288 110317 0917 AM The proposers of perpetualmotion machines generally have innovative minds but they usually lack formal engineering training which is very unfor tunate No one is immune from being deceived by an innovative perpetual motion machine As the saying goes however if something sounds too good to be true it probably is 66 REVERSIBLE AND IRREVERSIBLE PROCESSES The second law of thermodynamics states that no heat engine can have an efficiency of 100 percent Then one may ask what is the highest efficiency that a heat engine can possibly have Before we can answer this question we need to define an idealized process first which is called the reversible process The processes that were discussed at the beginning of this chapter occurred in a certain direction Once having taken place these processes cannot reverse themselves spontaneously and restore the system to its initial state For this reason they are classified as irreversible processes Once a cup of hot coffee cools it will not heat up by retrieving the heat it lost from the surroundings If it could the surroundings as well as the system coffee would be restored to their original condition and this would be a reversible process A reversible process is defined as a process that can be reversed with out leaving any trace on the surroundings Fig 629 That is both the system and the surroundings are returned to their initial states at the end of the reverse process This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined original and reverse process Processes that are not reversible are called irreversible processes It should be pointed out that a system can be restored to its initial state fol lowing a process regardless of whether the process is reversible or irrevers ible But for reversible processes this restoration is made without leaving any net change on the surroundings whereas for irreversible processes the sur roundings usually do some work on the system and therefore do not return to their original state Reversible processes actually do not occur in nature They are merely ide alizations of actual processes Reversible processes can be approximated by actual devices but they can never be achieved That is all the processes occurring in nature are irreversible You may be wondering then why we are bothering with such fictitious processes There are two reasons First they are easy to analyze since a system passes through a series of equilibrium states during a reversible process Second they serve as idealized models to which actual processes can be compared In daily life the concepts of Mr Right and Ms Right are also idealizations just like the concept of a reversible perfect process People who insist on finding Mr or Ms Right to settle down with are bound to remain Mr or Ms Single for the rest of their lives The possibility of finding the perfect prospec tive mate is no higher than the possibility of finding a perfect reversible process Likewise a person who insists on perfection in friends is bound to have no friends FIGURE 629 Two familiar reversible processes a Frictionless pendulum b Quasiequilibrium expansion and compression of a gas Final PDF to printer 289 CHAPTER 6 cen22672ch06271322indd 289 110317 0917 AM Engineers are interested in reversible processes because workproducing devices such as car engines and gas or steam turbines deliver the most work and workconsuming devices such as compressors fans and pumps consume the least work when reversible processes are used instead of irreversible ones Fig 630 Reversible processes can be viewed as theoretical limits for the correspond ing irreversible ones Some processes are more irreversible than others We may never be able to have a reversible process but we can certainly approach it The more closely we approximate a reversible process the more work delivered by a workproducing device or the less work required by a work consuming device The concept of reversible processes leads to the definition of the secondlaw efficiency for actual processes which is the degree of approxi mation to the corresponding reversible processes This enables us to compare the performance of different devices that are designed to do the same task on the basis of their efficiencies The better the design the lower the irrevers ibilities and the higher the secondlaw efficiency Irreversibilities The factors that cause a process to be irreversible are called irreversibilities They include friction unrestrained expansion mixing of two fluids heat transfer across a finite temperature difference electric resistance inelastic deformation of solids and chemical reactions The presence of any of these effects renders a process irreversible A reversible process involves none of these Some of the frequently encountered irreversibilities are discussed briefly below Friction is a familiar form of irreversibility associated with bodies in motion When two bodies in contact are forced to move relative to each other a piston in a cylinder for example as shown in Fig 631 a friction force that opposes the motion develops at the interface of these two bodies and some work is needed to overcome this friction force The energy supplied as work is eventually converted to heat during the process and is transferred to the bodies in contact as evidenced by a temperature rise at the interface When the direction of the motion is reversed the bodies are restored to their original position but the interface does not cool and heat is not converted back to work Instead more of the work is converted to heat while overcom ing the friction forces that also oppose the reverse motion Since the system the moving bodies and the surroundings cannot be returned to their original states this process is irreversible Therefore any process that involves friction is irreversible The larger the friction forces involved the more irreversible the process is Friction does not always involve two solid bodies in contact It is also encountered between a fluid and a solid and even between the layers of a fluid moving at different velocities A considerable fraction of the power produced by a car engine is used to overcome the friction the drag force between the air and the external surfaces of the car and it eventually becomes part of the internal energy of the air It is not possible to reverse this process and recover that lost power even though doing so would not violate the conservation of energy principle FIGURE 630 Reversible processes deliver the most and consume the least work Water Pressure distribution Water Water Water a Slow reversible process b Fast irreversible process Expansion Compression Expansion Compression FIGURE 631 Friction renders a process irreversible Gas Friction Final PDF to printer 290 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 290 110317 0917 AM Another example of irreversibility is the unrestrained expansion of a gas separated from a vacuum by a membrane as shown in Fig 632 When the membrane is ruptured the gas fills the entire tank The only way to restore the system to its original state is to compress it to its initial volume while transferring heat from the gas until it reaches its initial temperature From the conservation of energy considerations it can easily be shown that the amount of heat transferred from the gas equals the amount of work done on the gas by the surroundings The restoration of the surroundings involves conversion of this heat completely to work which would violate the second law Therefore unrestrained expansion of a gas is an irreversible process A third form of irreversibility familiar to us all is heat transfer through a finite temperature difference Consider a can of cold soda left in a warm room Fig 633 Heat is transferred from the warmer room air to the cooler soda The only way this process can be reversed and the soda restored to its original temperature is to provide refrigeration which requires some work input At the end of the reverse process the soda will be restored to its initial state but the surroundings will not be The internal energy of the surround ings will increase by an amount equal in magnitude to the work supplied to the refrigerator The restoration of the surroundings to the initial state can be done only by converting this excess internal energy completely to work which is impossible to do without violating the second law Since only the system not both the system and the surroundings can be restored to its initial condition heat transfer through a finite temperature difference is an irreversible process Heat transfer can occur only when there is a temperature difference between a system and its surroundings Therefore it is physically impossible to have a reversible heat transfer process But a heat transfer process becomes less and less irreversible as the temperature difference between the two bodies approaches zero Then heat transfer through a differential temperature dif ference dT can be considered to be reversible As dT approaches zero the process can be reversed in direction at least theoretically without requiring any refrigeration Notice that reversible heat transfer is a conceptual process and cannot be duplicated in the real world The smaller the temperature difference between two bodies the smaller the heat transfer rate will be Any significant heat transfer through a small temperature difference requires a very large surface area and a very long time Therefore even though approaching reversible heat transfer is desirable from a thermodynamic point of view it is impractical and not economically feasible Internally and Externally Reversible Processes A typical process involves interactions between a system and its surround ings and a reversible process involves no irreversibilities associated with either of them A process is called internally reversible if no irreversibilities occur within the boundaries of the system during the process During an internally revers ible process a system proceeds through a series of equilibrium states and when the process is reversed the system passes through exactly the same FIGURE 632 Irreversible compression and expansion processes a Fast compression b Fast expansion 50 kPa 700 kPa c Unrestrained expansion FIGURE 633 a Heat transfer through a temperature difference is irreversible and b the reverse process is impossible 5C 20C 20C 2C 5C 20C a An irreversible heat transfer process Heat b An impossible heat transfer process Heat Final PDF to printer 291 CHAPTER 6 cen22672ch06271322indd 291 110317 0917 AM equilibrium states while returning to its initial state That is the paths of the forward and reverse processes coincide for an internally reversible pro cess The quasiequilibrium process is an example of an internally reversible process A process is called externally reversible if no irreversibilities occur outside the system boundaries during the process Heat transfer between a reservoir and a system is an externally reversible process if the outer surface of the system is at the temperature of the reservoir A process is called totally reversible or simply reversible if it involves no irreversibilities within the system or its surroundings Fig 634 A totally reversible process involves no heat transfer through a finite temperature difference no nonquasiequilibrium changes and no friction or other dissipa tive effects As an example consider the transfer of heat to two identical systems that are undergoing a constantpressure thus constanttemperature phasechange process as shown in Fig 635 Both processes are internally reversible since both take place isothermally and both pass through exactly the same equilib rium states The first process shown is externally reversible also since heat transfer for this process takes place through an infinitesimal temperature dif ference dT The second process however is externally irreversible since it involves heat transfer through a finite temperature difference ΔT 67 THE CARNOT CYCLE We mentioned earlier that heat engines are cyclic devices and that the work ing fluid of a heat engine returns to its initial state at the end of each cycle Work is done by the working fluid during one part of the cycle and on the working fluid during another part The difference between these two is the net work delivered by the heat engine The efficiency of a heatengine cycle greatly depends on how the individual processes that make up the cycle are executed The net work thus the cycle efficiency can be maximized by using processes that require the least amount of work and deliver the most that is by using reversible processes Therefore it is no surprise that the most efficient cycles are reversible cycles that is cycles that consist entirely of reversible processes Reversible cycles cannot be achieved in practice because the irreversibili ties associated with each process cannot be eliminated However reversible cycles provide upper limits on the performance of real cycles Heat engines and refrigerators that work on reversible cycles serve as models to which actual heat engines and refrigerators can be compared Reversible cycles also serve as starting points in the development of actual cycles and are modified as needed to meet certain requirements Probably the best known reversible cycle is the Carnot cycle first proposed in 1824 by French engineer Sadi Carnot The theoretical heat engine that operates on the Carnot cycle is called the Carnot heat engine The Carnot cycle is composed of four reversible processestwo isother mal and two adiabaticand it can be executed either in a closed or a steadyflow system FIGURE 634 A reversible process involves no internal and external irreversibilities No irreversibilities inside the system No irreversibilities outside the system FIGURE 635 Totally and internally reversible heat transfer processes 20C Heat Thermal energy reservoir at 200001C Boundary at 20C a Totally reversible 20C Heat Thermal energy reservoir at 30C b Internally reversible Final PDF to printer 292 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 292 110317 0917 AM Consider a closed system that consists of a gas contained in an adiabatic pistoncylinder device as shown in Fig 636 The insulation of the cylinder head is such that it may be removed to bring the cylinder into contact with reservoirs to provide heat transfer The four reversible processes that make up the Carnot cycle are as follows Reversible Isothermal Expansion process 12 TH constant Initially state 1 the temperature of the gas is TH and the cylinder head is in close contact with a source at temperature TH The gas is allowed to expand slowly doing work on the surroundings As the gas expands the tempera ture of the gas tends to decrease But as soon as the temperature drops by an infinitesimal amount dT some heat is transferred from the reservoir into the gas raising the gas temperature to TH Thus the gas temperature is kept constant at TH Since the temperature difference between the gas and the reservoir never exceeds a differential amount dT this is a reversible heat transfer process It continues until the piston reaches position 2 The amount of total heat transferred to the gas during this process is QH Reversible Adiabatic Expansion process 23 temperature drops from TH to TL At state 2 the reservoir that was in contact with the cylinder head is removed and replaced by insulation so that the system becomes adiabatic The gas continues to expand slowly doing work on the surroundings until its temperature drops from TH to TL state 3 The piston is assumed to be frictionless and the process to be quasiequilibrium so the process is reversible as well as adiabatic Reversible Isothermal Compression process 34 TL constant At state 3 the insulation at the cylinder head is removed and the cylinder is brought into contact with a sink at temperature TL Now the piston is pushed inward by an external force doing work on the gas As the gas is compressed its temperature tends to rise But as soon as it rises by an infinitesimal amount dT heat is transferred from the gas to the sink causing the gas temperature to drop to TL Thus the gas temperature remains constant at TL Since the temperature difference between the gas and the sink never exceeds a differential amount dT this is a reversible heat transfer process It continues until the piston reaches state 4 The amount of heat rejected from the gas during this process is QL Reversible Adiabatic Compression process 41 temperature rises from TL to TH State 4 is such that when the lowtemperature reservoir is removed the insulation is put back on the cylinder head and the gas is compressed in a reversible manner so the gas returns to its initial state state 1 The temperature rises from TL to TH during this reversible adiabatic compres sion process which completes the cycle The PV diagram of this cycle is shown in Fig 637 Remembering that on a PV diagram the area under the process curve represents the boundary work for quasiequilibrium internally reversible processes we see that the area under curve 123 is the work done by the gas during the expansion part of the cycle and the area under curve 341 is the work done on the gas during the compression part of the cycle The area enclosed by the path of the cycle area 12341 is the difference between these two and represents the net work done during the cycle FIGURE 636 Execution of the Carnot cycle in a closed system 1 2 a Process 12 TH Energy sink at TL TL Insulation Insulation TH const 1 4 d Process 41 2 3 b Process 23 3 4 c Process 34 TL const Energy source at TH QH QL TH TL FIGURE 637 PV diagram of the Carnot cycle 1 QH TH const TL const QL 2 4 3 Wnetout P V Final PDF to printer 293 CHAPTER 6 cen22672ch06271322indd 293 110317 0917 AM Notice that if we acted stingily and compressed the gas at state 3 adiabati cally instead of isothermally in an effort to save QL we would end up back at state 2 retracing the process path 32 By doing so we would save QL but we would not be able to obtain any net work output from this engine This illustrates once more the necessity of a heat engine exchanging heat with at least two reservoirs at different temperatures to operate in a cycle and produce a net amount of work The Carnot cycle can also be executed in a steadyflow system It is dis cussed in later chapters in conjunction with other power cycles Being a reversible cycle the Carnot cycle is the most efficient cycle operat ing between two specified temperature limits Even though the Carnot cycle cannot be achieved in reality the efficiency of actual cycles can be improved by attempting to approximate the Carnot cycle more closely The Reversed Carnot Cycle The Carnot heatengine cycle just described is a totally reversible cycle There fore all the processes that comprise it can be reversed in which case it becomes the Carnot refrigeration cycle This time the cycle remains exactly the same except that the directions of any heat and work interactions are reversed Heat in the amount of QL is absorbed from the lowtemperature reservoir heat in the amount of QH is rejected to a hightemperature reservoir and a work input of Wnetin is required to accomplish all this The PV diagram of the reversed Carnot cycle is the same as the one given for the Carnot cycle except that the directions of the processes are reversed as shown in Fig 638 68 THE CARNOT PRINCIPLES The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the KelvinPlanck and Clausius statements A heat engine cannot operate by exchanging heat with a single reservoir and a refrig erator cannot operate without a net energy input from an external source We can draw valuable conclusions from these statements Two conclusions pertain to the thermal efficiency of reversible and irreversible ie actual heat engines and they are known as the Carnot principles Fig 639 expressed as follows 1 The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs 2 The efficiencies of all reversible heat engines operating between the same two reservoirs are the same These two statements can be proved by demonstrating that the violation of either statement results in the violation of the second law of thermody namics To prove the first statement consider two heat engines operating between the same reservoirs as shown in Fig 640 One engine is reversible and the other is irreversible Now each engine is supplied with the same amount of heat QH The amount of work produced by the reversible heat engine is Wrev and the amount produced by the irreversible one is Wirrev FIGURE 638 PV diagram of the reversed Carnot cycle 1 QH TH const TL const QL 4 2 3 Wnetin P V FIGURE 639 The Carnot principles ηth1 ηth2 ηth2 ηth3 1 Irrev HE 2 Rev HE 3 Rev HE Hightemperature reservoir at TH Lowtemperature reservoir at TL Final PDF to printer 294 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 294 110317 0917 AM In violation of the first Carnot principle we assume that the irreversible heat engine is more efficient than the reversible one that is ηthirrev ηthrev and thus delivers more work than the reversible one Now let the reversible heat engine be reversed and operate as a refrigerator This refrigerator will receive a work input of Wrev and reject heat to the hightemperature reser voir Since the refrigerator is rejecting heat in the amount of QH to the high temperature reservoir and the irreversible heat engine is receiving the same amount of heat from this reservoir the net heat exchange for this reservoir is zero Thus it could be eliminated by having the refrigerator discharge QH directly into the irreversible heat engine Now considering the refrigerator and the irreversible engine together we have an engine that produces a net work in the amount of Wirrev Wrev while exchanging heat with a single reservoira violation of the KelvinPlanck statement of the second law Therefore our initial assumption that ηthirrev ηthrev is incorrect Then we conclude that no heat engine can be more efficient than a reversible heat engine operating between the same reservoirs The second Carnot principle can also be proved in a similar manner This time let us replace the irreversible engine with another reversible engine that is more efficient and thus delivers more work than the first reversible engine By following through the same reasoning we end up having an engine that produces a net amount of work while exchanging heat with a single reservoir which is a violation of the second law Therefore we conclude that no revers ible heat engine can be more efficient than a reversible one operating between the same two reservoirs regardless of how the cycle is completed or the kind of working fluid used FIGURE 640 Proof of the first Carnot principle QH QH Wirrev Wrev QLirrev QLrev assumed QLrev a A reversible and an irreversible heat engine operating between the same two reservoirs the reversible heat engine is then reversed to run as a refrigerator Irreversible HE Wirrev Wrev QLrev QLirrev Combined HE R Hightemperature reservoir at TH Lowtemperature reservoir at TL Reversible HE or R Lowtemperature reservoir at TL b The equivalent combined system Final PDF to printer 295 CHAPTER 6 cen22672ch06271322indd 295 110317 0917 AM 69 THE THERMODYNAMIC TEMPERATURE SCALE A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale Such a temperature scale offers great conveniences in thermodynamic calcu lations and its derivation is given below using some reversible heat engines The second Carnot principle discussed in Sec 68 states that all reversible heat engines have the same thermal efficiency when operating between the same two reservoirs Fig 641 That is the efficiency of a reversible engine is independent of the working fluid employed and its properties the way the cycle is executed or the type of reversible engine used Since energy reser voirs are characterized by their temperatures the thermal efficiency of revers ible heat engines is a function of the reservoir temperatures only That is η threv g T H T L or Q H Q L f T H T L 613 since ηth 1 QLQH In these relations TH and TL are the temperatures of the high and lowtemperature reservoirs respectively The functional form of f TH TL can be developed with the help of the three reversible heat engines shown in Fig 642 Engines A and C are supplied with the same amount of heat Q1 from the hightemperature reservoir at T1 Engine C rejects Q3 to the lowtemperature reservoir at T3 Engine B receives the heat Q2 rejected by engine A at temperature T2 and rejects heat in the amount of Q3 to a reservoir at T3 The amounts of heat rejected by engines B and C must be the same since engines A and B can be combined into one reversible engine operating between the same reservoirs as engine C and thus the combined engine will have the same efficiency as engine C Since the heat input to engine C is the same as the heat input to the combined engines A and B both systems must reject the same amount of heat Applying Eq 613 to all three engines separately we obtain Q 1 Q 2 f T 1 T 2 Q 2 Q 3 f T 2 T 3 and Q 1 Q 3 f T 1 T 3 Now consider the identity Q 1 Q 3 Q 1 Q 2 Q 2 Q 3 which corresponds to f T 1 T 3 f T 1 T 2 f T 2 T 3 A careful examination of this equation reveals that the lefthand side is a func tion of T1 and T3 and therefore the righthand side must also be a function of FIGURE 641 All reversible heat engines operating between the same two reservoirs have the same efficiency the second Carnot principle A reversible HE ηthA Another reversible HE ηthB ηthA ηthB 70 Lowtemperature reservoir at TL 300 K Hightemperature reservoir at TH 1000 K FIGURE 642 The arrangement of heat engines used to develop the thermodynamic temperature scale WA Q2 Q1 Q2 Q3 Q3 Q1 T2 WC Thermal energy reservoir at T3 Rev HE B Rev HE A WB Rev HE C Thermal energy reservoir at T1 Final PDF to printer 296 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 296 110317 0917 AM T1 and T3 only and not T2 That is the value of the product on the righthand side of this equation is independent of the value of T2 This condition will be satisfied only if the function f has the following form f T 1 T 2 ϕ T 1 ϕ T 2 and f T 2 T 3 ϕ T 2 ϕ T 3 so that ϕT2 will cancel from the product of f T1 T2 and f T2 T3 yielding Q 1 Q 2 f T 1 T 3 ϕ T 1 ϕ T 3 614 This relation is much more specific than Eq 613 for the functional form of Q1Q3 in terms of T1 and T3 For a reversible heat engine operating between two reservoirs at tempera tures TH and TL Eq 614 can be written as Q H Q L ϕ T H ϕ T L 615 This is the only requirement that the second law places on the ratio of heat transfers to and from the reversible heat engines Several functions ϕT satisfy this equation and the choice is completely arbitrary Lord Kelvin first proposed taking ϕT T to define a thermodynamic temperature scale as Fig 643 Q H Q L rev T H T L 616 This temperature scale is called the Kelvin scale and the temperatures on this scale are called absolute temperatures On the Kelvin scale the temperature ratios depend on the ratios of heat transfer between a reversible heat engine and the reservoirs and are independent of the physical properties of any sub stance On this scale temperatures vary between zero and infinity The thermodynamic temperature scale is not completely defined by Eq 616 since it gives us only a ratio of absolute temperatures We also need to know the magnitude of a kelvin At the International Conference on Weights and Measures held in 1954 the triple point of water the state at which all three phases of water exist in equilibrium was assigned the value 27316 K Fig 644 The magnitude of a kelvin is defined as 127316 of the temperature interval between absolute zero and the triplepoint tempera ture of water The magnitudes of temperature units on the Kelvin and Celsius scales are identical 1 K 1C The temperatures on these two scales differ by a constant 27315 T C T K 27315 617 Even though the thermodynamic temperature scale is defined with the help of the reversible heat engines it is not possible nor is it practical to actually operate such an engine to determine numerical values on the absolute tem perature scale Absolute temperatures can be measured accurately by other means such as the constantvolume idealgas thermometer together with FIGURE 644 A conceptual experimental setup to determine thermodynamic temperatures on the Kelvin scale by measuring heat transfers QH and QL T 27316 QH QL W QH QL Carnot HE Heat reservoir T 27316 K assigned Water at triple point FIGURE 643 For reversible cycles the heat transfer ratio QHQL can be replaced by the absolute temperature ratio THTL Lowtemperature reservoir at TL QH QL TH TL QH Wnet QL Hightemperature reservoir at TH Reversible heat engine or refrigerator Final PDF to printer 297 CHAPTER 6 cen22672ch06271322indd 297 110317 0917 AM extrapolation techniques as discussed in Chap 1 The validity of Eq 616 can be demonstrated from physical considerations for a reversible cycle using an ideal gas as the working fluid 610 THE CARNOT HEAT ENGINE The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat engine The thermal efficiency of any heat engine reversible or irreversible is given by Eq 66 as η th 1 Q L Q H where QH is heat transferred to the heat engine from a hightemperature res ervoir at TH and QL is heat rejected to a lowtemperature reservoir at TL For reversible heat engines the heat transfer ratio in the preceding relation can be replaced by the ratio of the absolute temperatures of the two reservoirs as given by Eq 616 Then the efficiency of a Carnot engine or any revers ible heat engine becomes η threv 1 T L T H 618 This relation is often referred to as the Carnot efficiency since the Carnot heat engine is the best known reversible engine This is the highest efficiency a heat engine operating between the two thermal energy reservoirs at temperatures TL and TH can have Fig 645 All irreversible ie actual heat engines operating between these temperature limits TL and TH have lower efficiencies An actual heat engine cannot reach this maximum theo retical efficiency value because it is impossible to completely eliminate all the irreversibilities associated with the actual cycle Note that TL and TH in Eq 618 are absolute temperatures Using C or F for temperatures in this relation gives results grossly in error The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows Fig 646 η th η threv irreversible heat engine η threv reversible heat engine η threv impossible heat engine 619 Most workproducing devices heat engines in operation today have effi ciencies under 40 percent which appear low relative to 100 percent How ever when the performance of actual heat engines is assessed the efficiencies should not be compared to 100 percent instead they should be compared to the efficiency of a reversible heat engine operating between the same temper ature limitsbecause this is the true theoretical upper limit for the efficiency not 100 percent The maximum efficiency of a steam power plant operating between TH 1000 K and TL 300 K is 70 percent as determined from Eq 618 Compared with this value an actual efficiency of 40 percent does not seem so bad even though there is still plenty of room for improvement FIGURE 645 The Carnot heat engine is the most efficient of all heat engines operat ing between the same high and low temperature reservoirs Hightemperature reservoir at TH 1000 K Wnetout QH QL Carnot HE ηth 70 Lowtemperature reservoir at TL 300 K FIGURE 646 No heat engine can have a higher effi ciency than a reversible heat engine operating between the same high and lowtemperature reservoirs Lowtemperature reservoir at TL 300 K Rev HE ηth 70 Irrev HE ηth 45 Impossible HE ηth 80 Hightemperature reservoir at TH 1000 K Final PDF to printer 298 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 298 110317 0917 AM It is obvious from Eq 618 that the efficiency of a Carnot heat engine increases as TH is increased or as TL is decreased This is to be expected since as TL decreases so does the amount of heat rejected and as TL approaches zero the Carnot efficiency approaches unity This is also true for actual heat engines The thermal efficiency of actual heat engines can be maximized by supplying heat to the engine at the highest possible temperature limited by material strength and rejecting heat from the engine at the lowest possible temperature limited by the temperature of the cooling medium such as rivers lakes or the atmosphere FIGURE 647 Schematic for Example 65 Carnot HE QL Lowtemperature reservoir at TL 30C Wnetout QH 500 kJ Hightemperature reservoir at TH 652C FIGURE 648 The fraction of heat that can be converted to work as a function of source temperature for TL 303 K Lowtemperature reservoir at TL 303 K Rev HE ηth TH K ηth 925 800 700 500 350 672 621 567 394 134 Hightemperature reservoir at TH EXAMPLE 65 Analysis of a Carnot Heat Engine A Carnot heat engine shown in Fig 647 receives 500 kJ of heat per cycle from a hightemperature source at 652C and rejects heat to a lowtemperature sink at 30C Determine a the thermal efficiency of this Carnot engine and b the amount of heat rejected to the sink per cycle SOLUTION The heat supplied to a Carnot heat engine is given The thermal effi ciency and the heat rejected are to be determined Analysis a The Carnot heat engine is a reversible heat engine and so its efficiency can be determined from Eq 618 to be η threv 1 T L T H 1 30 273 K 652 273 K 0672 That is this Carnot heat engine converts 672 percent of the heat it receives to work b The amount of heat rejected QL by this reversible heat engine is easily determined from Eq 616 to be Q Lrev T L T H Q Hrev 30 273 K 652 273 K 500 kJ 164 kJ Discussion Note that this Carnot heat engine rejects to a lowtemperature sink 164 kJ of the 500 kJ of heat it receives during each cycle The Quality of Energy The Carnot heat engine in Example 65 receives heat from a source at 925 K and converts 672 percent of it to work while rejecting the rest 328 percent to a sink at 303 K Now let us examine how the thermal efficiency varies with the source temperature when the sink temperature is held constant The thermal efficiency of a Carnot heat engine that rejects heat to a sink at 303 K is evaluated at various source temperatures using Eq 618 and is listed in Fig 648 Clearly the thermal efficiency decreases as the source temperature is lowered When heat is supplied to the heat engine at 500 instead of 925 K for example the thermal efficiency drops from 672 to 394 percent That is the fraction of heat that can be converted to work drops to 394 percent when the temperature of the source drops to 500 K When the source tempera ture is 350 K this fraction becomes a mere 134 percent These efficiency values show that energy has quality as well as quantity It is clear from the thermal efficiency values in Fig 648 that more of the Final PDF to printer 299 CHAPTER 6 cen22672ch06271322indd 299 110317 0917 AM hightemperature thermal energy can be converted to work Therefore the higher the temperature the higher the quality of the energy Fig 649 Large quantities of solar energy for example can be stored in large bod ies of water called solar ponds at about 350 K This stored energy can then be supplied to a heat engine to produce work electricity However the efficiency of solar pond power plants is very low under 5 percent because of the low quality of the energy stored in the source and the construction and maintenance costs are relatively high Therefore they are not competitive even though the energy supply of such plants is free The temperature and thus the quality of the solar energy stored could be raised by utilizing con centrating collectors but the equipment cost in that case becomes very high Work is a more valuable form of energy than heat since 100 percent of work can be converted to heat but only a fraction of heat can be converted to work When heat is transferred from a hightemperature body to a lowertemperature one it is degraded since less of it now can be converted to work For example if 100 kJ of heat is transferred from a body at 1000 K to a body at 300 K at the end we will have 100 kJ of thermal energy stored at 300 K which has no practical value But if this conversion is made through a heat engine up to 1 3001000 70 percent of it could be converted to work which is a more valuable form of energy Thus 70 kJ of work potential is wasted as a result of this heat transfer and energy is degraded Quantity versus Quality in Daily Life At times of energy crisis we are bombarded with speeches and articles on how to conserve energy Yet we all know that the quantity of energy is already conserved What is not conserved is the quality of energy or the work potential of energy Wasting energy is synonymous with converting it to a less useful form One unit of highquality energy can be more valuable than three units of lowerquality energy For example a finite amount of thermal energy at high temperature is more attractive to power plant engineers than a vast amount of thermal energy at low temperature such as the energy stored in the upper layers of the oceans in tropical climates As part of our culture we seem to be fascinated by quantity and little attention is given to quality However quantity alone cannot give the whole picture and we need to consider quality as well That is we need to look at something from both the first and secondlaw points of view when evaluat ing something even in nontechnical areas Below we present some ordinary events and show their relevance to the second law of thermodynamics Consider two students Andy and Wendy Andy has 10 friends who never miss his parties and are always around during fun times However they seem to be busy when Andy needs their help Wendy on the other hand has five friends They are never too busy for her and she can count on them at times of need Let us now try to answer the question Who has more friends From the firstlaw point of view which considers quantity only it is obvi ous that Andy has more friends However from the secondlaw point of view which considers quality as well there is no doubt that Wendy is the one with more friends Another example with which most people will identify is the multibillion dollar diet industry which is primarily based on the first law of FIGURE 649 The higher the temperature of the thermal energy the higher its quality 2000 1500 1000 500 T K Thermal energy Quality Final PDF to printer 300 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 300 110317 0917 AM thermodynamics However the fact that 90 percent of the people who lose weight gain it back quickly with interest suggests that the first law alone does not give the whole picture People who seem to be eating whatever they want whenever they want without gaining weight are living proof that the caloriecounting technique the first law leaves many questions on dieting unanswered Obviously more research focused on the secondlaw effects of dieting is needed before we can fully understand the weightgain and weight loss process It is tempting to judge things on the basis of their quantity instead of their quality since assessing quality is much more difficult than assessing quantity However assessments made on the basis of quantity only the first law may be grossly inadequate and misleading 611 THE CARNOT REFRIGERATOR AND HEAT PUMP A refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump The coefficient of performance of any refrigerator or heat pump reversible or irreversible is given by Eqs 69 and 611 as COP R 1 Q H Q L 1 and COP HP 1 1 Q L Q H where QL is the amount of heat absorbed from the lowtemperature medium and QH is the amount of heat rejected to the hightemperature medium The COPs of all reversible refrigerators or heat pumps can be determined by replacing the heat transfer ratios in the preceding relations with the ratios of the absolute temperatures of the high and lowtemperature reservoirs as expressed by Eq 616 Then the COP relations for reversible refrigerators and heat pumps become COP Rrev 1 T H T L 1 620 and COP HPrev 1 1 T L T H 621 These are the highest coefficients of performance that a refrigerator or a heat pump operating between the temperature limits of TL and TH can have All actual refrigerators or heat pumps operating between these temperature limits TL and TH have lower coefficients of performance Fig 650 The coefficients of performance of actual and reversible refrigerators operating between the same temperature limits can be compared as follows COP R COP Rrev irreversible refrigerator COP Rrev reversible refrigerator COP Rrev impossible refrigerator 622 Final PDF to printer 301 CHAPTER 6 cen22672ch06271322indd 301 110317 0917 AM A similar relation can be obtained for heat pumps by replacing all COPRs in Eq 622 with COPHP The COP of a reversible refrigerator or heat pump is the maximum theo retical value for the specified temperature limits Actual refrigerators or heat pumps may approach these values as their designs are improved but they can never reach them As a final note the COPs of both the refrigerators and the heat pumps decrease as TL decreases That is it requires more work to absorb heat from lowertemperature media As the temperature of the refrigerated space approaches zero the amount of work required to produce a finite amount of refrigeration approaches infinity and COPR approaches zero FIGURE 650 No refrigerator can have a higher COP than a reversible refrigerator operating between the same temperature limits Reversible refrigerator COPR 11 Irreversible refrigerator COPR 7 Impossible refrigerator COPR 13 Cool refrigerated space at TL 275 K Warm environment at TH 300 K EXAMPLE 66 A Carnot Refrigeration Cycle Operating in the Saturation Dome A Carnot refrigeration cycle is executed in a closed system in the saturated liquidvapor mixture region using 08 kg of refrigerant134a as the working fluid Fig 651 The maximum and the minimum temperatures in the cycle are 20 and 8C respectively It is known that the refrigerant is saturated liquid at the end of the heat rejection process and the net work input to the cycle is 15 kJ Determine the fraction of the mass of the refrigerant that vaporizes during the heat addition process and the pressure at the end of the heat rejection process SOLUTION A Carnot refrigeration cycle is executed in a closed system The mass fraction of the refrigerant that vaporizes during the heat addition process and the pres sure at the end of the heat rejection process are to be determined Assumptions The refrigerator operates on the ideal Carnot cycle Analysis Knowing the high and low temperatures the coefficient of performance of the cycle is COP R 1 T H T L 1 1 20 273 K 8 273 K 1 9464 FIGURE 651 Schematic for Example 66 T 20C 8C 1 4 3 QH QL 2 V Final PDF to printer 302 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 302 110317 0917 AM EXAMPLE 67 Heating a House with a Carnot Heat Pump A heat pump is to be used to heat a house during the winter as shown in Fig 652 The house is to be maintained at 21C at all times The house is estimated to be losing heat at a rate of 135000 kJh when the outside temperature drops to 5C Determine the minimum power required to drive this heat pump SOLUTION A heat pump maintains a house at a constant temperature The required minimum power input to the heat pump is to be determined Assumptions Steady operating conditions exist Analysis The heat pump must supply heat to the house at a rate of Q H 135000 kJ h 375 kW The power requirements are minimum when a reversible heat pump is used to do the job The COP of a reversible heat pump operating between the house and the outside air is COP HPrev 1 1 T L T H 1 1 5 273 K 21 273 K 113 Then the required power input to this reversible heat pump becomes W netin Q H COP HP 375 kW 113 332 kW Discussion This reversible heat pump can meet the heating requirements of this house by consuming electric power at a rate of 332 kW only If this house were to be heated by electric resistance heaters instead the power consumption would jump up 113 times to 375 kW This is because in resistance heaters the electric energy is converted to heat at FIGURE 652 Schematic for Example 67 House TH 21C HP Wnetin QH QL Cold outside air TL 5C 135000 kJh Heat loss The amount of cooling is determined from the definition of the coefficient of perfor mance to be Q L COP R W in 9464 15 kJ 142 kJ The enthalpy of vaporization of R134a at 8C is hfg 20459 kJkg Table A11 Then the amount of refrigerant that vaporizes during heat absorption becomes Q L m evap h fg 8C m evap 142 kJ 20459 kJ kg 0694 kg Therefore the fraction of mass that vaporized during the heat addition process to the refrigerant is Mass fraction m evap m total 0694 kg 08 kg 0868 or 868 percent The pressure at the end of the heat rejection process is simply the saturation pressure at heat rejection temperature P 4 P sat 20C 5721 kPa Discussion The reversed Carnot cycle is an idealized refrigeration cycle thus it cannot be achieved in practice Practical refrigeration cycles are analyzed in Chap 11 Final PDF to printer 303 CHAPTER 6 cen22672ch06271322indd 303 110317 0917 AM Refrigerators to preserve perishable foods have long been essential appliances for households They have proven to be highly durable and reliable providing satisfactory service for over 15 years A typical household refrigerator is actu ally a combination refrigeratorfreezer since it has a freezer compartment to make ice and to store frozen food Todays refrigerators use much less energy than earlier models because they use smaller and higherefficiency motors and compressors better insulation materials larger coil surface areas and better door seals Fig 653 At a typ ical electricity rate of 115 cents per kWh an average refrigerator costs about 100 a year to run which is half the annual operating cost of a refrigerator 25 years ago Replacing a 25yearold 18ft3 refrigerator with a new energy efficient model will save over 1000 kWh of electricity per year For the envi ronment this means a reduction of over 1 ton of CO2 which causes global climate change and over 10 kg of SO2 which causes acid rain Despite the improvements made in several areas during the past 100 years in household refrigerators the basic vaporcompression refrigeration cycle has remained unchanged The alternative absorption refrigeration and thermoelec tric refrigeration systems are currently more expensive and less efficient but they have found limited use in some specialized applications Table 61 A household refrigerator is designed to maintain the freezer section at 18C 0F and the refrigerator section at 3C 37F Lower freezer tem peratures increase energy consumption without improving the storage life of frozen foods significantly Different temperatures for the storage of specific foods can be maintained in the refrigerator section by using specialpurpose compartments Practically all fullsize refrigerators have a large airtight drawer for leafy veg etables and fresh fruits to seal in moisture and to protect them from the drying effect of cool air circulating in the refrigerator A covered egg compartment in the lid extends the life of eggs by slowing down moisture loss from the eggs It is common for refrigerators to have a special warmer compartment for butter in the door to maintain butter at spreading temperature The compartment also isolates butter and prevents it from absorbing odors and tastes from other food items Some upscale models have a temperaturecontrolled meat compartment maintained at 05C 31F which keeps meat at the lowest safe temperature without freezing it thus extending its storage life The more expensive models come with an automatic icemaker located in the freezer section that is con nected to the water line as well as automatic ice and chilledwater dispensers TOPIC OF SPECIAL INTEREST Household Refrigerators This section can be skipped without a loss in continuity TABLE 61 Typical operating efficiencies of some refrigeration systems for a freezer temperature of 18C and ambient temperature of 32C Type of refrigeration system Coefficient of performance Vaporcompression 13 Absorption refrigeration 04 Thermoelectric refrigeration 01 a onetoone ratio With a heat pump however energy is absorbed from the outside and carried to the inside using a refrigeration cycle that consumes only 332 kW Notice that the heat pump does not create energy It merely transports it from one medium the cold outdoors to another the warm indoors FIGURE 653 Todays refrigerators are much more efficient because of the improvements in technology and manufacturing Refrigerator Better door seals Better insulation materials More efficient motors and compressors Final PDF to printer 304 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 304 110317 0917 AM A typical icemaker can produce 2 to 3 kg of ice per day and store 3 to 5 kg of ice in a removable ice storage container Household refrigerators consume from about 90 to 600 W of electrical energy when running and are designed to perform satisfactorily in environ ments at up to 43C 110F Refrigerators run intermittently as you may have noticed running about 30 percent of the time under normal use in a house at 25C 77F For specified external dimensions a refrigerator is desired to have maximum food storage volume minimum energy consumption and the lowest possible cost to the consumer The total food storage volume has been increased over the years without an increase in the external dimensions by using thinner but more effective insulation and minimizing the space occupied by the compressor and the condenser Switching from the fiberglass insulation thermal conductiv ity k 0032 0040 WmC to expandedinplace urethane foam insulation k 0019 WmC made it possible to reduce the wall thickness of the refrig erator by almost half from about 90 to 48 mm for the freezer section and from about 70 to 40 mm for the refrigerator section The rigidity and bonding action of the foam also provide additional structural support However the entire shell of the refrigerator must be carefully sealed to prevent any water leakage or moisture migration into the insulation since moisture degrades the effectiveness of insulation The size of the compressor and the other components of a refrigeration system are determined on the basis of the anticipated heat load or refrigera tion load which is the rate of heat flow into the refrigerator The heat load consists of the predictable part such as heat transfer through the walls and door gaskets of the refrigerator fan motors and defrost heaters Fig 654 and the unpredictable part which depends on user habits such as opening the door making ice and loading the refrigerator The amount of energy con sumed by the refrigerator can be minimized by practicing good conservation measures as discussed in the list that follows FIGURE 654 The cross section of a refrigerator showing the relative magnitudes of various effects that constitute the predictable heat load Thermal insulation 6 Fan motor 6 External heater 52 Wall insulation 30 Door gasket region 6 Defrost heater Plastic door liner Plastic breaker strips Final PDF to printer 305 CHAPTER 6 cen22672ch06271322indd 305 110317 0917 AM 1 Open the refrigerator door the fewest times possible for the shortest duration possible Each time the refrigerator door is opened the cool air inside is replaced by the warmer air outside which needs to be cooled Keeping the refrigerator or freezer full will save energy by reducing the amount of cold air that can escape each time the door is opened 2 Cool the hot foods to room temperature first before putting them into the refrigerator Moving a hot pan from the oven directly into the refrigerator not only wastes energy by making the refrigerator work longer but it also causes the nearby perishable foods to spoil by creating a warm environ ment in its immediate surroundings Fig 655 3 Clean the condenser coils located behind or beneath the refrigerator The dust and grime that collect on the coils act as insulation that slows down heat dissipation through them Cleaning the coils a couple of times a year with a damp cloth or a vacuum cleaner will improve the cooling ability of the refrigerator while cutting down the power consumption by a few percent Sometimes a fan is used to forcecool the condensers of large or builtin refrigerators and the strong air motion keeps the coils clean 4 Check the door gasket for air leaks This can be done by placing a flashlight into the refrigerator turning off the kitchen lights and looking for light leaks Heat transfer through the door gasket region accounts for almost onethird of the regular heat load of refrigerators and thus any defective door gaskets must be repaired immediately 5 Avoid unnecessarily low temperature settings The recommended temperatures for freezers and refrigerators are 18C 0F and 3C 37F respectively Setting the freezer temperature below 18C adds significantly to the energy consumption but does not add much to the storage life of frozen foods Keeping temperatures 6C or 10F below recommended levels can increase energy use by as much as 25 percent 6 Avoid excessive ice buildup on the interior surfaces of the evaporator The ice layer on the surface acts as insulation and slows down heat transfer from the freezer section to the refrigerant The refrigerator should be defrosted by manually turning off the temperature control switch when the ice thickness exceeds a few millimeters Defrosting is done automatically in nofrost refrigerators by supplying heat to the evaporator with a 300W to 1000W resistance heater or with hot refrigerant gas periodically for short periods The water is then drained to a pan outside where it is evaporated using the heat dissipated by the condenser The nofrost evaporators are basically finned tubes subjected to airflow circulated by a fan Practically all the frost collects on fins which are the coldest surfaces leaving the exposed surfaces of the freezer section and the frozen food frostfree 7 Use the powersaver switch that controls the heating coils and prevents condensation on the outside surfaces in humid environments The low wattage heaters are used to raise the temperature of the outer surfaces of the refrigerator at critical locations above the dew point in order to avoid FIGURE 655 Putting hot foods into the refrigerator without cooling them first not only wastes energy but also could spoil the foods nearby Hot food 80C Warm air 30C 5C Final PDF to printer 306 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 306 110317 0917 AM EXAMPLE 68 Malfunction of a Refrigerator Light Switch The interior lighting of refrigerators is provided by incandescent lamps whose switches are actuated by the opening of the refrigerator door Consider a refrigerator whose 40W lightbulb remains on continuously as a result of a malfunction of the switch Fig 657 If the refrigerator has a coefficient of performance of 13 and the cost of electricity is 12 cents per kWh determine the increase in the energy consumption of the refrigerator and its cost per year if the switch is not fixed SOLUTION The lightbulb of a refrigerator malfunctions and remains on The increases in the electricity consumption and cost are to be determined Assumptions The life of the lightbulb is more than 1 year Analysis The lightbulb consumes 40 W of power when it is on and thus adds 40 W to the heat load of the refrigerator Noting that the COP of the refrigerator is 13 the power consumed by the refrigerator to remove the heat generated by the lightbulb is W refrig Q refrig COP R 40 W 13 308 W Therefore the total additional power consumed by the refrigerator is W totaladditional W light W refrig 40 308 708 W The total number of hours in a year is Annual hours 365 days yr 24 h day 8760 h yr FIGURE 657 Schematic for Example 68 Light bulb 40 W water droplets forming on the surfaces and sliding down Condensation is most likely to occur in summer in hot and humid climates in homes without air conditioning The moisture formation on the surfaces is undesirable since it may cause the painted finish of the outer surface to deteriorate and it may wet the kitchen floor About 10 percent of the total energy consumed by the refrigerator can be saved by turning this heater off and keeping it off unless there is visible condensation on the outer surfaces 8 Do not block the airflow passages to and from the condenser coils of the refrigerator The heat dissipated by the condenser to the air is carried away by air that enters through the bottom and sides of the refrigerator and leaves through the top Any blockage of this natural convection air circulation path by large objects such as several cereal boxes on top of the refrigerator will degrade the performance of the condenser and thus the refrigerator Fig 656 These and other commonsense conservation measures will result in a reduction in the energy and maintenance costs of a refrigerator as well as an extended troublefree life of the device FIGURE 656 The condenser coils of a refrigerator must be cleaned periodically and the airflow passages must not be blocked to maintain high performance Cabinet Refrigerator Warm air Cool air Coils Final PDF to printer 307 CHAPTER 6 cen22672ch06271322indd 307 110317 0917 AM Assuming the refrigerator is opened 20 times a day for an average of 30 s the light would normally be on for Normal operating hours 20 times day 30 s time 1 h 3600 s 365 days yr 61 h yr Then the additional hours the light remains on as a result of the malfunction becomes Additional operating hours Annual hours Normal operating hours 8760 61 8699 h yr Therefore the additional electric power consumption and its cost per year are Additional power consumption W total additional Additional operating hours 00708 kW 8699 h yr 616 kWh yr and Additional power cost Additional power consumptionUnit cost 616 kWh yr 012 kWh 739 yr Discussion Note that not repairing the switch will cost the homeowner about 75 a year This is alarming when we consider that at 012kWh a typical refrigerator con sumes about 100 worth of electricity a year SUMMARY The second law of thermodynamics states that processes occur in a certain direction not in any direction A process does not occur unless it satisfies both the first and the sec ond laws of thermodynamics Bodies that can absorb or reject finite amounts of heat isothermally are called thermal energy reservoirs or heat reservoirs Work can be converted to heat directly but heat can be con verted to work only by some devices called heat engines The thermal efficiency of a heat engine is defined as η th W netout Q H 1 Q L Q H where Wnetout is the net work output of the heat engine QH is the amount of heat supplied to the engine and QL is the amount of heat rejected by the engine Refrigerators and heat pumps are devices that absorb heat from lowtemperature media and reject it to higher temperature ones The performance of a refrigerator or a heat pump is expressed in terms of the coefficient of performance which is defined as COP R Q L W netin 1 Q H Q L 1 COP HP Q H W netin 1 1 Q L Q H The KelvinPlanck statement of the second law of thermo dynamics states that no heat engine can produce a net amount of work while exchanging heat with a single reservoir only The Clausius statement of the second law states that no device can transfer heat from a cooler body to a warmer one without leaving an effect on the surroundings Any device that violates the first or the second law of ther modynamics is called a perpetualmotion machine A process is said to be reversible if both the system and the surroundings can be restored to their origi nal conditions Any other process is irreversible The effects such as friction nonquasiequilibrium expansion or compression and heat transfer through a finite tem perature difference render a process irreversible and are called irreversibilities Final PDF to printer 308 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 308 110317 0917 AM REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Refrigeration SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1994 2 D Stewart Wheels Go Round and Round but Always Run Down November 1986 Smithsonian pp 193208 3 J T Amann A Wilson and K Ackerly Consumer Guide to Home Energy Saving 9th ed American Council for an EnergyEfficient Economy Washington D C 2007 PROBLEMS Second Law of Thermodynamics and Thermal Energy Reservoirs 61C A mechanic claims to have developed a car engine that runs on water instead of gasoline What is your response to this claim 62C Describe an imaginary process that violates both the first and the second laws of thermodynamics 63C Describe an imaginary process that satisfies the first law but violates the second law of thermodynamics 64C Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics 65C An experimentalist claims to have raised the temperature of a small amount of water to 150C by transferring heat from highpressure steam at 120C Is this a reasonable claim Why Assume no refrigerator or heat pump is used in the process 66C Consider the process of baking potatoes in a conven tional oven Can the hot air in the oven be treated as a thermal energy reservoir Explain 67C Consider the energy generated by a TV set What is a suitable choice for a thermal energy reservoir Heat Engines and Thermal Efficiency 68C What are the characteristics of all heat engines 69C What is the KelvinPlanck expression of the second law of thermodynamics The Carnot cycle is a reversible cycle that is composed of four reversible processes two isothermal and two adia batic The Carnot principles state that the thermal efficien cies of all reversible heat engines operating between the same two reservoirs are the same and that no heat engine is more efficient than a reversible one operating between the same two reservoirs These statements form the basis for establish ing a thermodynamic temperature scale related to the heat transfers between a reversible device and the high and low temperature reservoirs by Q H Q L rev T H T L Therefore the QHQL ratio can be replaced by THTL for revers ible devices where TH and TL are the absolute temperatures of the high and lowtemperature reservoirs respectively A heat engine that operates on the reversible Carnot cycle is called a Carnot heat engine The thermal efficiency of a Carnot heat engine as well as all other reversible heat engines is given by η threv 1 T L T H This is the maximum efficiency a heat engine operating between two reservoirs at temperatures TH and TL can have The COPs of reversible refrigerators and heat pumps are given in a similar manner as COP Rrev 1 T H T L 1 and COP HPrev 1 1 T L T H Again these are the highest COPs a refrigerator or a heat pump operating between the temperature limits of TH and TL can have Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer 309 CHAPTER 6 cen22672ch06271322indd 309 110317 0917 AM 610C Is it possible for a heat engine to operate with out rejecting any waste heat to a lowtemperature reservoir Explain 611C Does a heat engine that has a thermal efficiency of 100 percent necessarily violate a the first law and b the second law of thermodynamics Explain 612C In the absence of any friction and other irrevers ibilities can a heat engine have an efficiency of 100 percent Explain 613C Are the efficiencies of all the workproducing devices including the hydroelectric power plants limited by the KelvinPlanck statement of the second law Explain 614C Baseboard heaters are basically electric resistance heaters and are frequently used in space heating A homeowner claims that her 5yearold baseboard heaters have a conversion efficiency of 100 percent Is this claim in violation of any ther modynamic laws Explain 615C Consider a pan of water being heated a by placing it on an electric range and b by placing a heating element in the water Which method is a more efficient way of heating water Explain 616 A heat engine has a total heat input of 13 kJ and a thermal efficiency of 35 percent How much work will it produce 617 A steam power plant receives heat from a furnace at a rate of 280 GJh Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJh If the waste heat is transferred to the cooling water at a rate of 165 GJh determine a net power output and b the thermal efficiency of this power plant Answers a 297 MW b 382 percent 619 A 600MW steam power plant which is cooled by a nearby river has a thermal efficiency of 40 percent Determine the rate of heat transfer to the river water Will the actual heat transfer rate be higher or lower than this value Why 620 A heat engine with a thermal efficiency of 45 percent rejects 500 kJkg of heat How much heat does it receive Answer 909 kJkg 621E A heat engine that propels a ship produces 500 Btulbm of work while rejecting 300 Btulbm of heat What is its ther mal efficiency 622 A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tonsh If the heating value of the coal is 30000 kJkg determine the overall efficiency of this plant Answer 300 percent 623 An automobile engine consumes fuel at a rate of 22 Lh and delivers 55 kW of power to the wheels If the fuel has a heating value of 44000 kJkg and a density of 08 gcm3 deter mine the efficiency of this engine Answer 256 percent 624E Solar energy stored in large bodies of water called solar ponds is being used to generate electricity If such a solar power plant has an efficiency of 3 percent and a net power output of 150 kW determine the average value of the required solar energy collection rate in Btuh 625 A coalburning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent The actual gravimetric airfuel ratio in the furnace is calcu lated to be 12 kg airkg fuel The heating value of the coal is 28000 kJkg Determine a the amount of coal consumed dur ing a 24hour period and b the rate of air flowing through the furnace Answers a 289 106 kg b 402 kgs 626E An Ocean Thermal Energy Conversion OTEC power plant built in Hawaii in 1987 was designed to operate FIGURE P616 Furnace Wout sink 13 kJ HE FIGURE P618E Sink ηth 40 3 104 Btuh Wnet HE Source 618E A heat engine has a heat input of 3 104 Btuh and a thermal efficiency of 40 percent Calculate the power it will produce in hp Final PDF to printer 310 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 310 110317 0917 AM between the temperature limits of 86F at the ocean surface and 41F at a depth of 2100 ft About 13300 gpm of cold seawater was to be pumped from deep ocean through a 40in diameter pipe to serve as the cooling medium or heat sink If the cooling water experiences a temperature rise of 6F and the thermal efficiency is 25 percent determine the amount of power generated Take the density of seawater to be 64 lbmft3 627 A country needs to build new power plants to meet the increasing demand for electric power One possibility is to build coalfired power plants which cost 1300 per kW to construct and have an efficiency of 40 percent Another possibility is to build cleanburning Integrated Gasification Combined Cycle IGCC plants where the coal is subjected to heat and pressure to gasify it while removing sulfur and par ticulate matter from it The gaseous coal is then burned in a gas turbine and part of the waste heat from the exhaust gases is recovered to generate steam for the steam turbine The con struction of IGCC plants costs about 1500 per kW but their efficiency is about 48 percent The average heating value of the coal is about 28000000 kJ per ton that is 28000000 kJ of heat is released when 1 ton of coal is burned If the IGCC plant is to recover its cost difference from fuel savings in five years determine what the price of coal should be in per ton 628 Reconsider Prob 627 Using appropriate software investigate the price of coal for varying simple pay back periods plant construction costs and operating efficiency 629 Repeat Prob 627 for a simple payback period of three years instead of five years Refrigerators and Heat Pumps 630C What is the difference between a refrigerator and a heat pump 631C What is the difference between a refrigerator and an air conditioner 632C Define the coefficient of performance of a refrigera tor in words Can it be greater than unity 633C Define the coefficient of performance of a heat pump in words Can it be greater than unity 634C A heat pump that is used to heat a house has a COP of 25 That is the heat pump delivers 25 kWh of energy to the house for each 1 kWh of electricity it consumes Is this a viola tion of the first law of thermodynamics Explain 635C A refrigerator has a COP of 15 That is the refrigera tor removes 15 kWh of energy from the refrigerated space for each 1 kWh of electricity it consumes Is this a violation of the first law of thermodynamics Explain 636C In a refrigerator heat is transferred from a lower temperature medium the refrigerated space to a higher temperature one the kitchen air Is this a violation of the second law of thermodynamics Explain 637C A heat pump is a device that absorbs energy from the cold outdoor air and transfers it to the warmer indoors Is this a violation of the second law of thermodynamics Explain 638C What is the Clausius expression of the second law of thermodynamics 639C Show that the KelvinPlanck and the Clausius expressions of the second law are equivalent FIGURE P640 2 kW QH QL Reservoir Reservoir HP 640 The coefficient of performance of a residential heat pump is 16 Calculate the heating effect in kJs this heat pump will produce when it consumes 2 kW of electrical power 641 A food freezer is to produce a 5kW cooling effect and its COP is 13 How many kW of power will this refrigerator require for operation 642 An automotive air conditioner produces a 1kW cool ing effect while consuming 075 kW of power What is the rate at which heat is rejected from this air conditioner 643 A food refrigerator is to provide a 15000kJh cooling effect while rejecting 22000 kJh of heat Calculate the COP of this refrigerator Answer 214 FIGURE P643 15000 kJh 22000 kJh Win Reservoir R Reservoir Final PDF to printer 311 CHAPTER 6 cen22672ch06271322indd 311 110317 0917 AM 644 Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJh for each kW of power it consumes Also determine the rate of heat rejec tion to the outside air 645 Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJh for each kW of electric power it draws Also determine the rate of energy absorption from the outdoor air Answers 222 4400 kJh 646 A heat pump has a COP of 17 Determine the heat transferred to and from this heat pump when 50 kJ of work is supplied 647E A heat pump with a COP of 14 is to produce a 100000 Btuh heating effect How much power does this device require in hp 648 An air conditioner removes heat steadily from a house at a rate of 750 kJmin while drawing electric power at a rate of 525 kW Determine a the COP of this air conditioner and b the rate of heat transfer to the outside air Answers a 238 b 1065 kJmin 649 A household refrigerator that has a power input of 450 W and a COP of 15 is to cool 5 large watermelons 10 kg each to 8C If the watermelons are initially at 28C determine how long it will take for the refrigerator to cool them The watermelons can be treated as water whose specific heat is 42 kJkgC Is your answer realistic or optimistic Explain Answer 104 min 650 When a man returns to his wellsealed house on a sum mer day he finds that the house is at 35C He turns on the air conditioner which cools the entire house to 20C in 30 min If the COP of the airconditioning system is 28 determine the power drawn by the air conditioner Assume the entire mass within the house is equivalent to 800 kg of air for which cv 072 kJkgC and cp 10 kJkgC results and include representative costs of airconditioning units in the EER rating range 652E Water enters an ice machine at 55F and leaves as ice at 25F If the COP of the ice machine is 24 during this opera tion determine the required power input for an ice production rate of 28 lbmh 169 Btu of energy needs to be removed from each lbm of water at 55F to turn it into ice at 25F 653 A refrigerator is used to cool water from 23 to 5C in a continuous manner The heat rejected in the condenser is 570 kJmin and the power is 265 kW Determine the rate at which water is cooled in Lmin and the COP of the refrigerator The specific heat of water is 418 kJkgC and its density is 1 kgL Answers 546 Lmin 258 654 A household refrigerator runs onefourth of the time and removes heat from the food compartment at an average rate of 800 kJh If the COP of the refrigerator is 22 determine the power the refrigerator draws when running 655E Consider an office room that is being cooled ade quately by a 12000 Btuh window air conditioner Now it is decided to convert this room into a computer room by install ing several computers terminals and printers with a total rated power of 84 kW The facility has several 7000 Btuh air conditioners in storage that can be installed to meet the addi tional cooling requirements Assuming a usage factor of 04 ie only 40 percent of the rated power will be consumed at any given time and additional occupancy of seven people each generating heat at a rate of 100 W determine how many of these air conditioners need to be installed in the room 656 A house that was heated by electric resistance heaters consumed 1200 kWh of electric energy in a winter month If this house were heated instead by a heat pump that has an average COP of 24 determine how much money the homeowner would have saved that month Assume a price of 012kWh for electricity 657 Refrigerant134a enters the condenser of a residen tial heat pump at 800 kPa and 35C at a rate of 0018 kgs and leaves at 800 kPa as a saturated liquid If the compressor FIGURE P650 35C 20C Win AC FIGURE P654 Win Refrigerator 800 kJh COP 22 651 Reconsider Prob 650 Using appropriate software determine the power input required by the air conditioner to cool the house as a function for air conditioner EER ratings in the range 5 to 15 Discuss your Final PDF to printer 312 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 312 110317 0917 AM consumes 12 kW of power determine a the COP of the heat pump and b the rate of heat absorption from the outside air PerpetualMotion Machines 659C An inventor claims to have developed a resistance heater that supplies 12 kWh of energy to a room for each kWh of electricity it consumes Is this a reasonable claim or has the inventor developed a perpetualmotion machine Explain 660C It is common knowledge that the temperature of air rises as it is compressed An inventor thought about using this hightemperature air to heat buildings He used a compres sor driven by an electric motor The inventor claims that the compressed hotair system is 25 percent more efficient than a resistance heating system that provides an equivalent amount of heating Is this claim valid or is this just another perpetual motion machine Explain Reversible and Irreversible Processes 661C Why are engineers interested in reversible processes even though they can never be achieved 662C A cold canned drink is left in a warmer room where its temperature rises as a result of heat transfer Is this a revers ible process Explain 663C A block slides down an inclined plane with friction and no restraining force Is this process reversible or irrevers ible Justify your answer 664C How do you distinguish between internal and exter nal irreversibilities 665C Show that processes involving rapid chemical reac tions are irreversible by considering the combustion of a natu ral gas eg methane and air mixture in a rigid container 666C Show that processes that use work for mixing are irreversible by considering an adiabatic system whose contents are stirred by turning a paddle wheel inside the system eg stirring a cake mix with an electric mixer 667C Why does a nonquasiequilibrium compression process require a larger work input than the corresponding quasiequilibrium one 668C Why does a nonquasiequilibrium expansion process deliver less work than the corresponding quasiequilibrium one 669C Is a reversible expansion or compression process nec essarily quasiequilibrium Is a quasiequilibrium expansion or compression process necessarily reversible Explain The Carnot Cycle and Carnot Principles 670C What are the four processes that make up the Carnot cycle 671C What are the two statements known as the Carnot principles 672C Is it possible to develop a an actual and b a revers ible heatengine cycle that is more efficient than a Carnot cycle operating between the same temperature limits Explain 658 Refrigerant134a enters the evaporator coils placed at the back of the freezer section of a household refrigerator at 100 kPa with a quality of 20 percent and leaves at 100 kPa and 26C If the compressor consumes 600 W of power and the COP of the refrigerator is 12 determine a the mass flow rate of the refrigerant and b the rate of heat rejected to the kitchen air Answers a 000414 kgs b 1320 W FIGURE P657 800 kPa x 0 Win QH QL Expansion valve 800 kPa 35C Compressor Evaporator Condenser FIGURE P658 100 kPa 26C 100 kPa x 02 Win QH QL Expansion valve Compressor Evaporator Condenser Final PDF to printer 313 CHAPTER 6 cen22672ch06271322indd 313 110317 0917 AM 673C Somebody claims to have developed a new reversible heatengine cycle that has a higher theoretical efficiency than the Carnot cycle operating between the same temperature lim its How do you evaluate this claim 674C Somebody claims to have developed a new reversible heatengine cycle that has the same theoretical efficiency as the Carnot cycle operating between the same temperature lim its Is this a reasonable claim Carnot Heat Engines 675C Is there any way to increase the efficiency of a Carnot heat engine other than by increasing TH or decreasing TL 676C Consider two actual power plants operating with solar energy Energy is supplied to one plant from a solar pond at 80C and to the other from concentrating collectors that raise the water temperature to 600C Which of these power plants will have a higher efficiency Explain 677 You are an engineer in an electricgeneration station You know that the flames in the boiler reach a temperature of 1200 K and that cooling water at 300 K is available from a nearby river What is the maximum efficiency your plant will ever achieve 678 Reconsider Prob 677 You also know that the metal lurgical temperature limit for the blades in the turbine is 1000 K before they will incur excessive creep Now what is the maxi mum efficiency for this plant 679E A thermodynamicist claims to have developed a heat engine with 50 percent thermal efficiency when operating with thermal energy reservoirs at 1260 R and 510 R Is this claim valid 680E A heat engine is operating on a Carnot cycle and has a thermal efficiency of 47 percent The waste heat from this engine is rejected to a nearby lake at 60F at a rate of 800 Btumin Determine a the power output of the engine and b the tem perature of the source Answers a 167 hp b 981 R 681E A completely reversible heat engine operates with a source at 1500 R and a sink at 500 R At what rate must heat be supplied to this engine in Btuh for it to produce 5 hp of power Answer 19100 Btuh 682 An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and pro duces 300 kJ of net work while rejecting the waste heat to a sink at 290 K Is this a reasonable claim Why 683 A Carnot heat engine operates between a source at 1000 K and a sink at 300 K If the heat engine is supplied with heat at a rate of 800 kJmin determine a the thermal efficiency and b the power output of this heat engine Answers a 70 percent b 933 kW 684E A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75 percent The waste heat from this engine is rejected to a nearby lake at 60F at a rate of 800 Btumin Determine a the power output of the engine and b the tem perature of the source Answers a 566 hp b 2080 R 685 A heat engine operates between a source at 477C and a sink at 25C If heat is supplied to the heat engine at a steady rate of 65000 kJmin determine the maximum power output of this heat engine 686 Reconsider Prob 685 Using appropriate soft ware study the effects of the temperatures of the heat source and the heat sink on the power produced and the cycle thermal efficiency Let the source temperature vary from 300 to 1000C and the sink temperature to vary from 0 to 50C Plot the power produced and the cycle efficiency against the source temperature for sink temperatures of 0C 25C and 50C and discuss the results 687E An experimentalist claims that based on his measure ments a heat engine receives 300 Btu of heat from a source of 900 R converts 160 Btu of it to work and rejects the rest as waste heat to a sink at 540 R Are these measurements reasonable Why FIGURE P680E Carnot HE 800 Btumin Wnet out Sink 60F Source TH FIGURE P684E Wnetout 800 Btumin Carnot HE Source TH Sink 60F Final PDF to printer 314 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 314 110317 0917 AM 688 In tropical climates the water near the surface of the ocean remains warm throughout the year as a result of solar energy absorption In the deeper parts of the ocean how ever the water remains at a relatively low temperature since the suns rays cannot penetrate very far It is proposed to take advantage of this temperature difference and construct a power plant that will absorb heat from the warm water near the sur face and reject the waste heat to the cold water a few hundred meters below Determine the maximum thermal efficiency of such a plant if the water temperatures at the two respective locations are 24 and 3C 689 It is claimed that the efficiency of a completely revers ible heat engine can be doubled by doubling the temperature of the energy source Justify the validity of this claim Carnot Refrigerators and Heat Pumps 690C What is the highest COP that a refrigerator operating between temperature levels TL and TH can have 691C A homeowner buys a new refrigerator and a new air conditioner Which one of these devices would you expect to have a higher COP Why 692C A homeowner buys a new refrigerator with no freezer compartment and a deep freezer for the new kitchen Which of these devices would you expect to have a lower COP Why 693C How can we increase the COP of a Carnot refrigerator 694C In an effort to conserve energy in a heatengine cycle somebody suggests incorporating a refrigerator that will absorb some of the waste energy QL and transfer it to the energy source of the heat engine Is this a smart idea Explain 695C It is well established that the thermal efficiency of a heat engine increases as the temperature TL at which heat is rejected from the heat engine decreases In an effort to increase the efficiency of a power plant somebody suggests refrigerating the cooling water before it enters the condenser where heat rejection takes place Would you be in favor of this idea Why 696C It is well known that the thermal efficiency of heat engines increases as the temperature of the energy source increases In an attempt to improve the efficiency of a power plant somebody suggests transferring heat from the available energy source to a highertemperature medium by a heat pump before energy is supplied to the power plant What do you think of this suggestion Explain 697 A thermodynamicist claims to have developed a heat pump with a COP of 17 when operating with thermal energy reservoirs at 273 K and 293 K Is this claim valid 698 Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with thermal energy reservoirs at 460 K and 535 K 699 A completely reversible refrigerator is driven by a 10kW compressor and operates with thermal energy res ervoirs at 250 K and 300 K Calculate the rate of cooling provided by this refrigerator Answer 50 kW 6100 An airconditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 750 kJmin to maintain its temperature at 24C If the outdoor air temperature is 35C determine the power required to operate this airconditioning system Answer 0463 kW 6101 A heat pump operates on a Carnot heat pump cycle with a COP of 125 It keeps a space at 24C by consuming 215 kW of power Determine the temperature of the reservoir from which the heat is absorbed and the heating load provided by the heat pump Answers 273 K 269 kW 6102E An airconditioning system is used to maintain a house at 70F when the temperature outside is 100F The house is gaining heat through the walls and the windows at a rate of 800 Btumin and the heat generation rate within the house from people lights and appliances amounts to 100 Btumin Determine the minimum power input required for this air conditioning system Answer 120 hp 6103 A Carnot refrigerator absorbs heat from a space at 15C at a rate of 16000 kJh and rejects heat to a reservoir at FIGURE P688 Boiler Turbine Condenser Pump 24C Ocean 3C Final PDF to printer 315 CHAPTER 6 cen22672ch06271322indd 315 110317 0917 AM 6104E A completely reversible refrigerator operates between thermal energy reservoirs at 450 R and 540 R How many kilowatts of power are required for this device to produce a 15000Btuh cooling effect 6105 A Carnot refrigerator operates in a room in which the temperature is 25C The refrigerator consumes 500 W of power when operating and has a COP of 45 Determine a the rate of heat removal from the refrigerated space and b the temperature of the refrigerated space Answers a 135 kJmin b 292C 6106 A heat pump is used to heat a house and maintain it at 24C On a winter day when the outdoor air temperature is 5C the house is estimated to lose heat at a rate of 80000 kJh Determine the minimum power required to operate this heat pump 6107 A commercial refrigerator with refrigerant134a as the working fluid is used to keep the refrigerated space at 35C by rejecting waste heat to cooling water that enters the condenser at 18C at a rate of 025 kgs and leaves at 26C The refrigerant enters the condenser at 12 MPa and 50C and leaves at the same pressure subcooled by 5C If the compres sor consumes 33 kW of power determine a the mass flow rate of the refrigerant b the refrigeration load c the COP 6108 The performance of a heat pump degrades ie its COP decreases as the temperature of the heat source decreases This makes using heat pumps at locations with severe weather conditions unattractive Consider a house that is heated and maintained at 20C by a heat pump during the winter What is the maximum COP for this heat pump if heat is extracted from the outdoor air at a 10C b 5C and c 30C 6109E A heat pump is to be used for heating a house in win ter The house is to be maintained at 78F at all times When the temperature outdoors drops to 25F the heat losses from the house are estimated to be 70000 Btuh Determine the minimum power required to run this heat pump if heat is extracted from a the outdoor air at 25F and b the well water at 50F 6110 A completely reversible heat pump has a COP of 16 and a sink temperature of 300 K Calculate a the temperature of the source and b the rate of heat transfer to the sink when 15 kW of power is supplied to this heat pump FIGURE P6103 Win 16000 kJh 36C R 15C FIGURE P6107 Expansion valve 12 MPa 5C subcooled 12 MPa 50C Compressor Evaporator Condenser Win QL Water 18C 26C 36C Determine the COP of the refrigerator the power input in kW and the rate of heat rejected to hightemperature reser voir in kJh FIGURE P6110 15 kW QH QL HP 300 K TL and d the minimum power input to the compressor for the same refrigeration load Final PDF to printer 316 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 316 110317 0917 AM 6111 A Carnot heat pump is to be used to heat a house and maintain it at 25C in winter On a day when the average out door temperature remains at about 2C the house is estimated to lose heat at a rate of 55000 kJh If the heat pump consumes 48 kW of power while operating determine a how long the heat pump ran on that day b the total heating costs assuming an average price of 011kWh for electricity and c the heat ing cost for the same day if resistance heating is used instead of a heat pump Answers a 590 h b 311 c 403 Special Topic Household Refrigerators 6116C Why are todays refrigerators much more efficient than those built in the past 6117C Why is it important to clean the condenser coils of a household refrigerator a few times a year Also why is it important not to block airflow through the condenser coils 6118C Someone proposes that the refrigeration system of a supermarket be overdesigned so that the entire air conditioning needs of the store can be met by refrigerated air without installing any airconditioning system What do you think of this proposal 6119C Someone proposes that the entire refrigerator freezer requirements of a store be met using a large freezer that supplies sufficient cold air at 20C instead of installing separate refrigerators and freezers What do you think of this proposal 6120 Explain how you can reduce the energy consumption of your household refrigerator 6121 The Energy Guide label of a refrigerator states that the refrigerator will consume 170 worth of electricity per year under normal use if the cost of electricity is 0125kWh If the electricity consumed by the lightbulb is negligible and the refrigerator consumes 400 W when running determine the fraction of the time the refrigerator will run 6122 The interior lighting of refrigerators is usually pro vided by incandescent lamps whose switches are actuated by the opening of the refrigerator door Consider a refrig erator whose 40W lightbulb remains on about 60 h per year It is proposed to replace the lightbulb with an energy efficient bulb that consumes only 18 W but costs 25 to purchase and install If the refrigerator has a coefficient of performance of 13 and the cost of electricity is 013kWh determine if the energy savings of the proposed lightbulb justify its cost 6123 It is commonly recommended that hot foods be cooled first to room temperature by simply waiting a while before they are put into the refrigerator to save energy Despite this com monsense recommendation a person keeps cooking a large pan of stew three times a week and putting the pan into the refrigerator while it is still hot thinking that the money saved is probably too little But he says he can be convinced if you can show that the money saved is significant The average mass of the pan and its contents is 5 kg The average temperature of the kitchen is 23C and the average temperature of the food is 95C when it is taken off the stove The refrigerated space is maintained at 3C and the average specific heat of the food and the pan can be taken to be 39 kJkgC If the refrigerator has a coefficient of performance of 15 and the cost of electric ity is 0125kWh determine how much this person will save a FIGURE P6111 25C 55000 kJh HP 48 kW 2C 6112 A Carnot heat engine receives heat from a reservoir at 900C at a rate of 800 kJmin and rejects the waste heat to the ambient air at 27C The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrig erated space at 5C and transfers it to the same ambient air at 27C Determine a the maximum rate of heat removal from the refrigerated space and b the total rate of heat rejection to the ambient air Answers a 4982 kJmin b 5782 kJmin 6113 The structure of a house is such that it loses heat at a rate of 4500kJh per C difference between the indoors and outdoors A heat pump that requires a power input of 4 kW is used to maintain this house at 24C Determine the lowest out door temperature for which the heat pump can meet the heat ing requirements of this house Answer 68C 6114 Derive an expression for the COP of a completely reversible refrigerator in terms of the thermal energy reservoir temperatures TL and TH 6115 Calculate and plot the COP of a completely reversible refrigerator as a function of the tem perature of the sink up to 500 K with the temperature of the source fixed at 250 K Final PDF to printer 317 CHAPTER 6 cen22672ch06271322indd 317 110317 0917 AM 6124 It is often stated that the refrigerator door should be opened as few times as possible for the shortest duration of time to save energy Consider a household refrigerator whose interior volume is 09 m3 and average internal temperature is 4C At any given time onethird of the refrigerated space is occupied by food items and the remaining 06 m3 is filled with air The average temperature and pressure in the kitchen are 20C and 95 kPa respectively Also the moisture con tents of the air in the kitchen and the refrigerator are 0010 and 0004 kg per kg of air respectively and thus 0006 kg of water vapor is condensed and removed for each kg of air that enters The refrigerator door is opened an average of 20 times a day and each time half of the air volume in the refrigerator is replaced by the warmer kitchen air If the refrigerator has a coefficient of performance of 14 and the cost of electricity is 0115 kWh determine the cost of the energy wasted per year as a result of opening the refrigerator door What would your answer be if the kitchen air were very dry and thus a negligible amount of water vapor condensed in the refrigerator Review Problems 6125 A manufacturer of ice cream freezers claims that its product has a coefficient of performance of 13 while freezing ice cream at 250 K when the surrounding environment is at 300 K Is this claim valid 6126 A heat pump designer claims to have an airsource heat pump whose coefficient of performance is 18 when heat ing a building whose interior temperature is 300 K and when the atmospheric air surrounding the building is at 260 K Is this claim valid 6127 An airconditioning system is used to maintain a house at a constant temperature of 20C The house is gaining heat from outdoors at a rate of 20000 kJh and the heat generated in the house from the people lights and appliances amounts to 8000 kJh For a COP of 25 determine the required power input to this airconditioning system Answer 311 kW 6128E A Carnot heat pump is used to heat and maintain a residential building at 75F An energy analysis of the house reveals that it loses heat at a rate of 2500 Btuh per F tem perature difference between the indoors and the outdoors For an outdoor temperature of 35F determine a the coefficient of performance and b the required power input to the heat pump Answers a 134 b 293 hp 6129E A refrigeration system uses a watercooled con denser for rejecting the waste heat The system absorbs heat from a space at 25F at a rate of 21000 Btuh Water enters the condenser at 65F at a rate of 145 lbms The COP of the system is estimated to be 19 Determine a the power input to the system in kW b the temperature of the water at the exit of the condenser in F and c the maximum possible COP of the system The specific heat of water is 10 BtubmF 6130 A refrigeration system is to cool bread loaves with an average mass of 350 g from 30 to 10C at a rate of 1200 loaves per hour with refrigerated air at 30C Taking the average specific and latent heats of bread to be 293 kJkgC and 1093 kJkg respectively determine a the rate of heat removal from the breads in kJh b the required volume flow rate of air in m3h if the temperature rise of air is not to exceed 8C and c the size of the compressor of the refrig eration system in kW for a COP of 12 for the refrigeration system 6131 A heat pump with a COP of 28 is used to heat an airtight house When running the heat pump consumes 5 kW of power If the temperature in the house is 7C when the heat pump is turned on how long will it take for the heat pump to raise the temperature of the house to 22C Is this answer realistic or optimistic Explain Assume the entire mass within the house air furniture etc is equivalent to 1500 kg of air Answer 192 min 6132 A promising method of power generation involves collecting and storing solar energy in large artificial lakes a few meters deep called solar ponds Solar energy is absorbed by all parts of the pond and the water temperature rises everywhere The top part of the pond however loses to the atmosphere much of the heat it absorbs and as a result its temperature drops This cool water serves as insulation for the bottom part of the pond and helps trap the energy there Usually salt is planted at the bottom of the pond to prevent the rise of this hot water to the top A power plant that uses an organic fluid such as alcohol as the working fluid can be operated between the top and the bottom portions of the FIGURE P6123 23C 3C Hot food 95C year by waiting for the food to cool to room temperature before putting it into the refrigerator Final PDF to printer 318 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 318 110317 0917 AM 6133 Consider a Carnot heatengine cycle executed in a closed system using 0025 kg of steam as the working fluid It is known that the maximum absolute temperature in the cycle is twice the minimum absolute temperature and the net work output of the cycle is 60 kJ If the steam changes from saturated vapor to saturated liquid during heat rejection determine the temperature of the steam during the heat rejec tion process 6134 Reconsider Prob 6133 Using appropriate software investigate the effect of the net work output on the required temperature of the steam during the heat rejection process Let the work output vary from 40 to 60 kJ 6135 Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquidvapor mixture region using 096 kg of refrigerant134a as the working fluid It is known that the maximum absolute temperature in the cycle is 12 times the minimum absolute temperature and the net work input to the cycle is 22 kJ If the refrigerant changes from satu rated vapor to saturated liquid during the heat rejection pro cess determine the minimum pressure in the cycle 6136 Reconsider Prob 6135 Using appropriate software investigate the effect of the net work input on the minimum pressure Let the work input vary from 10 to 30 kJ Plot the minimum pressure in the refrig eration cycle as a function of net work input and discuss the results 6137 Consider a Carnot heatengine cycle executed in a steadyflow system using steam as the working fluid The cycle has a thermal efficiency of 30 percent and steam changes from saturated liquid to saturated vapor at 275C during the heat addition process If the mass flow rate of the steam is 3 kgs determine the net power output of this engine in kW 6138 Consider two Carnot heat engines operating in series The first engine receives heat from the reservoir at 1400 K and rejects the waste heat to another reservoir at tem perature T The second engine receives this energy rejected by the first one converts some of it to work and rejects the rest to a reservoir at 300 K If the thermal efficiencies of both engines are the same determine the temperature T Answer 648 K 6139 A heat engine operates between two reservoirs at 800 and 20C Onehalf of the work output of the heat engine is used to drive a Carnot heat pump that removes heat from the cold surroundings at 2C and transfers it to a house maintained at 22C If the house is losing heat at a rate of 62000 kJh determine the minimum rate of heat supply to the heat engine required to keep the house at 22C 6140 An old gas turbine has an efficiency of 21 percent and develops a power output of 6000 kW Determine the fuel con sumption rate of this gas turbine in Lmin if the fuel has a heating value of 42000 kJkg and a density of 08 gcm3 6141 Consider a Carnot heatpump cycle executed in a steadyflow system in the saturated liquidvapor mixture region using refrigerant134a flowing at a rate of 018 kgs as the working fluid It is known that the maximum absolute temperature in the cycle is 12 times the minimum absolute temperature and the net power input to the cycle is 5 kW If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process determine the ratio of the maximum to minimum pressures in the cycle 6142 The cargo space of a refrigerated truck whose inner dimensions are 12 m 23 m 35 m is to be precooled from 25C to an average temperature of 5C The construction of the FIGURE P6132 Turbine Condenser Pump 80C 35C Solar pond Boiler pond If the water temperature is 35C near the surface and 80C near the bottom of the pond determine the maximum thermal efficiency that this power plant can have Is it real istic to use 35 and 80C for temperatures in the calculations Explain Answer 127 percent Final PDF to printer 319 CHAPTER 6 cen22672ch06271322indd 319 110317 0917 AM FIGURE P6144 Water reservoir Refrigeration system Water fountain Water inlet 22C 04 Lhperson Cold water 8C 25C truck is such that a transmission heat gain occurs at a rate of 120 WC If the ambient temperature is 25C determine how long it will take for a system with a refrigeration capacity of 11 kW to precool this truck 6145 A typical electric water heater has an efficiency of 95 percent and costs 350 a year to operate at a unit cost of electricity of 011kWh A typical heat pumppowered water heater has a COP of 33 but costs about 800 more to install Determine how many years it will take for the heat pump water heater to pay for its cost differential from the energy it saves FIGURE P6142 25C 120 WC Refrigerated truck 12 m 23 m 35 m 25 to 5C 6143 The maximum flow rate of a standard shower head is about 35 gpm 133 Lmin and can be reduced to 275 gpm 105 Lmin by switching to a lowflow shower head that is equipped with flow controllers Consider a family of four with each person taking a 6minute shower every morning City water at 15C is heated to 55C in an oil water heater whose efficiency is 65 percent and then tempered to 42C by cold water at the Telbow of the shower before being routed to the shower head The price of heating oil is 280gal and its heat ing value is 146300 kJgal Assuming a constant specific heat of 418 kJkgC for water determine the amount of oil and money saved per year by replacing the standard shower heads with the lowflow ones 6144 The drinking water needs of a production facility with 20 employees is to be met by a bubblertype water foun tain The refrigerated water fountain is to cool water from 22 to 8C and supply cold water at a rate of 04 L per hour per person Heat is transferred to the reservoir from the surround ings at 25C at a rate of 45 W If the COP of the refrigeration system is 29 determine the size of the compressor in W FIGURE P6145 Water heater McGrawHill EducationChristopher Kerrigan that will be suitable for the refrigeration system of this water cooler Final PDF to printer 320 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 320 110317 0917 AM 6146 Reconsider Prob 6145 Using appropriate software investigate the effect of the heat pump COP on the yearly operation costs and the number of years required to break even Let the COP vary from 2 to 5 Plot the payback period against the COP and discuss the results 6147 A homeowner is trying to decide between a high efficiency natural gas furnace with an efficiency of 97 percent and a groundsource heat pump with a COP of 35 The unit costs of electricity and natural gas are 0115kWh and 075therm 1 therm 105500 kJ Determine which system will have a lower energy cost 6148 Replacing incandescent lights with energyefficient fluorescent lights can reduce the lighting energy consumption to onefourth of what it was before The energy consumed by the lamps is eventually converted to heat and thus switching to energyefficient lighting also reduces the cooling load in sum mer but increases the heating load in winter Consider a build ing that is heated by a natural gas furnace with an efficiency of 80 percent and cooled by an air conditioner with a COP of 35 If electricity costs 012kWh and natural gas costs 140therm 1 therm 105500 kJ determine if efficient lighting will increase or decrease the total energy cost of the building a in summer and b in winter 6149 A heat pump supplies heat energy to a house at the rate of 140000 kJh when the house is maintained at 25C Over a period of one month the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house Consider a heat pump receiving heat from two different outside energy sources In one application the heat pump receives heat from the outside air at 0C In a second application the heat pump receives heat from a lake having a water temperature of 10C If electricity costs 012 kWh determine the maximum amount of money saved by using the lake water rather than the outside air as the outside energy source 6150 The kitchen bath and other ventilation fans in a house should be used sparingly since these fans can discharge a houseful of warmed or cooled air in just one hour Consider a 200m2 house whose ceiling height is 28 m The house is heated by a 96 percent efficient gas heater and is maintained at 22C and 92 kPa If the unit cost of natural gas is 120therm 1 therm 105500 kJ determine the cost of energy vented out by the fans in 1 h Assume the average outdoor tempera ture during the heating season to be 5C 6151 Repeat Prob 6150 for the airconditioning cost in a dry climate for an outdoor temperature of 33C Assume the COP of the airconditioning system to be 21 and the unit cost of electricity to be 012kWh 6152 A heat pump with refrigerant134a as the work ing fluid is used to keep a space at 25C by absorbing heat from geothermal water that enters the evaporator at 60C at a rate of 0065 kgs and leaves at 40C Refrigerant enters the evaporator at 12C with a quality of 15 percent and leaves at the same pressure as saturated vapor If the compressor con sumes 16 kW of power determine a the mass flow rate of the refrigerant b the rate of heat supply c the COP and d the minimum power input to the compressor for the same rate of heat supply Answers a 00338 kgs b 704 kW c 440 d 0740 kW FIGURE P6152 12C x 015 Sat vapor Geo water 60C 40C Expansion valve Compressor Evaporator Condenser Win FIGURE P6153 HP Water inlet Water heater Water exit Surroundings 0C Win QH QL 6153 Cold water at 10C enters a water heater at the rate of 002 m3min and leaves the water heater at 50C The water heater receives heat from a heat pump that receives heat from a heat source at 0C a Assuming the water to be an incompressible liquid that does not change phase during heat addition determine the rate of heat supplied to the water in kJs b Assuming the water heater acts as a heat sink having an average temperature of 30C determine the minimum power supplied to the heat pump in kW Final PDF to printer 321 CHAPTER 6 cen22672ch06271322indd 321 110317 0917 AM 6154 Using appropriate software determine the max imum work that can be extracted from a pond containing 105 kg of water at 350 K when the temperature of the surroundings is 300 K Notice that the temperature of water in the pond will be gradually decreasing as energy is extracted from it therefore the efficiency of the engine will be decreas ing Use temperature intervals of a 5 K b 2 K and c 1 K until the pond temperature drops to 300 K Also solve this problem exactly by integration and compare the results 6155 A Carnot heat engine is operating between a source at TH and a sink at TL If we wish to double the thermal efficiency of this engine what should the new source temperature be Assume the sink temperature is held constant 6156E Calculate and plot the thermal efficiency of a completely reversible heat engine as a function of the source temperature up to 2000 R with the sink tempera ture fixed at 500 R 6157 Show that COPHP COPR 1 when both the heat pump and the refrigerator have the same QL and QH values 6158 Prove that a refrigerators COP cannot exceed that of a completely reversible refrigerator that shares the same ther mal energy reservoirs Fundamentals of Engineering FE Exam Problems 6159 Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reser voirs If the COP of the refrigerator is 34 the COP of the heat pump is a 17 b 24 c 34 d 44 e 50 6160 A 24mhigh 200m2 house is maintained at 22C by an airconditioning system whose COP is 32 It is estimated that the kitchen bath and other ventilating fans of the house discharge a houseful of conditioned air once every hour If the average outdoor temperature is 32C the density of air is 120 kgm3 and the unit cost of electricity is 010kWh the amount of money vented out by the fans in 10 hours is a 050 b 16 c 50 d 110 e 160 6161 A window air conditioner that consumes 1 kW of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room and is plugged in The rate of cooling or heating this air conditioner will provide to the air in the room when running is a 3 kJs cooling b 1 kJs cooling c 033 kJs heating d 1 kJs heating e 3 kJs heating 6162 The drinking water needs of an office are met by cool ing tap water in a refrigerated water fountain from 23 to 6C at an average rate of 18 kgh If the COP of this refrigerator is 31 the required power input to this refrigerator is a 1100 W b 355 W c 195 W d 115 W e 35 W 6163 The label on a washing machine indicates that the washer will use 85 worth of hot water if the water is heated by a 90 percent efficient electric heater at an electricity rate of 0125kWh If the water is heated from 18 to 45C the amount of hot water an average family uses per year is a 195 tons b 217 tons c 241 tons d 272 tons e 304 tons 6164 A heat pump is absorbing heat from the cold out doors at 5C and supplying heat to a house at 25C at a rate of 18000 kJh If the power consumed by the heat pump is 19 kW the coefficient of performance of the heat pump is a 13 b 26 c 30 d 38 e 139 6165 A heat engine cycle is executed with steam in the saturation dome The pressure of steam is 1 MPa during heat addition and 04 MPa during heat rejection The highest pos sible efficiency of this heat engine is a 80 b 156 c 202 d 798 e 100 6166 A heat pump cycle is executed with R134a under the saturation dome between the pressure limits of 12 and 016 MPa The maximum coefficient of performance of this heat pump is a 57 b 52 c 48 d 45 e 41 6167 A refrigeration cycle is executed with R134a under the saturation dome between the pressure limits of 16 and 02 MPa If the power consumption of the refrigerator is 3 kW the maxi mum rate of heat removal from the cooled space of this refrig erator is a 045 kJs b 078 kJs c 30 kJs d 116 kJs e 146 kJs 6168 A heat pump with a COP of 32 is used to heat a per fectly sealed house no air leaks The entire mass within the house air furniture etc is equivalent to 1200 kg of air When running the heat pump consumes electric power at a rate of 5 kW The temperature of the house was 7C when the heat pump was turned on If heat transfer through the envelope of the house walls roof etc is negligible the length of time the heat pump must run to raise the temperature of the entire contents of the house to 22C is a 135 min b 431 min c 138 min d 188 min e 808 min 6169 A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 and 2 MPa If heat is supplied to the heat engine at a rate of 150 kJs the maximum power output of this heat engine is a 81 kW b 197 kW c 386 kW d 107 kW e 130 kW 6170 A heat engine receives heat from a source at 1000C and rejects the waste heat to a sink at 50C If heat is supplied to this engine at a rate of 100 kJs the maximum power this heat engine can produce is a 254 kW b 554 kW c 746 kW d 950 kW e 100 kW 6171 An airconditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate Final PDF to printer 322 THE SECOND LAW OF THERMODYNAMICS cen22672ch06271322indd 322 110317 0917 AM of 32 kJs to keep its temperature constant at 20C If the tem perature of the outdoors is 35C the power required to operate this airconditioning system is a 058 kW b 320 kW c 156 kW d 226 kW e 164 kW 6172 A refrigerator is removing heat from a cold medium at 3C at a rate of 5400 kJh and rejecting the waste heat to a medium at 30C If the coefficient of performance of the refrigerator is 2 the power consumed by the refrigerator is a 05 kW b 075 kW c 10 kW d 15 kW e 30 kW 6173 Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same the temperature of the intermediate reservoir is a 625 K b 800 K c 860 K d 453 K e 758 K 6174 A typical new household refrigerator consumes about 680 kWh of electricity per year and has a coefficient of perfor mance of 14 The amount of heat removed by this refrigerator from the refrigerated space per year is a 952 MJyr b 1749 MJyr c 2448 MJyr d 3427 MJyr e 4048 MJyr Design and Essay Problems 6175 Show that the work produced by a reversible process exceeds that produced by an equivalent irreversible process by considering a weight moving down a plane both with and with out friction 6176 Devise a Carnot heat engine using steadyflow com ponents and describe how the Carnot cycle is executed in that engine What happens when the directions of heat and work interactions are reversed 6177 The sun supplies electromagnetic energy to the earth It appears to have an effective temperature of approximately 5800 K On a clear summer day in North America the energy incident on a surface facing the sun is approximately 095 kWm2 The electromagnetic solar energy can be con verted into thermal energy by being absorbed on a darkened surface How might you characterize the work potential of the suns energy when it is to be used to produce work 6178 Using a thermometer measure the temperature of the main food compartment of your refrigerator and check if it is between 1 and 4C Also measure the temperature of the freezer compartment and check if it is at the recommended value of 18C 6179 Using a timer or watch and a thermometer conduct the following experiment to determine the rate of heat gain of your refrigerator First make sure that the door of the refrig erator is not opened for at least a few hours so that steady operating conditions are established Start the timer when the refrigerator stops running and measure the time ΔT1 it stays off before it kicks in Then measure the time ΔT2 it stays on Noting that the heat removed during ΔT2 is equal to the heat gain of the refrigerator during ΔT1 ΔT2 and using the power consumed by the refrigerator when it is running determine the average rate of heat gain for your refrigerator in W Take the COP coefficient of performance of your refrigerator to be 13 if it is not available 6180 Design a hydrocooling unit that can cool fruits and vegetables from 30 to 5C at a rate of 20000 kgh under the following conditions The unit will be of flood type which will cool the products as they are conveyed into the channel filled with water The products will be dropped into the channel filled with water at one end and be picked up at the other end The channel can be as wide as 3 m and as high as 90 cm The water is to be cir culated and cooled by the evaporator section of a refrigeration system The refrigerant temperature inside the coils is to be 2C and the water temperature is not to drop below 1C and not to exceed 6C Assuming reasonable values for the average product density specific heat and porosity the fraction of air volume in a box recommend reasonable values for a the water velocity through the channel and b the refrigeration capacity of the refrigeration system 6181 In the search to reduce thermal pollution and take advantage of renewable energy sources some people have proposed that we take advantage of such sources as discharges from electric power plants geothermal energy and ocean ther mal energy Although many of these sources contain an enor mous amount of energy the amount of work they are capable of producing is limited How might you use the work potential to assign an energy quality to these proposed sources Test your proposed energy quality measure by applying it to the ocean thermal source where the temperature 30 m below the surface is perhaps 5C lower than at the surface Apply it also to the geothermal water source where the temperature 2 to 3 km below the surface is perhaps 150C hotter than at the surface Final PDF to printer cen22672ch07323412indd 323 110617 0852 AM 323 OBJECTIVES The objectives of Chapter 7 are to Apply the second law of thermodynamics to processes Define a new property called entropy to quantify the second law effects Establish the increase of entropy principle Calculate the entropy changes that take place during processes for pure substances incompressible substances and ideal gases Examine a special class of idealized processes called isentropic processes and develop the property relations for these processes Derive the reversible steadyflow work relations Develop the isentropic efficiencies for various steadyflow devices Introduce and apply the entropy balance to various systems EN T R O PY I n Chap 6 we introduced the second law of thermodynamics and applied it to cycles and cyclic devices In this chapter we apply the second law to processes The first law of thermodynamics deals with the property energy and the conservation of it The second law leads to the definition of a new property called entropy Entropy is a somewhat abstract property and it is dif ficult to give a physical description of it without considering the microscopic state of the system Entropy is best understood and appreciated by studying its uses in commonly encountered engineering processes and this is what we intend to do This chapter starts with a discussion of the Clausius inequality which forms the basis for the definition of entropy and continues with the increase of entropy principle Unlike energy entropy is a nonconserved property and there is no such thing as conservation of entropy Next the entropy changes that take place during processes for pure substances incompress ible substances and ideal gases are discussed and a special class of ideal ized processes called isentropic processes is examined Then the reversible steadyflow work and the isentropic efficiencies of various engineering devices such as turbines and compressors are considered Finally entropy bal ance is introduced and applied to various systems 7 CHAPTER Final PDF to printer 324 ENTROPY cen22672ch07323412indd 324 110617 0852 AM 71 ENTROPY The second law of thermodynamics often leads to expressions that involve inequalities An irreversible ie actual heat engine for example is less effi cient than a reversible one operating between the same two thermal energy reservoirs Likewise an irreversible refrigerator or a heat pump has a lower coefficient of performance COP than a reversible one operating between the same temperature limits Another important inequality that has major conse quences in thermodynamics is the Clausius inequality It was first stated by the German physicist R J E Clausius 18221888 one of the founders of thermodynamics and is expressed in 1865 as δQ T 0 That is the cyclic integral of δQT is always less than or equal to zero This inequality is valid for all cycles reversible or irreversible The symbol integral symbol with a circle in the middle is used to indicate that the integration is to be performed over the entire cycle Any heat transfer to or from a system can be considered to consist of differential amounts of heat transfer Then the cyclic integral of δQT can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature at the boundary To demonstrate the validity of the Clausius inequality consider a sys tem connected to a thermal energy reservoir at a constant thermodynamic ie absolute temperature of TR through a reversible cyclic device Fig 71 The cyclic device receives heat δQR from the reservoir and supplies heat δQ to the system whose temperature at that part of the boundary is T a variable while producing work δWrev The system produces work δWsys as a result of this heat transfer Applying the energy balance to the combined system identi fied by dashed lines yields δ W C δ Q R d E C where δWC is the total work of the combined system δWrev δWsys and dEC is the change in the total energy of the combined system Considering that the cyclic device is a reversible one we have δ Q R T R δQ T where the sign of δQ is determined with respect to the system positive if to the system and negative if from the system and the sign of δQR is determined with respect to the reversible cyclic device Eliminating δQR from the two preceding relations yields δ W C T R δQ L d E C We now let the system undergo a cycle while the cyclic device undergoes an integral number of cycles Then the preceding relation becomes W C T R δQ T FIGURE 71 The system considered in the development of the Clausius inequality Combined system system and cyclic device Reversible cyclic device δWsys δWrev T System δQ δQR Thermal reservoir TR Final PDF to printer 325 CHAPTER 7 cen22672ch07323412indd 325 110617 0852 AM since the cyclic integral of energy the net change in the energy which is a property during a cycle is zero Here WC is the cyclic integral of δWC and it represents the net work for the combined cycle It appears that the combined system is exchanging heat with a single ther mal energy reservoir while involving producing or consuming work WC dur ing a cycle On the basis of the KelvinPlanck statement of the second law which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir we rea son that WC cannot be a work output and thus it cannot be a positive quantity Considering that TR is the thermodynamic temperature and thus a positive quantity we must have δQ T 0 71 which is the Clausius inequality This inequality is valid for all thermody namic cycles reversible or irreversible including the refrigeration cycles If no irreversibilities occur within the system as well as the revers ible cyclic device then the cycle undergone by the combined system is internally reversible As such it can be reversed In the reversed cycle case all the quantities have the same magnitude but the opposite sign Therefore the work WC which could not be a positive quantity in the regular case cannot be a negative quantity in the reversed case Then it follows that WCint rev 0 since it cannot be a positive or negative quantity and therefore δQ T int rev 0 72 for internally reversible cycles Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones To develop a relation for the definition of entropy let us examine Eq 72 more closely Here we have a quantity whose cyclic integral is zero Let us think for a moment what kinds of quantities can have this characteristic We know that the cyclic integral of work is not zero It is a good thing that it is not Otherwise heat engines that work on a cycle such as steam power plants would produce zero net work Neither is the cyclic integral of heat Now consider the volume occupied by a gas in a pistoncylinder device undergoing a cycle as shown in Fig 72 When the piston returns to its initial position at the end of a cycle the volume of the gas also returns to its initial value Thus the net change in volume during a cycle is zero This is also expressed as dV 0 73 FIGURE 72 The net change in volume a property during a cycle is always zero 1 m3 3 m3 dV ΔVcycle 0 1 m3 Final PDF to printer 326 ENTROPY cen22672ch07323412indd 326 110617 0852 AM That is the cyclic integral of volume or any other property is zero Con versely a quantity whose cyclic integral is zero depends on the state only and not the process path and thus it is a property Therefore the quantity δQTint rev must represent a property in the differential form Clausius realized in 1865 that he had discovered a new thermodynamic property and he chose to name this property entropy It is designated S and is defined as dS δQ T int rev kJ K 74 Entropy is an extensive property of a system and sometimes is referred to as total entropy Entropy per unit mass designated s is an intensive property and has the unit kJkgK The term entropy is generally used to refer to both total entropy and entropy per unit mass since the context usually clarifies which one is meant The entropy change of a system during a process can be determined by inte grating Eq 74 between the initial and the final states ΔS S 2 S 1 1 2 δQ T int rev kJ K 75 Notice that we have actually defined the change in entropy instead of entropy itself just as we defined the change in energy instead of the energy itself when we developed the firstlaw relation Absolute values of entropy are determined on the basis of the third law of thermodynamics which is discussed later in this chapter Engineers are usually concerned with the changes in entropy Therefore the entropy of a substance can be assigned a zero value at some arbitrarily selected reference state and the entropy values at other states can be determined from Eq 75 by choosing state 1 to be the reference state S 0 and state 2 to be the state at which entropy is to be determined To perform the integration in Eq 75 one needs to know the relation between Q and T during a process This relation is often not available and the integral in Eq 75 can be performed for a few cases only For the majority of cases we have to rely on tabulated data for entropy Note that entropy is a property and like all other properties it has fixed val ues at fixed states Therefore the entropy change ΔS between two specified states is the same no matter what path reversible or irreversible is followed during a process Fig 73 Also note that the integral of δQT gives us the value of entropy change only if the integration is carried out along an internally reversible path between the two states The integral of δQT along an irreversible path is not a prop erty and in general different values will be obtained when the integration is carried out along different irreversible paths Therefore even for irreversible processes the entropy change should be determined by carrying out this inte gration along some convenient imaginary internally reversible path between the specified states FIGURE 73 The entropy change between two specified states is the same whether the process is reversible or irreversible Irreversible process Reversible process 1 2 03 07 S kJK ΔS S2 S1 04 kJK T Final PDF to printer 327 CHAPTER 7 cen22672ch07323412indd 327 110617 0852 AM A Special Case Internally Reversible Isothermal Heat Transfer Processes Recall that isothermal heat transfer processes are internally reversible There fore the entropy change of a system during an internally reversible isother mal heat transfer process can be determined by performing the integration in Eq 75 ΔS 1 2 δQ T int rev 1 2 δQ T 0 int rev 1 T 0 1 2 δQ int rev which reduces to ΔS Q T 0 kJ K 76 where T0 is the constant temperature of the system and Q is the heat transfer for the internally reversible process Equation 76 is particularly useful for determining the entropy changes of thermal energy reservoirs that can absorb or supply heat indefinitely at a constant temperature Notice that the entropy change of a system during an internally reversible isothermal process can be positive or negative depending on the direction of heat transfer Heat transfer to a system increases the entropy of a system whereas heat transfer from a system decreases it In fact losing heat is the only way the entropy of a system can be decreased EXAMPLE 71 Entropy Change During an Isothermal Process A pistoncylinder device contains a liquidvapor mixture of water at 300 K During a constantpressure process 750 kJ of heat is transferred to the water As a result part of the liquid in the cylinder vaporizes Determine the entropy change of the water during this process SOLUTION Heat is transferred to a liquidvapor mixture of water in a piston cylinder device at constant pressure The entropy change of water is to be determined Assumptions No irreversibilities occur within the system boundaries during the process Analysis We take the entire water liquid vapor in the cylinder as the system Fig 74 This is a closed system since no mass crosses the system boundary during the process We note that the temperature of the system remains constant at 300 K during this process since the temperature of a pure substance remains constant at the saturation value during a phasechange process at constant pressure The system undergoes an internally reversible isothermal process and thus its entropy change can be determined directly from Eq 76 to be Δ S sysisothermal Q T sys 750 kJ 300 K 25 kJ K Discussion Note that the entropy change of the system is positive as expected since heat transfer is to the system FIGURE 74 Schematic for Example 71 Q 750 kJ T T 300 K const ΔSsys Q 25 kJ K Final PDF to printer 328 ENTROPY cen22672ch07323412indd 328 110617 0852 AM 72 THE INCREASE OF ENTROPY PRINCIPLE Consider a cycle that is made up of two processes process 12 which is arbi trary reversible or irreversible and process 21 which is internally revers ible as shown in Fig 75 From the Clausius inequality δQ T 0 or 1 2 δQ T 2 1 δQ T int rev 0 The second integral in the previous relation is recognized as the entropy change S1 S2 Therefore 1 2 δQ T S 1 S 2 0 which can be rearranged as S 2 S 1 1 2 δQ T 77 It can also be expressed in differential form as dS δQ T 78 where the equality holds for an internally reversible process and the inequality for an irreversible process We may conclude from these equations that the entropy change of a closed system during an irreversible process is greater than the integral of δQT evaluated for that process In the limiting case of a reversible process these two quantities become equal We again empha size that T in these relations is the thermodynamic temperature at the bound ary where the differential heat δQ is transferred between the system and the surroundings The quantity ΔS S2 S1 represents the entropy change of the system For a reversible process it becomes equal to 1 2 δQ T which represents the entropy transfer with heat The inequality sign in the preceding relations is a constant reminder that the entropy change of a closed system during an irreversible process is always greater than the entropy transfer That is some entropy is generated or cre ated during an irreversible process and this generation is due entirely to the presence of irreversibilities The entropy generated during a process is called entropy generation and is denoted by Sgen Noting that the difference between the entropy change of a closed system and the entropy transfer is equal to entropy generation Eq 77 can be rewritten as an equality as Δ S sys S 2 S 1 1 2 δQ T S gen 79 FIGURE 75 A cycle composed of a reversible and an irreversible process Process 12 reversible or irreversible 1 2 Process 21 internally reversible Final PDF to printer 329 CHAPTER 7 cen22672ch07323412indd 329 110617 0852 AM Note that the entropy generation Sgen is always a positive quantity or zero Its value depends on the process and thus it is not a property of the system Also in the absence of any entropy transfer the entropy change of a system is equal to the entropy generation Equation 77 has farreaching implications in thermodynamics For an iso lated system or simply an adiabatic closed system the heat transfer is zero and Eq 77 reduces to Δ S isolated 0 710 This equation can be expressed as the entropy of an isolated system during a process always increases or in the limiting case of a reversible process remains constant In other words it never decreases This is known as the increase of entropy principle Note that in the absence of any heat transfer entropy change is due to irreversibilities only and their effect is always to increase entropy Entropy is an extensive property thus the total entropy of a system is equal to the sum of the entropies of the parts of the system An isolated system may consist of any number of subsystems Fig 76 A system and its surroundings for example constitute an isolated system since both can be enclosed by a sufficiently large arbitrary boundary across which there is no heat work or mass transfer Fig 77 Therefore a system and its sur roundings can be viewed as the two subsystems of an isolated system and the entropy change of this isolated system during a process is the sum of the entropy changes of the system and its surroundings which is equal to the entropy generation since an isolated system involves no entropy transfer That is S gen Δ S total Δ S sys Δ S surr 0 711 where the equality holds for reversible processes and the inequality for irre versible ones Note that ΔSsurr refers to the change in the entropy of the sur roundings as a result of the occurrence of the process under consideration Since no actual process is truly reversible we can conclude that some entropy is generated during a process and therefore the entropy of the uni verse which can be considered to be an isolated system is continuously increasing The more irreversible a process the larger the entropy gener ated during that process No entropy is generated during reversible processes Sgen 0 Entropy increase of the universe is a major concern not only to engineers but also to philosophers theologians economists and environmentalists since entropy is viewed as a measure of the disorder or mixedupness in the universe The increase of entropy principle does not imply that the entropy of a sys tem cannot decrease The entropy change of a system can be negative during a process Fig 78 but entropy generation cannot The increase of entropy principle can be summarized as follows S gen 0 irreversible process 0 reversible process 0 impossible process FIGURE 76 The entropy change of an isolated system is the sum of the entropy changes of its components and is never less than zero Subsystem 1 Subsystem 2 Subsystem 3 Subsystem N Isolated ΔStotal ΔSi 0 i1 N Σ FIGURE 78 The entropy change of a system can be negative but the entropy generation cannot Sgen ΔStotal ΔSsys ΔSsurr 1 kJK Surroundings Q ΔSsys 2 kJK System ΔSsurr 3 kJK FIGURE 77 A system and its surroundings form an isolated system Isolated system boundary m 0 Q 0 W 0 Q W System Surroundings m Final PDF to printer 330 ENTROPY cen22672ch07323412indd 330 110617 0852 AM This relation serves as a criterion in determining whether a process is revers ible irreversible or impossible Things in nature have a tendency to change until they attain a state of equi librium The increase of entropy principle dictates that the entropy of an iso lated system increases until the entropy of the system reaches a maximum value At that point the system is said to have reached an equilibrium state since the increase of entropy principle prohibits the system from undergoing any change of state that results in a decrease in entropy Some Remarks About Entropy In light of the preceding discussions we draw the following conclusions 1 Processes can occur in a certain direction only not in any direction A process must proceed in the direction that complies with the increase of entropy principle that is Sgen 0 A process that violates this principle is impossible This principle often forces chemical reactions to come to a halt before reaching completion 2 Entropy is a nonconserved property and there is no such thing as the conservation of entropy principle Entropy is conserved during the idealized reversible processes only and increases during all actual processes 3 The performance of engineering systems is degraded by the presence of irreversibilities and entropy generation is a measure of the magnitudes of the irreversibilities present during that process The greater the extent of irreversibilities the greater the entropy generation Therefore entropy generation can be used as a quantitative measure of irreversibilities associated with a process It is also used to establish criteria for the performance of engineering devices This point is illustrated further in Example 72 FIGURE 79 Schematic for Example 72 Source 800 K Sink A 500 K 2000 kJ a Source 800 K Sink B 750 K b EXAMPLE 72 Entropy Generation During Heat Transfer Processes A heat source at 800 K loses 2000 kJ of heat to a sink at a 500 K and b 750 K Determine which heat transfer process is more irreversible SOLUTION Heat is transferred from a heat source to two heat sinks at different temperatures The heat transfer process that is more irreversible is to be determined Analysis A sketch of the reservoirs is shown in Fig 79 Both cases involve heat transfer through a finite temperature difference and therefore both are irreversible The magnitude of the irreversibility associated with each process can be determined by calculating the total entropy change for each case The total entropy change for a heat transfer process involving two reservoirs a source and a sink is the sum of the entropy changes of each reservoir since the two reservoirs form an adiabatic system Or do they The problem statement gives the impression that the two reservoirs are in direct contact during the heat transfer process But this cannot be the case since the temperature at a point can have only one value and thus it cannot be 800 K on one side of the point of contact and 500 K on the other side In other words the temperature function cannot have a jump discontinuity Therefore it is reason able to assume that the two reservoirs are separated by a partition through which the Final PDF to printer 331 CHAPTER 7 cen22672ch07323412indd 331 110617 0852 AM 73 ENTROPY CHANGE OF PURE SUBSTANCES Entropy is a property and thus the value of entropy of a system is fixed once the state of the system is fixed Specifying two intensive independent prop erties fixes the state of a simple compressible system and thus the value of entropy and of other properties at that state Starting with its defining relation the entropy change of a substance can be expressed in terms of other proper ties see Sec 77 But in general these relations are too complicated and are not practical to use for hand calculations Therefore using a suitable reference temperature drops from 800 K on one side to 500 K or 750 K on the other In that case the entropy change of the partition should also be considered when evaluating the total entropy change for this process However considering that entropy is a prop erty and the values of properties depend on the state of a system we can argue that the entropy change of the partition is zero since the partition appears to have undergone a steady process and thus experienced no change in its properties at any point We base this argument on the fact that the temperature on both sides of the partition and thus throughout remains constant during this process Therefore we are justified to assume that ΔSpartition 0 since the entropy as well as the energy content of the parti tion remains constant during this process The entropy change for each reservoir can be determined from Eq 76 since each reservoir undergoes an internally reversible isothermal process a For the heat transfer process to a sink at 500 K Δ S source Q source T source 2000 kJ 800 K 25 kJ K Δ S sink Q sink T sink 2000 kJ 500 K 40 kJ K and S gen Δ S total Δ S source Δ S sink 25 40 kJ K 15 kJ K Therefore 15 kJK of entropy is generated during this process Noting that both res ervoirs have undergone internally reversible processes the entire entropy generation took place in the partition b Repeating the calculations in part a for a sink temperature of 750 K we obtain Δ S source 25 kJ K Δ S sink 27 kJ K and S gen Δ S total 25 27 kJ K 02 kJ K The total entropy change for the process in part b is smaller and therefore it is less irreversible This is expected since the process in b involves a smaller temperature difference and thus a smaller irreversibility Discussion The irreversibilities associated with both processes could be elimi nated by operating a Carnot heat engine between the source and the sink For this case it can be shown that ΔStotal 0 Final PDF to printer 332 ENTROPY cen22672ch07323412indd 332 110617 0852 AM state the entropies of substances are evaluated from measurable property data following rather involved computations and the results are tabulated in the same manner as the other properties such as v u and h Fig 710 The entropy values in the property tables are given relative to an arbitrary reference state In steam tables the entropy of saturated liquid sf at 001C is assigned the value of zero For refrigerant134a the zero value is assigned to saturated liquid at 40C The entropy values become negative at tempera tures below the reference value The value of entropy at a specified state is determined just like any other property In the compressed liquid and superheated vapor regions it can be obtained directly from the tables at the specified state In the saturated mix ture region it is determined from s s f x s fg kJ kgK where x is the quality and sf and sfg values are listed in the saturation tables In the absence of compressed liquid data the entropy of the compressed liq uid can be approximated by the entropy of the saturated liquid at the given temperature s TP s f T kJ kgK The entropy change of a specified mass m a closed system during a process is simply ΔS m Δs m s 2 s 1 kJ K 712 which is the difference between the entropy values at the final and initial states When studying the secondlaw aspects of processes entropy is commonly used as a coordinate on diagrams such as the Ts and hs diagrams The general characteristics of the Ts diagram of pure substances are shown in Fig 711 using data for water Notice from this diagram that the constantvolume lines are steeper than the constantpressure lines and the constantpressure lines are FIGURE 710 The entropy of a pure substance is determined from the tables like other properties Superheated vapor T s 1 3 2 Saturated liquidvapor mixture T3 s3 P3 P1 s1 sf T1 T1 Compressed liquid T2 s2 sf x2sfg x2 FIGURE 711 Schematic of the Ts diagram for water T C 0 8 s kJkgK 7 6 5 4 3 2 1 100 200 300 400 500 Saturated liquid line Critical state Saturated vapor line P 10 MPa P 1 MPa v 01 m3kg v 05 m3kg Final PDF to printer 333 CHAPTER 7 cen22672ch07323412indd 333 110617 0852 AM parallel to the constanttemperature lines in the saturated liquidvapor mix ture region Also the constantpressure lines almost coincide with the satu rated liquid line in the compressed liquid region EXAMPLE 73 Entropy Change of a Substance in a Tank A rigid tank contains 5 kg of refrigerant134a initially at 20C and 140 kPa The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa Determine the entropy change of the refrigerant during this process SOLUTION The refrigerant in a rigid tank is cooled while being stirred The entropy change of the refrigerant is to be determined Assumptions The volume of the tank is constant and thus v2 v1 Analysis We take the refrigerant in the tank as the system Fig 712 This is a closed system since no mass crosses the system boundary during the process We note that the change in entropy of a substance during a process is simply the differ ence between the entropy values at the final and initial states The initial state of the refrigerant is completely specified Recognizing that the specific volume remains constant during this process the properties of the refrigerant at both states are State 1 P 1 140 kPa T 1 20C s 1 10625 kJ kgK v 1 016544 m 3 kg State 2 P 2 100 kPa v 2 v 1 v f 00007258 m 3 kg v g 019255 m 3 kg The refrigerant is a saturated liquidvapor mixture at the final state since vf v2 vg at 100 kPa pressure Therefore we need to determine the quality first x 2 v 2 v f v fg 016544 00007258 019255 00007258 0859 Thus s 2 s f x 2 s fg 007182 0859088008 08278 kJ kgK Then the entropy change of the refrigerant during this process is ΔS m s 2 s 1 5 kg08278 10625 kJ kgK 1173 kJ K Discussion The negative sign indicates that the entropy of the system is decreasing during this process This is not a violation of the second law however since it is the entropy generation Sgen that cannot be negative FIGURE 712 Schematic and Ts diagram for Example 73 T s s2 1 2 s1 v const Heat m 5 kg Refrigerant134a T1 20C P1 140 kPa ΔS EXAMPLE 74 Entropy Change During a ConstantPressure Process A pistoncylinder device initially contains 3 lbm of liquid water at 20 psia and 70F The water is now heated at constant pressure by the addition of 3450 Btu of heat Determine the entropy change of the water during this process Final PDF to printer 334 ENTROPY cen22672ch07323412indd 334 110617 0852 AM 74 ISENTROPIC PROCESSES We mentioned earlier that the entropy of a fixed mass can be changed by 1 heat transfer and 2 irreversibilities Then it follows that the entropy of a fixed mass does not change during a process that is internally reversible and adiabatic Fig 714 A process during which the entropy remains constant is called an isentropic process It is characterized by Isentropic process Δs 0 or s 2 s 1 kJ kgK 713 That is a substance will have the same entropy value at the end of the process as it does at the beginning if the process is carried out in an isentropic manner FIGURE 714 During an internally reversible adiabatic isentropic process the entropy remains constant No heat transfer adiabatic Steam s1 s2 s1 No irreversibilities internally reversible SOLUTION Liquid water in a pistoncylinder device is heated at constant pres sure The entropy change of water is to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero ΔKE ΔPE 0 2 The process is quasiequilibrium 3 The pres sure remains constant during the process and thus P2 P1 Analysis We take the water in the cylinder as the system Fig 713 This is a closed system since no mass crosses the system boundary during the process We note that a pistoncylinder device typically involves a moving boundary and thus bound ary work Wb Also heat is transferred to the system Water exists as a compressed liquid at the initial state since its pressure is greater than the saturation pressure of 03632 psia at 70F By approximating the com pressed liquid as a saturated liquid at the given temperature the properties at the initial state are State 1 P 1 20 psia T 1 70F s 1 s f 70F 007459 Btu lbmR h 1 h f 70F 3808 Btu lbm At the final state the pressure is still 20 psia but we need one more property to fix the state This property is determined from the energy balance E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q in W b ΔU Q in ΔH m h 2 h 1 3450 Btu 3 lbm h 2 3808 Btu lbm h 2 11881 Btu lbm since ΔU Wb ΔH for a constantpressure quasiequilibrium process Then State 2 P 2 20 psia h 2 11881 Btu lbm s 2 17761 Btu lbmR Table A6E interpolation Therefore the entropy change of water during this process is ΔS m s 2 s 1 3 lbm17761 007459 Btu lbmR 5105 Btu R FIGURE 713 Schematic and Ts diagram for Example 74 T s 1 2 s2 s1 P const Qin H2O P1 20 psia T1 70F Final PDF to printer 335 CHAPTER 7 cen22672ch07323412indd 335 110617 0852 AM Many engineering systems or devices such as pumps turbines nozzles and diffusers are essentially adiabatic in their operation and they perform best when the irreversibilities such as the friction associated with the process are minimized Therefore an isentropic process can serve as an appropriate model for actual processes Also isentropic processes enable us to define efficiencies for processes to compare the actual performance of these devices to the performance under idealized conditions It should be recognized that a reversible adiabatic process is necessarily isentropic s2 s1 but an isentropic process is not necessarily a reversible adiabatic process The entropy increase of a substance during a process as a result of irreversibilities may be offset by a decrease in entropy as a result of heat losses for example However the term isentropic process is customarily used in thermodynamics to imply an internally reversible adiabatic process FIGURE 715 Schematic and Ts diagram for Example 75 wout T s 1 2 s2 s1 Isentropic expansion 14 MPa 5 MPa P1 5 MPa T1 450C P2 14 MPa s2 s1 Steam turbine EXAMPLE 75 Isentropic Expansion of Steam in a Turbine Steam enters an adiabatic turbine at 5 MPa and 450C and leaves at a pressure of 14 MPa Determine the work output of the turbine per unit mass of steam if the pro cess is reversible SOLUTION Steam is expanded in an adiabatic turbine to a specified pressure in a reversible manner The work output of the turbine is to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 ΔECV 0 and ΔSCV 0 2 The process is reversible 3 Kinetic and potential energies are negligible 4 The turbine is adiabatic and thus there is no heat transfer Analysis We take the turbine as the system Fig 715 This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one exit and thus m 1 m 2 m The power output of the turbine is determined from the rate form of the energy balance E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 E in E out m h 1 W out m h 2 since Q 0 ke pe 0 W out m h 1 h 2 The inlet state is completely specified since two properties are given But only one property pressure is given at the final state and we need one more property to fix it The second property comes from the observation that the process is reversible and adiabatic and thus isentropic Therefore s2 s1 and State 1 P 1 5 MPa T 1 450C h 1 33172 kJ kg s 1 68210 kJ kgK State 2 P 2 14 MPa s 2 s 1 h 2 29674 kJ kg 0 steady Final PDF to printer 336 ENTROPY cen22672ch07323412indd 336 110617 0852 AM 75 PROPERTY DIAGRAMS INVOLVING ENTROPY Property diagrams serve as great visual aids in the thermodynamic analysis of processes We have used Pv and Tv diagrams extensively in previous chap ters in conjunction with the first law of thermodynamics In the secondlaw analysis it is very helpful to plot the processes on diagrams for which one of the coordinates is entropy The two diagrams commonly used in the second law analysis are the temperatureentropy and the enthalpyentropy diagrams Consider the defining equation of entropy Eq 74 It can be rearranged as δ Q int rev T dS kJ 714 As shown in Fig 716 δQint rev corresponds to a differential area on a TS dia gram The total heat transfer during an internally reversible process is deter mined by integration to be Q int rev 1 2 T dS kJ 715 which corresponds to the area under the process curve on a TS diagram There fore we conclude that the area under the process curve on a TS diagram rep resents heat transfer during an internally reversible process This is somewhat analogous to reversible boundary work being represented by the area under the process curve on a PV diagram Note that the area under the process curve rep resents heat transfer for processes that are internally or totally reversible The area has no meaning for irreversible processes Equations 714 and 715 can also be expressed on a unitmass basis as δ q int rev T ds kJ kg 716 and q int rev 1 2 T ds kJ kg 717 To perform the integrations in Eqs 715 and 717 one needs to know the relationship between T and s during a process One special case for which these integrations can be performed easily is the internally reversible isother mal process It yields Q int rev T 0 ΔS kJ 718 or q int rev T 0 ΔS kJ kg 719 Then the work output of the turbine per unit mass of the steam becomes w out h 1 h 2 33172 29674 3498 kJ kg FIGURE 716 On a TS diagram the area under the process curve represents the heat transfer for internally reversible processes Internally reversible process T S dA T dS δQ Area T dS Q 1 2 Final PDF to printer 337 CHAPTER 7 cen22672ch07323412indd 337 110617 0852 AM where T0 is the constant temperature and ΔS is the entropy change of the sys tem during the process An isentropic process on a Ts diagram is easily recognized as a verticalline segment This is expected since an isentropic process involves no heat transfer and therefore the area under the process path must be zero Fig 717 The Ts diagrams serve as valuable tools for visualizing the secondlaw aspects of processes and cycles and thus they are frequently used in thermodynamics The Ts diagram of water is given in the appendix in Fig A9 Another diagram commonly used in engineering is the enthalpyentropy diagram which is quite valuable in the analysis of steadyflow devices such as turbines compressors and nozzles The coordinates of an hs diagram represent two properties of major interest enthalpy which is a primary property in the firstlaw analysis of the steadyflow devices and entropy which is the property that accounts for irreversibilities during adiabatic processes In analyzing the steady flow of steam through an adiabatic tur bine for example the vertical distance between the inlet and the exit states Δh is a measure of the work output of the turbine and the horizontal dis tance Δs is a measure of the irreversibilities associated with the process Fig 718 The hs diagram is also called a Mollier diagram after the German sci entist R Mollier 18631935 An hs diagram is given in the appendix for steam in Fig A10 FIGURE 719 The TS diagram of a Carnot cycle Example 76 T TH TL S1 S4 S2 S3 S Wnet A B 4 3 1 2 EXAMPLE 76 The TS Diagram of the Carnot Cycle Show the Carnot cycle on a TS diagram and indicate the areas that represent the heat supplied QH heat rejected QL and the net work output Wnetout on this diagram SOLUTION The Carnot cycle is to be shown on a TS diagram and the areas that represent QH QL and Wnetout are to be indicated Analysis Recall that the Carnot cycle is made up of two reversible isothermal T constant processes and two isentropic s constant processes These four pro cesses form a rectangle on a TS diagram as shown in Fig 719 On a TS diagram the area under the process curve represents the heat transfer for that process Thus the area A12B represents QH the area A43B represents QL and the difference between these two the area in color represents the net work since W netout Q H Q L Therefore the area enclosed by the path of a cycle area 1234 on a TS diagram represents the net work Recall that the area enclosed by the path of a cycle also rep resents the net work on a PV diagram 76 WHAT IS ENTROPY It is clear from the previous discussion that entropy is a useful property and serves as a valuable tool in the secondlaw analysis of engineering devices But this does not mean that we know and understand entropy well Because we do not In fact we cannot even give an adequate answer to the question FIGURE 717 The isentropic process appears as a vertical line segment on a Ts diagram T s s2 s1 2 Isentropic process 1 FIGURE 718 For adiabatic steadyflow devices the vertical distance Δh on an hs diagram is a measure of work and the horizontal distance Δs is a measure of irreversibilities h s 1 2 Δs Δh Final PDF to printer 338 ENTROPY cen22672ch07323412indd 338 110617 0852 AM what is entropy Not being able to describe entropy fully however does not take anything away from its usefulness We could not define energy either but it did not interfere with our understanding of energy transformations and the conservation of energy principle Granted entropy is not a house hold word like energy But with continued use our understanding of entropy will deepen and our appreciation of it will grow The next discussion should shed some light on the physical meaning of entropy by considering the micro scopic nature of matter Entropy can be viewed as a measure of molecular disorder or molecular randomness As a system becomes more disordered the positions of the molecules become less predictable and the entropy increases Thus it is not surprising that the entropy of a substance is lowest in the solid phase and highest in the gas phase Fig 720 In the solid phase the molecules of a substance continually oscillate about their equilibrium positions but they cannot move relative to each other and their position at any instant can be predicted with good certainty In the gas phase however the molecules move about at random collide with each other and change direction mak ing it extremely difficult to predict accurately the microscopic state of a system at any instant Associated with this molecular chaos is a high value of entropy When viewed microscopically from a statistical thermodynamics point of view an isolated system that appears to be at a state of equilibrium actu ally exhibits a high level of activity because of the continual motion of the molecules To each state of macroscopic equilibrium there corresponds a large number of molecular microscopic states or molecular configurations Boltzmann first hypothesized that the entropy of a system at a specified macrostate is related to the total number of possible relevant microstates of that system W from Wahrscheinlichkeit the German word for probabil ity This thought was formulated later by Planck using a constant k with the entropy unit of JK named after Boltzmann and inscribed on Boltzmanns tombstone as S k ln W 720a which is known as the Boltzmann relation The thermal motion randomness or disorder as related to entropy was later generalized by Gibbs as a measure of the sum of all microstates uncertainties ie probabilities as S k p i log p i 720b Gibbs formulation is more general since it allows for nonuniform prob ability pi of microstates With an increase of particle momenta or thermal disorder and volume occupied more information is required for the charac terization of the system relative to more ordered systems Gibbs formulation reduces to Boltzmann relation for equiprobable uniform probability of all W microstates since pi 1W constant 1 From a microscopic point of view the entropy of a system increases when ever the thermal randomness or disorder ie the number of possible relevant molecular microstates corresponding to a given bulk macrostate of a system increases Thus entropy can be viewed as a measure of thermal randomness FIGURE 720 The level of molecular disorder entropy of a substance increases as it melts or evaporates Liquid Solid Gas Entropy kJkgK Final PDF to printer 339 CHAPTER 7 cen22672ch07323412indd 339 110617 0852 AM or molecular disorder which increases any time an isolated system undergoes a process As mentioned earlier the molecules of a substance in solid phase con tinually oscillate creating an uncertainty about their position These oscil lations however fade as the temperature is decreased and the molecules supposedly become motionless at absolute zero This represents a state of ultimate molecular order and minimum energy Therefore the entropy of a pure crystalline substance at absolute zero temperature is zero since there is no uncertainty about the state of the molecules at that instant Fig 721 This statement is known as the third law of thermodynamics The third law of thermodynamics provides an absolute reference point for the deter mination of entropy The entropy determined relative to this point is called absolute entropy and it is extremely useful in the thermodynamic analysis of chemical reactions Notice that the entropy of a substance that is not pure crystalline such as a solid solution is not zero at absolute zero temperature This is because more than one molecular configuration exists for such sub stances which introduces some uncertainty about the microscopic state of the substance Molecules in the gas phase possess a considerable amount of kinetic energy However we know that no matter how large their kinetic energies are the gas molecules do not rotate a paddle wheel inserted into the container and pro duce work This is because the gas molecules and the energy they possess are disorganized Probably the number of molecules trying to rotate the wheel in one direction at any instant is equal to the number of molecules that are try ing to rotate it in the opposite direction causing the wheel to remain motion less Therefore we cannot extract any useful work directly from disorganized energy Fig 722 Now consider a rotating shaft shown in Fig 723 This time the energy of the molecules is completely organized since the molecules of the shaft are rotating in the same direction together This organized energy can readily be used to perform useful tasks such as raising a weight or generating electric ity Being an organized form of energy work is free of disorder or random ness and thus free of entropy There is no entropy transfer associated with energy transfer as work Therefore in the absence of any friction the pro cess of raising a weight by a rotating shaft or a flywheel does not produce any entropy Any process that does not produce a net entropy is reversible and thus the process just described can be reversed by lowering the weight Therefore energy is not degraded during this process and no potential to do work is lost Instead of raising a weight let us operate the paddle wheel in a container filled with a gas as shown in Fig 724 The paddlewheel work in this case is converted to the internal energy of the gas as evidenced by a rise in gas temperature creating a higher level of molecular disorder in the container This process is quite different from raising a weight since the organized paddlewheel energy is now converted to a highly disorganized form of energy which cannot be converted back to the paddle wheel as the rotational kinetic energy Only a portion of this energy can be con verted to work by partially reorganizing it through the use of a heat engine Therefore energy is degraded during this process the ability to do work is FIGURE 721 A pure crystalline substance at absolute zero temperature is in perfect order and its entropy is zero the third law of thermodynamics Pure crystal T 0 K Entropy 0 FIGURE 722 Disorganized energy does not create much useful effect no matter how large it is Load FIGURE 723 In the absence of friction raising a weight by a rotating shaft does not create any disorder entropy and thus energy is not degraded during this process Weight Wsh Final PDF to printer 340 ENTROPY cen22672ch07323412indd 340 110617 0852 AM reduced molecular disorder is produced and associated with all this is an increase in entropy The quantity of energy is always preserved during an actual process the first law but the quality is bound to decrease the second law This decrease in quality is always accompanied by an increase in entropy As an example consider the transfer of 10 kJ of energy as heat from a hot medium to a cold one At the end of the process we still have the 10 kJ of energy but at a lower temperature and thus at a lower quality Heat is in essence a form of disorganized energy and some disorganiza tion entropy flows with heat Fig 725 As a result the entropy and the level of molecular disorder or randomness of the hot body decreases with the entropy and the level of molecular disorder of the cold body increases The second law requires that the increase in entropy of the cold body be greater than the decrease in entropy of the hot body and thus the net entropy of the combined system the cold body and the hot body increases That is the combined system is at a state of greater disorder at the final state Thus we can conclude that processes can occur only in the direction of increased overall entropy or molecular disorder That is the entire universe is getting more and more chaotic every day Entropy and Entropy Generation in Daily Life The concept of entropy can also be applied to other areas Entropy can be viewed as a measure of disorder or disorganization in a system Likewise entropy generation can be viewed as a measure of disorder or disorganization generated during a process The concept of entropy is not used in daily life nearly as extensively as the concept of energy even though entropy is readily applicable to various aspects of daily life The extension of the entropy con cept to nontechnical fields is not a novel idea It has been the topic of several articles and even some books Next we present several ordinary events and show their relevance to the concept of entropy and entropy generation Efficient people lead lowentropy highly organized lives They have a place for everything minimum uncertainty and it takes minimum energy for them to locate something Inefficient people on the other hand are disorga nized and lead highentropy lives It takes them minutes if not hours to find something they need and they are likely to create a bigger disorder as they are searching since they will probably conduct the search in a disorganized man ner People leading highentropy lifestyles are always on the run and never seem to catch up Consider two identical buildings each containing one million books In the first building the books are piled on top of each other whereas in the sec ond building they are highly organized shelved and indexed for easy refer ence There is no doubt about which building a student will prefer to go to for checking out a certain book Yet some may argue from the firstlaw point of view that these two buildings are equivalent since the mass and knowledge content of the two buildings are identical despite the high level of disorgani zation entropy in the first building This example illustrates that any realistic comparisons should involve the secondlaw point of view Two textbooks that seem to be identical because both cover basically the same topics and present the same information may actually be very different FIGURE 724 The paddlewheel work done on a gas increases the level of disorder entropy of the gas and thus energy is degraded during this process Gas T Wsh FIGURE 725 During a heat transfer process the net entropy increases The increase in the entropy of the cold body more than offsets the decrease in the entropy of the hot body Entropy increases Hot body Heat 80C 20C Entropy decreases Cold body Final PDF to printer 341 CHAPTER 7 cen22672ch07323412indd 341 110617 0852 AM depending on how they cover the topics After all two seemingly identical cars are not so identical if one goes only half as many miles as the other one on the same amount of fuel Likewise two seemingly identical books are not so identical if it takes twice as long to learn a topic from one of them as it does from the other Thus comparisons made on the basis of the first law only may be highly misleading Having a disorganized highentropy army is like having no army at all It is no coincidence that the command centers of any armed forces are among the primary targets during a war One army that consists of 10 divi sions is 10 times more powerful than 10 armies each consisting of a single division Likewise one country that consists of 10 states is more powerful than 10 countries each consisting of a single state The United States would not be such a powerful country if there were 50 independent countries in its place instead of a single country with 50 states The old cliché divide and conquer can be rephrased as increase the entropy and conquer We know that mechanical friction is always accompanied by entropy gen eration and thus reduced performance We can generalize this to daily life friction in the workplace with fellow workers is bound to generate entropy and thus adversely affect performance Fig 726 It results in reduced productivity We also know that unrestrained expansion or explosion and uncontrolled electron exchange chemical reactions generate entropy and are highly irre versible Likewise unrestrained opening of the mouth to scatter angry words is highly irreversible since this generates entropy and it can cause consider able damage A person who gets up in anger is bound to sit down at a loss Hopefully someday we will be able to come up with some procedures to quantify entropy generated during nontechnical activities and maybe even pinpoint its primary sources and magnitude 77 THE T ds RELATIONS Recall that the quantity δQTint rev corresponds to a differential change in the property entropy The entropy change for a process then can be evaluated by integrating δQT along some imaginary internally reversible path between the actual end states For isothermal internally reversible processes this integra tion is straightforward But when the temperature varies during the process we have to have a relation between δQ and T to perform this integration Find ing such relations is what we intend to do in this section The differential form of the conservation of energy equation for a closed stationary system a fixed mass containing a simple compressible substance can be expressed for an internally reversible process as δ Q int rev δ W int revout dU 721 But δ Q int rev T dS δ W int revout P dV FIGURE 726 As in mechanical systems friction in the workplace is bound to generate entropy and reduce performance PurestockSuperStock RF Final PDF to printer 342 ENTROPY cen22672ch07323412indd 342 110617 0852 AM Thus T dS dU P dV kJ 722 or T ds du P dv kJ kg 723 This equation is known as the first T ds or Gibbs equation Notice that the only type of work interaction a simple compressible system may involve as it undergoes an internally reversible process is the boundary work The second T ds equation is obtained by eliminating du from Eq 723 by using the definition of enthalpy h u Pv h u Pv Eq 723 dh du P dv v dP T ds du P dv T ds dh v dP 724 Equations 723 and 724 are extremely valuable since they relate entropy changes of a system to the changes in other properties Unlike Eq 74 they are property relations and therefore are independent of the type of the processes These T ds relations are developed with an internally reversible process in mind since the entropy change between two states must be evaluated along a reversible path However the results obtained are valid for both reversible and irreversible processes since entropy is a property and the change in a property between two states is independent of the type of process the system under goes Equations 723 and 724 are relations between the properties of a unit mass of a simple compressible system as it undergoes a change of state and they are applicable whether the change occurs in a closed or an open system Fig 727 Explicit relations for differential changes in entropy are obtained by solving for ds in Eqs 723 and 724 ds du T P dv T 725 and ds dh T v dP T 726 The entropy change during a process can be determined by integrating either of these equations between the initial and the final states To perform these integrations however we must know the relationship between du or dh and the temperature such as du cv dT and dh cp dT for ideal gases as well as the equation of state for the substance such as the idealgas equation of state Pv RT For substances for which such relations exist the integration of Eq 725 or 726 is straightforward For other substances we have to rely on tabulated data The T ds relations for nonsimple systems that is systems that involve more than one mode of quasiequilibrium work can be obtained in a similar man ner by including all the relevant quasiequilibrium work modes FIGURE 727 The T ds relations are valid for both reversible and irreversible processes and for both closed and open systems Closed system T ds du P dv T ds dh v dP CV Final PDF to printer 343 CHAPTER 7 cen22672ch07323412indd 343 110617 0852 AM 78 ENTROPY CHANGE OF LIQUIDS AND SOLIDS Recall that liquids and solids can be approximated as incompressible sub stances since their specific volumes remain nearly constant during a process Thus dv 0 for liquids and solids and Eq 725 for this case reduces to ds du T c dT T 727 since cp cv c and du c dT for incompressible substances Then the entropy change during a process is determined by integration to be Liquids solids s 2 s 1 1 2 c T dT T c avg ln T 2 T 1 kJ kgK 728 where cavg is the average specific heat of the substance over the given tem perature interval Note that the entropy change of a truly incompressible sub stance depends on temperature only and is independent of pressure Equation 728 can be used to determine the entropy changes of solids and liquids with reasonable accuracy However for liquids that expand consid erably with temperature it may be necessary to consider the effects of vol ume change in calculations This is especially the case when the temperature change is large A relation for isentropic processes of liquids and solids is obtained by set ting the entropy change relation above equal to zero It gives Isentropic s 2 s 1 c avg ln T 2 T 1 0 T 2 T 1 729 That is the temperature of a truly incompressible substance remains constant during an isentropic process Therefore the isentropic process of an incom pressible substance is also isothermal This behavior is closely approximated by liquids and solids EXAMPLE 77 Effect of Density of a Liquid on Entropy Liquid methane is commonly used in various cryogenic applications The critical temperature of methane is 191 K or 82C and thus methane must be maintained below 191 K to keep it in liquid phase The properties of liquid methane at various temperatures and pressures are given in Table 71 Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa a using tabulated properties and b approximating liquid methane as an incom pressible substance What is the error involved in the latter case SOLUTION Liquid methane undergoes a process between two specified states The entropy change of methane is to be determined by using actual data and by assuming methane to be incompressible Analysis a We consider a unit mass of liquid methane Fig 728 The properties of the methane at the initial and final states are FIGURE 728 Schematic for Example 77 Methane pump P1 1 MPa T1 110 K P2 5 MPa T2 120 K Final PDF to printer 344 ENTROPY cen22672ch07323412indd 344 110617 0852 AM State 1 P 1 1 MPa T 1 110 K s 1 4875 kJ kgK c p1 3471 kJ kgK State 2 P 2 5 MPa T 2 120 K s 2 5145 kJ kgK c p2 3486 kJ kgK Therefore Δs s 2 s 1 5145 4875 0270 kJ kgK b Approximating liquid methane as an incompressible substance its entropy change is determined to be Δs c avg ln T 2 T 1 34785 kJ kgK ln 120 K 110 K 0303 kJ kgK since c avg c p1 c p2 2 3471 3486 2 34785 kJ kgK Therefore the error involved in approximating liquid methane as an incompressible substance is Error Δ s actual Δ s ideal Δ s actual 0270 0303 0270 0122 or 122 Discussion This result is not surprising since the density of liquid methane changes during this process from 4258 to 4152 kgm3 about 3 percent which makes us question the validity of the incompressible substance assumption Still this assump tion enables us to obtain reasonably accurate results with less effort which proves to be very convenient in the absence of compressed liquid data TABLE 71 Properties of liquid methane Temp T K Pressure P MPa Density ρ kgm3 Enthalpy h kJkg Entropy s kJkgK Specific heat cp kJkgK 110 05 10 20 50 4253 4258 4266 4291 2083 2090 2105 2150 4878 4875 4867 4844 3476 3471 3460 3432 120 05 10 20 50 4104 4110 4120 4152 2434 2441 2454 2496 5185 5180 5171 5145 3551 3543 3528 3486 Final PDF to printer 345 CHAPTER 7 cen22672ch07323412indd 345 110617 0852 AM FIGURE 729 Liquefied natural gas LNG turbine after being removed from an LNG tank Courtesy of Ebara International Corp Cryodynamics Division Sparks Nevada EXAMPLE 78 Economics of Replacing a Valve with a Turbine A cryogenic manufacturing facility handles liquid methane at 115 K and 5 MPa at a rate of 0280 m3s A process requires dropping the pressure of liquid methane to 1 MPa which is done by throttling the liquid methane by passing it through a flow resistance such as a valve A recently hired engineer proposes to replace the throt tling valve with a turbine in order to produce power while dropping the pressure to 1 MPa Using data from Table 71 determine the maximum amount of power that can be produced by such a turbine Also determine how much this turbine will save the facility from electricity usage costs per year if the turbine operates continuously 8760 hyr and the facility pays 0075kWh for electricity SOLUTION Liquid methane is expanded in a turbine to a specified pressure at a specified rate The maximum power that this turbine can produce and the amount of money it can save per year are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 ΔECV 0 and ΔSCV 0 2 The turbine is adiabatic and thus there is no heat transfer 3 The process is reversible 4 Kinetic and potential energies are negligible Analysis We take the turbine as the system Fig 729 This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one exit and thus m 1 m 2 m The preceding assumptions are reasonable since a turbine is normally well insu lated and it must involve no irreversibilities for best performance and thus maximum power production Therefore the process through the turbine must be reversible adia batic or isentropic Then s2 s1 and State 1 P 1 5 MPa T 1 115 K h 1 2323 kJ kg s 1 49945 kJ kgK ρ 1 42215 kg m 3 State 2 P 2 1 MPa s 2 s 1 h 2 2228 kJ kg Also the mass flow rate of liquid methane is m ρ 1 V 1 42215 kg m 3 0280 m 3 s 1182 kg s Then the power output of the turbine is determined from the rate form of the energy balance to be E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 E in E out m h 1 W out m h 2 since Q 0 ke pe 0 W out m h 1 h 2 1182 kg s2323 2228 kJ kg 1123 kW For continuous operation 365 24 8760 h the amount of power produced per year is 0 steady Final PDF to printer 346 ENTROPY cen22672ch07323412indd 346 110617 0852 AM Annual power production W out Δt 1123 kW 8760 h yr 09837 10 7 kWh yr At 0075kWh the amount of money this turbine can save the facility is Annual money savings Annual power production Unit cost of power 09837 10 7 kWh yr0075 kWh 737800 yr That is this turbine can save the facility 737800 a year by simply taking advantage of the potential that is currently being wasted by a throttling valve and the engineer who made this observation should be rewarded Discussion This example shows the importance of the property entropy since it enabled us to quantify the work potential that is being wasted In practice the tur bine will not be isentropic and thus the power produced will be less The analysis above gave us the upper limit An actual turbinegenerator assembly can utilize about 80 percent of the potential and produce more than 900 kW of power while saving the facility more than 600000 a year It can also be shown that the temperature of methane drops to 1139 K a drop of 11 K during the isentropic expansion process in the turbine instead of remaining constant at 115 K as would be the case if methane were assumed to be an incompress ible substance The temperature of methane would rise to 1166 K a rise of 16 K during the throttling process FIGURE 730 A broadcast from channel IG Tony CardozaGetty Images RF Pv RT du cv dT dh cp dT 79 THE ENTROPY CHANGE OF IDEAL GASES An expression for the entropy change of an ideal gas can be obtained from Eq 725 or 726 by employing the property relations for ideal gases Fig 730 By substituting du cv dT and P RTv into Eq 725 the differential entropy change of an ideal gas becomes ds c v dT T R dv v 730 The entropy change for a process is obtained by integrating this relation between the end states s 2 s 1 1 2 c v T dT T R ln v 2 v 1 731 A second relation for the entropy change of an ideal gas is obtained in a simi lar manner by substituting dh cp dT and v RTP into Eq 726 and inte grating The result is s 2 s 1 1 2 c p T dT T R ln P 2 P 1 732 The specific heats of ideal gases with the exception of monatomic gases depend on temperature and the integrals in Eqs 731 and 732 cannot be Final PDF to printer 347 CHAPTER 7 cen22672ch07323412indd 347 110617 0852 AM performed unless the dependence of cv and cp on temperature is known Even when the cvT and cpT functions are available performing long integrations every time entropy change is calculated is not practical Then two reasonable choices are left either perform these integrations by simply assuming constant specific heats or evaluate those integrals once and tabulate the results Both approaches are presented next Constant Specific Heats Approximate Analysis Assuming constant specific heats for ideal gases is a common approximation and we used this assumption before on several occasions It usually simplifies the analysis greatly and the price we pay for this convenience is some loss in accuracy The magnitude of the error introduced by this assumption depends on the situation at hand For example for monatomic ideal gases such as helium the specific heats are independent of temperature and therefore the constant specificheat assumption introduces no error For ideal gases whose specific heats vary almost linearly in the temperature range of interest the possible error is minimized by using specific heat values evaluated at the average temperature Fig 731 The results obtained in this way usually are sufficiently accurate if the temperature range is not greater than a few hundred degrees The entropychange relations for ideal gases under the constantspecificheat assumption are easily obtained by replacing cvT and cpT in Eqs 731 and 732 with cvavg and cpavg respectively and performing the integrations We obtain s 2 s 1 c vavg ln T 2 T 1 R ln v 2 v 1 kJ kgK 733 and s 2 s 1 c pavg ln T 2 T 1 R ln P 2 P 1 kJ kgK 734 Entropy changes can also be expressed on a unitmole basis by multiplying these relations by molar mass s 2 s 1 c vavg ln T 2 T 1 R u ln v 2 v 1 kJ kmolK 735 and s 2 s 1 c pavg ln T 2 T 1 R u ln P 2 P 1 kJ kmolK 736 Variable Specific Heats Exact Analysis When the temperature change during a process is large and the specific heats of the ideal gas vary nonlinearly within the temperature range the assumption of constant specific heats may lead to considerable errors in entropychange calculations For those cases the variation of specific heats with tempera ture should be properly accounted for by utilizing accurate relations for the FIGURE 731 Under the constantspecificheat assumption the specific heat is assumed to be constant at some average value T1 T T2 cpavg Tavg Average cp cp Actual cp Final PDF to printer 348 ENTROPY cen22672ch07323412indd 348 110617 0852 AM specific heats as a function of temperature The entropy change during a process is then determined by substituting these cvT or cpT relations into Eq 731 or 732 and performing the integrations Instead of performing these laborious integrals each time we have a new process it is convenient to perform these integrals once and tabulate the results For this purpose we choose absolute zero as the reference tempera ture and define a function s as s 0 T c p T dT T 737 Obviously s is a function of temperature alone and its value is zero at abso lute zero temperature The values of s are calculated at various temperatures and the results are tabulated in the appendix as a function of temperature for air Given this definition the integral in Eq 732 becomes 1 2 c p T dT T s2 s1 738 where s2 is the value of s at T2 and s1 is the value at T1 Thus s 2 s 1 s2 s1 R ln P 2 P 1 kJ kgK 739 It can also be expressed on a unitmole basis as s 2 s 1 s 2 s 1 R u ln P 2 P 1 kJ kmolK 740 Note that unlike internal energy and enthalpy the entropy of an ideal gas varies with specific volume or pressure as well as the temperature Therefore entropy cannot be tabulated as a function of temperature alone The s values in the tables account for the temperature dependence of entropy Fig 732 The variation of entropy with pressure is accounted for by the last term in Eq 739 Another relation for entropy change can be developed based on Eq 731 but this would require the definition of another function and tabula tion of its values which is not practical FIGURE 732 The entropy of an ideal gas depends on both T and P The function s repre sents only the temperaturedependent part of entropy Table A17 T K 300 310 320 s kJkgK 170203 173498 176690 EXAMPLE 79 Entropy Change of an Ideal Gas Air is compressed from an initial state of 100 kPa and 17C to a final state of 600 kPa and 57C Determine the entropy change of air during this compression process by using a property values from the air table and b average specific heats SOLUTION Air is compressed between two specified states The entropy change of air is to be determined by using tabulated property values and also by using average specific heats Assumptions Air is an ideal gas since it is at a high temperature and low pres sure relative to its criticalpoint values Therefore entropy change relations developed under the idealgas assumption are applicable Final PDF to printer 349 CHAPTER 7 cen22672ch07323412indd 349 110617 0852 AM Analysis A sketch of the system and the Ts diagram for the process are given in Fig 733 We note that both the initial and the final states of air are completely specified a The properties of air are given in the air table Table A17 Reading s values at given temperatures and substituting we find s 2 s 1 s2 s1 R ln P 2 P 1 179783 166802 kJ kgK 0287 kJ kgK ln 600 kPa 100 kPa 03844 kJ kgK b The entropy change of air during this process can also be determined approxi mately from Eq 734 by using a cp value at the average temperature of 37C Table A2b and treating it as a constant s 2 s 1 c pavg ln T 2 T 1 R ln P 2 P 1 1006 kJ kgK ln 330 K 290 K 0287 kJ kgK ln 600 kPa 100 kPa 03842 kJ kgK Discussion The two results above are almost identical since the change in tempera ture during this process is relatively small Fig 734 When the temperature change is large however they may differ significantly For those cases Eq 739 should be used instead of Eq 734 since it accounts for the variation of specific heats with temperature FIGURE 733 Schematic and Ts diagram for Example 79 T s P2 600 kPa P1 100 kPa 1 2 P2 600 kPa T2 330 K P1 100 kPa T1 290 K Air Compressor FIGURE 734 For small temperature differences the exact and approximate relations for entropy changes of ideal gases give almost identical results Air T1 290 K T2 330 K s2 s1 s2 s1 R ln 03844 kJkgK 03842 kJkgK P2 P1 P2 P1 T2 T1 s2 s1 cpavg ln R ln Isentropic Processes of Ideal Gases Several relations for the isentropic processes of ideal gases can be obtained by setting the entropychange relations developed previously equal to zero Again this is done first for the case of constant specific heats and then for the case of variable specific heats Constant Specific Heats Approximate Analysis When the constantspecificheat assumption is valid the isentropic relations for ideal gases are obtained by setting Eqs 733 and 734 equal to zero From Eq 733 ln T 2 T 1 R c v ln v 2 v 1 which can be rearranged as ln T 2 T 1 ln v 1 v 2 R c v 741 Final PDF to printer 350 ENTROPY cen22672ch07323412indd 350 110617 0852 AM or T 2 T 1 s const v 1 v 2 k 1 ideal gas 742 since R cp cv k cpcv and thus Rcv k 1 Equation 742 is the first isentropic relation for ideal gases under the constantspecificheat assumption The second isentropic relation is obtained in a similar manner from Eq 734 with the following result T 2 T 1 s const P 2 P 1 k 1 k ideal gas 743 The third isentropic relation is obtained by substituting Eq 743 into Eq 742 and simplifying P 2 P 1 s const v 1 v 2 k ideal gas 744 Equations 742 through 744 can also be expressed in a compact form as T v k 1 constant 745 T P 1 k k constant 746 P v k constant 747 The specific heat ratio k in general varies with temperature and thus an average k value for the given temperature range should be used Note that the preceding idealgas isentropic relations as the name implies are strictly valid for isentropic processes only when the constantspecificheat assumption is appropriate Fig 735 Variable Specific Heats Exact Analysis When the constantspecificheat assumption is not appropriate the isentro pic relations developed previously yield results that are not quite accurate For such cases we should use an isentropic relation obtained from Eq 739 that accounts for the variation of specific heats with temperature Setting this equation equal to zero gives 0 s2 s1 R ln P 2 P 1 or s2 s1 R ln P 2 P 1 748 where s 2 o is the s value at the end of the isentropic process Relative Pressure and Relative Specific Volume Equation 748 provides an accurate way of evaluating property changes of ideal gases during isentropic processes since it accounts for the variation FIGURE 735 The isentropic relations of ideal gases are valid for the isentropic processes of ideal gases only Valid for ideal gas isentropic process constant specific heats T2 T1 sconst P2 P1 k1k v1 k1 v2 Final PDF to printer 351 CHAPTER 7 cen22672ch07323412indd 351 110617 0852 AM of specific heats with temperature However it involves tedious iterations when the volume ratio is given instead of the pressure ratio This is quite an inconvenience in optimization studies which usually require many repetitive calculations To remedy this deficiency we define two new dimensionless quantities associated with isentropic processes The definition of the first is based on Eq 748 which can be rearranged as P 2 P 1 exp s2 s1 R or P 2 P 1 exp s2 R exp s1 R The quantity C1 expsR is defined as the relative pressure Pr where C1 is a constant With this definition the last relation becomes P 2 P 1 s const P r2 P r1 749 Note that the relative pressure Pr is a dimensionless quantity that is a function of temperature only since s depends on temperature alone Therefore values of Pr can be tabulated against temperature This is done for air in Table A17 The use of Pr data is illustrated in Fig 736 Sometimes specific volume ratios are given instead of pressure ratios This is particularly the case when automotive engines are analyzed In such cases one needs to work with volume ratios Therefore we define another quantity related to specific volume ratios for isentropic processes This is done by uti lizing the idealgas relation and Eq 749 P 1 v 1 T 1 P 2 v 2 T 2 v 2 v 1 T 2 P 1 T 1 P 2 T 2 P r1 T 1 P r2 T 2 P r2 T 1 P r1 The quantity C2TPr is a function of temperature only and is defined as relative specific volume vr where C2 is a constant Thus v 2 v 1 s const v r2 v r1 750 Equations 749 and 750 are strictly valid for isentropic processes of ideal gases only They account for the variation of specific heats with temperature and therefore give more accurate results than Eqs 742 through 747 The values of Pr and vr are listed for air in Table A17 EXAMPLE 710 Isentropic Compression of Air in a Car Engine Air is compressed in a car engine from 22C and 95 kPa in a reversible and adiabatic manner If the compression ratio V1V2 of this engine is 8 determine the final tem perature of the air FIGURE 736 The use of Pr data for calculating the final temperature during an isentropic process Process isentropic Given P1 T1 and P2 Find T2 Pr2 Pr1 Pr P2 P1 Pr1 T T2 T1 read read Final PDF to printer 352 ENTROPY cen22672ch07323412indd 352 110617 0852 AM SOLUTION Air is compressed in a car engine isentropically For a given com pression ratio the final air temperature is to be determined Assumptions At specified conditions air can be treated as an ideal gas Therefore the isentropic relations for ideal gases are applicable Analysis A sketch of the system and the Ts diagram for the process are given in Fig 737 This process is easily recognized as being isentropic since it is both reversible and adi abatic The final temperature for this isentropic process can be determined from Eq 750 with the help of relative specific volume data Table A17 as illustrated in Fig 738 For closed systems V 2 V 1 v 2 v 1 At T 1 295 K v r1 6479 From Eq 750 v r2 v r1 v 2 v 1 6479 1 8 8099 T 2 6627 K Therefore the temperature of air will increase by 3677C during this process ALTERNATIVE SOLUTION The final temperature could also be determined from Eq 742 by assuming constant specific heats for air T 2 T 1 s const v 1 v 2 k 1 The specific heat ratio k also varies with temperature and we need to use the value of k corresponding to the average temperature However the final temperature is not given and so we cannot determine the average temperature in advance For such cases calculations can be started with a k value at the initial or the anticipated aver age temperature This value could be refined later if necessary and the calculations can be repeated We know that the temperature of the air will rise considerably dur ing this adiabatic compression process so we guess the average temperature to be about 450 K The k value at this anticipated average temperature is determined from Table A2b to be 1391 Then the final temperature of air becomes T 2 295 K 8 1391 1 6652 K This gives an average temperature value of 4801 K which is sufficiently close to the assumed value of 450 K Therefore it is not necessary to repeat the calculations by using the k value at this average temperature The result obtained by assuming constant specific heats for this case is in error by about 04 percent which is rather small This is not surprising since the temperature change of air is relatively small only a few hundred degrees and the specific heats of air vary almost linearly with temperature in this temperature range FIGURE 737 Schematic and Ts diagram for Example 710 T K s 1 2 Air P1 95 kPa T1 295 K V1 V2 8 295 Isentropic compression v1 const v2 const FIGURE 738 The use of vr data for calculating the final temperature during an isentropic process Example 710 Process isentropic Given v1 T1 and v2 Find T2 vr2 vr vr1 v2 v1 vr1 T T2 T1 read read EXAMPLE 711 Isentropic Expansion of an Ideal Gas Air enters an isentropic turbine at 150 psia and 900F through a 05ft2 inlet section with a velocity of 500 fts Fig 739 It leaves at 15 psia with a velocity of 100 fts Calculate the air temperature at the turbine exit and the power produced in hp by this turbine Final PDF to printer 353 CHAPTER 7 cen22672ch07323412indd 353 110617 0852 AM FIGURE 739 Schematic for Example 711 150 psia 900F 500 fts 15 psia 100 fts Wout Air turbine T s 2 1 150 psia 15 psia SOLUTION Air is expanded in an isentropic turbine The exit temperature of the air and the power produced are to be determined Assumptions 1 This is a steadyflow process since there is no change with time 2 The process is isentropic ie reversibleadiabatic 3 Air is an ideal gas with con stant specific heats Properties The properties of air at an anticipated average temperature of 600F are c p 0250 Btu lbm R and k 1377 Table A2Eb The gas constant of air is R 03704 psia ft 3 lbm R Table A1E Analysis There is only one inlet and one exit and thus m 1 m 2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steadyflow system can be expressed in the rate form as E in E out Rate of net energy transfer by heat work and mass Δ E system dt Rate of change in internal kinetic potential etc energies 0 E in E out m h1 V 1 2 2 m h 2 V 2 2 2 W out W out m h 1 h 2 V 1 2 V 2 2 2 m c p T 1 T 2 V 1 2 V 2 2 2 The exit temperature of the air for this isentropic process is T 2 T 1 P 2 P 1 k 1 k 900 460 R 15 psia 150 psia 03771377 724 R The specific volume of air at the inlet and the mass flow rate are v 1 R T 1 P 1 03704 psia ft 3 lbm R900 460 R 150 psia 3358 ft 3 lbm m A 1 V 1 v 1 05 ft 2 500 ft s 3358 ft 3 lbm 7445 lbm s Substituting into the energy balance equation gives W out m c p T 1 T 2 V 1 2 V 2 2 2 7445 lbm s 0250 Btu lbm R 1360 724 R 500 ft s 2 100 ft s 2 2 1 Btu lbm 25037 ft 2 s 2 12194 Btu s 1 hp 07068 Btu s 17250 hp Discussion An actual adiabatic turbine will produce less power due to irrevers ibilities Also the exit temperature of the air will be higher in the actual turbine cor responding to a smaller enthalpy change between the inlet and the exit 0 steady Final PDF to printer 354 ENTROPY cen22672ch07323412indd 354 110617 0852 AM 710 REVERSIBLE STEADYFLOW WORK The work done during a process depends on the path followed as well as on the properties at the end states Recall that reversible quasiequilibrium moving boundary work associated with closed systems is expressed in terms of the fluid properties as W b 1 2 P dV We mentioned that the quasiequilibrium work interactions lead to the maxi mum work output for workproducing devices and the minimum work input for workconsuming devices It would also be very insightful to express the work associated with steady flow devices in terms of fluid properties Taking the positive direction of work to be from the system work output the energy balance for a steadyflow device undergoing an internally revers ible process can be expressed in differential form as δ q rev δ w rev dh d ke d pe But δ q rev Tds Tds dh v dP Eq 716 Eq 724 δ q rev dh v dP Substituting this into the relation above and canceling dh yield δ w rev v dP d ke d pe Integrating we find w rev 1 2 v dP Δke Δpe kJ kg 751 When the changes in kinetic and potential energies are negligible this equa tion reduces to w rev 1 2 v dP kJ kg 752 Equations 751 and 752 are relations for the reversible work output associ ated with an internally reversible process in a steadyflow device They will give a negative result when work is done on the system To avoid the negative sign Eq 751 can be written for work input to steadyflow devices such as compressors and pumps as w revin 1 2 v dP Δke Δpe 753 The resemblance between the v dP in these relations and P dv is striking They should not be confused with each other however since P dv is associ ated with reversible boundary work in closed systems Fig 740 FIGURE 740 Reversible work relations for steadyflow and closed systems wrev wrev a Steadyflow system b Closed system wrev v dP 1 2 wrev P dv 1 2 Final PDF to printer 355 CHAPTER 7 cen22672ch07323412indd 355 110617 0852 AM Obviously one needs to know v as a function of P for the given process to perform the integration When the working fluid is incompressible the spe cific volume v remains constant during the process and can be taken out of the integration Then Eq 751 simplifies to w rev v P 2 P 1 Δke Δpe kJ kg 754 For the steady flow of a liquid through a device that involves no work inter actions such as a nozzle or a pipe section the work term is zero and the preceding equation can be expressed as v P 2 P 1 V 2 2 V 1 2 2 g z 2 z 1 0 755 which is known as the Bernoulli equation in fluid mechanics It is developed for an internally reversible process and thus is applicable to incompressible fluids that involve no irreversibilities such as friction or shock waves This equation can be modified however to incorporate these effects Equation 752 has farreaching implications in engineering for devices that produce or consume work steadily such as turbines compressors and pumps It is obvious from this equation that the reversible steadyflow work is closely associated with the specific volume of the fluid flowing through the device The larger the specific volume the larger the reversible work produced or consumed by the steadyflow device Fig 741 This conclusion is equally valid for actual steadyflow devices Therefore every effort should be made to keep the specific volume of a fluid as small as possible during a compres sion process to minimize the work input and as large as possible during an expansion process to maximize the work output In steam or gas power plants the pressure rise in the pump or compressor is equal to the pressure drop in the turbine if we disregard the pressure losses in various other components In steam power plants the pump handles liquid which has a very small specific volume and the turbine handles vapor whose specific volume is many times larger Therefore the work output of the tur bine is much larger than the work input to the pump This is one of the reasons for the widespread use of steam power plants in electric power generation If we were to compress the steam exiting the turbine back to the turbine inlet pressure before cooling it first in the condenser in order to save the heat rejected we would have to supply all the work produced by the turbine back to the compressor In reality the required work input would be even greater than the work output of the turbine because of the irreversibilities present in both processes In gas power plants the working fluid typically air is compressed in the gas phase and a considerable portion of the work output of the turbine is con sumed by the compressor As a result a gas power plant delivers less net work per unit mass of the working fluid EXAMPLE 712 Compressing a Substance in the Liquid Versus Gas Phases Determine the compressor work input required to compress steam isentropically from 100 kPa to 1 MPa assuming that the steam exists as a saturated liquid and b saturated vapor at the inlet state FIGURE 741 The larger the specific volume the greater the work produced or consumed by a steadyflow device w w w v dP 1 2 v dP 1 2 v dP 1 2 Final PDF to printer 356 ENTROPY cen22672ch07323412indd 356 110617 0852 AM Proof that SteadyFlow Devices Deliver the Most and Consume the Least Work When the Process Is Reversible We have shown in Chap 6 that cyclic devices heat engines refrigerators and heat pumps deliver the most work and consume the least when reversible SOLUTION Steam is to be compressed from a given pressure to a specified pres sure isentropically The work input is to be determined for the cases of steam being a saturated liquid and saturated vapor at the inlet Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 The process is given to be isentropic Analysis We take first the pump and then the compressor as the system Both are control volumes since mass crosses the boundary Sketches of the pump and the tur bine together with the Ts diagram are given in Fig 742 a In this case steam is a saturated liquid initially and its specific volume is v 1 v f 100 kPa 0001043 m 3 kg Table A5 which remains essentially constant during the process Thus w rev 1 2 v dP v 1 P 2 P 1 0001043 m 3 kg 1000 100 kPa 1 kJ 1 kPa m 3 094 kJ kg b This time steam is a saturated vapor initially and remains a vapor during the entire compression process Since the specific volume of a gas changes considerably during a compression process we need to know how v varies with P to perform the integra tion in Eq 753 This relation in general is not readily available But for an isentro pic process it is easily obtained from the second T ds relation by setting ds 0 T ds dh v dP Eq 724 ds 0 isentropic process v dP dh Thus w revin 1 2 v dP 1 2 dh h 2 h 1 This result could also be obtained from the energy balance relation for an isentropic steadyflow process Next we determine the enthalpies State 1 P 1 100 kPa sat vapor h 1 26750 kJ kg s 1 73589 kJ kgK Table A5 State 2 P 2 1 MPa s 2 s 1 h 2 31945 kJ kg Table A6 Thus w revin 31945 26750 kJ kg 5195 kJ kg Discussion Note that compressing steam in the vapor form would require over 500 times more work than compressing it in the liquid form between the same pressure limits T s 1 2 1 2 1 MPa 100 kPa a b P2 1 MPa P1 100 kPa Compressor P1 100 kPa a Compressing a liquid b Compressing a vapor P2 1 MPa Pump FIGURE 742 Schematic and Ts diagram for Example 712 Final PDF to printer 357 CHAPTER 7 cen22672ch07323412indd 357 110617 0852 AM processes are used Now we demonstrate that this is also the case for indi vidual devices such as turbines and compressors in steady operation Consider two steadyflow devices one reversible and the other irreversible operating between the same inlet and exit states Again taking heat transfer to the system and work done by the system to be positive quantities the energy balance for each of these devices can be expressed in the differential form as Actual δ q act δ w act dh d ke d pe Reversible δ q rev δ w rev dh d ke d pe The righthand sides of these two equations are identical since both devices are operating between the same end states Thus δ q act δ w act δ q rev δ w rev or δ w rev δ w act δ q rev δ q act However δ q rev T ds Substituting this relation into the preceding equation and dividing each term by T we obtain δ w rev δ w act T ds δ q act T 0 since ds δ q act T Also T is the absolute temperature which is always positive Thus δ w rev δ w act or w rev w act Therefore workproducing devices such as turbines w is positive deliver more work and workconsuming devices such as pumps and compressors w is negative require less work when they operate reversibly Fig 743 711 MINIMIZING THE COMPRESSOR WORK We have just shown that the work input to a compressor is minimized when the compression process is executed in an internally reversible manner When the changes in kinetic and potential energies are negligible the compressor work is given by Eq 753 w revin 1 2 v dP 756 Obviously one way of minimizing the compressor work is to approximate an internally reversible process as much as possible by minimizing the irrevers ibilities such as friction turbulence and nonquasiequilibrium compression FIGURE 743 A reversible turbine delivers more work than an irreversible one if both operate between the same end states wrev wact P1 T1 P2 T2 Turbine Final PDF to printer 358 ENTROPY cen22672ch07323412indd 358 110617 0852 AM The extent to which this can be accomplished is limited by economic con siderations A second and more practical way of reducing the compressor work is to keep the specific volume of the gas as small as possible during the compression process This is done by keeping the temperature of the gas as low as possible during compression since the specific volume of a gas is pro portional to temperature Therefore reducing the work input to a compressor requires that the gas be cooled as it is compressed To have a better understanding of the effect of cooling during the com pression process we compare the work input requirements for three kinds of processes an isentropic process involves no cooling a polytropic process involves some cooling and an isothermal process involves maximum cool ing Assuming all three processes are executed between the same pressure levels P1 and P2 in an internally reversible manner and the gas behaves as an ideal gas Pv RT with constant specific heats we see that the compression work is determined by performing the integration in Eq 756 for each case with the following results Isentropic Pv k constant w compin kR T 2 T 1 k 1 kR T 1 k 1 P 2 P 1 k 1 k 1 757a Polytropic Pv n constant w compin nR T 2 T 1 n 1 nR T 1 n 1 P 2 P 1 n 1 n 1 757b Isothermal Pv constant w compin RT ln P 2 P 1 757c The three processes are plotted on a Pv diagram in Fig 744 for the same inlet state and exit pressure On a Pv diagram the area to the left of the process curve is the integral of v dP Thus it is a measure of the steadyflow compres sion work It is interesting to observe from this diagram that of the three inter nally reversible cases considered the adiabatic compression Pvk constant requires the maximum work and the isothermal compression T constant or Pv constant requires the minimum The work input requirement for the poly tropic case Pvn constant is between these two and decreases as the polytropic exponent n is decreased by increasing the heat rejection during the compres sion process If sufficient heat is removed the value of n approaches unity and the process becomes isothermal One common way of cooling the gas during compression is to use cooling jackets around the casing of the compressors Multistage Compression with Intercooling It is clear from these arguments that cooling a gas as it is compressed is desir able since this reduces the required work input to the compressor However often it is not possible to have adequate cooling through the casing of the com pressor and it becomes necessary to use other techniques to achieve effective cooling One such technique is multistage compression with intercooling FIGURE 744 Pv diagrams of isentropic polytropic and isothermal compression processes between the same pressure limits P 1 P2 P1 Isentropic n k Polytropic 1 n k Isothermal n 1 v Final PDF to printer 359 CHAPTER 7 cen22672ch07323412indd 359 110617 0852 AM where the gas is compressed in stages and cooled between stages by passing it through a heat exchanger called an intercooler Ideally the cooling process takes place at constant pressure and the gas is cooled to the initial tempera ture T1 at each intercooler Multistage compression with intercooling is espe cially attractive when a gas is to be compressed to very high pressures The effect of intercooling on compressor work is graphically illustrated on Pv and Ts diagrams in Fig 745 for a twostage compressor The gas is com pressed in the first stage from P1 to an intermediate pressure Px cooled at con stant pressure to the initial temperature T1 and compressed in the second stage to the final pressure P2 The compression processes in general can be modeled as polytropic Pvn constant where the value of n varies between k and 1 The colored area on the Pv diagram represents the work saved as a result of twostage compression with intercooling The process paths for single stage isothermal and polytropic processes are also shown for comparison The size of the colored area the saved work input varies with the value of the intermediate pressure Px and it is of practical interest to determine the conditions under which this area is maximized The total work input for a twostage compressor is the sum of the work inputs for each stage of compres sion as determined from Eq 757b w compin w comp Iin w comp IIin nR T 1 n 1 P x P 1 n 1 n 1 nR T 1 n 1 P 2 P x n 1 n 1 758 The only variable in this equation is Px The Px value that minimizes the total work is determined by differentiating this expression with respect to Px and setting the resulting expression equal to zero It yields P x P 1 P 2 12 or P x P 1 P 2 P x 759 That is to minimize compression work during twostage compression the pressure ratio across each stage of the compressor must be the same When this condition is satisfied the compression work at each stage becomes identi cal that is wcomp Iin wcomp IIin EXAMPLE 713 Work Input for Various Compression Processes Air is compressed steadily by a reversible compressor from an inlet state of 100 kPa and 300 K to an exit pressure of 900 kPa Determine the compressor work per unit mass for a isentropic compression with k 14 b polytropic compression with n 13 c isothermal compression and d ideal twostage compression with inter cooling with a polytropic exponent of 13 SOLUTION Air is compressed reversibly from a specified state to a specified pressure The compressor work is to be determined for the cases of isentropic poly tropic isothermal and twostage compression Assumptions 1 Steady operating conditions exist 2 At specified conditions air can be treated as an ideal gas 3 Kinetic and potential energy changes are negligible FIGURE 745 Pv and Ts diagrams for a twostage steadyflow compression process P 1 P2 P1 Isothermal Px Polytropic Work saved Intercooling 2 T s 1 T1 Intercooling 2 P2 P1 Px v Final PDF to printer 360 ENTROPY cen22672ch07323412indd 360 110617 0852 AM FIGURE 746 Schematic and Pv diagram for Example 713 P kPa 900 100 Isentropic k 14 Polytropic n 13 Twostage Isothermal 1 v P2 900 kPa P1 100 kPa T1 300 K Air compressor wcomp Analysis We take the compressor to be the system This is a control volume since mass crosses the boundary A sketch of the system and the Ts diagram for the process are given in Fig 746 The steadyflow compression work for all these four cases is determined by using the relations developed earlier in this section a Isentropic compression with k 14 w compin kR T 1 k 1 P 2 P 1 k 1 k 1 14 0287 kJ kgK 300 K 14 1 900 kPa 100 kPa 14 1 14 1 2632 kJ kg b Polytropic compression with n 13 w compin nR T 1 n 1 P 2 P 1 n 1 n 1 13 0287 kJ kgK 300 K 13 1 900 kPa 100 kPa 13 1 13 1 2464 kJ kg c Isothermal compression w compin RT ln P 2 P 1 0287 kJ kgK300 K ln 900 kPa 100 kPa 1892 kJ kg d Ideal twostage compression with intercooling n 13 In this case the pressure ratio across each stage is the same and the value of the intermediate pressure is P x P 1 P 2 12 100 kPa 900 kPa 12 300 kPa The compressor work across each stage is also the same Thus the total compressor work is twice the compression work for a single stage w compin 2 w comp Iin 2 nR T 1 n 1 P x P 1 n 1 n 1 2 13 0287 kJ kgK 300 K 13 1 300 kPa 100 kPa 13 1 13 1 2153 kJ kg Discussion Of all four cases considered the isothermal compression requires the minimum work and the isentropic compression the maximum The compressor work is decreased when two stages of polytropic compression are utilized instead of just one As the number of compressor stages is increased the compressor work approaches the value obtained for the isothermal case Final PDF to printer 361 CHAPTER 7 cen22672ch07323412indd 361 110617 0852 AM 712 ISENTROPIC EFFICIENCIES OF STEADYFLOW DEVICES We have said repeatedly that irreversibilities accompany all actual processes and that their effect is always to downgrade the performance of devices In engineering analysis it would be very useful to have some parameters that would enable us to quantify the degree of degradation of energy in these devices In Chap 6 we did this for cyclic devices such as heat engines and refrigerators by comparing the actual cycles to the idealized ones such as the Carnot cycle A cycle that was composed entirely of reversible processes served as the model cycle to which the actual cycles could be compared This idealized model cycle enabled us to determine the theoretical limits of perfor mance for cyclic devices under specified conditions and to examine how the performance of actual devices suffered as a result of irreversibilities Now we extend the analysis to discrete engineering devices working under steadyflow conditions such as turbines compressors and nozzles and we examine the degree of degradation of energy in these devices as a result of irreversibilities However first we need to define an ideal process that serves as a model for the actual processes Although some heat transfer between these devices and the surrounding medium is unavoidable many steadyflow devices are intended to operate under adiabatic conditions Therefore the model process for these devices should be an adiabatic one Furthermore an ideal process should involve no irreversibilities since the effect of irreversibilities is always to downgrade the performance of engineering devices Thus the ideal process that can serve as a suitable model for adiabatic steadyflow devices is the isentropic process Fig 747 The more closely the actual process approximates the idealized isen tropic process the better the device performs Thus it would be desirable to have a parameter that expresses quantitatively how efficiently an actual device approximates an idealized one This parameter is the isentropic or adiabatic efficiency which is a measure of the deviation of actual processes from the corresponding idealized ones Isentropic efficiencies are defined differently for different devices since each device is set up to perform different tasks Next we define the isentropic efficiencies of turbines compressors and nozzles by comparing the actual performance of these devices to their performance under isentropic condi tions for the same inlet state and exit pressure Isentropic Efficiency of Turbines For a turbine under steady operation the inlet state of the working fluid and the exhaust pressure are fixed Therefore the ideal process for an adia batic turbine is an isentropic process between the inlet state and the exhaust pressure The desired output of a turbine is the work produced and the isentropic efficiency of a turbine is defined as the ratio of the actual work output of the turbine to the work output that would be achieved if the process between the inlet state and the exit pressure were isentropic η T Actual turbine work Isentropic turbine work w a w s 760 FIGURE 747 The isentropic process involves no irreversibilities and serves as the ideal process for adiabatic devices P1 T1 P2 Actual irreversible P1 T1 P2 s2 s1 Ideal reversible Final PDF to printer 362 ENTROPY cen22672ch07323412indd 362 110617 0852 AM Usually the changes in kinetic and potential energies associated with a fluid stream flowing through a turbine are small relative to the change in enthalpy and can be neglected Then the work output of an adiabatic turbine simply becomes the change in enthalpy and Eq 760 becomes η T h 1 h 2a h 1 h 2s 761 where h2a and h2s are the enthalpy values at the exit state for actual and isen tropic processes respectively Fig 748 The value of ηT greatly depends on the design of the individual components that make up the turbine Welldesigned large turbines have isentropic effi ciencies above 90 percent For small turbines however isentropic efficiency may drop below 70 percent The value of the isentropic efficiency of a turbine is determined by measuring the actual work output of the turbine and by cal culating the isentropic work output for the measured inlet conditions and the exit pressure This value can then be used conveniently in the design of power plants FIGURE 748 The hs diagram for the actual and isentropic processes of an adiabatic turbine h s 1 h2s h2a h1 2s 2a P2 P1 Inlet state Actual process Isentropic process Exit pressure s2s s1 wa ws EXAMPLE 714 Isentropic Efficiency of a Steam Turbine Steam enters an adiabatic turbine steadily at 3 MPa and 400C and leaves at 50 kPa and 100C If the power output of the turbine is 2 MW determine a the isentropic efficiency of the turbine and b the mass flow rate of the steam flowing through the turbine SOLUTION Steam flows steadily in a turbine between inlet and exit states For a specified power output the isentropic efficiency and the mass flow rate are to be determined Assumptions 1 Steady operating conditions exist 2 The changes in kinetic and potential energies are negligible Analysis A sketch of the system and the Ts diagram of the process are given in Fig 749 a The enthalpies at various states are State 1 P 1 3 MPa T 1 400C h 1 32317 kJ kg s 1 69235 kJ kgK Table A6 State 2a P 2a 50 kPa T 2a 100C h 2a 26824 kJ kg Table A6 The exit enthalpy of the steam for the isentropic process h2s is determined from the requirement that the entropy of the steam remain constant s2s s1 State 2s P 2s 50 kPa s 2s s 1 s f 10912 kJ kgK sg 75931 kJ kgK Table A5 Obviously at the end of the isentropic process steam exists as a saturated mixture since sf s2s sg Thus we need to find the quality at state 2s first x 2s s 2s s f s fg 69235 10912 65019 0897 FIGURE 749 Schematic and Ts diagram for Example 714 T C s 1 2s s2s s1 50 kPa 3 MPa 400 100 2a Actual process Isentropic process P1 3 MPa T1 400C P2 50 kPa T2 100C 2 MW Steam turbine Final PDF to printer 363 CHAPTER 7 cen22672ch07323412indd 363 110617 0852 AM Isentropic Efficiencies of Compressors and Pumps The isentropic efficiency of a compressor is defined as the ratio of the work input required to raise the pressure of a gas to a specified value in an isentro pic manner to the actual work input η C Isentropic compressor work Actual compressor work w s w a 762 Notice that the isentropic compressor efficiency is defined with the isentropic work input in the numerator instead of in the denominator This is because ws is a smaller quantity than wa and this definition prevents ηC from becoming greater than 100 percent which would falsely imply that the actual compres sors performed better than the isentropic ones Also notice that the inlet con ditions and the exit pressure of the gas are the same for both the actual and the isentropic compressor When the changes in kinetic and potential energies of the gas being com pressed are negligible the work input to an adiabatic compressor becomes equal to the change in enthalpy and Eq 762 for this case becomes η C h 2s h 1 h 2a h 1 763 where h2a and h2s are the enthalpy values at the exit state for actual and isentropic compression processes respectively as illustrated in Fig 750 Again the value of ηC greatly depends on the design of the compressor Welldesigned compressors have isentropic efficiencies that range from 80 to 90 percent and h 2s h f x 2s h fg 34054 0897 23047 24079 kJ kg By substituting these enthalpy values into Eq 761 the isentropic efficiency of this turbine is determined to be η T h 1 h 2a h 1 h 2s 32317 26824 32317 24079 0667 or 667 b The mass flow rate of steam through this turbine is determined from the energy balance for steadyflow systems E in E out m h 1 W aout m h 2a W aout m h 1 h 2a 2 MW 1000 kJ s 1 MW m 32317 26824 kJ kg m 364 kg s FIGURE 750 The hs diagram of the actual and isentropic processes of an adiabatic compressor h2s h s 1 h1 h2a 2s 2a P2 P1 Inlet state s2s s1 wa Exit pressure ws Actual process Isentropic process Final PDF to printer 364 ENTROPY cen22672ch07323412indd 364 110617 0852 AM When the changes in potential and kinetic energies of a liquid are negli gible the isentropic efficiency of a pump is defined similarly as η P w s w a v P 2 P 1 h 2a h 1 764 When no attempt is made to cool the gas as it is compressed the actual compression process is nearly adiabatic and the reversible adiabatic ie isen tropic process serves well as the ideal process However sometimes compres sors are cooled intentionally by utilizing fins or a water jacket placed around the casing to reduce the work input requirements Fig 751 In this case the isentropic process is not suitable as the model process since the device is no longer adiabatic and the isentropic compressor efficiency defined above is meaningless A realistic model process for compressors that are intentionally cooled during the compression process is the reversible isothermal process Then we can conveniently define an isothermal efficiency for such cases by comparing the actual process to a reversible isothermal one η C w t w a 765 where wt and wa are the required work inputs to the compressor for the revers ible isothermal and actual cases respectively EXAMPLE 715 Effect of Efficiency on Compressor Power Input Air is compressed by an adiabatic compressor from 100 kPa and 12C to a pressure of 800 kPa at a steady rate of 02 kgs If the isentropic efficiency of the compressor is 80 percent determine a the exit temperature of air and b the required power input to the compressor SOLUTION Air is compressed to a specified pressure at a specified rate For a given isentropic efficiency the exit temperature and the power input are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas 3 The changes in kinetic and potential energies are negligible Analysis A sketch of the system and the Ts diagram of the process are given in Fig 752 a We know only one property pressure at the exit state and we need to know one more to fix the state and thus determine the exit temperature The property that can be determined with minimal effort in this case is h2a since the isentropic efficiency of the compressor is given At the compressor inlet T 1 285 K h 1 28514 kJ kg Table A17 P r1 11584 The enthalpy of the air at the end of the isentropic compression process is determined by using one of the isentropic relations of ideal gases P r2 P r1 P 2 P 1 11584 800 kPa 100 kPa 92672 FIGURE 752 Schematic and Ts diagram for Example 715 T K s 1 2s s2s s1 Isentropic process 285 Actual process T2a T2s 2a 800 kPa 100 kPa P2 800 kPa P1 100 kPa T1 285 K Air compressor m 02 kgs FIGURE 751 Compressors are sometimes intentionally cooled to minimize the work input Cooling water Air Compressor Final PDF to printer 365 CHAPTER 7 cen22672ch07323412indd 365 110617 0852 AM and P r2 92672 h 2s 51705 kJ kg Substituting the known quantities into the isentropic efficiency relation we have η C h 2s h 1 h 2a h 1 080 51705 28514 kJ kg h 2a 28514 kJ kg Thus h 2a 57503 kJ kg T 2a 5695 K b The required power input to the compressor is determined from the energy balance for steadyflow devices E in E out m h 1 W ain m h 2a W ain m h 2a h 1 02 kg s 57503 28514 kJ kg 580 kW Discussion Notice that in determining the power input to the compressor we used h2a instead of h2s since h2a is the actual enthalpy of the air as it exits the compressor The quantity h2s is a hypothetical enthalpy value that the air would have if the process were isentropic Isentropic Efficiency of Nozzles Nozzles are essentially adiabatic devices and are used to accelerate a fluid Therefore the isentropic process serves as a suitable model for nozzles The isentropic efficiency of a nozzle is defined as the ratio of the actual kinetic energy of the fluid at the nozzle exit to the kinetic energy value at the exit of an isentropic nozzle for the same inlet state and exit pressure That is η N Actual KE at nozzle exit Isentropic KE at nozzle exit V 2a 2 V 2s 2 766 Note that the exit pressure is the same for both the actual and isentropic pro cesses but the exit state is different Nozzles involve no work interactions and the fluid experiences little or no change in its potential energy as it flows through the device If in addition the inlet velocity of the fluid is small relative to the exit velocity the energy balance for this steadyflow device reduces to h 1 h 2a V 2a 2 2 Final PDF to printer 366 ENTROPY cen22672ch07323412indd 366 110617 0852 AM Then the isentropic efficiency of the nozzle can be expressed in terms of enthalpies as η N h 1 h 2a h 1 h 2s 767 where h2a and h2s are the enthalpy values at the nozzle exit for the actual and isentropic processes respectively Fig 753 Isentropic efficiencies of noz zles are typically above 90 percent and nozzle efficiencies above 95 percent are not uncommon FIGURE 753 The hs diagram of the actual and isentropic processes of an adiabatic nozzle h2s h s 1 h1 h2a 2s 2a P2 P1 Inlet state Actual process Isentropic process s2s s1 Exit pressure 2 V2 2a V2 2a 2 EXAMPLE 716 Effect of Efficiency on Nozzle Exit Velocity Air at 200 kPa and 950 K enters an adiabatic nozzle at low velocity and is discharged at a pressure of 110 kPa If the isentropic efficiency of the nozzle is 92 percent deter mine a the maximum possible exit velocity b the exit temperature and c the actual exit velocity of the air Assume constant specific heats for air SOLUTION The acceleration of air in a nozzle is considered For specified exit pressure and isentropic efficiency the maximum and actual exit velocities and the exit temperature are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas 3 The inlet kinetic energy is negligible Analysis A sketch of the system and the Ts diagram of the process are given in Fig 754 The temperature of air will drop during this acceleration process because some of its internal energy is converted to kinetic energy This problem can be solved accu rately by using property data from the air table But we will assume constant specific heats thus sacrifice some accuracy to demonstrate their use Let us guess the aver age temperature of the air to be about 850 K Then the average values of cp and k at this anticipated average temperature are determined from Table A2b to be cp 111 kJkgK and k 1349 a The exit velocity of the air will be a maximum when the process in the nozzle involves no irreversibilities The exit velocity in this case is determined from the steadyflow energy equation However first we need to determine the exit tempera ture For the isentropic process of an ideal gas we have T 2s T 1 P 2s P 1 k 1 k or T 2s T 1 P 2s P 1 k 1 k 950 K 110 kPa 200 kPa 03491349 814 K This gives an average temperature of 882 K which is somewhat higher than the assumed average temperature 850 K This result could be refined by reevaluating the k value at 882 K and repeating the calculations but it is not warranted since the two average temperatures are sufficiently close doing so would change the tempera ture by only 06 K which is not significant FIGURE 754 Schematic and Ts diagram for Example 716 Air nozzle ηN 092 T2s T K s 1 950 T2a 2s 2a Actual process Isentropic process s2s s1 P1 200 kPa T1 950 K V1 V2 P2 110 kPa 200 kPa 110 kPa Final PDF to printer 367 CHAPTER 7 cen22672ch07323412indd 367 110617 0852 AM 713 ENTROPY BALANCE The property entropy is a measure of molecular disorder or randomness of a system and the second law of thermodynamics states that entropy can be created but it cannot be destroyed Therefore the entropy change of a system during a process is greater than the entropy transfer by an amount equal to the entropy generated during the process within the system and the increase of entropy principle for any system is expressed as Fig 756 Total entropy entering Total entropy leaving Total entropy generated Change in the total entropy of the system or S in S out S gen Δ S system 768 FIGURE 755 A substance leaves actual nozzles at a higher temperature thus a lower velocity as a result of friction 825 K 527 ms Actual nozzle Isentropic nozzle 814 K 549 ms 950 K Air Now we can determine the isentropic exit velocity of the air from the energy bal ance for this isentropic steadyflow process e in e out h 1 V 1 2 2 h 2s V 2s 2 2 or V 2s 2 h 1 h 2s 2 c pavg T 1 T 2s 2 111 kJ kgK 950 814 K 1000 m 2 s 2 1 kJ kg 549 m s b The actual exit temperature of the air is higher than the isentropic exit temperature evaluated above and is determined from η N h 1 h 2a h 1 h 2s c pavg T 1 T 2a c pavg T 1 T 2s or 092 950 T 2a 950 814 T 2a 825 K That is the temperature is 11 K higher at the exit of the actual nozzle as a result of irreversibilities such as friction It represents a loss since this rise in temperature comes at the expense of kinetic energy Fig 755 c The actual exit velocity of air can be determined from the definition of isentropic efficiency of a nozzle η N V 2a 2 V 2s 2 V 2a η N V 2s 2 092 549 m s 2 527 m s FIGURE 756 Energy and entropy balances for a system System ΔEsystem ΔEsystem Ein Eout ΔSsystem Sin Sout Sgen Eout Ein Sout Sin ΔSsystem Sgen 0 Final PDF to printer 368 ENTROPY cen22672ch07323412indd 368 110617 0852 AM which is a verbal statement of Eq 79 This relation is often referred to as the entropy balance and is applicable to any system undergoing any process The entropy balance relation above can be stated as the entropy change of a system during a process is equal to the net entropy transfer through the system boundary plus the entropy generated within the system Next we discuss the various terms in that relation Entropy Change of a System ΔSsystem Despite the reputation of entropy as being vague and abstract and the intimi dation associated with it entropy balance is actually easier to deal with than energy balance since unlike energy entropy does not exist in various forms Therefore the determination of entropy change of a system during a process involves evaluating the entropy of the system at the beginning and at the end of the process and taking their difference That is Entropy change Entropy at final state Entropy at initial state or Δ S system S final S intial S 2 S 1 769 Note that entropy is a property and the value of a property does not change unless the state of the system changes Therefore the entropy change of a system is zero if the state of the system does not change during the process For example the entropy change of steadyflow devices such as nozzles compressors turbines pumps and heat exchangers is zero during steady operation When the properties of the system are not uniform the entropy of the sys tem can be determined by integration from S system s δm V sρ dV 770 where V is the volume of the system and ρ is density Mechanisms of Entropy Transfer Sin and Sout Entropy can be transferred to or from a system by two mechanisms heat transfer and mass flow in contrast energy is transferred by work also Entropy transfer is recognized at the system boundary as it crosses the boundary and it represents the entropy gained or lost by a system during a process The only form of entropy interaction associated with a fixed mass or closed system is heat transfer and thus the entropy transfer for an adiabatic closed system is zero 1 Heat Transfer Heat is in essence a form of disorganized energy and some disorganization entropy will flow with heat Heat transfer to a system increases the entropy of that system and thus the level of molecular disorder or randomness and heat transfer from a system decreases it In fact heat rejection is the only way the entropy of a fixed mass can be decreased The ratio of the heat transfer Q Final PDF to printer 369 CHAPTER 7 cen22672ch07323412indd 369 110617 0852 AM at a location to the absolute temperature T at that location is called the entropy flow or entropy transfer and is expressed as Fig 757 Entropy transfer by heat transfer S heat Q T T constant 771 The quantity QT represents the entropy transfer accompanied by heat trans fer and the direction of entropy transfer is the same as the direction of heat transfer since thermodynamic temperature T is always a positive quantity When the temperature T is not constant the entropy transfer during a process 12 can be determined by integration or by summation if appropri ate as S heat 1 2 δQ T Q k T k 772 where Qk is the heat transfer through the boundary at temperature Tk at loca tion k When two systems are in contact the entropy transfer from the warmer system is equal to the entropy transfer into the cooler one at the point of con tact That is no entropy can be created or destroyed at the boundary since the boundary has no thickness and occupies no volume Note that work is entropyfree and no entropy is transferred by work Energy is transferred by both heat and work whereas entropy is transferred only by heat That is Entropy transfer by work S work 0 773 The first law of thermodynamics makes no distinction between heat trans fer and work it considers them as equals The distinction between heat transfer and work is brought out by the second law an energy interaction that is accompanied by entropy transfer is heat transfer and an energy interaction that is not accompanied by entropy transfer is work That is no entropy is exchanged during a work interaction between a system and its surroundings Thus only energy is exchanged during work interac tion whereas both energy and entropy are exchanged during heat transfer Fig 758 2 Mass Flow Mass contains entropy as well as energy and the entropy and energy con tents of a system are proportional to the mass When the mass of a system is doubled so are the entropy and energy contents of the system Both entropy and energy are carried into or out of a system by streams of mat ter and the rates of entropy and energy transport into or out of a system are proportional to the mass flow rate Closed systems do not involve any mass flow and thus any entropy transfer by mass When a mass in the amount of m enters or leaves a system entropy in the amount of ms where s is the specific entropy entropy per unit mass entering or leaving accompanies it Fig 759 That is Entropy transfer by mass flow S mass ms 774 FIGURE 757 Heat transfer is always accompanied by entropy transfer in the amount of QT where T is the boundary temperature Surroundings System Q 500 kJ Sheat Q Tb 125 kJK Tb 400 K FIGURE 758 No entropy accompanies work as it crosses the system boundary But entropy may be generated within the system as work is dissipated into a less useful form of energy Entropy is not transferred with work Entropy generation via friction FIGURE 759 Mass contains entropy as well as energy and thus mass flow into or out of a system is always accompanied by energy and entropy transfer Control volume mh ms h s m Final PDF to printer 370 ENTROPY cen22672ch07323412indd 370 110617 0852 AM Therefore the entropy of a system increases by ms when mass in the amount of m enters and decreases by the same amount when the same amount of mass at the same state leaves the system When the properties of the mass change during the process the entropy transfer by mass flow can be determined by integration from S mass A c sρ V n d A c and S mass s δm Δt S mass dt 775 where Ac is the crosssectional area of the flow and Vn is the local velocity normal to dAc Entropy Generation Sgen Irreversibilities such as friction mixing chemical reactions heat transfer through a finite temperature difference unrestrained expansion nonquasiequilibrium compression or expansion always cause the entropy of a system to increase and entropy generation is a measure of the entropy created by such effects during a process For a reversible process a process that involves no irreversibilities the entropy generation is zero and thus the entropy change of a system is equal to the entropy transfer Therefore the entropy balance relation in the revers ible case becomes analogous to the energy balance relation which states that the energy change of a system during a process is equal to the energy transfer during that process However note that the energy change of a system equals the energy transfer for any process but the entropy change of a system equals the entropy transfer only for a reversible process The entropy transfer by heat QT is zero for adiabatic systems and the entropy transfer by mass ms is zero for systems that involve no mass flow across their boundary ie closed systems Entropy balance for any system undergoing any process can be expressed more explicitly as S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy kJ K 776 or in the rate form as S in S out Rate of net entropy transfer by heat and mass S gen Rate of entropy generation d S system dt Rate of entropy generation kW K 777 where the rates of entropy transfer by heat transferred at a rate of Q and mass flowing at a rate of m are S heat Q T and S mass m s The entropy balance can also be expressed on a unitmass basis as s in s out s gen Δ s system kJ kgK 778 where all the quantities are expressed per unit mass of the system Note that for a reversible process the entropy generation term Sgen drops out from all of the relations above Final PDF to printer 371 CHAPTER 7 cen22672ch07323412indd 371 110617 0852 AM The term Sgen represents the entropy generation within the system bound ary only Fig 760 and not the entropy generation that may occur outside the system boundary during the process as a result of external irreversibili ties Therefore a process for which Sgen 0 is internally reversible but not necessarily totally reversible The total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes the system itself and its immediate surroundings where exter nal irreversibilities might be occurring Fig 761 Also the entropy change in this case is equal to the sum of the entropy change of the system and the entropy change of the immediate surroundings Note that under steady condi tions the state and thus the entropy of the immediate surroundings let us call it the buffer zone at any point does not change during the process and the entropy change of the buffer zone is zero The entropy change of the buffer zone if any is usually small relative to the entropy change of the system and thus it is usually disregarded When evaluating the entropy transfer between an extended system and the surroundings the boundary temperature of the extended system is simply taken to be the environment temperature Closed Systems A closed system involves no mass flow across its boundaries and its entropy change is simply the difference between the initial and final entropies of the system The entropy change of a closed system is due to the entropy trans fer accompanying heat transfer and the entropy generation within the system boundaries Taking the positive direction of heat transfer to be to the system the general entropy balance relation Eq 776 can be expressed for a closed system as Closed system Q k T k S gen Δ S system S 2 S 1 kJ K 779 The entropy balance relation above can be stated as The entropy change of a closed system during a process is equal to the sum of the net entropy transferred through the system boundary by heat transfer and the entropy generated within the system boundaries For an adiabatic process Q 0 the entropy transfer term in the preced ing relation drops out and the entropy change of the closed system becomes equal to the entropy generation within the system boundaries That is Adiabatic closed system S gen Δ S adiabatic system 780 Noting that any closed system and its surroundings can be treated as an adia batic system and the total entropy change of a system is equal to the sum of the entropy changes of its parts the entropy balance for a closed system and its surroundings can be written as System Surroundings S gen ΔS Δ S system Δ S surroundings 781 where ΔSsystem ms2 s1 and the entropy change of the surroundings can be determined from ΔSsurr QsurrTsurr if its temperature is constant At initial stages of studying entropy and entropy transfer it is more instructive to start FIGURE 760 Mechanisms of entropy transfer for a general system System Mass Heat Mass Heat ΔSsystem Sgen 0 Sin Sout FIGURE 761 Entropy generation outside system boundaries can be accounted for by writing an entropy balance on an extended system that includes the system and its immediate surroundings Immediate surroundings System Q Tsurr Final PDF to printer 372 ENTROPY cen22672ch07323412indd 372 110617 0852 AM with the general form of the entropy balance Eq 776 and to simplify it for the problem under consideration The specific relations above are convenient to use after a certain degree of intuitive understanding of the material is achieved Control Volumes The entropy balance relations for control volumes differ from those for closed systems in that they involve one more mechanism of entropy exchange mass flow across the boundaries As mentioned earlier mass possesses entropy as well as energy and the amounts of these two extensive properties are propor tional to the amount of mass Fig 762 Taking the positive direction of heat transfer to be to the system the general entropy balance relations Eqs 776 and 777 can be expressed for control volumes as Q k T k m i s i m e s e S gen S 2 S 1 CV kJ K 782 or in the rate form as Q k T k m i s i m e s e S gen d S CV dt kW K 783 This entropy balance relation can be stated as The rate of entropy change within the control volume during a process is equal to the sum of the rate of entropy transfer through the control volume boundary by heat transfer the net rate of entropy transfer into the control volume by mass flow and the rate of entropy generation within the boundaries of the control volume as a result of irreversibilities Most control volumes encountered in practice such as turbines compres sors nozzles diffusers heat exchangers pipes and ducts operate steadily and thus they experience no change in their entropy Therefore the entropy balance relation for a general steadyflow process can be obtained from Eq 783 by setting dSCVdt 0 and rearranging to give Steadyflow S gen m e s e m i s i Q k T k 784 For singlestream one inlet and one exit steadyflow devices the entropy balance relation simplifies to Steadyflow singlestream S gen m s e s i Q k T k 785 For the case of an adiabatic singlestream device the entropy balance relation further simplifies to Steadyflow singlestream adiabatic S gen m s e s i 786 which indicates that the specific entropy of the fluid must increase as it flows through an adiabatic device since S gen 0 Fig 763 If the flow through the device is reversible and adiabatic then the entropy remains constant se si regardless of the changes in other properties FIGURE 762 The entropy of a control volume changes as a result of mass flow as well as heat transfer Control volume Surroundings Entropy transfer by heat Entropy transfer by mass ΔSCV Q T misi mese Sgen mi si me se Q T FIGURE 763 The entropy of a substance always increases or remains constant in the case of a reversible process as it flows through a singlestream adiabatic steadyflow device se si si Steadyflow device Final PDF to printer 373 CHAPTER 7 cen22672ch07323412indd 373 110617 0852 AM FIGURE 764 Schematic for Example 717 Q Brick wall 30 cm 20C 5C 0C 27C EXAMPLE 717 Entropy Generation in a Wall Consider steady heat transfer through a 5m 7m brick wall of a house of thickness 30 cm On a day when the temperature of the outdoors is 0C the house is maintained at 27C The temperatures of the inner and outer surfaces of the brick wall are mea sured to be 20C and 5C respectively and the rate of heat transfer through the wall is 1035 W Determine the rate of entropy generation in the wall and the rate of total entropy generation associated with this heat transfer process SOLUTION Steady heat transfer through a wall is considered For specified heat transfer rate wall temperatures and environment temperatures the entropy genera tion rate within the wall and the total entropy generation rate are to be determined Assumptions 1 The process is steady and thus the rate of heat transfer through the wall is constant 2 Heat transfer through the wall is onedimensional Analysis We first take the wall as the system Fig 764 This is a closed sys tem since no mass crosses the system boundary during the process We note that the entropy change of the wall is zero during this process since the state and thus the entropy of the wall do not change anywhere in the wall Heat and entropy are entering from one side of the wall and leaving from the other side The rate form of the entropy balance for the wall simplifies to S in S out Rate of net entropy transfer by heat and mass S gen Rate of entropy generation d S system dt Rate of change in entropy Q T in Q T out S gen 0 1035 W 293 K 1035 W 278 K S gen 0 Therefore the rate of entropy generation in the wall is S gen 0191 W K Note that entropy transfer by heat at any location is QT at that location and the direc tion of entropy transfer is the same as the direction of heat transfer To determine the rate of total entropy generation during this heat transfer process we extend the system to include the regions on both sides of the wall that experience a temperature change Then one side of the system boundary becomes room tempera ture while the other side becomes the temperature of the outdoors The entropy bal ance for this extended system system immediate surroundings is the same as that given above except the two boundary temperatures are now 300 and 273 K instead of 293 and 278 K respectively Then the rate of total entropy generation becomes 1035 W 300 K 1035 W 273 K S gentotal 0 S gentotal 0341 W K Discussion Note that the entropy change of this extended system is also zero since the state of air does not change at any point during the process The difference between the two entropy generations is 0150 WK and it represents the entropy gen erated in the air layers on both sides of the wall The entropy generation in this case is entirely due to irreversible heat transfer through a finite temperature difference 0 steady Final PDF to printer 374 ENTROPY cen22672ch07323412indd 374 110617 0852 AM FIGURE 765 Schematic and Ts diagram for Example 718 T C s 1 2 s1 h const 450 s2 Throttling process P1 7 MPa T1 450 C P2 3 MPa EXAMPLE 718 Entropy Generation During a Throttling Process Steam at 7 MPa and 450C is throttled in a valve to a pressure of 3 MPa during a steadyflow process Determine the entropy generated during this process and check if the increase of entropy principle is satisfied SOLUTION Steam is throttled to a specified pressure The entropy generated dur ing this process is to be determined and the validity of the increase of entropy prin ciple is to be verified Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 ΔECV 0 and ΔSCV 0 2 Heat transfer to or from the valve is negligible 3 The kinetic and potential energy changes are negligible Δke Δpe 0 Analysis We take the throttling valve as the system Fig 765 This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one exit and thus m 1 m 2 m Also the enthalpy of a fluid remains nearly constant during a throttling process and thus h2 h1 The entropy of the steam at the inlet and the exit states is determined from the steam tables to be State 1 P 1 7 MPA T 1 450C h 1 32883 kJ kg s 1 66353 kJ kgK State 2 P 2a 3 MPa h 2 h 1 s 2 70046 kJ kgK Then the entropy generation per unit mass of the steam is determined from the entropy balance applied to the throttling valve S in S out Rate of net entropy transfer by heat and mass S gen Rate of entropy generation d S system dt Rate of change in entropy m s 1 m s 2 S gen 0 S gen m s 2 s 1 Dividing by mass flow rate and substituting gives s gen s 2 s 1 70046 66353 03693 kJ kgK This is the amount of entropy generated per unit mass of steam as it is throttled from the inlet state to the final pressure and it is caused by unrestrained expansion The increase of entropy principle is obviously satisfied during this process since the entropy generation is positive 0 steady Final PDF to printer 375 CHAPTER 7 cen22672ch07323412indd 375 110617 0852 AM EXAMPLE 719 Entropy Generated when a Hot Block Is Dropped in a Lake A 50kg block of iron casting at 500 K is thrown into a large lake that is at a tem perature of 285 K The iron block eventually reaches thermal equilibrium with the lake water Assuming an average specific heat of 045 kJkgK for the iron determine a the entropy change of the iron block b the entropy change of the lake water and c the entropy generated during this process SOLUTION A hot iron block is thrown into a lake and cools to the lake tempera ture The entropy changes of the iron and of the lake as well as the entropy generated during this process are to be determined Assumptions 1 Both the water and the iron block are incompressible substances 2 Constant specific heats can be used for the water and the iron 3 The kinetic and potential energy changes of the iron are negligible ΔKE ΔPE 0 and thus ΔE ΔU Properties The specific heat of the iron is 045 kJkgK Table A3 Analysis We take the iron casting as the system Fig 766 This is a closed system since no mass crosses the system boundary during the process To determine the entropy change for the iron block and for the lake first we need to know the final equilibrium temperature Given that the thermal energy capacity of the lake is very large relative to that of the iron block the lake will absorb all the heat rejected by the iron block without experiencing any change in its temperature There fore the iron block will cool to 285 K during this process while the lake temperature remains constant at 285 K a The entropy change of the iron block can be determined from Δ S iron m s 2 s 1 m c avg ln T 2 T 1 50 kg045 kJ kgK ln 285 K 500 K 1265 kJ K b The temperature of the lake water remains constant during this process at 285 K Also the amount of heat transfer from the iron block to the lake is determined from an energy balance on the iron block to be E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q out ΔU m c avg T 2 T 1 or Q out m c avg T 1 T 2 50 kg 045 kJ kgK 500 285 K 4838 kJ Then the entropy change of the lake becomes Δ S lake Q lake T lake 4838 kJ 285 K 1697 kJ K c The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the iron block and its immediate FIGURE 766 Schematic for Example 719 Lake 285 K Iron casting m 50 kg T1 500 K Final PDF to printer 376 ENTROPY cen22672ch07323412indd 376 110617 0852 AM FIGURE 767 Schematic for Example 720 Steam 35C 10000 kgh 32C 30C 3 2 1 4 Air 20C surroundings so that the boundary temperature of the extended system is at 285 K at all times S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy Q out T b S gen Δ S system or S gen Q out T b Δ S system 4838 kJ 285 K 1265 kJ K 432 kJ K Discussion The entropy generated can also be determined by taking the iron block and the entire lake as the system which is an isolated system and applying an entropy balance An isolated system involves no heat or entropy transfer and thus the entropy generation in this case becomes equal to the total entropy change S gen Δ S total Δ S iron Δ S lake 1265 1697 432 kJ K which is the same result obtained above EXAMPLE 720 Entropy Generation in a Heat Exchanger Air in a large building is kept warm by heating it with steam in a heat exchanger Fig 767 Saturated water vapor enters this unit at 35C at a rate of 10000 kgh and leaves as saturated liquid at 32C Air at 1atm pressure enters the unit at 20C and leaves at 30C at about the same pressure Determine the rate of entropy generation associated with this process SOLUTION Air is heated by steam in a heat exchanger The rate of entropy gen eration associated with this process is to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid 3 Changes in the kinetic and potential energies of fluid streams are negligible 4 Air is an ideal gas with constant specific heats at room temperature 5 The pressure of air remains constant Analysis The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger S in S out Rate of net entropy transfer by heat and mass S gen Rate of entropy generation d S system dt Rate of change in entropy m steam s 1 m air s 3 m steam s 2 m air s 4 S gen 0 S gen m steam s 2 s 1 m air s 4 s 3 The specific heat of air at room temperature is cp 1005 kJkgC Table A2a The properties of the steam at the inlet and exit states are 0 steady Final PDF to printer 377 CHAPTER 7 cen22672ch07323412indd 377 110617 0852 AM T 1 35C x 1 1 h 1 25646 kJ kg s 1 83517 kJ kgK Table A4 T 2 32C x 2 0 h 2 13410 kJ kg s 2 04641 kJ kgK Table A4 From an energy balance the heat transferred from steam is equal to the heat trans ferred to the air Then the mass flow rate of air is determined to be Q m steam h 2 h 1 10000 3600 kg s 25646 13410 kJ kg 6751 kW m air Q c p T 4 T 3 6751 kW 1005 kJ kgC 30 20 C 6717 kg s Substituting into the entropy balance relation the rate of entropy generation becomes S gen m steam s 2 s 1 m air s 4 s 3 m steam s 2 s 1 m air c p ln T 4 T 3 100003600 kgs04641 83517kJkgK 6711 kg s 1005 kJ kgK ln 303 K 293 K 0745 kW K Discussion Note that the pressure of air remains nearly constant as it flows through the heat exchanger and thus the pressure term is not included in the entropy change expression for air EXAMPLE 721 Entropy Generation Associated with Heat Transfer A frictionless pistoncylinder device contains a saturated liquidvapor mixture of water at 100C During a constantpressure process 600 kJ of heat is transferred to the surrounding air at 25C As a result part of the water vapor contained in the cylin der condenses Determine a the entropy change of the water and b the total entropy generation during this heat transfer process SOLUTION A saturated liquidvapor mixture of water loses heat to its surround ings and some of the vapor condenses The entropy change of the water and the total entropy generation are to be determined Assumptions 1 There are no irreversibilities involved within the system boundar ies and thus the process is internally reversible 2 The water temperature remains constant at 100C everywhere including the boundaries Analysis We first take the water in the cylinder as the system Fig 768 This is a closed system since no mass crosses the system boundary during the process We note that the pressure and thus the temperature of water in the cylinder remain constant during this process Also the entropy of the system decreases during the process because of heat loss FIGURE 768 Schematic for Example 721 T 100C H2O Tsurr 25C 600 kJ Final PDF to printer 378 ENTROPY cen22672ch07323412indd 378 110617 0852 AM a Noting that water undergoes an internally reversible isothermal process its entropy change can be determined from Δ S system Q T system 600 kJ 100 273 K 161 kJ K b To determine the total entropy generation during this process we consider the extended system which includes the water the pistoncylinder device and the region immediately outside the system that experiences a temperature change so that the entire boundary of the extended system is at the surrounding temperature of 25C The entropy balance for this extended system system immediate surroundings yields S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy Q out T b S gen Δ S system or S gen Q out T b Δ S system 600 kJ 25 273 K 161 kJ K 040 kJ K The entropy generation in this case is entirely due to irreversible heat transfer through a finite temperature difference Note that the entropy change of this extended system is equivalent to the entropy change of water since the pistoncylinder device and the immediate surroundings do not experience any change of state at any point and thus any change in any property including entropy Discussion For the sake of argument consider the reverse process ie the transfer of 600 kJ of heat from the surrounding air at 25C to saturated water at 100C and see if the increase of entropy principle can detect the impossibility of this process This time heat transfer will be to the water heat gain instead of heat loss and thus the entropy change of water will be 161 kJK Also the entropy transfer at the boundary of the extended system will have the same magnitude but opposite direc tion This will result in an entropy generation of 04 kJK The negative sign for the entropy generation indicates that the reverse process is impossible To complete the discussion let us consider the case where the surrounding air tem perature is a differential amount below 100C say 99999 9C instead of being 25C This time heat transfer from the saturated water to the surrounding air will take place through a differential temperature difference rendering this process reversible It can be shown that Sgen 0 for this process Remember that reversible processes are idealized processes and they can be approached but never reached in reality Entropy Generation Associated with a Heat Transfer Process In Example 721 it is determined that 04 kJK of entropy is generated during the heat transfer process but it is not clear where exactly the entropy genera tion takes place and how To pinpoint the location of entropy generation we need to be more precise about the description of the system its surroundings and the system boundary Final PDF to printer 379 CHAPTER 7 cen22672ch07323412indd 379 110617 0852 AM In that example we assumed both the system and the surrounding air to be isothermal at 100 and 25C respectively This assumption is reasonable if both fluids are well mixed The inner surface of the wall must also be at 100C while the outer surface is at 25C since two bodies in physical con tact must have the same temperature at the point of contact Considering that entropy transfer with heat transfer Q through a surface at constant tempera ture T is QT the entropy transfer from the water into the wall is QTsys 161 kJK Likewise entropy transfer from the outer surface of the wall into the surrounding air is QTsurr 201 kJK Obviously entropy in the amount of 201 161 04 kJK is generated in the wall as illustrated in Fig 769b Identifying the location of entropy generation enables us to determine whether a process is internally reversible A process is internally reversible if no entropy is generated within the system boundaries Therefore the heat trans fer process discussed in Example 721 is internally reversible if the inner sur face of the wall is taken as the system boundary and thus the system excludes the container wall If the system boundary is taken to be the outer surface of the container wall then the process is no longer internally reversible since the wall which is the site of entropy generation is now part of the system For thin walls it is very tempting to ignore the mass of the wall and to regard the wall as the boundary between the system and the surroundings This seemingly harmless choice hides the site of the entropy generation from view and is a source of confusion The temperature in this case drops suddenly from Tsys to Tsurr at the boundary surface and confusion arises as to which temperature to use in the relation QT for entropy transfer at the boundary Note that if the system and the surrounding air are not isothermal as a result of insufficient mixing then part of the entropy generation will occur in both the system and the surrounding air in the vicinity of the wall as shown in Fig 769c FIGURE 769 Graphical representation of entropy generation during a heat transfer process through a finite temperature difference System Surrounding Heat transfer Entropy transfer Wall Tsys Tsurr Location of entropy generation Q Tsys Q Tsurr Wall Tsys Tsurr Q Tsys Q Tsurr Boundary Tsys Tsurr Q Tsys Q Tsurr Sgen a The wall is ignored b The wall is considered c The wall as well as the variations of temperature in the system and the surroundings are considered Q Q Q Q Q Q Final PDF to printer 380 ENTROPY cen22672ch07323412indd 380 110617 0852 AM FIGURE 770 A large compressor assembly Photo courtesy of the DresserRand business part of Siemens Power Gas TOPIC OF SPECIAL INTEREST Reducing the Cost of Compressed Air Compressed air at gage pressures of 550 to 1000 kPa 80 to 150 psig is com monly used in industrial facilities to perform a wide variety of tasks such as cleaning operating pneumatic equipment and even refrigeration It is often referred to as the fourth utility after electricity water and natural gas or oil In production facilities there is a widespread waste of energy associated with compressedair systems and a general lack of awareness about the opportuni ties to conserve energy A considerable portion of the energy waste associated with compressedair systems can be avoided by following some common sense measures In this section we discuss the energy losses associated with compressed air systems and their costs to manufacturers We also show how to reduce the cost of compressed air in existing facilities by making some modi fications with attractive payback periods With the exception of a few com pressors that are driven by natural gas engines all compressors are driven by electric motors Fig 770 Some primitive methods of producing an air blast to keep the fire in fur naces alive such as forge bellows and the Chinese wind box date back at least to 2000 BC The water trompe that compresses air by the fall of water in a tube to blow forges metal heat shops is believed to have been in use by 150 BC In 1650 Otto van Guericke made great improvements in both the compressor and the vacuum pump In 1683 Papin proposed using compressed air to transmit power over long distances In 1829 William Mann received a patent for multistage compression of air In 1830 Thilorier was recognized for compressing gases to high pressures in stages In 1890 Edward Rix transmit ted power with air several miles to operate lifting machines in the North Star mine near Grass Valley California by using a compressor driven by Pelton wheels In 1872 cooling was adapted to increase efficiency by spraying water directly into the cylinder through the air inlet valves This wet compression was abandoned later because of the problems it caused The cooling then was accomplished externally by water jacketing the cylinders The first largescale compressor used in the United States was a fourcylinder unit built in 1866 for use in the Hoosac tunnel The cooling was first accomplished by water injection into the cylinder and later by running a stream of water over the cylinder Major advances in recent compressor technology are due to Burleigh Ingersoll Sergeant Rand and Clayton among others The compressors used range from a few horsepower to more than 10000 hp in size and they are among the major energyconsuming equipment in most manufacturing facilities Manufacturers are quick to identify energy and thus money losses from hot surfaces and to insulate those surfaces However somehow they are not so sensitive when it comes to saving compressed air since they view air as being free and the only time the air leaks and dirty air filters get some attention is when the air and pressure losses interfere with the normal operation of the plant However paying attention to the compressedair system and practicing some simple conservation measures can result in con siderable energy and cost savings for the plants This section can be skipped without a loss in continuity Final PDF to printer 381 CHAPTER 7 cen22672ch07323412indd 381 110617 0852 AM The hissing of air leaks can sometimes be heard even in highnoise manu facturing facilities Pressure drops at enduse points on the order of 40 percent of the compressordischarged pressure are not uncommon Yet a common response to such a problem is to install a larger compressor instead of check ing the system and finding out what the problem is The latter corrective action is usually taken only after the larger compressor also fails to eliminate the problem The energy wasted in compressedair systems because of poor installation and maintenance can account for up to 50 percent of the energy consumed by the compressor and about half of this amount can be saved by simple measures The cost of electricity to operate a compressor for one year can exceed the purchase price of the compressor This is especially the case for larger com pressors operating during two or three daily shifts For example operating a 125hp compressor powered by a 90percentefficient electric motor at full load for 6000 hours a year at 0085kWh will cost 52820 a year in electric ity cost which greatly exceeds the purchase and installation cost of a typical unit Fig 771 Next we describe some procedures to reduce the cost of compressed air in industrial facilities and to quantify the energy and cost savings associated with them Once the compressor power wasted is determined the annual energy usually electricity and cost savings can be determined from Energy savings Power saved Operating hours η motor 787 and Cost savings Energy savings Unit cost of energy 788 where ηmotor is the efficiency of the motor driving the compressor and the unit cost of energy is usually expressed in dollars per kilowatthour 1 kWh 3600 kJ 1 Repairing Air Leaks on CompressedAir Lines Air leaks are the greatest single cause of energy loss from compressedair systems in manufacturing facilities It takes energy to compress the air and thus the loss of compressed air is a loss of energy for the facility A compres sor must work harder and longer to make up for the lost air and must use more energy in the process Several studies at plants have revealed that up to 40 percent of the compressed air is lost through leaks Eliminating the air leaks totally is impractical and a leakage rate of 10 percent is considered acceptable Air leaks in general occur at the joints flange connections elbows reduc ing bushes sudden expansions valve systems filters hoses check valves relief valves extensions and the equipment connected to the compressedair lines Fig 772 Expansion and contraction as a result of thermal cycling and vibration are common causes of loosening at the joints and thus air leaks Therefore it is a good practice to check the joints for tightness and to tighten them periodically Air leaks also commonly occur at the points of end use or where the compressedair lines are connected to the equipment that oper ates on compressed air Because of the frequent opening and closing of the FIGURE 771 The cost of electricity to operate a compressor for one year can exceed the purchase price of the compressor Compressor 125 hp 9321 kW Operating hours 6000 hyr Unit cost of electricity 0085kWh Motor efficiency 090 Annual energy usage 621417 kWhyr Annual electricity cost 52820yr FIGURE 772 Air leaks commonly occur at joints and connections Compressed air Joint Air leak Final PDF to printer 382 ENTROPY cen22672ch07323412indd 382 110617 0852 AM FIGURE 773 The energy wasted as compressed air escapes through the leaks is equivalent to the energy it takes to compress it Air compressor Motor 120 kW Air inlet 1 atm m Air leak 20 02 m 24 kW compressedair lines at these points the gaskets wear out quickly and they need to be replaced periodically There are many ways of detecting air leaks in a compressedair system Per haps the simplest way of detecting a large air leak is to listen for it The high velocity of the air escaping the line produces a hissing sound that is difficult not to notice except in environments with a high noise level Another way of detecting air leaks especially small ones is to test the suspected area with soap water and to watch for soap bubbles This method is obviously not practi cal for a large system with many connections A modern way of checking for air leaks is to use an acoustic leak detector which consists of a directional microphone amplifiers audio filters and digital indicators A practical way of quantifying the air leaks in a production facility in its entirety is to conduct a pressure drop test The test is conducted by stopping all the operations that use compressed air and by shutting down the compressors and closing the pressure relief valve which relieves pressure automatically if the compressor is equipped with one This way any pressure drop in the compressedair lines is due to the cumulative effects of air leaks The drop in pressure in the system with time is observed and the test is conducted until the pressure drops by an amount that can be measured accurately usually 05 atm The time it takes for the pressure to drop by this amount is measured and the decay of pressure as a function of time is recorded The total volume of the compressedair system including the compressedair tanks the headers accumulators and the primary compressedair lines is calculated Ignoring the small lines will make the job easier and will cause the result to be more conservative The rate of air leak can be determined using the ideal gas equa tion of state The amount of mechanical energy wasted as a unit mass of air escapes through the leaks is equivalent to the actual amount of energy it takes to com press it and is determined from Eq 757 modified as Fig 773 w compin w reversible compin η comp nR T 1 η comp n 1 P 2 P 1 n 1 n 1 789 where n is the polytropic compression exponent n 14 when the compres sion is isentropic and 1 n 14 when there is intercooling and ηcomp is the compressor efficiency whose value usually ranges between 07 and 09 Using compressibleflow theory see Chap 17 it can be shown that when ever the line pressure is above 2 atm which is usually the case the velocity of air at the leak site must be equal to the local speed of sound Then the mass flow rate of air through a leak of minimum crosssectional area A becomes m air C discharge 2 k 1 1 k 1 P line R T line A kR 2 k 1 T line 790 where k is the specific heat ratio k 14 for air and Cdischarge is a discharge or loss coefficient that accounts for imperfections in flow at the leak site Its value ranges from about 060 for an orifice with sharp edges to 097 for a Final PDF to printer 383 CHAPTER 7 cen22672ch07323412indd 383 110617 0852 AM wellrounded circular hole The airleak sites are imperfect in shape and thus the discharge coefficient can be taken to be 065 in the absence of actual data Also Tline and Pline are the temperature and pressure in the compressedair line respectively Once m air and wcompin are available the power wasted by the leaking com pressed air or the power saved by repairing the leak is determined from Power saved Power wasted m air w compin 791 FIGURE 774 Schematic for Example 722 700 kPa 24C Air leak D 3 mm Air compressor Motor Air inlet 101 kPa 20C EXAMPLE 722 Energy and Cost Savings by Fixing Air Leaks The compressors of a production facility maintain the compressedair lines at a gauge pressure of 700 kPa at sea level where the atmospheric pressure is 101 kPa Fig 774 The average temperature of air is 20C at the compressor inlet and 24C in the compressedair lines The facility operates 4200 hours a year and the average price of electricity is 0078kWh Taking the compressor efficiency to be 08 the motor efficiency to be 092 and the discharge coefficient to be 065 determine the energy and money saved per year by sealing a leak equivalent to a 3mmdiameter hole on the compressedair line SOLUTION An air leak in the compressed air lines of a facility is considered The energy and money saved per year by sealing the leak are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas 3 Pressure losses in the compressed air lines are negligible Analysis We note that the absolute pressure is the sum of the gauge and atmospheric pressures The work needed to compress a unit mass of air at 20C from the atmospheric pressure of 101 kPa to 700 101 801 kPa is w compin nR T 1 η comp n 1 P 2 P 1 n 1 n 1 14 0287 kJ kgK 293 K 08 14 1 801 kPa 101 kPa 0414 1 2969 kJ kg The crosssectional area of the 3mmdiameter hole is A π D 2 4 π 3 10 3 m 2 4 7069 10 6 m 2 Noting that the line conditions are 297 K and 801 kPa the mass flow rate of the air leaking through the hole is determined to be m air C discharge 2 k 1 1 k 1 P line R T line A kR 2 k 1 T line 065 2 14 1 1 14 1 801 kPa 0287 kPa m 3 kgK297 K 7069 10 6 m 2 14 0287 kJ kgK 1000 m 2 s 2 1 kJ kg 2 14 1 297 K 0008632 kg s Final PDF to printer 384 ENTROPY cen22672ch07323412indd 384 110617 0852 AM 2 Installing HighEfficiency Motors Practically all compressors are powered by electric motors and the electrical energy a motor draws for a specified power output is inversely proportional to its efficiency Electric motors cannot convert the electrical energy they consume into mechanical energy completely and the ratio of the mechanical power supplied to the electrical power consumed during operation is called the motor efficiency ηmotor Therefore the electric power consumed by the motor and the mechanical shaft power supplied to the compressor are related to each other by Fig 775 W electric W comp η motor 792 For example assuming no transmission losses a motor that is 80 percent effi cient will draw 108 125 kW of electric power for each kW of shaft power it delivers to the compressor whereas a motor that is 95 percent efficient will draw only 1095 105 kW to deliver 1 kW Therefore highefficiency motors cost less to operate than their standard counterparts but they also usu ally cost more to purchase However the energy savings usually make up for the price differential during the first few years This is especially true for large compressors that operate more than one regular shift The electric power saved by replacing the existing standard motor of efficiency ηstandard with a high efficiency one of efficiency ηefficient is determined from W electricsaved W electricstandard W electricefficient W comp 1 η standard 1 η efficient Rated powerLoad factor1 η standard 1 η efficient 793 Then the power wasted by the leaking compressed air becomes Power wasted m air w compin 0008632 kg s 2969 kJ kg 2563 kW The compressor operates 4200 hyr and the motor efficiency is 092 Then the annual energy and cost savings resulting from repairing this leak are determined to be Energy savings Power saved operatings hours η motor 2563 kW 4200 h yr 092 11700 kWh yr Cost savings Energy savings Unit cost of energy 11700 kWh yr 0078 kWh 913 yr Discussion Note that the facility will save 11700 kWh of electricity worth 913 a year when this air leak is fixed This is a substantial amount for a single leak whose equivalent diameter is 3 mm FIGURE 775 The electrical energy consumed by a motor is inversely proportional to its efficiency ηmotor Welectric Wshaftηmotor Motor efficiency Electrical power consumed per kW of mechanical shaft power output 100 90 80 70 60 50 40 30 20 10 100 kW 111 125 143 167 200 250 333 500 1000 ηmotor Welectric Wshaft Final PDF to printer 385 CHAPTER 7 cen22672ch07323412indd 385 110617 0852 AM where rated power is the nominal power of the motor listed on its label the power the motor delivers at full load and the load factor is the fraction of the rated power at which the motor normally operates Then the annual energy savings as a result of replacing a motor with a highefficiency motor instead of a comparable standard one is Energy savings W electricsaved Annual operating hours 794 The efficiencies of motors used to power compressors usually range from about 70 percent to over 96 percent The portion of electrical energy not converted to mechanical energy is converted to heat The amount of heat generated by the motors may reach high levels especially at part load and it may cause overheating if not dissipated effectively It may also cause the air temperature in the compressor room to rise to undesirable levels For example a 90percentefficient 100kW motor generates as much heat as a 10kW resistance heater in the confined space of the compressor room and it contributes greatly to the heating of the air in the room If this heated air is not vented properly and the air into the compressor is drawn from inside the compressor room the performance of the compressor will also decline as explained later Important considerations in the selection of a motor for a compressor are the operating profile of the compressor ie the variation of the load with time and the efficiency of the motor at partload conditions The partload efficiency of a motor is as important as the fullload efficiency if the compres sor is expected to operate at part load during a significant portion of the total operating time A typical motor has a nearly flat efficiency curve between half load and full load and peak efficiency is usually at about 75 percent load Efficiency falls off pretty steeply below half load and thus operation below 50 percent load should be avoided as much as possible For example the effi ciency of a motor may drop from 90 percent at full load to 87 percent at half load and 80 percent at quarter load Fig 776 The efficiency of another motor of similar specifications on the other hand may drop from 91 percent at full load to 75 percent at quarter load The first motor is obviously better suited for a situation in which a compressor is expected to operate at quarter load during a significant portion of the time The efficiency at partload con ditions can be improved greatly by installing variable voltage controllers if it is economical to do so Also oversizing a motor just to be on the safe side and to have some excess power just in case is a bad practice since this will cause the motor to operate almost always at part load and thus at a lower efficiency Besides oversized motors have a higher initial cost However oversized motors waste little energy as long as they operate at loads above 50 percent of design Using a Smaller Motor at High Capacity We tend to purchase larger equipment than needed for reasons like having a safety margin or anticipated future expansion and compressors are no excep tion The uncertainties in plant operation are partly responsible for opting for a larger compressor since it is better to have an oversized compressor than FIGURE 776 The efficiency of an electric motor decreases at part load ηmotor 10 20 30 40 Motor efficiency 50 60 70 80 90 100 0 20 40 60 80 100 Load Final PDF to printer 386 ENTROPY cen22672ch07323412indd 386 110617 0852 AM an undersized one Sometimes compressors with several times the required capacity are purchased with the perception that the extra capacity may be needed someday The result is a compressor that runs intermittently at full load or one that runs continuously at part load A compressor that operates at part load also causes the motor to oper ate less efficiently since the efficiency of an electric motor decreases as the point of operation shifts down from its rated power as explained above The result is a motor that consumes more electricity per unit power delivered and thus a more expensive operation The operating costs can be reduced by switching to a smaller motor that runs at rated power and thus at a higher efficiency 3 Using Outside Air for Compressor Intake We pointed out earlier that the power consumed by a compressor is pro portional to the specific volume which is proportional to the absolute tem perature of the gas at a given pressure It is also clear from Eq 789 that the compressor work is directly proportional to the inlet temperature of air Therefore the lower the inlet temperature of the air the smaller the compressor work Then the power reduction factor which is the fraction of compressor power reduced as a result of taking intake air from the outside becomes f reduction w compinside w compinside w compinside T inside T outside T inside 1 T outside T inside 795 where Tinside and Toutside are the absolute temperatures in K or R of the ambi ent air inside and outside the facility respectively Thus reducing the absolute inlet temperature by 5 percent for example will reduce the compressor power input by 5 percent As a rule of thumb for a specified amount of compressed air the power consumption of the compressor decreases or for a fixed power input the amount of compressed air increases by 1 percent for each 3C drop in the temperature of the inlet air to the compressor Compressors are usually located inside the production facilities or in adjacent shelters specifically built outside these facilities The intake air is normally drawn from inside the building or the shelter However in many locations the air temperature in the building is higher than the outside air temperature because of space heaters in the winter and the heat given up by a large number of mechanical and electrical equipment as well as the furnaces year round The temperature rise in the shelter is also due to the heat dissipation from the compressor and its motor The outside air is gener ally cooler and thus denser than the air in the compressor room even on hot summer days Therefore it is advisable to install an intake duct to the com pressor inlet so that the air is supplied directly from the outside of the build ing instead of the inside as shown in Fig 777 This will reduce the energy consumption of the compressor since it takes less energy to compress a specified amount of cool air than the same amount of warm air Compress ing the warm air in a building in winter also wastes the energy used to heat the air FIGURE 777 The power consumption of a compres sor can be reduced by taking in air from the outside Wall Outside air Air intake duct Air filter Compressor Final PDF to printer 387 CHAPTER 7 cen22672ch07323412indd 387 110617 0852 AM 4 Reducing the Air Pressure Setting Another source of energy waste in compressedair systems is compressing the air to a higher pressure than required by the airdriven equipment since it takes more energy to compress air to a higher pressure In such cases considerable energy savings can be realized by determining the minimum required pressure and then reducing the air pressure control setting on the compressor accord ingly This can be done on both screwtype and reciprocating compressors by simply adjusting the pressure setting to match the needs The amount of energy it takes to compress a unit mass of air is determined from Eq 789 We note from that relation that the higher the pressure P2 at the compressor exit the larger the work required for compression Reducing the exit pressure of the compressor to P2reduced will reduce the power input require ments of the compressor by a factor of f reduction w compcurrent w compreduced w compcurrent 1 P 2reduced P 1 n 1 n 1 P 2 P 1 n 1 n 1 796 A power reduction or savings factor of freduction 008 for example indicates that the power consumption of the compressor is reduced by 8 percent as a result of reducing the pressure setting Some applications require slightly compressed air In such cases the need can be met by a blower instead of a compressor Considerable energy can be saved in this manner since a blower requires a small fraction of the power needed by a compressor for a specified mass flow rate EXAMPLE 723 Reducing the Pressure Setting to Reduce Cost The compressedair requirements of a plant located at 1400m elevation are being met by a 75hp compressor that takes in air at the local atmospheric pressure of 856 kPa and the average temperature of 15C and compresses it to 900 kPa gauge Fig 778 The plant is currently paying 12000 a year in electricity costs to run the compressor An investigation of the compressedair system and the equipment using the compressed air reveals that compressing the air to 800 kPa is sufficient for this plant Determine how much money will be saved by reducing the pressure of the compressed air SOLUTION It is observed that the compressor of a facility compresses the air to higher pressures than needed The cost savings associated with a pressure reduction are to be determined Assumptions 1 Air is an ideal gas 2 Compression process is isentropic and thus n k 14 Analysis The fraction of energy saved as a result of reducing the pressure setting of the compressor is f reduction 1 P 2reduced P 1 n 1 n 1 P 2 P 1 n 1 n 1 1 8856 856 14 1 14 1 9856 856 14 1 14 1 0060 FIGURE 778 Schematic for Example 723 800 kPa 900 kPa Compressed air Air compressor Motor Air inlet 856 kPa Final PDF to printer 388 ENTROPY cen22672ch07323412indd 388 110617 0852 AM There are also other ways to reduce the cost of compressed air in industrial facilities An obvious way is turning the compressor off during nonproduc tion periods such as lunch hours nights and even weekends A consider able amount of power can be wasted by a compressor in standby mode This is especially the case for screwtype compressors since they consume up to 85 percent of their rated power in this mode Reciprocating compressors are not immune to this deficiency however since they also must cycle on and off because of air leaks in the compressedair lines The system can be shut down manually during nonproduction periods to save energy but it is better to install a timer with manual override to do this automatically since it is human nature to put things off when the benefits are not obvious or immediate The compressed air is sometimes cooled considerably below its dew point in refrigerated dryers in order to condense and remove a large fraction of the water vapor in the air as well as other condensable gases such as oil vapors The temperature of air rises considerably as it is compressed sometimes exceeding 250C at compressor exit when compressed adiabatically to just 700 kPa Therefore it is desirable to cool air after compression in order to minimize the amount of power consumed by the refrigeration system just as it is desirable to let the hot food in a pan cool to the ambient temperature before putting it into the refrigerator The cooling can be done by either ambi ent air or water and the heat picked up by the cooling medium can be used for space heating feedwater heating or processrelated heating Compressors are commonly cooled directly by air or by circulating a liquid such as oil or water through them in order to minimize the power consump tion The heat picked up by the oil or water is usually rejected to the ambient air in a liquidtoair heat exchanger This heat rejected usually amounts to 60 to 90 percent of the power input and thus it represents a huge amount of energy that can be used for space heating in winter preheating the air or water in a furnace or other processrelated purposes Fig 779 For exam ple assuming 80 percent of the power input is converted to heat a 150hp compressor can reject as much heat as a 90kW electric resistance heater or a 400000Btuh natural gas heater when operating at full load Thus the proper utilization of the waste heat from a compressor can result in significant energy and cost savings That is reducing the pressure setting will reduce the energy consumed by the com pressor by about 6 percent Then Cost savings Current cost f reduction 12000 yr 006 720 yr Therefore reducing the pressure setting by 100 kPa will result in annual savings of 720 in this case FIGURE 779 Waste heat from a compressor can be used to heat a building in winter Air Cooling liquid from compressor Liquidtoair heat exchanger Damper summer mode Damper winter mode Heated air Outside Inside facility Final PDF to printer 389 CHAPTER 7 cen22672ch07323412indd 389 110617 0852 AM SUMMARY The second law of thermodynamics leads to the definition of a new property called entropy which is a quantitative measure of microscopic disorder for a system Any quantity whose cyclic integral is zero is a property and entropy is defined as dS dQ T int rev For the special case of an internally reversible isothermal process it gives ΔS Q T 0 The inequality part of the Clausius inequality combined with the definition of entropy yields an inequality known as the increase of entropy principle expressed as S gen 0 where Sgen is the entropy generated during the process Entropy change is caused by heat transfer mass flow and irreversibili ties Heat transfer to a system increases the entropy and heat transfer from a system decreases it The effect of irreversibili ties is always to increase the entropy The entropychange and isentropic relations for a process can be summarized as follows 1 Pure substances Any process Δs s 2 s 1 Isentropic process s 2 s 1 2 Incompressible substances Any process s 2 s 1 c avg ln T 2 T 1 Isentropic process T 2 T 1 3 Ideal gases a Constant specific heats approximate treatment Any process s 2 s 1 c vavg ln T 2 T 1 R ln v 2 v 1 s 2 s 1 c pavg ln T 2 T 1 R ln P 2 P 1 Isentropic process T 2 T 1 s const v 1 v 2 k 1 T 2 T 1 s const P 1 P 2 k 1k P 2 P 1 s const v 1 v 2 k b Variable specific heats exact treatment Any process s 2 s 1 s 2 s 1 R ln P 2 P 1 Isentropic process s 2 s 1 R ln P 2 P 1 P 2 P 1 s const P r2 P r1 v 2 v 1 s const v r2 v r1 where Pr is the relative pressure and vr is the relative specific volume The function s depends on temperature only The steadyflow work for a reversible process can be expressed in terms of the fluid properties as w rev 1 2 v dP Δke Δpe For incompressible substances v constant it simplifies to w rev v P 2 P 1 Δke Δpe The work done during a steadyflow process is proportional to the specific volume Therefore v should be kept as small as possible during a compression process to minimize the work input and as large as possible during an expansion process to maximize the work output The reversible work inputs to a compressor compressing an ideal gas from T1 P1 to P2 in an isentropic Pvk constant polytropic Pvn constant or isothermal Pv constant manner are determined by integration for each case with the following results Isentropic w compin kR T 2 T 1 k 1 kR T 1 k 1 P 2 P 1 k 1 k 1 Polytropic w compin nR T 2 T 1 n 1 nR T 1 n 1 P 2 P 1 n 1 n 1 Isothermal w compin RT ln P 2 P 1 The work input to a compressor can be reduced by using multistage compression with intercooling For maximum sav ings from the work input the pressure ratio across each stage of the compressor must be the same Most steadyflow devices operate under adiabatic condi tions and the ideal process for these devices is the isentropic Final PDF to printer cen22672ch07323412indd 390 110617 0852 AM 390 ENTROPY process The parameter that describes how efficiently a device approximates a corresponding isentropic device is called isentropic or adiabatic efficiency It is expressed for turbines compressors and nozzles as follows η T Actual turbine work Isentropic turbine work w a w s h 1 h 2a h 1 h 2s η C Isentropic compressor work Actual compressor work w s w a h 2s h 1 h 2a h 1 η N Actual KE at nozzle exit Isentropic KE at nozzle exit V 2a 2 V 2s 2 h 1 h 2a h 1 h 2s In the preceding relations h2a and h2s are the enthalpy val ues at the exit state for actual and isentropic processes respectively The entropy balance for any system undergoing any process can be expressed in the general form as S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy or in the rate form as S in S out Rate of net entropy transfer by heat and mass S gen Rate of entropy generation d S system dt Rate of change in entropy For a general steadyflow process it simplifies to S gen m e s e m i s i Q k T k Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software PROBLEMS Entropy and the Increase of Entropy Principle 71C Does a cycle for which δQ 0 violate the Clausius inequality Why 72C Does the cyclic integral of heat have to be zero ie does a system have to reject as much heat as it receives to com plete a cycle Explain 73C Is a quantity whose cyclic integral is zero necessarily a property 74C Is an isothermal process necessarily internally revers ible Explain your answer with an example 75C Is the value of the integral 1 2 δQ T the same for all reversible processes between states 1 and 2 Why 76C How do the values of the integral 1 2 δQ T compare for a reversible and an irreversible process between the same end states 77C Is it possible to create entropy Is it possible to destroy it 78C The entropy of a hot baked potato decreases as it cools Is this a violation of the increase of entropy principle Explain 79C When a system is adiabatic what can be said about the entropy change of the substance in the system REFERENCES AND SUGGESTED READINGS 1 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley Interscience 2006 2 A Bejan Entropy Generation through Heat and Fluid Flow New York Wiley Interscience 1982 3 Y A Çengel and H Kimmel Optimization of Expansion in Natural Gas Liquefaction Processes LNG Journal UK MayJune 1998 4 Y Çerci Y A Çengel and R H Turner Reducing the Cost of Compressed Air in Industrial Facilities International Mechanical Engineering Congress and Exposition San Francisco CA November 1217 1995 5 W F E Feller Air Compressors Their Installation Operation and Maintenance New York McGrawHill 1944 6 D W Nutter A J Britton and W M Heffington Conserve Energy to Cut Operating Costs Chemical Engineering September 1993 pp 127137 7 J Rifkin Entropy New York The Viking Press 1980 8 M Kostic Revisiting the Second Law of Energy Degradation and Entropy Generation From Sadi Carnots Ingenious Reasoning to Holistic Generalization AIP Conf Proc 1411 pp 327350 2011 doi 10106313665247 Final PDF to printer cen22672ch07323412indd 391 110617 0852 AM 391 CHAPTER 7 710C Is it possible for the entropy change of a closed sys tem to be zero during an irreversible process Explain 711C A pistoncylinder device contains helium gas During a reversible isothermal process the entropy of the helium will never sometimes always increase 712C A pistoncylinder device contains nitrogen gas During a reversible adiabatic process the entropy of the nitro gen will never sometimes always increase 713C A pistoncylinder device contains superheated steam During an actual adiabatic process the entropy of the steam will never sometimes always increase 714C The entropy of steam will increase decrease remain the same as it flows through an actual adiabatic turbine 715C During a heat transfer process the entropy of a system always sometimes never increases 716C Steam is accelerated as it flows through an actual adia batic nozzle The entropy of the steam at the nozzle exit will be greater than equal to less than the entropy at the nozzle inlet 717 Heat is transferred at a rate of 2 kW from a hot reser voir at 800 K to a cold reservoir at 300 K Calculate the rate at which the entropy of the two reservoirs changes and determine if the second law is satisfied Answer 000417 kWK 718E A completely reversible air conditioner provides 36000 Btuh of cooling for a space maintained at 70F while rejecting heat to the environmental air at 110F Calculate the rate at which the entropies of the two reservoirs change and verify that this air conditioner satisfies the increase of entropy principle 719 Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold reservoir at 600 K Calculate the entropy change of the two reservoirs and deter mine if the increase of entropy principle is satisfied 722 Reconsider Prob 721 Using appropriate soft ware study the effects of the varying heat added to the working fluid and the source temperature on the entropy change of the working fluid the entropy change of the source and the total entropy change for the process Let the source temperature vary from 100 to 1000C Plot the entropy changes of the source and of the working fluid against the source tem perature for heat transfer amounts of 500 kJ 900 kJ and 1300 kJ and discuss the results 723E During the isothermal heat rejection process of a Carnot cycle the working fluid experiences an entropy change of 07 BtuR If the temperature of the heat sink is 95F deter mine a the amount of heat transfer b the entropy change of the sink and c the total entropy change for this process Answers a 389 Btu b 07 BtuR c 0 FIGURE P723E Carnot heat engine 95F Heat Sink 95F FIGURE P726 Ideal gas 40C 30C 200 kJ Heat FIGURE P719 100 kJ 600 K 1200 K 720 In Prob 719 assume that the heat is transferred from the cold reservoir to the hot reservoir contrary to the Clausius statement of the second law Prove that this violates the increase of entropy principleas it must according to Clausius 721 During the isothermal heat addition process of a Car not cycle 900 kJ of heat is added to the working fluid from a source at 400C Determine a the entropy change of the working fluid b the entropy change of the source and c the total entropy change for the process 724 Air is compressed by a 40kW compressor from P1 to P2 The air temperature is maintained constant at 25C dur ing this process as a result of heat transfer to the surround ing medium at 20C Determine the rate of entropy change of the air State the assumptions made in solving this problem Answer 0134 KWK 725 Refrigerant134a enters the coils of the evaporator of a refrigeration system as a saturated liquidvapor mixture at a pressure of 140 kPa The refrigerant absorbs 180 kJ of heat from the cooled space which is maintained at 10C and leaves as saturated vapor at the same pressure Determine a the entropy change of the refrigerant b the entropy change of the cooled space and c the total entropy change for this process 726 A rigid tank contains an ideal gas at 40C that is being stirred by a paddle wheel The paddle wheel does 200 kJ of work on the ideal gas It is observed that the temperature of the ideal gas remains constant during this process as a result of heat transfer between the system and the surroundings at 30C Determine the entropy change of the ideal gas Final PDF to printer cen22672ch07323412indd 392 110617 0852 AM 392 ENTROPY 727 A rigid vessel is filled with a fluid from a source whose properties remain constant How does the entropy of the sur roundings change if the vessel is filled such that the specific entropy of the vessel contents remains constant 728 A rigid vessel filled with a fluid is allowed to leak some fluid out through an opening During this process the specific entropy of the remaining fluid remains constant How does the entropy of the environment change during this process Entropy Changes of Pure Substances 729C Is a process that is internally reversible and adiabatic necessarily isentropic Explain 730E One lbm of R134a is expanded isentropically in a closed system from 100 psia and 100F to 10 psia Determine the total heat transfer and work production for this process 731E Two lbm of water at 300 psia fill a weighted piston cylinder device whose volume is 25 ft3 The water is then heated at constant pressure until the temperature reaches 500F Determine the resulting change in the waters total entropy Answer 0474 BtuR 732 A wellinsulated rigid tank contains 3 kg of a satu rated liquidvapor mixture of water at 200 kPa Initially three quarters of the mass is in the liquid phase An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized Determine the entropy change of the steam during this process Answer 111 kJK water at 400 kPa and 60C while the other part is evacuated The partition is now removed and the water expands to fill the entire tank Determine the entropy change of water during this process if the final pressure in the tank is 40 kPa Answer 0492 kJK FIGURE P732 733 Using the relation ds δQTint rev for the definition of entropy calculate the change in the specific entropy of R134a as it is heated at a constant pressure of 200 kPa from a saturated liquid to a saturated vapor Use the R134a tables to verify your answer 734 The radiator of a steam heating system has a volume of 20 L and is filled with superheated water vapor at 200 kPa and 150C At this moment both the inlet and the exit valves to the radiator are closed After a while the temperature of the steam drops to 40C as a result of heat transfer to the room air Determine the entropy change of the steam during this process Answer 0132 kJK 735 A rigid tank is divided into two equal parts by a parti tion One part of the tank contains 25 kg of compressed liquid FIGURE P735 736 An insulated pistoncylinder device contains 005 m3 of saturated refrigerant 134a vapor at 08MPa pressure The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 04 MPa Determine a the final tempera ture in the cylinder and b the work done by the refrigerant FIGURE P736 R134a 005m3 08 MPa 737 Reconsider Prob 736 Using appropriate soft ware evaluate and plot the work done by the refrigerant as a function of final pressure as it varies from 08 to 04 MPa Compare the work done for this process to one for which the temperature is constant over the same pressure range Discuss your results 738 An insulated pistoncylinder device contains 5 L of satu rated liquid water at a constant pressure of 150 kPa An elec tric resistance heater inside the cylinder is now turned on and 1700 kJ of energy is transferred to the steam Determine the entropy change of the water during this process Answer 442 kJK 739 Onekg of R134a initially at 600 kPa and 25C under goes a process during which the entropy is kept constant until the pressure drops to 100 kPa Determine the final temperature of the R134a and the final specific internal energy 740 Refrigerant134a is expanded isentropically from 600 kPa and 70C at the inlet of a steadyflow turbine to 100 kPa at the outlet The outlet area is 1 m2 and the inlet area is 05 m2 Calculate the inlet and outlet velocities when the mass flow rate is 075 kgs Answers 00646 ms 0171 ms Final PDF to printer cen22672ch07323412indd 393 110617 0852 AM 393 CHAPTER 7 741 Refrigerant134a at 320 kPa and 40C undergoes an iso thermal process in a closed system until its quality is 45 percent On a perunitmass basis determine how much work and heat transfer are required Answers 406 kJkg 130 kJkg 747E R134a vapor enters into a turbine at 250 psia and 175F The temperature of R134a is reduced to 20F in this turbine while its specific entropy remains constant Determine the change in the enthalpy of R134a as it passes through the turbine FIGURE P741 R134a 320 kPa 40C 742 A rigid tank contains 5 kg of saturated vapor steam at 100C The steam is cooled to the ambient temperature of 25C a Sketch the process with respect to the saturation lines on a Tv diagram b Determine the entropy change of the steam in kJK c For the steam and its surroundings determine the total entropy change associated with this process in kJK 743 A 05m3 rigid tank contains refrigerant134a initially at 200 kPa and 40 percent quality Heat is transferred now to the refrigerant from a source at 35C until the pressure rises to 400 kPa Determine a the entropy change of the refriger ant b the entropy change of the heat source and c the total entropy change for this process 744 Reconsider Prob 743 Using appropriate soft ware investigate the effects of the source tempera ture and final pressure on the total entropy change for the process Let the source temperature vary from 30 to 210C and let the final pressure vary from 250 to 500 kPa Plot the total entropy change for the process as a function of the source tem perature for final pressures of 250 kPa 400 kPa and 500 kPa and discuss the results 745 Steam enters a steadyflow adiabatic nozzle with a low inlet velocity as a saturated vapor at 6 MPa and expands to 12 MPa a Under the conditions that the exit velocity is to be the maximum possible value sketch the Ts diagram with respect to the saturation lines for this process b Determine the maximum exit velocity of the steam in ms Answer 764 ms 746 Steam enters an adiabatic diffuser at 150 kPa and 120C with a velocity of 550 ms Determine the minimum velocity that the steam can have at the outlet when the outlet pressure is 300 kPa FIGURE P747E 250 psia 175F 20F R134a turbine 748 Water vapor enters a compressor at 35 kPa and 160C and leaves at 300 kPa with the same specific entropy as at the inlet What are the temperature and the specific enthalpy of water at the compressor exit 749 Refrigerant134a enters an adiabatic compressor as saturated vapor at 160 kPa at a rate of 2 m3min and is com pressed to a pressure of 900 kPa Determine the minimum power that must be supplied to the compressor 750E The compressor in a refrigerator compresses satu rated R134a vapor at 0F to 200 psia Calculate the work required by this compressor in Btulbm when the compres sion process is isentropic 751 An isentropic steam turbine processes 2 kgs of steam at 3 MPa which is exhausted at 50 kPa and 100C Five per cent of this flow is diverted for feedwater heating at 500 kPa FIGURE P748 35 kPa 160C Wout 300 kPa Steam compressor Final PDF to printer cen22672ch07323412indd 394 110617 0852 AM 394 ENTROPY 752 Water at 10C and 814 percent quality is compressed isentropically in a closed system to 3 MPa How much work does this process require in kJkg 753 Twokg of saturated water vapor at 600 kPa are con tained in a pistoncylinder device The water expands adiabati cally until the pressure is 100 kPa and is said to produce 700 kJ of work output a Determine the entropy change of the water in kJkgK b Is this process realistic Using the Ts diagram for the process and the concepts of the second law support your answer 754 A pistoncylinder device contains 5 kg of steam at 100C with a quality of 50 percent This steam undergoes two processes as follows 12 Heat is transferred to the steam in a reversible manner while the temperature is held constant until the steam exists as a saturated vapor 23 The steam expands in an adiabatic reversible process until the pressure is 15 kPa a Sketch these processes with respect to the saturation lines on a single Ts diagram b Determine the heat transferred to the steam in process 12 in kJ c Determine the work done by the steam in process 23 in kJ 755 A rigid 20L steam cooker is arranged with a pres sure relief valve set to release vapor and maintain the pressure once the pressure inside the cooker reaches 150 kPa Initially this cooker is filled with water at 175 kPa with a quality of 10 percent Heat is now added until the quality inside the cooker is 40 percent Determine the minimum entropy change of the thermal energy reservoir supplying this heat 756 In Prob 755 the water is stirred at the same time that it is being heated Determine the minimum entropy change of the heatsupplying source if 100 kJ of work is done on the water as it is being heated 757E A 055ft3 wellinsulated rigid can initially contains refrigerant134a at 90 psia and 30F Now a crack develops in the can and the refrigerant starts to leak out slowly Assuming the refrigerant remaining in the can has undergone a reversible adiabatic process determine the final mass in the can when the pressure drops to 20 psia FIGURE P751 500 kPa 3 MPa 2 kgs 50 kPa 100C Steam turbine FIGURE P757E R134a 90 psia 30F 758 Determine the total heat transfer for the reversible process 13 shown in Fig P758 759 Determine the total heat transfer for the reversible process 12 shown in Fig P759 FIGURE P758 2 3 1 55 360 T C S kJK 2 3 1 FIGURE P759 02 10 500 100 S kJK 2 1 T C Determine the power produced by this turbine in kW Answer 2285 kW Final PDF to printer cen22672ch07323412indd 395 110617 0852 AM 395 CHAPTER 7 Entropy Change of Incompressible Substances 761C Consider two solid blocks one hot and the other cold brought into contact in an adiabatic container After a while thermal equilibrium is established in the container as a result of heat transfer The first law requires that the amount of energy lost by the hot solid be equal to the amount of energy gained by the cold one Does the second law require that the decrease in entropy of the hot solid be equal to the increase in entropy of the cold one 762 An adiabatic pump is to be used to compress saturated liquid water at 10 kPa to a pressure to 15 MPa in a reversible manner Determine the work input using a entropy data from the compressed liquid table b inlet specific volume and pres sure values c average specific volume and pressure values Also determine the errors involved in parts b and c Presuming that the pressure remains constant while the chips are being cooled determine the entropy change of a the chips b the R134a and c the entire system Is this process possible Why 764 A 25kg iron block initially at 280C is quenched in an insulated tank that contains 100 kg of water at 18C Assum ing the water that vaporizes during the process condenses back in the tank determine the total entropy change during this process 765 A 30kg aluminum block initially at 140C is brought into contact with a 40kg block of iron at 60C in an insulated enclosure Determine the final equilibrium temperature and the total entropy change for this process Answers 109C 0251 kJK 766 Reconsider Prob 765 Using appropriate soft ware study the effect of the mass of the iron block on the final equilibrium temperature and the total entropy change for the process Let the mass of the iron vary from 10 to 100 kg Plot the equilibrium temperature and the total entropy change as a function of iron mass and discuss the results 767 A 50kg copper block initially at 140C is dropped into an insulated tank that contains 90 L of water at 10C Deter mine the final equilibrium temperature and the total entropy change for this process 760 Calculate the heat transfer in kJkg for the reversible steadyflow process 13 shown on a Ts diagram in Fig P760 Answer 341 kJkg FIGURE P760 002 10 30 120 100 s kJkg K 2 3 1 T C FIGURE P762 Pump 15 MPa 10 kPa 763 Ten grams of computer chips with a specific heat of 03 kJkgK are initially at 20C These chips are cooled by placement in 5 grams of saturated liquid R134a at 40C 768 A 30kg iron block and a 40kg copper block both ini tially at 80C are dropped into a large lake at 15C Thermal equilibrium is established after a while as a result of heat trans fer between the blocks and the lake water Determine the total entropy change for this process FIGURE P767 Water 90 L Copper 50 kg FIGURE P768 Lake 15C Copper 40 kg Iron 30 kg Final PDF to printer cen22672ch07323412indd 396 110617 0852 AM 396 ENTROPY Entropy Change of Ideal Gases 769C What are Pr and vr called Is their use limited to isen tropic processes Explain 770C Some properties of ideal gases such as internal energy and enthalpy vary with temperature only that is u uT and h hT Is this also the case for entropy 771C Can the entropy of an ideal gas change during an iso thermal process 772C An ideal gas undergoes a process between two speci fied temperatures first at constant pressure and then at con stant volume For which case will the ideal gas experience a larger entropy change Explain 773 What is the difference between entropies of oxygen at 150 kPa and 39C and oxygen at 150 kPa and 337C on a per unitmass basis 774E Air is expanded from 200 psia and 500F to 100 psia and 50F Assuming constant specific heats determine the change in the specific entropy of air Answer 0106 BtulbmR 775 Determine the final temperature when air is expanded isentropically from 1000 kPa and 477C to 100 kPa in a piston cylinder device 776E Air is expanded isentropically from 100 psia and 500F to 20 psia in a closed system Determine its final temperature 777 Which of the two gaseshelium or nitrogenhas the higher final temperature as it is compressed isentropically from 100 kPa and 25C to 1 MPa in a closed system 778 Which of the two gasesneon or airhas the lower final temperature as it is expanded isentropically from 1000 kPa and 500C to 100 kPa in a pistoncylinder device 779 A 15m3 insulated rigid tank contains 27 kg of carbon dioxide at 100 kPa Now paddlewheel work is done on the system until the pressure in the tank rises to 150 kPa Deter mine the entropy change of carbon dioxide during this process Assume constant specific heats Answer 0719 kJK 780 An insulated pistoncylinder device initially contains 300 L of air at 120 kPa and 17C Air is now heated for 15 min by a 200W resistance heater placed inside the cylinder The pressure of air is kept constant during this process Determine the entropy change of air assuming a constant specific heats and b variable specific heats 781 A pistoncylinder device contains 075 kg of nitrogen gas at 140 kPa and 37C The gas is now compressed slowly in a polytropic process during which PV13 constant The pro cess ends when the volume is reduced by onehalf Determine the entropy change of nitrogen during this process Answer 00385 kJK 782 Reconsider Prob 781 Using appropriate soft ware investigate the effect of varying the poly tropic exponent from 1 to 14 on the entropy change of the nitrogen Show the processes on a common Pv diagram 783E A mass of 25 lbm of helium undergoes a process from an initial state of 50 ft3lbm and 60F to a final state of 20 ft3lbm and 240F Determine the entropy change of helium during this process assuming a the process is reversible and b the process is irreversible 784 One kg of air at 200 kPa and 127C is contained in a pistoncylinder device Air is now allowed to expand in a reversible isothermal process until its pressure is 100 kPa Determine the amount of heat transferred to the air during this expansion 785 An insulated rigid tank is divided into two equal parts by a partition Initially one part contains 12 kmol of an ideal gas at 330 kPa and 50C and the other side is evacu ated The partition is now removed and the gas fills the entire tank Determine the total entropy change during this process Answer 692 kJK 786 Air at 27C and 100 kPa is contained in a piston cylinder device When the air is compressed adiabatically a minimum work input of 1000 kJ will increase the pressure to 600 kPa Assuming air has constant specific heats evaluated at 300 K determine the mass of air in the device 787 Air at 35 MPa and 500C is expanded in an adiabatic gas turbine to 02 MPa Calculate the maximum work that this turbine can produce in kJkg 788 Air is compressed in a pistoncylinder device from 90 kPa and 20C to 600 kPa in a reversible isothermal pro cess Determine a the entropy change of air and b the work done 789 Helium gas is compressed from 90 kPa and 30C to 450 kPa in a reversible adiabatic process Determine the final temperature and the work done assuming the process takes place a in a pistoncylinder device and b in a steadyflow compressor FIGURE P779 CO2 15 m3 100 kPa 27 kg Final PDF to printer cen22672ch07323412indd 397 110617 0852 AM 397 CHAPTER 7 791 Five kg of air at 427C and 600 kPa are contained in a pistoncylinder device The air expands adiabatically until the pressure is 100 kPa and produces 600 kJ of work output Assume air has constant specific heats evaluated at 300 K a Determine the entropy change of the air in kJkgK b Since the process is adiabatic is the process realistic Using concepts of the second law support your answer 792 A container filled with 45 kg of liquid water at 95C is placed in a 90m3 room that is initially at 12C Thermal equilibrium is established after a while as a result of heat trans fer between the water and the air in the room Using constant specific heats determine a the final equilibrium temperature b the amount of heat transfer between the water and the air in the room and c the entropy generation Assume the room is well sealed and heavily insulated 795E The wellinsulated container shown in Fig P795E is initially evacuated The supply line contains air that is main tained at 150 psia and 140F The valve is opened until the pressure in the container is the same as the pressure in the sup ply line Determine the minimum temperature in the container when the valve is closed FIGURE P790 600 kPa 120 kPa 30C Nitrogen compressor FIGURE P792 Room 90 m3 12C Water 45 kg 95C 793 Oxygen at 300 kPa and 90C flowing at an average velocity of 3 ms is expanded in an adiabatic nozzle What is the maximum velocity of the oxygen at the outlet of this nozzle when the outlet pressure is 120 kPa Answer 390 ms 794 Air at 800 kPa and 400C enters a steadyflow nozzle with a low velocity and leaves at 100 kPa If the air undergoes an adiabatic expansion process through the nozzle what is the maximum velocity of the air at the nozzle exit in ms FIGURE P795E Valve Supply line Vessel 796 An insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30C A valve is now opened and argon is allowed to escape until the pressure inside drops to 200 kPa Assuming the argon remaining inside the tank has undergone a reversible adiabatic process determine the final mass in the tank Answer 246 kg FIGURE P796 Argon 4 kg 30C 450 kPa 797 Reconsider Prob 796 Using appropriate soft ware investigate the effect of the final pressure on the final mass in the tank as the pressure varies from 450 to 150 kPa and plot the results Reversible SteadyFlow Work 798C In large compressors the gas is often cooled while being compressed to reduce the power consumed by the com pressor Explain how cooling the gas during a compression process reduces the power consumption 790 Nitrogen at 120 kPa and 30C is compressed to 600 kPa in an adiabatic compressor Calculate the minimum work needed for this process in kJkg Answer 184 kJkg Final PDF to printer cen22672ch07323412indd 398 110617 0852 AM 398 ENTROPY 799C The turbines in steam power plants operate essentially under adiabatic conditions A plant engineer suggests ending this practice She proposes to run cooling water through the outer surface of the casing to cool the steam as it flows through the turbine This way she reasons the entropy of the steam will decrease the performance of the turbine will improve and as a result the work output of the turbine will increase How would you evaluate this proposal 7100C It is well known that the power consumed by a com pressor can be reduced by cooling the gas during compression Inspired by this somebody proposes to cool the liquid as it flows through a pump in order to reduce the power consump tion of the pump Would you support this proposal Explain 7101 Calculate the work produced in kJkg for the revers ible steadyflow process 13 shown in Fig P7101 7104 Saturated water vapor at 150C is compressed in a reversible steadyflow device to 1000 kPa while its specific vol ume remains constant Determine the work required in kJkg 7105 Liquid water at 120 kPa enters a 7kW pump where its pressure is raised to 5 MPa If the elevation difference between the exit and the inlet levels is 10 m determine the highest mass flow rate of liquid water this pump can handle Neglect the kinetic energy change of water and take the specific volume of water to be 0001 m3kg 7106 Water enters the pump of a steam power plant as satu rated liquid at 20 kPa at a rate of 45 kgs and exits at 6 MPa Neglecting the changes in kinetic and potential energies and assuming the process to be reversible determine the power input to the pump 7107 Consider a steam power plant that operates between the pressure limits of 5 MPa and 10 kPa Steam enters the pump as saturated liquid and leaves the turbine as saturated vapor Determine the ratio of the work delivered by the turbine to the work consumed by the pump Assume the entire cycle to be reversible and the heat losses from the pump and the turbine to be negligible 7108 Reconsider Prob 7107 Using appropriate software investigate the effect of the quality of the steam at the turbine exit on the net work output Vary the quality from 05 to 10 and plot the net work output as a func tion of this quality 7109E Saturated refrigerant134a vapor at 15 psia is com pressed reversibly in an adiabatic compressor to 80 psia Determine the work input to the compressor What would your answer be if the refrigerant were first condensed at constant pressure before it was compressed 7110E Helium gas is compressed from 16 psia and 85F to 120 psia at a rate of 10 ft3s Determine the power input to the compressor assuming the compression process to be a isen tropic b polytropic with n 12 c isothermal and d ideal twostage polytropic with n 12 7111E Reconsider Prob 7110E Using appropriate software evaluate and plot the work of compres sion and entropy change of the helium as functions of the poly tropic exponent as it varies from 1 to 1667 Discuss your results 7112 Nitrogen gas is compressed from 80 kPa and 27C to 480 kPa by a 10kW compressor Determine the mass flow rate of nitrogen through the compressor assuming the compres sion process to be a isentropic b polytropic with n 13 c isothermal and d ideal twostage polytropic with n 13 Answers a 0048 kgs b 0051 kgs c 0063 kgs d 0056 kgs Isentropic Efficiencies of SteadyFlow Devices 7113C Describe the ideal process for an a adiabatic tur bine b adiabatic compressor and c adiabatic nozzle and define the isentropic efficiency for each device FIGURE P7101 1 50 400 500 100 v m3kg P kPa 2 3 1 7102E Calculate the work produced in Btulbm for the reversible steadyflow process 12 shown in Fig P7102E FIGURE P7102E 01 17 500 100 v ft3lbm P psia 2 1 7103E Air is compressed isothermally from 13 psia and 55F to 80 psia in a reversible steadyflow device Calculate the work required in Btulbm for this compression Answer 642 Btulbm Final PDF to printer cen22672ch07323412indd 399 110617 0852 AM 399 CHAPTER 7 7114C Is the isentropic process a suitable model for com pressors that are cooled intentionally Explain 7115C On a Ts diagram does the actual exit state state 2 of an adiabatic turbine have to be on the righthand side of the isentropic exit state state 2s Why 7116 Argon gas enters an adiabatic turbine at 800C and 15 MPa at a rate of 80 kgmin and exhausts at 200 kPa If the power output of the turbine is 370 kW determine the isentro pic efficiency of the turbine 7117E Steam at 100 psia and 650F is expanded adiabatically in a closed system to 10 psia Determine the work produced in Btulbm and the final temperature of steam for an isentropic expansion efficiency of 80 percent Answers 132 Btulbm 275F 7118E Combustion gases enter an adiabatic gas turbine at 1540F and 120 psia and leave at 60 psia with a low velocity Treating the combustion gases as air and assuming an isentro pic efficiency of 82 percent determine the work output of the turbine Answer 717 Btulbm 7119 Steam at 4 MPa and 350C is expanded in an adia batic turbine to 120 kPa What is the isentropic efficiency of this turbine if the steam is exhausted as a saturated vapor 7123 Reconsider Prob 7122 Using appropriate software redo the problem by including the effects of the kinetic energy of the flow by assuming an inlet toexit area ratio of 15 for the compressor when the compres sor exit pipe inside diameter is 2 cm 7124 The adiabatic compressor of a refrigeration system compresses saturated R134a vapor at 0C to 600 kPa and 50C What is the isentropic efficiency of this compressor FIGURE P7119 4 MPa 350C 120 kPa sat vapor Steam turbine 7120 Steam at 3 MPa and 400C is expanded to 30 kPa in an adiabatic turbine with an isentropic efficiency of 92 percent Determine the power produced by this turbine in kW when the mass flow rate is 2 kgs 7121 Repeat Prob 7120 for a turbine efficiency of 85 percent 7122 Refrigerant134a enters an adiabatic compressor as saturated vapor at 100 kPa at a rate of 07 m3min and exits at 1MPa pressure If the isentropic efficiency of the compressor is 87 percent determine a the temperature of the refrigerant at the exit of the compressor and b the power input in kW FIGURE P7122 1 MPa 100 kPa sat vapor R134a Compressor FIGURE P7124 600 kPa 50C 0C sat vapor R134a compressor 7125 Air is compressed by an adiabatic compressor from 95 kPa and 27C to 600 kPa and 277C Assuming variable specific heats and neglecting the changes in kinetic and poten tial energies determine a the isentropic efficiency of the compressor and b the exit temperature of air if the process were reversible Answers a 819 percent b 506 K 7126E Argon gas enters an adiabatic compressor at 14 psia and 75F with a velocity of 60 fts and it exits at 200 psia and 240 fts If the isentropic efficiency of the compressor is 87 percent determine a the exit temperature of the argon and b the work input to the compressor 7127 An adiabatic steadyflow device compresses argon at 200 kPa and 27C to 2 MPa If the argon leaves this compressor at 550C what is the isentropic efficiency of the compressor Also show the process on a Ts diagram with respect to satura tion lines Final PDF to printer cen22672ch07323412indd 400 110617 0852 AM 400 ENTROPY 7128E Air enters an adiabatic nozzle at 45 psia and 940F with low velocity and exits at 650 fts If the isentropic effi ciency of the nozzle is 85 percent determine the exit tempera ture and pressure of the air 7129E Reconsider Prob 7128E Using appropriate software study the effect of varying the nozzle isentropic efficiency from 08 to 10 on both the exit tempera ture and pressure of the air and plot the results 7130E An adiabatic diffuser at the inlet of a jet engine increases the pressure of the air that enters the diffuser at 11 psia and 30F to 20 psia What will the air velocity at the diffuser exit be if the diffuser isentropic efficiency defined as the ratio of the actual kinetic energy change to the isentro pic kinetic energy change is 82 percent and the diffuser inlet velocity is 1200 fts Answer 735 fts 7134E Refrigerant134a is expanded adiabatically from 100 psia and 100F to a pressure of 10 psia Determine the entropy generation for this process in BtulbmR FIGURE P7130E 11 psia 30F 1200 fts 20 psia Air 7131 Hot combustion gases enter the nozzle of a turbojet engine at 260 kPa 747C and 80 ms and they exit at a pressure of 85 kPa Assuming an isentropic efficiency of 92 percent and treating the combustion gases as air determine a the exit veloc ity and b the exit temperature Answers a 728 ms b 786 K FIGURE P7131 260 kPa 747C 80 ms 85 kPa Nozzle ηN 92 7132 The exhaust nozzle of a jet engine expands air at 300 kPa and 180C adiabatically to 100 kPa Determine the air velocity at the exit when the inlet velocity is low and the nozzle isentropic efficiency is 93 percent Entropy Balance 7133E An iron block of unknown mass at 185F is dropped into an insulated tank that contains 08 ft3 of water at 70F At the same time a paddle wheel driven by a 200W motor is activated to stir the water Thermal equilibrium is established after 10 min with a final temperature of 75F Determine a the mass of the iron block and b the entropy generated during this process 7135E A frictionless pistoncylinder device contains satu rated liquid water at 40psia pressure Now 600 Btu of heat is transferred to water from a source at 1000F and part of the liquid vaporizes at constant pressure Determine the total entropy generated during this process in BtuR 7136 Air enters a compressor steadily at the ambient condi tions of 100 kPa and 22C and leaves at 800 kPa Heat is lost from the compressor in the amount of 120 kJkg and the air experiences an entropy decrease of 040 kJkgK Using con stant specific heats determine a the exit temperature of the air b the work input to the compressor and c the entropy generation during this process 7137 Steam enters an adiabatic turbine steadily at 7 MPa 500C and 45 ms and leaves at 100 kPa and 75 ms If the power output of the turbine is 5 MW and the isentropic effi ciency is 77 percent determine a the mass flow rate of steam through the turbine b the temperature at the turbine exit and c the rate of entropy generation during this process FIGURE P7134E 10 psia R134a 100 psia 100F FIGURE P7137 Turbine Steam 7 MPa 500C 45 ms 100 kPa 75 ms 7138 In an icemaking plant water at 0C is frozen at atmo spheric pressure by evaporating saturated R134a liquid at 16C The refrigerant leaves this evaporator as a saturated vapor and the plant is sized to produce ice at 0C at a rate of 5500 kgh Deter mine the rate of entropy generation in this plant Answer 0115 kWK FIGURE P7138 R134a 16C 16C sat vapor Q Final PDF to printer cen22672ch07323412indd 401 110617 0852 AM 401 CHAPTER 7 7139 Oxygen enters an insulated 12cmdiameter pipe with a velocity of 70 ms At the pipe entrance the oxygen is at 240 kPa and 20C and at the exit it is at 200 kPa and 18C Calculate the rate at which entropy is generated in the pipe 7140E Water at 20 psia and 50F enters a mixing chamber at a rate of 300 lbmmin where it is mixed steadily with steam entering at 20 psia and 240F The mixture leaves the chamber at 20 psia and 130F and heat is lost to the surrounding air at 70F at a rate of 180 Btumin Neglecting the changes in kinetic and potential energies determine the rate of entropy generation during this process To save energy and money the plant installs a regenerator that has an effectiveness of 82 percent If the cost of natural gas is 130therm 1 therm 105500 kJ determine how much energy and money the regenerator will save this company per year and the annual reduction in entropy generation FIGURE P7143 Heat Pasteurizing section 72C 72C Hot milk Cold milk Regenerator 4C FIGURE P7140E Mixing chamber P 20 psia T3 130F T1 50F 300 lbmmin 180 Btumin T2 240F FIGURE P7141 70C 85C 16 kgs Hot water Cold water 095 kgs 10C 7141 Cold water cp 418 kJkgC leading to a shower enters a wellinsulated thinwalled doublepipe counterflow heat exchanger at 10C at a rate of 095 kgs and is heated to 70C by hot water cp 419 kJkgC that enters at 85C at a rate of 16 kgs Determine a the rate of heat transfer and b the rate of entropy generation in the heat exchanger 7142 Air cp 1005 kJkgC is to be preheated by hot exhaust gases in a crossflow heat exchanger before it enters the furnace Air enters the heat exchanger at 95 kPa and 20C at a rate of 16 m3s The combustion gases cp 110 kJkgC enter at 180C at a rate of 22 kgs and leave at 95C Deter mine a the rate of heat transfer to the air b the outlet tem perature of the air and c the rate of entropy generation 7143 In a dairy plant milk at 4C is pasteurized continu ously at 72C at a rate of 12 Ls for 24 hours a day and 365 days a year The milk is heated to the pasteurizing temperature by hot water heated in a naturalgasfired boiler that has an effi ciency of 82 percent The pasteurized milk is then cooled by cold water at 18C before it is finally refrigerated back to 4C 7144 Steam is to be condensed in the condenser of a steam power plant at a temperature of 60C with cooling water from a nearby lake which enters the tubes of the condenser at 18C at a rate of 75 kgs and leaves at 27C Assuming the condenser to be perfectly insulated determine a the rate of condensa tion of the steam and b the rate of entropy generation in the condenser Answers a 120 kgs b 106 kWK 7145 An ordinary egg can be approximated as a 55cmdiameter sphere The egg is initially at a uniform tem perature of 8C and is dropped into boiling water at 97C Taking the properties of the egg to be ρ 1020 kgm3 and cp 332 kJkgC determine a how much heat is transferred to the egg by the time the average temperature of the egg rises to 70C and b the amount of entropy generation associated with this heat transfer process FIGURE P7145 Egg Ti 8C Boiling water 97C 7146 Chickens with an average mass of 22 kg and average specific heat of 354 kJkgC are to be cooled by chilled water that enters a continuousflowtype immersion chiller at 05C and leaves at 25C Chickens are dropped into the chiller at a uniform temperature of 15C at a rate of 250 chickens per hour Final PDF to printer cen22672ch07323412indd 402 110617 0852 AM 402 ENTROPY and are cooled to an average temperature of 3C before they are taken out The chiller gains heat from the surroundings at 25C at a rate of 150 kJh Determine a the rate of heat removal from the chickens in kW and b the rate of entropy generation during this chilling process 7147E In a production facility 12inthick 2ft 2ft square brass plates ρ 5325 lbmft3 and cp 0091 BtulbmF that are initially at a uniform temperature of 75F are heated by passing them through an oven at 1300F at a rate of 450 per minute If the plates remain in the oven until their average tem perature rises to 1000F determine a the rate of heat transfer to the plates in the furnace and b the rate of entropy genera tion associated with this heat transfer process 7148 Long cylindrical steel rods ρ 7833 kgm3 and cp 0465 kJkgC of 10cm diameter are heat treated by drawing them at a velocity of 3 mmin through a 7mlong oven maintained at 900C If the rods enter the oven at 30C and leave at 700C determine a the rate of heat transfer to the rods in the oven and b the rate of entropy generation asso ciated with this heat transfer process 7152 Steam enters an adiabatic nozzle at 2 MPa and 350C with a velocity of 55 ms and exits at 08 MPa and 390 ms If the nozzle has an inlet area of 75 cm2 determine a the exit temperature and b the rate of entropy generation for this process Answers a 303C b 00854 kWK 7153 Steam expands in a turbine steadily at a rate of 40000 kgh entering at 8 MPa and 500C and leaving at 40 kPa as saturated vapor If the power generated by the tur bine is 82 MW determine the rate of entropy generation for this process Assume the surrounding medium is at 25C Answer 114 kWK 7149 Stainlesssteel ball bearings ρ 8085 kgm3 and cp 0480 kJkgC having a diameter of 18 cm are to be quenched in water at a rate of 1100 per minute The balls leave the oven at a uniform temperature of 900C and are exposed to air at 20C for a while before they are dropped into the water If the temperature of the balls drops to 850C prior to quench ing determine a the rate of heat transfer from the balls to the air and b the rate of entropy generation due to heat loss from the balls to the air 7150 The inner and outer surfaces of a 4m 10m brick wall of thickness 20 cm are maintained at temperatures of 16C and 4C respectively If the rate of heat transfer through the wall is 1800 W determine the rate of entropy generation within the wall 7151E Steam enters a diffuser at 20 psia and 240F with a velocity of 900 fts and exits as saturated vapor at 240F and 100 fts The exit area of the diffuser is 1 ft2 Determine a the mass flow rate of the steam and b the rate of entropy generation during this process Assume an ambient temperature of 77F FIGURE P7148 7 m Oven 900C 3 mmin Stainless steel 30C FIGURE P7153 8 MPa 500C 40 kPa sat vapor Steam turbine 82 MW 7154 Liquid water at 200 kPa and 15C is heated in a cham ber by mixing it with superheated steam at 200 kPa and 150C Liquid water enters the mixing chamber at a rate of 43 kgs and the chamber is estimated to lose heat to the surrounding air at 20C at a rate of 1200 kJmin If the mixture leaves the mixing chamber at 200 kPa and 80C determine a the mass flow rate of the superheated steam and b the rate of entropy generation during this mixing process Answers a 0481 kgs b 0746 kWK FIGURE P7154 Mixing chamber 200 kPa 80C 15C 43 kgs 1200 kJmin 150C 7155 A 018m3 rigid tank is filled with saturated liq uid water at 120C A valve at the bottom of the tank is now opened and onehalf of the total mass is withdrawn from the tank in the liquid form Heat is transferred to water from a Final PDF to printer cen22672ch07323412indd 403 110617 0852 AM 403 CHAPTER 7 source at 230C so that the temperature in the tank remains constant Determine a the amount of heat transfer and b the total entropy generation for this process 7156 A rigid tank contains 75 kg of saturated water mixture at 400 kPa A valve at the bottom of the tank is now opened and liquid is withdrawn from the tank Heat is transferred to the steam such that the pressure inside the tank remains con stant The valve is closed when no liquid is left in the tank If it is estimated that a total of 5 kJ of heat is transferred to the tank determine a the quality of steam in the tank at the ini tial state b the amount of mass that has escaped and c the entropy generation during this process if heat is supplied to the tank from a source at 500C Special Topic Reducing the Cost of Compressed Air 7157 The compressedair requirements of a plant at sea level are being met by a 90hp compressor that takes in air at the local atmospheric pressure of 1013 kPa and the average temperature of 15C and compresses it to 1100 kPa An investigation of the compressedair system and the equipment using the compressed air reveals that compressing the air to 750 kPa is sufficient for this plant The compressor operates 3500 hyr at 75 percent of the rated load and is driven by an electric motor that has an efficiency of 94 percent Taking the price of electricity to be 0105kWh determine the amount of energy and money saved as a result of reducing the pressure of the compressed air 7158 The compressedair requirements of a plant are being met by a 100hp screw compressor that runs at full load dur ing 40 percent of the time and idles the rest of the time during operating hours The compressor consumes 35 percent of the rated power when idling and 90 percent of the power when compressing air The annual operating hours of the facility are 3800 h and the unit cost of electricity is 0083kWh It is determined that the compressedair requirements of the facility during 60 percent of the time can be met by a 25hp reciprocating compressor that consumes 95 percent of the rated power when compressing air and no power when not compressing air It is estimated that the 25hp compressor will run 85 percent of the time The efficiencies of the motors of the large and the small compressors at or near full load are 090 and 088 respectively The efficiency of the large motor at 35 percent load is 082 Determine the amount of energy and money that can be saved by switching to the 25hp compressor during 60 percent of the time 7159 The compressedair requirements of a plant are being met by a 90hp screw compressor The facility stops produc tion for one hour every day including weekends for lunch break but the compressor is kept operating The compressor consumes 35 percent of the rated power when idling and the unit cost of electricity is 009kWh Determine the amount of energy and money that can be saved per year as a result of turn ing the compressor off during lunch break Take the efficiency of the motor at part load to be 84 percent 7160 Compressed air is one of the key utilities in manufac turing facilities and the total installed power of compressedair systems in the United States is estimated to be about 20 million horsepower Assuming the compressors operate at full load for onethird of the time on average and the average motor effi ciency is 90 percent determine how much energy and money will be saved per year if the energy consumed by compres sors is reduced by 5 percent as a result of implementing some conservation measures Take the unit cost of electricity to be 009kWh 7161 The compressedair requirements of a plant are met by a 150hp compressor equipped with an intercooler an after cooler and a refrigerated dryer The plant operates 6300 hyr but the compressor is estimated to be compressing air dur ing only onethird of the operating hours that is 2100 hours a year The compressor is either idling or is shut off the rest of the time Temperature measurements and calculations indi cate that 25 percent of the energy input to the compressor is removed from the compressed air as heat in the aftercooler The COP of the refrigeration unit is 25 and the cost of elec tricity is 0065kWh Determine the amount of energy and money saved per year as a result of cooling the compressed air before it enters the refrigerated dryer 7162 The 1800rpm 150hp motor of a compressor is burned out and is to be replaced by either a standard motor that has a fullload efficiency of 930 percent and costs 9031 or a highefficiency motor that has an efficiency of 962 percent and costs 10942 The compressor operates 4368 hyr at full load and its operation at part load is negligible If the cost of electricity is 0095kWh determine the amount of energy and money this facility will save by purchasing the highefficiency motor instead of the standard motor Also determine if the savings from the highefficiency motor justify the price differ ential if the expected life of the motor is 10 years Ignore any possible rebates from the local power company 7163 The space heating of a facility is accomplished by nat ural gas heaters that are 85 percent efficient The compressed air needs of the facility are met by a large liquidcooled compressor The coolant of the compressor is cooled by air in a liquidtoair heat exchanger whose airflow section is 10 m high and 10 m wide During typical operation the air is heated from 20 to 52C as it flows through the heat exchanger The average velocity of air on the inlet side is measured at 3 ms The compressor operates 20 hours a day and 5 days a week throughout the year Taking the heating season to be 6 months 26 weeks and the cost of the natural gas to be 125therm 1 therm 100000 Btu 105500 kJ determine how much money will be saved by diverting the compressor waste heat into the facility during the heating season 7164 The compressors of a production facility maintain the compressedair lines at a gage pressure of 700 kPa at 1400m elevation where the atmospheric pressure is 856 kPa The average temperature of air is 15C at the compressor inlet Final PDF to printer cen22672ch07323412indd 404 110617 0852 AM 404 ENTROPY and 25C in the compressedair lines The facility operates 4200 hyr and the average price of electricity is 010kWh Taking the compressor efficiency to be 08 the motor effi ciency to be 093 and the discharge coefficient to be 065 determine the energy and money saved per year by sealing a leak equivalent to a 3mmdiameter hole on the compressed air line 7165 The energy used to compress air in the United States is estimated to exceed onehalf quadrillion 05 1015 kJ per year It is also estimated that 10 to 40 percent of the compressed air is lost through leaks Assuming on average 20 percent of the compressed air is lost through air leaks and the unit cost of electricity is 011kWh determine the amount and cost of electricity wasted per year due to air leaks 7166 A 150hp compressor in an industrial facility is housed inside the production area where the average tem perature during operating hours is 25C The average outdoor temperature during the same hours is 10C The compressor operates 4500 hyr at 85 percent of rated load and is driven by an electric motor that has an efficiency of 90 percent Tak ing the price of electricity to be 0075kWh determine the amount of energy and money that can be saved by drawing outside air to the compressor instead of using the inside air Review Problems 7167 A proposed heat pump design creates a heating effect of 25 kW while using 5 kW of electrical power The thermal energy reservoirs are at 300 K and 260 K Is this possible according to the increase of entropy principle 7168 A refrigerator with a coefficient of performance of 4 transfers heat from a cold region at 20C to a hot region at 30C Calculate the total entropy change of the regions when 1 kJ of heat is transferred from the cold region Is the second law satisfied Will this refrigerator still satisfy the second law if its coefficient of performance is 6 7169 What is the minimum internal energy that steam can achieve as it is expanded adiabatically in a closed system from 1500 kPa and 320C to 100 kPa 7170E Is it possible to expand water at 30 psia and 70 percent quality to 10 psia in a closed system undergoing an isothermal reversible process while exchanging heat with an energy reser voir at 300F 7171 What is the maximum volume that 3 kg of oxygen at 950 kPa and 373C can be adiabatically expanded to in a pistoncylinder device if the final pressure is to be 100 kPa Answer 266 m3 7172E A 100lbm block of a solid material whose specific heat is 05 BtulbmR is at 80F It is heated with 10 lbm of saturated water vapor that has a constant pressure of 20 psia Determine the final temperature of the block and water and the entropy change of a the block b the water and c the entire system Is this process possible Why 7173 A pistoncylinder device contains air that undergoes a reversible thermodynamic cycle Initially air is at 400 kPa and 300 K with a volume of 03 m3 Air is first expanded isother mally to 150 kPa then compressed adiabatically to the initial pressure and finally compressed at the constant pressure to the initial state Accounting for the variation of specific heats with temperature determine the work and heat transfer for each process 7174E A pistoncylinder device initially contains 15 ft3 of helium gas at 25 psia and 70F Helium is now compressed in a polytropic process PVn constant to 70 psia and 300F Determine a the entropy change of helium b the entropy change of the surroundings and c whether this process is reversible irreversible or impossible Assume the surroundings are at 70F Answers a 0016 BtuR b 0019 BtuR c irreversible 7175 A pistoncylinder device contains steam that under goes a reversible thermodynamic cycle Initially the steam is at 400 kPa and 350C with a volume of 05 m3 The steam is first expanded isothermally to 150 kPa then compressed adia batically to the initial pressure and finally compressed at the constant pressure to the initial state Determine the net work and heat transfer for the cycle after you calculate the work and heat interaction for each process 7176 One hundred kg of saturated steam at 100 kPa is to be adiabatically compressed in a closed system to 1000 kPa How much work is required if the isentropic compression efficiency is 90 percent Answer 44160 kJ 7177E Ten lbm of R134a is expanded without any heat transfer in a closed system from 120 psia and 100F to 20 psia If the isentropic expansion efficiency is 95 percent what is the final volume of this steam 7178 Refrigerant134a at 700 kPa and 40C is expanded adiabatically in a closed system to 60 kPa Determine the work produced in kJkg and final enthalpy for an isentropic expan sion efficiency of 80 percent Answer 379 kJkg 2384 kJkg FIGURE P7168 QH Win 1 kJ R 30C 20C Final PDF to printer cen22672ch07323412indd 405 110617 0852 AM 405 CHAPTER 7 7179 A 08m3 rigid tank contains carbon dioxide CO2 gas at 250 K and 100 kPa A 500W electric resistance heater placed in the tank is now turned on and kept on for 40 min after which the pressure of CO2 is measured to be 175 kPa Assuming the surroundings to be at 300 K and using constant specific heats determine a the final temperature of CO2 b the net amount of heat transfer from the tank and c the entropy generation during this process 7182E Helium gas enters a nozzle whose isentropic effi ciency is 94 percent with a low velocity and it exits at 14 psia 180F and 1000 fts Determine the pressure and temperature at the nozzle inlet 7183 An inventor claims to have invented an adiabatic steadyflow device with a single inletoutlet that produces 230 kW when expanding 1 kgs of air from 1200 kPa and 300C to 100 kPa Is this claim valid 7184 An adiabatic capillary tube is used in some refrigera tion systems to drop the pressure of the refrigerant from the condenser level to the evaporator level R134a enters the cap illary tube as a saturated liquid at 70C and leaves at 20C Determine the rate of entropy generation in the capillary tube for a mass flow rate of 02 kgs Answer 00166 kWK FIGURE P7179 CO2 250 K 100 kPa We 7180 Air enters the evaporator section of a window air condi tioner at 100 kPa and 27C with a volume flow rate of 6 m3min The refrigerant134a at 120 kPa with a quality of 03 enters the evaporator at a rate of 2 kgmin and leaves as saturated vapor at the same pressure Determine the exit temperature of the air and the rate of entropy generation for this process assuming a the outer surfaces of the air conditioner are insulated and b heat is transferred to the evaporator of the air conditioner from the surrounding medium at 32C at a rate of 30 kJmin Answers a 159C 000196 kWK b 116C 000225 kWK FIGURE P7180 R134a 120 kPa x 03 Air 100 kPa 27C sat vapor 7181 Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 ms and leaves at 300 kPa and 350 K Using variable specific heats determine a the isentropic efficiency b the exit velocity and c the entropy generation FIGURE P7181 Air 500 kPa 400 K 30 ms 300 kPa 350 K FIGURE P7184 R134a 70C sat liq 20C Capillary tube 7185 Helium gas is throttled steadily from 400 kPa and 60C Heat is lost from the helium in the amount of 175 kJ kg to the surroundings at 25C and 100 kPa If the entropy of the helium increases by 034 kJkgK in the valve deter mine a the exit temperature and pressure and b the entropy generation during this process Answers a 597C 339 kPa b 0346 kJkgK 7186 Determine the work input and entropy generation dur ing the compression of steam from 100 kPa to 1 MPa in a an adiabatic pump and b an adiabatic compressor if the inlet state is saturated liquid in the pump and saturated vapor in the compressor and the isentropic efficiency is 85 percent for both devices FIGURE P7186 Pump 1 MPa 100 kPa Compressor 100 kPa 1 MPa 7187 Carbon dioxide is compressed in a reversible isother mal process from 100 kPa and 20C to 400 kPa using a steady flow device with one inlet and one outlet Determine the work required and the heat transfer both in kJkg for this compression 7188 Reconsider Prob 7187 Determine the change in the work and heat transfer when the compression process is isen tropic rather than isothermal Final PDF to printer cen22672ch07323412indd 406 110617 0852 AM 406 ENTROPY 7189 The compressor of a refrigerator compresses satu rated R134a vapor at 10C to 800 kPa How much work in kJkg does this process require when the process is isentropic 7190 Air enters a twostage compressor at 100 kPa and 27C and is compressed to 625 kPa The pressure ratio across each stage is the same and the air is cooled to the initial tem perature between the two stages Assuming the compression process to be isentropic determine the power input to the compressor for a mass flow rate of 015 kgs What would your answer be if only one stage of compression were used Answers 271 kW 311 kW 7193 Steam at 6000 kPa and 500C enters a steadyflow turbine The steam expands in the turbine while doing work until the pressure is 1000 kPa When the pressure is 1000 kPa 10 percent of the steam is removed from the turbine for other uses The remaining 90 percent of the steam continues to expand through the turbine while doing work and leaves the turbine at 10 kPa The entire expansion process by the steam through the turbine is reversible and adiabatic a Sketch the process on a Ts diagram with respect to the saturation lines Be sure to label the data states and the lines of constant pressure b If the turbine has an isentropic efficiency of 85 percent what is the work done by the steam as it flows through the turbine per unit mass of steam flowing into the turbine in kJkg 7194 Refrigerant134a at 140 kPa and 10C is compressed by an adiabatic 13kW compressor to an exit state of 700 kPa and 60C Neglecting the changes in kinetic and potential ener gies determine a the isentropic efficiency of the compressor b the volume flow rate of the refrigerant at the compressor inlet in Lmin and c the maximum volume flow rate at the inlet conditions that this adiabatic 13kW compressor can handle without violating the second law 7195 Refrigerant134a enters a compressor as a saturated vapor at 160 kPa at a rate of 003 m3s and leaves at 800 kPa The power input to the compressor is 10 kW If the surroundings at 20C experience an entropy increase of 0008 kWK determine a the rate of heat loss from the compressor b the exit tem perature of the refrigerant and c the rate of entropy generation 7196 Air is expanded in an adiabatic turbine of 90 percent isentropic efficiency from an inlet state of 2800 kPa and 400C to an outlet pressure of 100 kPa Calculate the outlet tempera ture of air the work produced by this turbine and the entropy generation Answers 303 K 375 kJkg 0148 kJkgK 7197 A steam turbine is equipped to bleed 6 percent of the inlet steam for feedwater heating It is operated with 4 MPa and 350C steam at the inlet a bleed pressure of 800 kPa and an exhaust pressure of 30 kPa Calculate the work produced by this turbine when the isentropic efficiency between the inlet and bleed point is 97 percent and the isentropic efficiency between the bleed point and exhaust is 95 percent What is the overall isentropic efficiency of the turbine Hint Treat this turbine as two separate turbines with one operating between the inlet and bleed conditions and the other operating between the bleed and exhaust conditions 7198E Work can be produced by passing the vapor phase of a twophase substance stored in a tank through a turbine as shown in Fig P7198E Consider such a system using R134a which is initially at 80F and a 10ft3 tank that initially is entirely filled with liquid R134a The turbine is isentropic the temperature in the storage tank remains constant as mass is removed from it and FIGURE P7190 100 kPa 27C Air compressor 1st stage 625 kPa 2nd stage 27C Px Px W Heat 7191 Three kg of helium gas at 100 kPa and 27C are adia batically compressed to 900 kPa If the isentropic compression efficiency is 80 percent determine the required work input and the final temperature of helium 7192 Steam at 6 MPa and 500C enters a twostage adia batic turbine at a rate of 15 kgs Ten percent of the steam is extracted at the end of the first stage at a pressure of 12 MPa for other use The remainder of the steam is further expanded in the second stage and leaves the turbine at 20 kPa Determine the power output of the turbine assuming a the process is reversible and b the turbine has an isentropic efficiency of 88 percent Answers a 16290 kW b 14335 kW FIGURE P7192 20 kPa 6 MPa 500C 12 MPa 10 90 Steam turbine 1st stage 2nd stage Final PDF to printer cen22672ch07323412indd 407 110617 0852 AM 407 CHAPTER 7 7199E An engineer has proposed that compressed air be used to level the load in an electricalgeneration and distribution system The proposed system is illustrated in Fig P7199E During those times when electricalgeneration capacity exceeds the demand for electrical energy the excess electrical energy is used to run the compressor and fill the storage tank When the demand exceeds the generation capacity compressed air in the tank is passed through the turbine to generate additional electrical energy Consider this system when the compressor and turbine are isentropic the tanks temperature stays constant at 70F air enters the compressor at 70F and 1 atm the tank volume is 1 million cubic feet and air leaves the turbine at 1 atm The compressor is activated when the tank pressure is 1 atm and it remains on until the tank pressure is 10 atm Calculate the total work required to fill the tank and the total heat transferred from the air in the tank as it is being filled drops to 200 kPa During this process 300 kJ of heat is trans ferred from tank B to the surroundings at 17C Assuming the steam remaining inside tank A to have undergone a revers ible adiabatic process determine a the final temperature in each tank and b the entropy generated during this process Answers a 1202C 1161C b 0498 kJK FIGURE P7198E Storage tank Turbine FIGURE P7199E 4 3 1 2 Storage tank Compressor Turbine 7200E Reconsider Prob 7199E The filled compressedair storage tank is discharged at a later time through the turbine until the pressure in the tank is 1 atm During this discharge the temperature of the air in the storage tank remains constant at 70F Calculate the total work produced by the turbine and the total heat transferred to the air in the tank during this discharge 7201 Two rigid tanks are connected by a valve Tank A is insulated and contains 03 m3 of steam at 400 kPa and 60 percent quality Tank B is uninsulated and contains 2 kg of steam at 200 kPa and 250C The valve is now opened and steam flows from tank A to tank B until the pressure in tank A FIGURE P7201 A 03 m3 steam 400 kPa x 06 B 2 kg steam 200 kPa 250C 300 kJ 7202 A 1200W electric resistance heating element whose diameter is 05 cm is immersed in 40 kg of water initially at 20C Assuming the water container is well insulated deter mine how long it will take for this heater to raise the water temperature to 50C Also determine the entropy generated during this process in kJK 7203E A 15ft3 steel container that has a mass of 75 lbm when empty is filled with liquid water Initially both the steel tank and the water are at 120F Now heat is transferred and the entire system cools to the surrounding air temperature of 70F Determine the total entropy generated during this process 7204 In order to cool 1 ton of water at 20C in an insu lated tank a person pours 140 kg of ice at 5C into the water Determine a the final equilibrium temperature in the tank and b the entropy generation during this process The melting temperature and the heat of fusion of ice at atmospheric pres sure are 0C and 3337 kJkg 7205 One ton of liquid water at 80C is brought into a well insulated and wellsealed 4m 5m 7m room initially at 22C and 100 kPa Assuming constant specific heats for both air and water at room temperature determine a the final equilibrium temperature in the room and b the total entropy change during this process in kJK FIGURE P7205 4 m 5 m 7 m Room 22C 100 kPa Water 80C Heat the R134a leaves the turbine at 10 psia How much work will be produced when the half of the liquid mass in the tank is used Final PDF to printer cen22672ch07323412indd 408 110617 0852 AM 408 ENTROPY 7206 A wellinsulated 4m 4m 5m room initially at 10C is heated by the radiator of a steam heating system The radiator has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200C At this moment both the inlet and the exit valves to the radiator are closed A 120W fan is used to distribute the air in the room The pressure of the steam is observed to drop to 100 kPa after 30 min as a result of heat transfer to the room Assuming constant specific heats for air at room temperature determine a the average temperature of air in 30 min b the entropy change of the steam c the entropy change of the air in the room and d the entropy gen erated during this process in kJK Assume the air pressure in the room remains constant at 100 kPa at all times 7207 A passive solar house that is losing heat to the out doors at 3C at an average rate of 50000 kJh is maintained at 22C at all times during a winter night for 10 h The house is heated by 50 glass containers each containing 20 L of water that is heated to 80C during the day by absorbing solar energy A thermostatcontrolled 15kW backup electric resis tance heater turns on whenever necessary to keep the house at 22C Determine how long the electric heating system is on during this night and the amount of entropy generated during the night 7208 An insulated pistoncylinder device initially contains 002 m3 of saturated liquidvapor mixture of water with a qual ity of 01 at 100C Now some ice at 18C is dropped into the cylinder If the cylinder contains saturated liquid at 100C when thermal equilibrium is established determine a the amount of ice added and b the entropy generation during this process The melting temperature and the heat of fusion of ice at atmospheric pressure are 0C and 3337 kJkg exchanger has an effectiveness of 050 that is it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water determine the electric power input required in this case and the reduction in the rate of entropy generation in the resistance heating section FIGURE P7208 002 m3 100C Ice 18C 7209 a Water flows through a shower head steadily at a rate of 10 Lmin An electric resistance heater placed in the water pipe heats the water from 16 to 43C Taking the density of water to be 1 kgL determine the electric power input to the heater in kW and the rate of entropy generation during this process in kWK b In an effort to conserve energy it is proposed to pass the drained warm water at a temperature of 39C through a heat exchanger to preheat the incoming cold water If the heat FIGURE P7209 Resistance heater 7210 Using appropriate software determine the work input to a multistage compressor for a given set of inlet and exit pressures for any number of stages Assume that the pressure ratio across each stage is identical and the compression process is polytropic List and plot the compres sor work against the number of stages for P1 100 kPa T1 25C P2 1000 kPa and n 135 for air Based on this chart can you justify using compressors with more than three stages 7211 The inner and outer glasses of a 2m 2m double pane window are at 18C and 6C respectively If the glasses are very nearly isothermal and the rate of heat transfer through the window is 110 W determine the rates of entropy transfer through both sides of the window and the rate of entropy gen eration within the window in WK FIGURE P7211 Air Q 18C 6C Final PDF to printer cen22672ch07323412indd 409 110617 0852 AM 409 CHAPTER 7 7212 A hotwater pipe at 80C is losing heat to the sur rounding air at 5C at a rate of 1600 W Determine the rate of entropy generation in the surrounding air in WK 7213 When the transportation of natural gas in a pipeline is not feasible for economic reasons it is first liquefied using nonconventional refrigeration techniques and then trans ported in superinsulated tanks In a natural gas liquefaction plant the liquefied natural gas LNG enters a cryogenic tur bine at 30 bar and 160C at a rate of 20 kgs and leaves at 3 bar If 115 kW power is produced by the turbine determine the efficiency of the turbine Take the density of LNG to be 4238 kgm3 Answer 903 percent 7215 Consider a 50L evacuated rigid bottle that is sur rounded by the atmosphere at 95 kPa and 27C A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle The air trapped in the bottle eventually reaches thermal equilibrium with the atmo sphere as a result of heat transfer through the wall of the bottle The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere Determine the net heat transfer through the wall of the bottle and the entropy generation during this filling process Answers 475 kJ 00158 kJK 7216 A 040m3 insulated pistoncylinder device initially contains 13 kg of air at 30C At this state the piston is free to move Now air at 500 kPa and 70C is allowed to enter the cylinder from a supply line until the volume increases by 50 percent Using constant specific heats at room tempera ture determine a the final temperature b the amount of mass that has entered c the work done and d the entropy generation FIGURE P7213 3 bar LNG 30 bar 160C 20 kgs Cryogenic turbine 7214 Consider the turbocharger of an internal combustion engine The exhaust gases enter the turbine at 450C at a rate of 002 kgs and leave at 400C Air enters the compressor at 70C and 95 kPa at a rate of 0018 kgs and leaves at 135 kPa The mechanical efficiency between the turbine and the com pressor is 95 percent 5 percent of turbine work is lost during its transmission to the compressor Using air properties for the exhaust gases determine a the air temperature at the com pressor exit and b the isentropic efficiency of the compressor Answers a 126C b 642 percent FIGURE P7214 400C Air 70C 95 kPa 0018 kgs 135 kPa Exhaust gases 450C 002 kgs Turbine Compressor 7217E A 5ft3 rigid tank initially contains refrigerant134a at 60 psia and 100 percent quality The tank is connected by a valve to a supply line that carries refrigerant134a at 140 psia and 80F The valve is now opened allowing the refrigerant to enter the tank and is closed when it is observed that the tank contains only saturated liquid at 100 psia Determine a the mass of the refrigerant that entered the tank b the amount of heat transfer with the surroundings at 70F and c the entropy generated during this process 7218 During a heat transfer process the entropy change of incompressible substances such as liquid water can be determined from ΔS mcavg lnT2T1 Show that for thermal energy reservoirs such as large lakes this relation reduces to ΔS QT 7219 Show that the difference between the reversible steadyflow work and reversible moving boundary work is equal to the flow energy 7220 Demonstrate the validity of the Clausius inequality using a reversible and an irreversible heat engine operating FIGURE P7216 Air 040 m3 13 kg 30C Air 500 kPa 70C Final PDF to printer cen22672ch07323412indd 410 110617 0852 AM 410 ENTROPY 7221 Consider two bodies of identical mass m and specific heat c used as thermal reservoirs source and sink for a heat engine The first body is initially at an absolute temperature T1 while the second one is at a lower absolute temperature T2 Heat is transferred from the first body to the heat engine which rejects the waste heat to the second body The process contin ues until the final temperatures of the two bodies Tf become equal Show that T f T 1 T 2 when the heat engine produces the maximum possible work 7223 For an ideal gas with constant specific heats show that the compressor and turbine isentropic efficiencies may be written as η C P 2 P 1 k 1 k T 2 T 1 1 and η T T 4 T 3 1 P 4 P 3 k 1k 1 The states 1 and 2 represent the compressor inlet and exit states and the states 3 and 4 represent the turbine inlet and exit states 7224 A rigid adiabatic container is filled through a sin gle opening from a source of working fluid whose proper ties remain fixed How does the final specific entropy of the singlephase contents of this container compare to the initial specific entropy 7225 The temperature of an ideal gas having constant spe cific heats is given as a function of specific entropy and pres sure as Ts P A P k 1 k exps c p where A is a constant Determine the TP relation for this ideal gas undergoing an isentropic process 7226 The polytropic or small stage efficiency of a tur bine ηT is defined as the ratio of the actual differential work done to the isentropic differential work done by the fluid flowing through the turbine ηT dhdhs Consider an ideal gas with constant specific heats as the working fluid in a turbine undergoing a process in which the poly tropic efficiency is constant Show that the temperature ratio across the turbine is related to the pressure ratio across the turbine by T 2 T 1 P 2 P 1 η T R c p P 2 P 1 η T k 1 k Fundamentals of Engineering FE Exam Problems 7227 Steam is condensed at a constant temperature of 30C as it flows through the condensor of a power plant by rejecting heat at a rate of 55 MW The rate of entropy change of steam as it flows through the condenser is a 183 MWK b 018 MWK c 0 MWK d 056 MWK e 122 MWK 7228 Steam is compressed from 6 MPa and 300C to 10 MPa isentropically The final temperature of the steam is a 290C b 300C c 311C d 371C e 422C 7229 An apple with a mass of 012 kg and average specific heat of 365 kJkgC is cooled from 25C to 5C The entropy change of the apple is a 0705 kJK b 0254 kJK c 00304 kJK d 0 kJK e 0348 kJK 7230 A pistoncylinder device contains 5 kg of saturated water vapor at 3 MPa Now heat is rejected from the cylinder at constant pressure until the water vapor completely condenses so that the cylinder contains saturated liquid at 3 MPa at the end of the process The entropy change of the system during this process is a 0 kJK b 35 kJK c 125 kJK d 177 kJK e 195 kJK FIGURE P7220 Lowtemperature reservoir at TL Rev HE QH Wnetrev QL Irrev HE QH Wnetirrev QL irrev TH Hightemperature reservoir at FIGURE P7221 HE m c T1 m c T2 W QH QL 7222 Consider a threestage isentropic compressor with two intercoolers that cool the gas to the initial temperature between the stages Determine the two intermediate pressures Px and Py in terms of inlet and exit pressures P1 and P2 that will minimize the work input to the compressor Answers P x P 1 2 P 2 13 P y P 1 P 2 2 13 between the same two thermal energy reservoirs at constant temperatures of TL and TH Final PDF to printer cen22672ch07323412indd 411 110617 0852 AM 411 CHAPTER 7 7231 Argon gas expands in an adiabatic turbine from 3 MPa and 750C to 03 MPa at a rate of 5 kgs The maximum power output of the turbine is a 064 MW b 112 MW c 160 MW d 195 MW e 240 MW 7232 A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining heat from the sur roundings at temperature T in the amount of q If the entropy of the substance is s1 at state 1 and s2 at state 2 the entropy change of the substance Δs during this process is a Δs s2 s1 b Δs s2 s1 c Δs s2 s1 d Δs s2 s1 qT e Δs s2 s1 qT 7233 A unit mass of an ideal gas at temperature T under goes a reversible isothermal process from pressure P1 to pres sure P2 while losing heat to the surroundings at temperature T in the amount of q If the gas constant of the gas is R the entropy change of the gas Δs during this process is a Δs R lnP2P1 b Δs R lnP2P1 qT c Δs R lnP1P2 d Δs R lnP1P2 qT e Δs 0 7234 Heat is lost through a plane wall steadily at a rate of 1500 W If the inner and outer surface temperatures of the wall are 20C and 5C respectively the rate of entropy generation within the wall is a 007 WK b 015 WK c 028 WK d 142 WK e 521 WK 7235 Air is compressed steadily and adiabatically from 17C and 90 kPa to 200C and 400 kPa Assuming constant specific heats for air at room temperature the isentropic effi ciency of the compressor is a 076 b 094 c 086 d 084 e 1 7236 Argon gas expands in an adiabatic turbine steadily from 600C and 800 kPa to 80 kPa at a rate of 25 kgs For isentropic efficiency of 88 percent the power produced by the turbine is a 240 kW b 361 kW c 414 kW d 602 kW e 777 kW 7237 Water enters a pump steadily at 100 kPa at a rate of 35 Ls and leaves at 800 kPa The flow velocities at the inlet and the exit are the same but the pump exit where the dis charge pressure is measured is 61 m above the inlet section The minimum power input to the pump is a 34 kW b 22 kW c 27 kW d 52 kW e 44 kW 7238 Air is to be compressed steadily and isentropically from 1 atm to 16 atm by a twostage compressor To minimize the total compression work the intermediate pressure between the two stages must be a 3 atm b 4 atm c 85 atm d 9 atm e 12 atm 7239 Helium gas enters an adiabatic nozzle steadily at 500C and 600 kPa with a low velocity and exits at a pressure of 90 kPa The highest possible velocity of helium gas at the nozzle exit is a 1475 ms b 1662 ms c 1839 ms d 2066 ms e 3040 ms 7240 Combustion gases with a specific heat ratio of 13 enter an adiabatic nozzle steadily at 800C and 800 kPa with a low velocity and exit at a pressure of 85 kPa The lowest pos sible temperature of combustion gases at the nozzle exit is a 43C b 237C c 367C d 477C e 640C 7241 Steam enters an adiabatic turbine steadily at 400C and 5 MPa and leaves at 20 kPa The highest possible percent age of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is a 4 b 8 c 12 d 18 e 0 7242 Liquid water enters an adiabatic piping system at 15C at a rate of 8 kgs If the water temperature rises by 02C during flow due to friction the rate of entropy generation in the pipe is a 23 WK b 55 WK c 68 WK d 220 WK e 443 WK 7243 Liquid water is to be compressed by a pump whose isentropic efficiency is 85 percent from 02 MPa to 5 MPa at a rate of 015 m3min The required power input to this pump is a 85 kW b 102 kW c 120 kW d 141 kW e 153 kW 7244 Steam enters an adiabatic turbine at 8 MPa and 500C at a rate of 18 kgs and exits at 02 MPa and 300C The rate of entropy generation in the turbine is a 0 kWK b 72 kWK c 21 kWK d 15 kWK e 17 kWK 7245 Helium gas is compressed steadily from 90 kPa and 25C to 800 kPa at a rate of 2 kgmin by an adiabatic compres sor If the compressor consumes 80 kW of power while operat ing the isentropic efficiency of this compressor is a 540 b 805 c 758 d 901 e 100 7246 Helium gas is compressed from 1 atm and 25C to a pressure of 10 atm adiabatically The lowest temperature of helium after compression is a 25C b 63C c 250C d 384C e 476C Design and Essay Problems 7247 Compressors powered by natural gas engines are increasing in popularity Several major manufacturing facili ties have already replaced the electric motors that drive their compressors with gasdriven engines in order to reduce their energy bills since the cost of natural gas is much lower than the cost of electricity Consider a facility that has a 130kW compressor that runs 4400 hyr at an average load factor of 06 Making reasonable assumptions and using unit costs for Final PDF to printer cen22672ch07323412indd 412 110617 0852 AM 412 ENTROPY natural gas and electricity at your location determine the potential cost savings per year by switching to gasdriven engines 7248 It is well known that the temperature of a gas rises while it is compressed as a result of the energy input in the form of compression work At high compression ratios the air tempera ture may rise above the autoignition temperature of some hydro carbons including some lubricating oil Therefore the presence of some lubricating oil vapor in highpressure air raises the pos sibility of an explosion creating a fire hazard The concentration of the oil within the compressor is usually too low to create a real danger However the oil that collects on the inner walls of exhaust piping of the compressor may cause an explosion Such explosions have largely been eliminated by using the proper lubricating oils carefully designing the equipment intercooling between compressor stages and keeping the system clean A compressor is to be designed for an industrial appli cation in Los Angeles If the compressor exit temperature is not to exceed 250C for safety reasons determine the maxi mum allowable compression ratio that is safe for all possible weather conditions for that area 7249 Identify the major sources of entropy generation in your house and propose ways of reducing them 7250 Obtain the following information about a power plant that is closest to your town the net power output the type and amount of fuel the power consumed by the pumps fans and other auxiliary equipment stack gas losses temperatures at several locations and the rate of heat rejection at the con denser Using these and other relevant data determine the rate of entropy generation in that power plant 7251 You are designing a closedsystem isentropic expansion process using an ideal gas that operates between the pressure limits of P1 and P2 The gases under consideration are hydrogen nitrogen air helium argon and carbon diox ide Which of these gases will produce the greatest amount of work Which will require the least amount of work in a com pression process 7252 In large gascompression stations for example on a natural gas pipeline the compression is done in several stages as in Fig P7252 At the end of each stage the compressed gas is cooled at constant pressure back to the temperature at the inlet of the compressor Consider a compression station that is to compress a gas say methane from P1 to P2 in N stages where each stage has an isentropic compressor coupled to a reversible isobaric cooling unit Determine the N1 interme diate pressures at the outlet of each stage of compression that minimize the total work required How does this work com pare to the work needed to do the entire compression with one isentropic compressor FIGURE P7252 P2 T1 P1 T1 P2 T2 Q P3 T1 P3 T3 Q P4 T4 Q Final PDF to printer cen22672ch08413474indd 413 110817 1156 AM 413 CHAPTER8 EX E R GY T he increased awareness that the worlds energy resources are limited has caused many countries to reexamine their energy policies and take drastic measures in eliminating waste It has also sparked inter est in the scientific community to take a closer look at the energy conversion devices and to develop new techniques to better utilize the existing limited resources The first law of thermodynamics deals with the quantity of energy and asserts that energy cannot be created or destroyed This law merely serves as a necessary tool for the bookkeeping of energy during a process and offers no challenges to the engineer The second law however deals with the quality of energy More specifically it is concerned with the degradation of energy during a process the entropy generation and the lost opportunities to do work and it offers plenty of room for improvement The second law of thermodynamics has proved to be a very powerful tool in the optimization of complex thermodynamic systems In this chapter we examine the performance of engineering devices in light of the second law of thermodynamics We start our discussions with the introduction of exergy also called availability which is the maximum useful work that could be obtained from the system at a given state in a specified environment and we continue with the reversible work which is the maximum useful work that can be obtained as a system undergoes a process between two specified states Next we discuss the irreversibility also called the exergy destruction or lost work which is the wasted work potential during a process as a result of irreversibilities and we define a secondlaw efficiency We then develop the exergy balance relation and apply it to closed systems and control volumes OBJECTIVES The objectives of Chapter 8 are to Examine the performance of engineering devices in light of the second law of thermodynamics Define exergy which is the maximum useful work that could be obtained from the system at a given state in a specified environment Define reversible work which is the maximum useful work that can be obtained as a system undergoes a process between two specified states Define the exergy destruction which is the wasted work potential during a process as a result of irreversibilities Define the secondlaw efficiency Develop the exergy balance relation Apply exergy balance to closed systems and control volumes Final PDF to printer 414 EXERGY cen22672ch08413474indd 414 110817 1156 AM 81 EXERGY WORK POTENTIAL OF ENERGY When a new energy source such as a geothermal well is discovered the first thing the explorers do is estimate the amount of energy contained in the source This information alone however is of little value in deciding whether to build a power plant on that site What we really need to know is the work potential of the sourcethat is the amount of energy we can extract as useful work The rest of the energy is eventually discarded as waste energy and is not worthy of our consideration Thus it would be very desirable to have a prop erty to enable us to determine the useful work potential of a given amount of energy at some specified state This property is exergy which is also called the availability or available energy The work potential of the energy contained in a system at a specified state is simply the maximum useful work that can be obtained from the system You will recall that the work done during a process depends on the initial state the final state and the process path That is Work f initial state process path final state In an exergy analysis the initial state is specified and thus it is not a vari able The work output is maximized when the process between two specified states is executed in a reversible manner as shown in Chap 7 Therefore all the irreversibilities are disregarded in determining the work potential Finally the system must be in the dead state at the end of the process to maximize the work output A system is said to be in the dead state when it is in thermodynamic equilibrium with the environment it is in Fig 81 At the dead state a system is at the temperature and pressure of its environment in thermal and mechanical equilibrium it has no kinetic or potential energy relative to the environment zero velocity and zero elevation above a reference level and it does not react with the environment chemically inert Also there are no unbalanced magnetic electrical or surface tension effects between the system and its surroundings if these are relevant to the situation at hand The properties of a system at the dead state are denoted by subscript zero for example P0 T0 h0 u0 and s0 Unless specified otherwise the deadstate temperature and pressure are taken to be T0 25C 77F and P0 1 atm 101325 kPa or 147 psia A system has zero exergy at the dead state Distinction should be made between the surroundings the immediate sur roundings and the environment By definition surroundings are everything outside the system boundaries The immediate surroundings refer to the portion of the surroundings that is affected by the process and environment refers to the region beyond the immediate surroundings whose properties are not affected by the process at any point Therefore any irreversibilities during a process occur within the system and its immediate surroundings and the environment is free of any irreversibilities When analyzing the cooling of a hot baked potato in a room at 25C for example the warm air that surrounds the potato is the immediate surroundings and the remaining part of the room air at 25C is the environment Note that the temperature of the immediate surroundings changes from the temperature of the potato at the boundary to the environment temperature of 25C Fig 82 FIGURE 81 A system that is in equilibrium with its environment is said to be at the dead state Air 25C 101 kPa V 0 z 0 T0 25C P0 101 kPa FIGURE 82 The immediate surroundings of a hot potato are simply the temperature gradient zone of the air next to the potato Hot potato 70C 25C 25C Environment Immediate surroundings Final PDF to printer 415 CHAPTER 8 cen22672ch08413474indd 415 110817 1156 AM The notion that a system must go to the dead state at the end of the pro cess to maximize the work output can be explained as follows If the system temperature at the final state is greater than or less than the temperature of the environment it is in we can always produce additional work by running a heat engine between these two temperature levels If the final pressure is greater than or less than the pressure of the environment we can still obtain work by letting the system expand to the pressure of the environment If the final velocity of the system is not zero we can catch that extra kinetic energy with a turbine and convert it to rotating shaft work and so on No work can be produced from a system that is initially at the dead state The atmosphere around us contains a tremendous amount of energy However the atmosphere is in the dead state and the energy it contains has no work potential Fig 83 Therefore we conclude that a system delivers the maximum possible work as it undergoes a reversible process from the specified initial state to the state of its environment that is the dead state This represents the use ful work potential of the system at the specified state and is called exergy It is important to realize that exergy does not represent the amount of work that a workproducing device will actually deliver upon installation Rather it represents the upper limit on the amount of work a device can deliver without violating any thermodynamic laws There will always be a difference large or small between exergy and the actual work deliv ered by a device This difference represents the room engineers have for improvement Note that the exergy of a system at a specified state depends on the conditions of the environment the dead state as well as the properties of the system Therefore exergy is a property of the systemenvironment combina tion and not of the system alone Altering the environment is another way of increasing exergy but it is definitely not an easy alternative The term availability was made popular in the United States by the MIT School of Engineering in the 1940s Today an equivalent term exergy introduced in Europe in the 1950s has found global acceptance partly because it is shorter it rhymes with energy and entropy and it can be adapted without requiring translation In this text the preferred term is exergy Exergy Work Potential Associated with Kinetic and Potential Energy Kinetic energy is a form of mechanical energy and thus it can be converted to work entirely Therefore the work potential or exergy of the kinetic energy of a system is equal to the kinetic energy itself regardless of the temperature and pressure of the environment That is Exergy of kinetic energy x ke ke V 2 2 kJkg 81 where V is the velocity of the system relative to the environment Potential energy is also a form of mechanical energy and thus it can be converted to work entirely Therefore the exergy of the potential energy of a FIGURE 83 The atmosphere contains a tremendous amount of energy but no exergy Design PicsDean MuzGetty Images RF Final PDF to printer 416 EXERGY cen22672ch08413474indd 416 110817 1156 AM system is equal to the potential energy itself regardless of the temperature and pressure of the environment Fig 84 That is Exergy of potential energy x pe pe gz kJkg 82 where g is the gravitational acceleration and z is the elevation of the system relative to a reference level in the environment Therefore the exergies of kinetic and potential energies are equal to themselves and they are entirely available for work However the internal energy u and enthalpy h of a system are not entirely available for work as shown later FIGURE 84 The work potential or exergy of potential energy is equal to the potential energy itself m z Wmax mgz EXAMPLE 81 Maximum Power Generation by a Wind Turbine A wind turbine with a 12mdiameter rotor as shown in Fig 85 is to be installed at a location where the wind is blowing steadily at an average velocity of 10 ms Determine the maximum power that can be generated by the wind turbine SOLUTION A wind turbine is being considered for a specified location The maximum power that can be generated by the wind turbine is to be determined Assumptions Air is at standard conditions of 1 atm and 25C and thus its density is 118 kgm3 Analysis The air flowing with the wind has the same properties as the stagnant atmo spheric air except that it possesses a velocity and thus some kinetic energy This air will reach the dead state when it is brought to a complete stop Therefore the exergy of the blowing air is simply the kinetic energy it possesses ke V 2 2 10 ms 2 2 1 kJkg 1000 m 2 s 2 005 kJkg That is every unit mass of air flowing at a velocity of 10 ms has a work potential of 005 kJkg In other words a perfect wind turbine will bring the air to a complete stop and capture that 005 kJkg of work potential To determine the maximum power we need to know the amount of air passing through the rotor of the wind turbine per unit time that is the mass flow rate which is determined to be m ρAV ρ π D 2 4 V 118 kgm 3 π 12 m 2 4 10 ms 1335 kgs Thus Maximum power m ke 1335 kgs005 kJkg 668 kW This is the maximum power available to the wind turbine Assuming a conversion efficiency of 30 percent an actual wind turbine will convert 200 kW to electricity Notice that the work potential for this case is equal to the entire kinetic energy of the air Discussion It should be noted that although the entire kinetic energy of the wind is available for power production Betzs law states that the power output of a wind machine is at maximum when the wind is slowed to onethird of its initial velocity Therefore for maximum power and thus minimum cost per installed power the highest efficiency of a FIGURE 85 Schematic for Example 81 10 ms Final PDF to printer 417 CHAPTER 8 cen22672ch08413474indd 417 110817 1156 AM wind turbine is about 59 percent In practice the actual efficiency ranges between 20 and 40 percent and is about 35 percent for many wind turbines Wind power is suitable for harvesting when there are steady winds with an average velocity of at least 6 ms or 13 mph Recent improvements in wind turbine design have brought the cost of generating wind power to about 5 cents per kWh which is competitive with electricity generated from other resources EXAMPLE 82 Exergy Transfer from a Furnace Consider a large furnace that can transfer heat at a temperature of 2000 R at a steady rate of 3000 Btus Determine the rate of exergy flow associated with this heat trans fer Assume an environment temperature of 77F SOLUTION Heat is being supplied by a large furnace at a specified temperature The rate of exergy flow is to be determined Analysis The furnace in this example can be modeled as a heat reservoir that supplies heat indefinitely at a constant temperature The exergy of this heat energy is its useful work potential that is the maximum possible amount of work that can be extracted from it This corresponds to the amount of work that a reversible heat engine operating between the furnace and the environment can produce The thermal efficiency of this reversible heat engine is η thmax η threv 1 T L T H 1 T 0 T H 1 537 R 2000 R 0732 or 732 That is a heat engine can convert at best 732 percent of the heat received from this furnace to work Thus the exergy of this furnace is equivalent to the power produced by the reversible heat engine W max W rev η threv Q in 07323000 Btus 2196 Btus Discussion Notice that 268 percent of the heat transferred from the furnace is not available for doing work The portion of energy that cannot be converted to work is called unavailable energy Fig 86 Unavailable energy is simply the difference between the total energy of a system at a specified state and the exergy of that energy FIGURE 86 Unavailable energy is the portion of energy that cannot be converted to work by even a reversible heat engine Total energy Exergy Unavailable energy 82 REVERSIBLE WORK AND IRREVERSIBILITY The property exergy serves as a valuable tool in determining the quality of energy and comparing the work potentials of different energy sources or systems The evaluation of exergy alone however is not sufficient for studying engineering devices operating between two fixed states This is because when evaluating exergy the final state is always assumed to be the dead state which is hardly ever the case for actual engineering systems The isentropic efficiencies discussed in Chap 7 are also of limited use because the exit state of the model isentropic process is not the same as the actual exit state and it is limited to adiabatic processes Final PDF to printer 418 EXERGY cen22672ch08413474indd 418 110817 1156 AM In this section we describe two quantities that are related to the actual initial and final states of processes and serve as valuable tools in the thermodynamic analysis of components or systems These two quantities are the reversible work and irreversibility or exergy destruction But first we examine the surroundings work which is the work done by or against the surroundings during a process The work done by workproducing devices is not always entirely in a usable form For example when a gas in a pistoncylinder device expands part of the work done by the gas is used to push the atmospheric air out of the way of the piston Fig 87 This work which cannot be recovered and utilized for any useful purpose is equal to the atmospheric pressure P0 times the volume change of the system W surr P 0 V 2 V 1 83 The difference between the actual work W and the surroundings work Wsurr is called the useful work Wu W u W W surr W P 0 V 2 V 1 84 When a system is expanding and doing work part of the work done is used to overcome the atmospheric pressure and thus Wsurr represents a loss When a system is compressed however the atmospheric pressure helps the com pression process and thus Wsurr represents a gain Note that the work done by or against the atmospheric pressure has sig nificance only for systems whose volume changes during the process ie systems that involve moving boundary work It has no significance for cyclic devices and systems whose boundaries remain fixed during a process such as rigid tanks and steadyflow devices turbines compressors nozzles heat exchangers etc as shown in Fig 88 Reversible work Wrev is defined as the maximum amount of useful work that can be produced or the minimum work that needs to be supplied as a system undergoes a process between the specified initial and final states This is the useful work output or input obtained or expended when the process between the initial and final states is executed in a totally reversible manner When the final state is the dead state the reversible work equals exergy For processes that require work reversible work represents the minimum amount of work necessary to carry out that process For convenience in presentation the term work is used to denote both work and power throughout this chapter Any difference between the reversible work Wrev and the useful work Wu is due to the irreversibilities present during the process and this difference is called irreversibility I It is expressed as Fig 89 I W revout W uout or I W uin W revin 85 The irreversibility is equivalent to the exergy destroyed which is discussed in Sec 86 For a totally reversible process the actual and reversible work terms are identical and thus the irreversibility is zero This is expected since totally reversible processes generate no entropy Irreversibility is a positive quan tity for all actual irreversible processes since Wrev Wu for workproducing devices and Wrev Wu for workconsuming devices FIGURE 87 As a closed system expands some work needs to be done to push the atmospheric air out of the way Wsurr Atmospheric air System V1 P0 Atmospheric air System V2 P0 FIGURE 88 For constantvolume systems the total actual and useful works are identical Wu W FIGURE 89 The difference between reversible work and actual useful work is the irreversibility Initial state Actual process Wu Wrev Reversible process Wrev Final state I Wrev Wu Final PDF to printer 419 CHAPTER 8 cen22672ch08413474indd 419 110817 1156 AM Irreversibility can be viewed as the wasted work potential or the lost oppor tunity to do work It represents the energy that could have been converted to work but was not The smaller the irreversibility associated with a process the greater the work that is produced or the smaller the work that is con sumed The performance of a system can be improved by minimizing the irreversibility associated with it EXAMPLE 83 The Rate of Irreversibility of a Heat Engine A heat engine receives heat from a source at 1200 K at a rate of 500 kJs and rejects the waste heat to a medium at 300 K Fig 810 The power output of the heat engine is 180 kW Determine the reversible power and the irreversibility rate for this process SOLUTION The operation of a heat engine is considered The reversible power and the irreversibility rate associated with this operation are to be determined Analysis The reversible power for this process is the amount of power that a reversible heat engine such as a Carnot heat engine would produce when operating between the same temperature limits and is determined to be W revout η threv Q in 1 T sink T source Q in 1 300 K 1200 K 500 kW 375 kW This is the maximum power that can be produced by a heat engine operating between the specified temperature limits and receiving heat at the specified rate This would also represent the available power if 300 K were the lowest temperature available for heat rejection The irreversibility rate is the difference between the reversible power maximum power that could have been produced and the useful power output I W revout W uout 375 180 195 kW Discussion Note that 195 kW of power potential is wasted during this process as a result of irreversibilities Also the 500 375 125 kW of heat rejected to the sink is not available for converting to work and thus is not part of the irreversibility FIGURE 810 Schematic for Example 83 HE W 180 kW Qin 500 kJs Sink 300 K Source 1200 K EXAMPLE 84 Irreversibility During the Cooling of an Iron Block A 500kg iron block shown in Fig 811 is initially at 200C and is allowed to cool to 27C by transferring heat to the surrounding air at 27C Determine the reversible work and the irreversibility for this process SOLUTION A hot iron block is allowed to cool in air The reversible work and irreversibility associated with this process are to be determined Assumptions 1 The kinetic and potential energies are negligible 2 The process involves no work interactions Analysis We take the iron block as the system This is a closed system since no mass crosses the system boundary We note that heat is lost from the system FIGURE 811 Schematic for Example 84 Surrounding air Iron 200C 27C T0 27C Heat Final PDF to printer 420 EXERGY cen22672ch08413474indd 420 110817 1156 AM It probably came as a surprise to you that we are asking to find the reversible work for a process that does not involve any work interactions Well even if no attempt is made to produce work during this process the potential to do work still exists and the reversible work is a quantitative measure of this potential The reversible work in this case is determined by considering a series of imaginary reversible heat engines operating between the source at a variable temperature T and the sink at a constant temperature T0 as shown in Fig 812 Summing their work output δ W rev η threv δ Q in 1 T sink T source δ Q in 1 T 0 T δ Q in and W rev 1 T 0 T δ Q in The source temperature T changes from T1 200C 473 K to T0 27C 300 K during this process A relation for the differential heat transfer from the iron block can be obtained from the differential form of the energy balance applied on the iron block δ E in δ E out Net energy transfer by heat work and mass d E system Change in internal kinetic potential etc energies δ Q out dU m c avg dT Then δ Q inheat engine δ Q outsystem m c avg dT since heat transfers from the iron and to the heat engine are equal in magnitude and opposite in direction Substituting and performing the integration the reversible work is determined to be W rev T 1 T 0 1 T 0 T m c avg dT m c avg T 1 T 0 m c avg T 0 ln T 1 T 0 500 kg045 kJkgK 476 300 K 300 K ln 473 K 300 K 8191 kJ where the specific heat value is obtained from Table A3 The first term in the preceding equation Q mcavgT1 T0 38925 kJ is the total heat transfer from the iron block to the heat engine The reversible work for this problem is found to be 8191 kJ which means that 8191 21 percent of the 38925 kJ of heat transferred from the iron block to the ambient air could have been converted to work If the specified ambient temperature of 27C is the lowest available environment temperature the reversible work determined above also represents the exergy which is the maximum work potential of the sensible energy contained in the iron block The irreversibility for this process is determined from its definition I W rev W u 8191 0 8191 kJ FIGURE 812 An irreversible heat transfer process can be made reversible by the use of a reversible heat engine Wrev Qin Surroundings 27C Rev HE Iron 200C 27C Final PDF to printer 421 CHAPTER 8 cen22672ch08413474indd 421 110817 1156 AM EXAMPLE 85 Heating Potential of a Hot Iron Block The iron block discussed in Example 84 is to be used to maintain a house at 27C when the outdoor temperature is 5C Determine the maximum amount of heat that can be supplied to the house as the iron cools to 27C SOLUTION The iron block is now reconsidered for heating a house The maxi mum amount of heating this block can provide is to be determined Analysis Probably the first thought that comes to mind to make the most use of the energy stored in the iron block is to take it inside and let it cool in the house as shown in Fig 813 transferring its sensible energy as heat to the indoors air provided that it meets the approval of the household of course The iron block can keep losing heat until its temperature drops to the indoor temperature of 27C transferring a total of 38925 kJ of heat Since we utilized the entire energy of the iron block available for heating without wasting a single kilojoule it seems like we have a 100percentefficient operation and nothing can beat this right Well not quite In Example 84 we determined that this process has an irreversibility of 8191 kJ which implies that things are not as perfect as they seem A perfect process is one that involves zero irreversibility The irreversibility in this process is associated with the heat transfer through a finite temperature difference that can be eliminated by running a reversible heat engine between the iron block and the indoor air This heat engine produces as determined in Example 84 8191 kJ of work and rejects the remaining 38925 8191 30734 kJ of heat to the house Now we managed to eliminate the irreversibility and ended up with 8191 kJ of work What can we do with this work Well at worst we can convert it to heat by running a paddle wheel for example creating an equal amount of irreversibility Or we can supply this work to a heat pump that transports heat from the outdoors at 5C to the indoors at 27C Such a heat pump if reversible has a coefficient of performance of COP HP 1 1 T L T H 1 1 278 K 300 K 136 That is this heat pump can supply the house with 136 times the energy it consumes as work In our case it will consume the 8191 kJ of work and deliver 8191 136 111398 kJ of heat to the house Therefore the hot iron block has the potential to supply 30734 111398 kJ 142132 kJ 142 MJ of heat to the house The irreversibility for this process is zero and this is the best we can do under the specified conditions A similar argument can be given for the electric heating of residential or commercial buildings Discussion Now try to answer the following question What would happen if the heat engine were operated between the iron block and the outside air instead of the house until the temperature of the iron block fell to 27C Would the amount of heat supplied to the house still be 142 MJ Here is a hint The initial and final states in both cases are the same and the irreversibility for both cases is zero FIGURE 813 Schematic for Example 85 5C 27C Heat Iron 200C Discussion Notice that the reversible work and irreversibility the wasted work poten tial are the same for this case since the entire work potential is wasted The source of irreversibility in this process is the heat transfer through a finite temperature difference Final PDF to printer 422 EXERGY cen22672ch08413474indd 422 110817 1156 AM 83 SECONDLAW EFFICIENCY In Chap 6 we defined the thermal efficiency and the coefficient of performance for devices as a measure of their performance They are defined on the basis of the first law only and they are sometimes referred to as the firstlaw effi ciencies The firstlaw efficiency however makes no reference to the best possible performance and thus it may be misleading Consider two heat engines both having a thermal efficiency of 30 percent as shown in Fig 814 One of the engines engine A is supplied with heat from a source at 600 K and the other one engine B from a source at 1000 K Both engines reject heat to a medium at 300 K At first glance both engines seem to convert to work the same fraction of heat that they receive thus they are performing equally well When we take a second look at these engines in light of the second law of thermodynamics however we see a totally different picture These engines at best can perform as reversible engines in which case their efficiencies would be η revA 1 T L T H A 1 300 K 600 K 050 or 50 η revB 1 T L T H B 1 300 K 1000 K 070 or 70 Now it appears that engine B has a greater work potential available to it 70 percent of the heat supplied as compared to 50 percent for engine A and thus it should do a lot better than engine A Therefore we can say that engine B is performing poorly relative to engine A even though both have the same thermal efficiency It is obvious from this example that the firstlaw efficiency alone is not a realistic measure of performance of engineering devices To overcome this deficiency we define a secondlaw efficiency ηII as the ratio of the actual thermal efficiency to the maximum possible reversible thermal efficiency under the same conditions Fig 815 η II η th η threv heat engines 86 Based on this definition the secondlaw efficiencies of the two heat engines discussed above are η IIA 030 050 060 and η IIB 030 070 043 That is engine A is converting 60 percent of the available work potential to useful work This ratio is only 43 percent for engine B The secondlaw efficiency can also be expressed as the ratio of the useful work output and the maximum possible reversible work output η II W u W rev workproducing devices 87 FIGURE 814 Two heat engines that have the same thermal efficiency but different maximum thermal efficiencies Sink 300 K Source 600 K A ηth 30 ηthmax 50 Sink 300 K Source 1000 K B ηth 30 ηthmax 70 FIGURE 815 Secondlaw efficiency is a measure of the performance of a device relative to its performance under reversible conditions 60 ηth 30 ηrev 50 ηII Final PDF to printer 423 CHAPTER 8 cen22672ch08413474indd 423 110817 1156 AM This definition is more general since it can be applied to processes in turbines pistoncylinder devices etc as well as to cycles Note that the secondlaw efficiency cannot exceed 100 percent Fig 816 We can also define a secondlaw efficiency for workconsuming noncyclic such as compressors and cyclic such as refrigerators devices as the ratio of the minimum reversible work input to the useful work input η II W rev W u workconsuming devices 88 For cyclic devices such as refrigerators and heat pumps it can also be expressed in terms of the coefficients of performance as η II COP COP rev refrigerators and heat pumps 89 Again because of the way we defined the secondlaw efficiency its value cannot exceed 100 percent In the preceding relations the reversible work Wrev should be determined by using the same initial and final states as in the actual process The definitions just given for the secondlaw efficiency do not apply to devices that are not intended to produce or consume work Therefore we need a more general definition However there is some disagreement on a general definition of the secondlaw efficiency and thus a person may encounter dif ferent definitions for the same device The secondlaw efficiency is intended to serve as a measure of approximation to reversible operation and thus its value should range from zero in the worst case complete destruction of exergy to 1 in the best case no destruction of exergy With this in mind we define the secondlaw efficiency of a system during a process as Fig 817 η II Exergy recovered Exergy expended 1 Exergy destroyed Exergy expended 810 Therefore when determining the secondlaw efficiency the first thing we need to do is determine how much exergy or work potential is expended or consumed during a process In a reversible operation we should be able to recover entirely the exergy expended during the process and the irreversibil ity in this case should be zero The secondlaw efficiency is zero when we recover none of the exergy expended by the system Note that the exergy can be supplied or recovered at various amounts in various forms such as heat work kinetic energy potential energy internal energy and enthalpy Some times there are differing though valid opinions on what constitutes expended exergy and this causes differing definitions for secondlaw efficiency At all times however the exergy recovered and the exergy destroyed the irrevers ibility must add up to the exergy expended Also we need to define the system precisely in order to identify correctly any interactions between the system and its surroundings For a heat engine the exergy expended is the decrease in the exergy of the heat transferred to the engine which is the difference between the exergy of FIGURE 816 Secondlaw efficiency of all reversible devices is 100 percent 100 ηth 70 ηrev 70 ηII Sink 300 K Source 1000 K FIGURE 817 The secondlaw efficiency of naturally occurring processes is zero if none of the work potential is recovered Hot water 80C Atmosphere 25C Heat Final PDF to printer 424 EXERGY cen22672ch08413474indd 424 110817 1156 AM the heat supplied and the exergy of the heat rejected The exergy of the heat rejected at the temperature of the surroundings is zero The net work output is the recovered exergy For a refrigerator or heat pump the exergy expended is the work input since the work supplied to a cyclic device is entirely consumed The recovered exergy is the exergy of the heat transferred to the hightemperature medium for a heat pump and the exergy of the heat transferred from the low temperature medium for a refrigerator For a heat exchanger with two unmixed fluid streams usually the exergy expended is the decrease in the exergy of the highertemperature fluid stream and the exergy recovered is the increase in the exergy of the lower temperature fluid stream This is discussed further in Sec 88 In the case of electric resistance heating the exergy expended is the electri cal energy the resistance heater consumes from the resource of the electric grid The exergy recovered is the exergy content of the heat supplied to the room which is the work that can be produced by a Carnot engine receiving this heat If the heater maintains the heated space at a constant temperature of TH in an environment at T0 the secondlaw efficiency for the electric heater becomes η IIelectric heater X recovered X expended X heat W e Q e 1 T 0 T H W e 1 T 0 T H 811 since from the firstlaw considerations Q e W e Note that the secondlaw efficiency of a resistance heater becomes zero when the heater is outdoors as in a radiant heater and thus the exergy of the heat supplied to the environ ment is not recoverable EXAMPLE 86 SecondLaw Efficiency of Resistance Heaters A dealer advertises that he has just received a shipment of electric resistance heaters for residential buildings that have an efficiency of 100 percent Fig 818 Assum ing an indoor temperature of 21C and outdoor temperature of 10C determine the secondlaw efficiency of these heaters SOLUTION Electric resistance heaters are being considered for residential build ings The secondlaw efficiency of these heaters is to be determined Analysis Obviously the efficiency that the dealer is referring to is the firstlaw effi ciency meaning that for each unit of electric energy work consumed the heater will sup ply the house with 1 unit of energy heat That is the advertised heater has a COP of 1 At the specified conditions a reversible heat pump would have a coefficient of performance of COP HPrev 1 1 T L T H 1 1 10 273 K 21 273 K 267 That is it would supply the house with 267 units of heat 257 of which is extracted from the cold outside air for each unit of electric energy it consumes FIGURE 818 Schematic for Example 86 10C 21C Resistance heater Final PDF to printer 425 CHAPTER 8 cen22672ch08413474indd 425 110817 1156 AM 84 EXERGY CHANGE OF A SYSTEM The property exergy is the work potential of a system in a specified environ ment and represents the maximum amount of useful work that can be obtained as the system is brought to equilibrium with the environment Unlike energy the value of exergy depends on the state of the environment as well as the state of the system Therefore exergy is a combination property The exergy of a system that is in equilibrium with its environment is zero The state of the environment is referred to as the dead state since the system is practically dead cannot do any work from a thermodynamic point of view when it reaches that state In this section we limit the discussion to thermomechanical exergy and thus disregard any mixing and chemical reactions Therefore a system at this restricted dead state is at the temperature and pressure of the environment and it has no kinetic or potential energies relative to the environment However it may have a different chemical composition than the environment Exergy associated with different chemical compositions and chemical reactions is discussed in later chapters Below we develop relations for the exergies and exergy changes for a fixed mass and a flow stream Exergy of a Fixed Mass Nonflow or Closed System Exergy In general internal energy consists of sensible latent chemical and nuclear energies However in the absence of any chemical or nuclear reactions the chemical and nuclear energies can be disregarded and the internal energy can be considered to consist of only sensible and latent energies that can be transferred to or from a system as heat whenever there is a temperature differ ence across the system boundary The second law of thermodynamics states that heat cannot be converted to work entirely and thus the work potential of internal energy must be less than the internal energy itself But how much less The secondlaw efficiency of this resistance heater is η II COP COP rev 10 267 0037 or 37 which does not look so impressive The dealer will not be happy to see this value Considering the high price of electricity a consumer will probably be better off with a less efficient gas heater Discussion The secondlaw efficiency of this electric heater can also be determined directly from Eq 811 to be η II 1 T 0 T H 1 10 273 K 21 273 K 0037 or 37 Therefore if we change our minds and decide to convert the heat back to electricity the best we can do is 37 percent That is 963 percent of the heat can never be con verted to electrical energy Final PDF to printer 426 EXERGY cen22672ch08413474indd 426 110817 1156 AM To answer that question we need to consider a stationary closed system at a specified state that undergoes a reversible process to the state of the envi ronment that is the final temperature and pressure of the system should be T0 and P0 respectively The useful work delivered during this process is the exergy of the system at its initial state Fig 819 Consider a pistoncylinder device that contains a fluid of mass m at tem perature T and pressure P The system the mass inside the cylinder has a volume V internal energy U and entropy S The system is now allowed to undergo a differential change of state during which the volume changes by a differential amount dV and heat is transferred in the differential amount of δQ Taking the direction of heat and work transfers to be from the system heat and work outputs the energy balance for the system during this differ ential process can be expressed as δ E in δ E out Net energy transfer by heat work and mass d E system Change in internal kinetic potential etc energies δQ δW dU 812 since the only form of energy the system contains is internal energy and the only forms of energy transfer a fixed mass can involve are heat and work Also the only form of work a simple compressible system can involve during a reversible process is the boundary work which is given to be δW P dV when the direc tion of work is taken to be from the system otherwise it would be P dV The pressure P in the P dV expression is the absolute pressure which is measured from absolute zero Any useful work delivered by a pistoncylinder device is due to the pressure above the atmospheric level Therefore δW P dV P P 0 dV P 0 dV δ W buseful P 0 dV 813 A reversible process cannot involve any heat transfer through a finite tempera ture difference and thus any heat transfer between the system at temperature T and its surroundings at T0 must occur through a reversible heat engine Noting that dS δQT for a reversible process in this case dS δQ T since δQ denotes the amount of heat leaving the system and the thermal efficiency of a reversible heat engine operating between the temperatures of T and T0 is ηth 1 T0T the differential work produced by the engine as a result of this heat transfer is δ W HE 1 T 0 T δQ δQ T 0 T δQ δQ T 0 dS δQ δ W HE T 0 dS 814 Substituting the δW and δQ expressions in Eqs 813 and 814 into the energy balance relation Eq 812 gives after rearranging δ W total useful δ W HE δ W buseful dU P 0 dV T 0 dS Integrating from the given state no subscript to the dead state 0 subscript we obtain W total useful U U 0 P 0 V V 0 T 0 S S 0 where Wtotal useful is the total useful work delivered as the system undergoes a revers ible process from the given state to the dead state which is exergy by definition FIGURE 819 The exergy of a specified mass at a specified state is the useful work that can be produced as the mass undergoes a reversible process to the state of the environment δWHE P0 P0 T0 P T δWbuseful δQ T0 Heat engine Final PDF to printer 427 CHAPTER 8 cen22672ch08413474indd 427 110817 1156 AM A closed system in general may possess kinetic and potential ener gies and the total energy of a closed system is equal to the sum of its internal kinetic and potential energies Noting that kinetic and potential energies themselves are forms of exergy the exergy of a closed system of mass m is X U U 0 P 0 V V 0 T 0 S S 0 m V 2 2 mgz 815 On a unit mass basis the closed system or nonflow exergy ϕ is expressed as ϕ u u 0 P 0 v v 0 T 0 s s 0 V 2 2 gz e e 0 P 0 v v 0 T 0 s s 0 816 where u0 v0 and s0 are the properties of the system evaluated at the dead state Note that the exergy of a system is zero at the dead state since e e0 v v0 and s s0 at that state The exergy change of a closed system during a process is simply the differ ence between the final and initial exergies of the system ΔX X 2 X 1 m ϕ 2 ϕ 1 E 2 E 1 P 0 V 2 V 1 T 0 S 2 S 1 U 2 U 1 P 0 V 2 V 1 T 0 S 2 S 1 m V 2 2 V 1 2 2 mg z 2 z 1 817 or on a unit mass basis Δϕ ϕ 2 ϕ 1 u 2 u 1 P 0 v 2 v 1 T 0 s 2 s 1 V 2 2 V 1 2 2 g z 2 z 1 e 2 e 1 P 0 v 2 v 1 T 0 s 2 s 1 For stationary closed systems the kinetic and potential energy terms drop out When the properties of a system are not uniform the exergy of the system can be determined by integration from X system ϕ δm V 1 ϕρ dV 819 where V is the volume of the system and ρ is density Note that exergy is a property and the value of a property does not change unless the state changes Therefore the exergy change of a system is zero if the state of the system or the environment does not change during the process For example the exergy change of steadyflow devices such as nozzles compressors turbines pumps and heat exchangers in a given environment is zero during steady operation The exergy of a closed system is either positive or zero It is never negative Even a medium at low temperature T T0 andor low pressure P P0 contains exergy since a cold medium can serve as the heat sink to a heat engine that absorbs heat from the environment at T0 and an evacuated space makes it possible for the atmospheric pressure to move a piston and do useful work Fig 820 818 FIGURE 820 The exergy of a cold medium is also a positive quantity since work can be produced by transferring heat to it Work output Heat engine Cold medium T 3C Atmosphere T0 25C Final PDF to printer 428 EXERGY cen22672ch08413474indd 428 110817 1156 AM Exergy of a Flow Stream Flow or Stream Exergy In Chap 5 it was shown that a flowing fluid has an additional form of energy called the flow energy which is the energy needed to maintain flow in a pipe or duct and this was expressed as wflow Pv where v is the specific volume of the fluid which is equivalent to the volume change of a unit mass of the fluid as it is displaced during flow The flow work is essentially the boundary work done by a fluid on the fluid downstream and thus the exergy associ ated with flow work is equivalent to the exergy associated with the boundary work which is the boundary work in excess of the work done against the atmospheric air at P0 to displace it by a volume v Fig 821 Noting that the flow work is Pv and the work done against the atmosphere is P0v the exergy associated with flow energy can be expressed as x flow Pv P 0 v P P 0 v 820 Therefore the exergy associated with flow energy is obtained by replacing the pressure P in the flow work relation with the pressure in excess of the atmospheric pressure P P0 Then the exergy of a flow stream is determined by simply adding the flow exergy relation above to the exergy relation in Eq 816 for a nonflowing fluid x flowing fluid x nonflowing fluid x flow u u 0 P 0 v v 0 T 0 s s 0 V 2 2 gz P P 0 v u Pv u 0 P 0 v 0 T 0 s s 0 V 2 2 gz h h 0 T 0 s s 0 V 2 2 gz The final expression is called flow or stream exergy and is denoted by ψ Fig 822 Flow exergy ψ h h 0 T 0 s s 0 V 2 2 gz 822 Then the exergy change of a fluid stream as it undergoes a process from state 1 to state 2 becomes Δψ ψ 2 ψ 1 h 2 h 1 T 0 s 2 s 1 V 2 2 V 1 2 2 g z 2 z 1 823 For fluid streams with negligible kinetic and potential energies the kinetic and potential energy terms drop out Note that the exergy change of a closed system or a fluid stream represents the maximum amount of useful work that can be done or the minimum amount of useful work that needs to be supplied if it is negative as the system changes from state 1 to state 2 in a specified environment and it represents the reversible work Wrev It is independent of the type of process executed the kind of system used and the nature of energy interactions with the surround ings Also note that the exergy of a closed system cannot be negative but the exergy of a flow stream can at pressures below the environment pressure P0 821 FIGURE 822 The energy and exergy contents of a a fixed mass and b a fluid stream V 2 2 ϕ u u0 P0v v0 T0s s0 gz V 2 2 ψ h h0 T0s s0 gz Energy Exergy a A fixed mass nonflowing b A fluid stream flowing e u gz V 2 2 Fixed mass Energy Exergy θ h gz V 2 2 Fluid stream FIGURE 821 The exergy associated with flow energy is the useful work that would be delivered by an imaginary piston in the flow section v P0 Pv P0v wshaft wshaft Flowing fluid Imaginary piston represents the fluid downstream Atmospheric air displaced P v Final PDF to printer 429 CHAPTER 8 cen22672ch08413474indd 429 110817 1156 AM EXAMPLE 87 Work Potential of Compressed Air in a Tank A 200m3 rigid tank contains compressed air at 1 MPa and 300 K Determine how much work can be obtained from this air if the environment conditions are 100 kPa and 300 K SOLUTION Compressed air stored in a large tank is considered The work potential of this air is to be determined Assumptions 1 Air is an ideal gas 2 The kinetic and potential energies are negligible Analysis We take the air in the rigid tank as the system Fig 823 This is a closed system since no mass crosses the system boundary during the process Here the question is the work potential of a fixed mass which is the nonflow exergy by definition Taking the state of the air in the tank to be state 1 and noting that T1 T0 300 K the mass of air in the tank is m 1 P 1 V R T 1 1000 kPa200 m 3 0287 kPam 3 kgK300 K 2323 kg The exergy content of the compressed air can be determined from X 1 m ϕ 1 m u 1 u 0 P 0 V 1 V 0 T 0 s 1 s 0 V 1 2 2 g z 1 m P 0 v 1 v 0 T 0 s 1 s 0 We note that P 0 v 1 v 0 P 0 R T 1 P 1 R T 0 P 0 R T 0 P 0 P 1 1 since T 1 T 0 T 0 s 2 s 0 T 0 c p ln T 1 T 0 R ln P 1 P 0 R T 0 ln P 1 P 0 since T 1 T 0 Therefore ϕ 1 R T 0 P 0 P 1 1 R T 0 ln P 1 P 0 R T 0 ln P 1 P 0 P 0 P 1 1 0287 kJ kgK 300 K ln 1000 kPa 100 kPa 100 kPa 1000 kPa 1 12076 kJ kg and X 1 m 1 ϕ 1 2323 kg 12076 kJ kg 280525 kJ 281 MJ Discussion The work potential of the system is 281 MJ and thus a maximum of 281 MJ of useful work can be obtained from the compressed air stored in the tank in the specified environment 0 0 0 FIGURE 823 Schematic for Example 87 Compressed air 1 MPa 100 kPa 300 K 300 K 200 m3 Final PDF to printer 430 EXERGY cen22672ch08413474indd 430 110817 1156 AM EXAMPLE 88 Exergy Change During a Compression Process Refrigerant134a is to be compressed from 014 MPa and 10C to 08 MPa and 50C steadily by a compressor Taking the environment conditions to be 20C and 95 kPa determine the exergy change of the refrigerant during this process and the minimum work input that needs to be supplied to the compressor per unit mass of the refrigerant SOLUTION Refrigerant134a is being compressed from a specified inlet state to a specified exit state The exergy change of the refrigerant and the minimum compres sion work per unit mass are to be determined Assumptions 1 Steady operating conditions exist 2 The kinetic and potential energies are negligible Analysis We take the compressor as the system Fig 824 This is a control volume since mass crosses the system boundary during the process Here the question is the exergy change of a fluid stream which is the change in the flow exergy ψ The properties of the refrigerant at the inlet and the exit states are Inlet state P 1 014 MPa T 1 10C h 1 24637 kJkg s 1 09724 kJkgK Exit state P 2 08 MPa T 2 50C h 2 28671 kJkg s 2 09803 kJkgK The exergy change of the refrigerant during this compression process is determined directly from Eq 823 to be Δψ ψ 2 ψ 1 h 2 h 1 T 0 s 2 s 1 V 2 2 V 1 2 2 g z 2 z 1 h 2 h 1 T 0 s 2 s 1 28671 24637 kJ kg 293 K 09803 09724 kJ kgK 380 kJ kg Therefore the exergy of the refrigerant increases during compression by 380 kJkg The exergy change of a system in a specified environment represents the revers ible work in that environment which is the minimum work input required for work consuming devices such as compressors Therefore the increase in exergy of the refrigerant is equal to the minimum work that needs to be supplied to the compressor w inmin ψ 2 ψ 1 380 kJ kg Discussion Note that if the compressed refrigerant at 08 MPa and 50C were to be expanded to 014 MPa and 10C in a turbine in the same environment in a reversible manner 380 kJkg of work would be produced 0 0 FIGURE 824 Schematic for Example 88 P1 014 MPa T1 10C T2 50C P2 08 MPa T0 20C Compressor 85 EXERGY TRANSFER BY HEAT WORK AND MASS Exergy like energy can be transferred to or from a system in three forms heat work and mass flow Exergy transfer is recognized at the system boundary as exergy crosses it and it represents the exergy gained or lost by a Final PDF to printer 431 CHAPTER 8 cen22672ch08413474indd 431 110817 1156 AM system during a process The only two forms of exergy interactions associated with a fixed mass or closed system are heat transfer and work Exergy Transfer by Heat Q Recall from Chap 6 that the work potential of the energy transferred from a heat source at temperature T is the maximum work that can be obtained from that energy in an environment at temperature T0 and is equivalent to the work produced by a Carnot heat engine operating between the source and the environment Therefore the Carnot efficiency ηCarnot 1 T0T represents the fraction of energy of a heat source at temperature T that can be converted to work Fig 825 For example only 70 percent of the energy transferred from a heat source at T 1000 K can be converted to work in an environment at T0 300 K Heat is a form of disorganized energy and thus only a portion of it can be converted to work which is a form of organized energy the second law We can always produce work from heat at a temperature above the environment temperature by transferring it to a heat engine that rejects the waste heat to the environment Therefore heat transfer is always accompanied by exergy trans fer Heat transfer Q at a location at thermodynamic temperature T is always accompanied by exergy transfer Xheat in the amount of Exergy transfer by heat X heat 1 T 0 T Q kJ 824 This relation gives the exergy transfer accompanying heat transfer Q whether T is greater than or less than T0 When T T0 heat transfer to a system increases the exergy of that system and heat transfer from a system decreases it But the opposite is true when T T0 In this case the heat transfer Q is the heat rejected to the cold medium the waste heat and it should not be con fused with the heat supplied by the environment at T0 The exergy transferred with heat is zero when T T0 at the point of transfer Perhaps you are wondering what happens when T T0 That is what if we have a medium that is at a lower temperature than the environment In this case it is conceivable that we can run a heat engine between the environment and the cold medium and thus a cold medium offers us an opportunity to produce work However this time the environment serves as the heat source and the cold medium as the heat sink In this case the relation above gives the negative of the exergy transfer associated with the heat Q transferred to the cold medium For example for T 100 K and a heat transfer of Q 1 kJ to the medium Eq 824 gives Xheat 1 3001001 kJ 2 kJ which means that the exergy of the cold medium decreases by 2 kJ It also means that this exergy can be recovered and the cold mediumenvironment combination has the potential to produce two units of work for each unit of heat rejected to the cold medium at 100 K That is a Carnot heat engine operating between T0 300 K and T 100 K produces two units of work while rejecting one unit of heat for each three units of heat it receives from the environment When T T0 the exergy and heat transfer are in the same direction That is both the exergy and energy content of the medium to which heat is transferred increase When T T0 cold medium however the exergy and heat transfer are in opposite directions That is the energy of the cold medium increases as a result of heat transfer but its exergy decreases The exergy of the cold FIGURE 825 The Carnot efficiency ηCarnot 1 T0T represents the fraction of the energy transferred from a heat source at temperature T that can be converted to work in an environment at temperature T0 Heat Source Temperature T Energy transferred E Exergy 1 T0 E T T0 Final PDF to printer 432 EXERGY cen22672ch08413474indd 432 110817 1156 AM medium eventually becomes zero when its temperature reaches T0 Equation 824 can also be viewed as the exergy associated with thermal energy Q at temperature T When the temperature T at the location where heat transfer is taking place is not constant the exergy transfer accompanying heat transfer is determined by integration to be X heat 1 1 1 T 0 T δQ 825 Note that heat transfer through a finite temperature difference is irreversible and some entropy is generated as a result The entropy generation is always accompanied by exergy destruction as illustrated in Fig 826 Also note that heat transfer Q at a location at temperature T is always accompanied by entropy transfer in the amount of QT and exergy transfer in the amount of 1 T0TQ Exergy Transfer by Work W Exergy is the useful work potential and the exergy transfer by work can sim ply be expressed as Exergy transfer by work X work W W surr for boundary work W for other forms of work 826 where Wsurr P0V2 V1 P0 is atmospheric pressure and V1 and V2 are the initial and final volumes of the system Therefore the exergy transfer with work such as shaft work and electrical work is equal to the work W itself In the case of a system that involves boundary work such as a piston cylinder device the work done to push the atmospheric air out of the way during expansion cannot be transferred and thus it must be sub tracted Also during a compression process part of the work is done by the atmospheric air and thus we need to supply less useful work from an external source To clarify this point further consider a vertical cylinder fitted with a weightless and frictionless piston Fig 827 The cylinder is filled with a gas that is maintained at the atmospheric pressure P0 at all times Heat is now transferred to the system and the gas in the cylinder expands As a result the piston rises and boundary work is done However this work cannot be used for any useful purpose since it is just enough to push the atmospheric air aside If we connect the piston to an external load to extract some useful work the pressure in the cylinder will have to rise above P0 to beat the resis tance offered by the load When the gas is cooled the piston moves down compressing the gas Again no work is needed from an external source to accomplish this compression process Thus we conclude that the work done by or against the atmosphere is not available for any useful purpose and it should be excluded from available work Exergy Transfer by Mass m Mass contains exergy as well as energy and entropy and the exergy energy and entropy contents of a system are proportional to mass Also the rates of exergy entropy and energy transport into or out of a system are proportional to FIGURE 826 The transfer and destruction of exergy during a heat transfer process through a finite temperature difference Medium 1 Medium 2 Wall Q Q Heat transfer T1 T2 Entropy transfer Entropy generated Q T1 Q T2 Exergy transfer Exergy destroyed 1 T0 Q T1 1 T0 Q T2 FIGURE 827 There is no useful work transfer associated with boundary work when the pressure of the system is maintained constant at atmospheric pressure Weightless piston P0 Heat P0 Final PDF to printer 433 CHAPTER 8 cen22672ch08413474indd 433 110817 1156 AM the mass flow rate Mass flow is a mechanism to transport exergy entropy and energy into or out of a system When mass in the amount of m enters or leaves a system exergy in the amount of mψ where ψ h h0 T0s s0 V 22 gz accompanies it That is Exergy transfer by mass X mass mψ 827 Therefore the exergy of a system increases by mψ when mass in the amount of m enters and it decreases by the same amount when the same amount of mass at the same state leaves the system Fig 828 Exergy flow associated with a fluid stream when the fluid properties are variable can be determined by integration from X mass A c ψρ V n d A c and X mass 1 1 ψ δm Δt X mass dt 828 where Ac is the crosssectional area of the flow and Vn is the local velocity normal to dAc Note that exergy transfer by heat Xheat is zero for adiabatic systems and the exergy transfer by mass Xmass is zero for systems that involve no mass flow across their boundaries ie closed systems The total exergy transfer is zero for isolated systems since they involve no heat work or mass transfer 86 THE DECREASE OF EXERGY PRINCIPLE AND EXERGY DESTRUCTION In Chap 2 we presented the conservation of energy principle and indicated that energy cannot be created or destroyed during a process In Chap 7 we established the increase of entropy principle which can be regarded as one of the statements of the second law and we indicated that entropy can be cre ated but cannot be destroyed That is entropy generation Sgen must be positive actual processes or zero reversible processes but it cannot be negative Now we are about to establish an alternative statement of the second law of thermodynamics called the decrease of exergy principle which is the counterpart of the increase of entropy principle Consider an isolated system shown in Fig 829 By definition no heat work or mass can cross the boundaries of an isolated system and thus there is no energy and entropy transfer Then the energy and entropy balances for an isolated system can be expressed as Energy balance E in E out Δ E system 0 E 2 E 1 Energy balance S in S out S gen Δ S system S gen S 2 S 1 Multiplying the second relation by T0 and subtracting it from the first one gives T 0 S gen E 2 E 1 T 0 S 2 S 1 829 From Eq 817 we have X 2 X 1 E 2 E 1 P 0 V 2 V 1 T 0 S 2 S 1 E 2 E 1 T 0 S 2 S 1 830 0 0 0 0 0 FIGURE 828 Mass contains energy entropy and exergy and thus mass flow into or out of a system is accompanied by energy entropy and exergy transfer Control volume h s m mh ms m ψ ψ FIGURE 829 The isolated system considered in the development of the decrease of exergy principle No heat work or mass transfer Isolated system ΔXisolated 0 or Xdestroyed 0 Final PDF to printer 434 EXERGY cen22672ch08413474indd 434 110817 1156 AM since V2 V1 for an isolated system it cannot involve any moving boundary and thus any boundary work Combining Eqs 829 and 830 gives T 0 S gen X 2 X 1 0 831 since T0 is the thermodynamic temperature of the environment and thus a positive quantity Sgen 0 and thus T0Sgen 0 Then we conclude that Δ X isolated X 2 X 1 isolated 0 832 This equation can be expressed as the exergy of an isolated system during a process always decreases or in the limiting case of a reversible process remains constant In other words it never increases and exergy is destroyed during an actual process This is known as the decrease of exergy principle For an isolated system the decrease in exergy equals exergy destroyed Exergy Destruction Irreversibilities such as friction mixing chemical reactions heat transfer through a finite temperature difference unrestrained expansion nonquasi equilibrium compression or expansion always generate entropy and any thing that generates entropy always destroys exergy The exergy destroyed is proportional to the entropy generated as can be seen from Eq 831 and is expressed as X destroyed T 0 S gen 0 833 Note that exergy destroyed is a positive quantity for any actual process and becomes zero for a reversible process Exergy destroyed represents the lost work potential and is also called the irreversibility or lost work Equations 832 and 833 for the decrease of exergy and the exergy destruc tion are applicable to any kind of system undergoing any kind of process since any system and its surroundings can be enclosed by a sufficiently large arbitrary boundary across which there is no heat work and mass transfer and thus any system and its surroundings constitute an isolated system No actual process is truly reversible and thus some exergy is destroyed dur ing a process Therefore the exergy of the universe which can be considered to be an isolated system is continuously decreasing The more irreversible a process is the larger the exergy destruction during that process No exergy is destroyed during a reversible process Xdestroyedrev 0 The decrease of exergy principle does not imply that the exergy of a system cannot increase The exergy change of a system can be positive or negative during a process Fig 830 but exergy destroyed cannot be negative The decrease of exergy principle can be summarized as follows X destroyed 0 irreversible process 0 reversible process 0 impossible process 834 This relation serves as an alternative criterion to determine whether a process is reversible irreversible or impossible FIGURE 830 The exergy change of a system can be negative but the exergy destruction cannot Surroundings System Xsys 2 kJ Xdest 1 kJ Q Final PDF to printer 435 CHAPTER 8 cen22672ch08413474indd 435 110817 1156 AM 87 EXERGY BALANCE CLOSED SYSTEMS The nature of exergy is opposite to that of entropy in that exergy can be destroyed but it cannot be created Therefore the exergy change of a system during a process is less than the exergy transfer by an amount equal to the exergy destroyed during the process within the system boundaries Then the decrease of exergy principle can be expressed as Fig 831 Total exergy entering Total exergy leaving Total exergy destroyed Change in the total exergy of the system or X in X out X destroyed Δ X system 835 This relation is referred to as the exergy balance and can be stated as the exergy change of a system during a process is equal to the difference between the net exergy transfer through the system boundary and the exergy destroyed within the system boundaries as a result of irreversibilities We mentioned earlier that exergy can be transferred to or from a system by heat work and mass transfer Then the exergy balance for any system undergoing any process can be expressed more explicitly as General X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy kJ 836 or in the rate form as General rate form X in X out Rate of net exergy transfer by heat work and mass X destroyed Rate of exergy destruction d X system dt Rate of change in exergy kW 837 where the rates of exergy transfer by heat work and mass are expressed as X heat 1 T 0 T Q X work W useful and X mass m ψ respectively The exergy balance can also be expressed per unit mass as General unitmass basis x in x out x destroyed Δ x system kJkg 838 where all the quantities are expressed per unit mass of the system Note that for a reversible process the exergy destruction term Xdestroyed drops out from all of the relations above Also it is usually more convenient to find the entropy generation Sgen first and then to evaluate the exergy destroyed directly from Eq 833 That is X destroyed T 0 S gen or X destroyed T 0 S gen 839 When the environment conditions P0 and T0 and the end states of the sys tem are specified the exergy change of the system ΔXsystem X2 X1 can be determined directly from Eq 817 regardless of how the process is executed However the determination of the exergy transfers by heat work and mass requires a knowledge of these interactions FIGURE 831 Mechanisms of exergy transfer System ΔXsystem Xdestroyed Xin Xout Mass Heat Work Mass Heat Work Final PDF to printer 436 EXERGY cen22672ch08413474indd 436 110817 1156 AM A closed system does not involve any mass flow and thus any exergy trans fer associated with mass flow Taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system the exergy balance for a closed system can be expressed more explic itly as Fig 832 Closed system X heat X work X destroyed Δ X system 840 or Closed system 1 T 0 T k Q k W P 0 V 2 V 1 T 0 S gen X 2 X 1 841 where Qk is the heat transfer through the boundary at temperature Tk at location k Dividing the previous equation by the time interval Δt and taking the limit as Δt 0 gives the rate form of the exergy balance for a closed system Rate form 1 T 0 T k Q k W P 0 d V system dt T 0 S gen d X system dt 842 Note that the relations above for a closed system are developed by taking the heat transfer to a system and work done by the system to be positive quanti ties Therefore heat transfer from the system and work done on the system should be taken to be negative quantities when using those relations The exergy balance relations presented above can be used to determine the reversible work Wrev by setting the exergy destruction term equal to zero The work W in that case becomes the reversible work That is W Wrev when Xdestroyed T0Sgen 0 Note that Xdestroyed represents the exergy destroyed within the system bound ary only and not the exergy destruction that may occur outside the system boundary during the process as a result of external irreversibilities Therefore a process for which Xdestroyed 0 is internally reversible but not necessarily totally reversible The total exergy destroyed during a process can be deter mined by applying the exergy balance to an extended system that includes the system itself and its immediate surroundings where external irreversibili ties might be occurring Fig 833 Also the exergy change in this case is equal to the sum of the exergy changes of the system and the exergy change of the immediate surroundings Note that under steady conditions the state and thus the exergy of the immediate surroundings the buffer zone at any point does not change during the process and thus the exergy change of the immediate surroundings is zero When evaluating the exergy transfer between an extended system and the environment the boundary temperature of the extended system is simply taken to be the environment temperature T0 For a reversible process the entropy generation and thus the exergy destruction are zero and the exergy balance relation in this case becomes analogous to the energy balance relation That is the exergy change of the system becomes equal to the exergy transfer Note that the energy change of a system equals the energy transfer for any process but the exergy change of a system equals the exergy transfer only for a reversible process The quantity of energy is always preserved during FIGURE 832 Exergy balance for a closed system when the direction of heat transfer is taken to be to the system and the direction of work from the system ΔXsystem Xdestroyed Q Xheat Xheat Xwork Xdestroyed ΔXsystem W Xwork FIGURE 833 Exergy destroyed outside system boundaries can be accounted for by writing an exergy balance on the extended system that includes the system and its immediate surroundings Immediate surroundings System Q T0 Outer surroundings environment T0 Final PDF to printer 437 CHAPTER 8 cen22672ch08413474indd 437 110817 1156 AM an actual process the first law but the quality is bound to decrease the second law This decrease in quality is always accompanied by an increase in entropy and a decrease in exergy When 10 kJ of heat is transferred from a hot medium to a cold one for example we still have 10 kJ of energy at the end of the process but at a lower temperature and thus at a lower quality and at a lower potential to do work EXAMPLE 89 General Exergy Balance for Closed Systems Starting with energy and entropy balances derive the general exergy balance relation for a closed system Eq 841 SOLUTION Starting with energy and entropy balance relations a general relation for exergy balance for a closed system is to be obtained Analysis We consider a general closed system a fixed mass that is free to exchange heat and work with its surroundings Fig 834 The system undergoes a process from state 1 to state 2 Taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system the energy and entropy balances for this closed system can be expressed as Energy balance E in E out Δ E system Q W E 2 E 1 S in S out S gen Δ S system 1 2 δQ T boundary S gen S 2 S 1 Multiplying the second relation by T0 and subtracting it from the first one gives Q T 0 1 2 δQ T boundary W T 0 S gen E 2 E 1 T 0 S 2 S 1 However the heat transfer for the process 12 can be expressed as Q 1 2 δQ and the right side of the preceding equation is from Eq 817 X2 X1 P0V2 V1 Thus 1 2 δQ T 0 1 2 δQ T boundary W T 0 S gen X 2 X 1 P 0 V 2 V 1 Letting Tb denote the boundary temperature and rearranging give 1 2 1 T 0 T b δQ W P 0 V 2 V 1 T 0 S gen X 2 X 1 843 which is equivalent to Eq 841 for the exergy balance except that the integration is replaced by summation in that equation for convenience This completes the proof Discussion Note that the exergy balance relation here is obtained by adding the energy and entropy balance relations and thus it is not an independent equation However it can be used in place of the entropy balance relation as an alternative secondlaw expression in exergy analysis Entropy balance FIGURE 834 A general closed system considered in Example 89 Closed system Q Tb W Final PDF to printer 438 EXERGY cen22672ch08413474indd 438 110817 1156 AM EXAMPLE 810 Exergy Destruction During Heat Conduction Consider steady heat transfer through a 5m 6m brick wall of a house of thickness 30 cm On a day when the temperature of the outdoors is 0C the house is maintained at 27C The temperatures of the inner and outer surfaces of the brick wall are mea sured to be 20C and 5C respectively and the rate of heat transfer through the wall is 1035 W Determine the rate of exergy destruction in the wall and the rate of total exergy destruction associated with this heat transfer process SOLUTION Steady heat transfer through a wall is considered For specified heat transfer rate wall surface temperatures and environmental conditions the rate of exergy destruction within the wall and the rate of total exergy destruction are to be determined Assumptions 1 The process is steady and thus the rate of heat transfer through the wall is constant 2 The exergy change of the wall is zero during this process since the state and thus the exergy of the wall do not change anywhere in the wall 3 Heat transfer through the wall is onedimensional Analysis We first take the wall as the system Fig 835 This is a closed system since no mass crosses the system boundary during the process We note that heat and exergy are entering from one side of the wall and leaving from the other side Applying the rate form of the exergy balance to the wall gives X in X out Rate of net exergy transfer by heat work and mass X destroyed Rate of exergy destruction d X system dt Rate of change in exergy 0 steady 0 Q 1 T 0 T in Q 1 T 0 T out X destroyed 0 1035 W 1 273 K 293 K 1035 W 1 273 K 298 K X destroyed 0 Solving the rate of exergy destruction in the wall is determined to be X destroyed 520 W Note that exergy transfer with heat at any location is 1 T0TQ at that location and the direction of exergy transfer is the same as the direction of heat transfer To determine the rate of total exergy destruction during this heat transfer process we extend the system to include the regions on both sides of the wall that experience a temperature change Then one side of the system boundary becomes room tempera ture while the other side becomes the temperature of the outdoors The exergy bal ance for this extended system system immediate surroundings is the same as that given above except the two boundary temperatures are 300 and 273 K instead of 293 and 278 K respectively Then the rate of total exergy destruction becomes X destroyedtotal 1035 W 1 273 K 300 K 1035 W 1 273 K 273 K 932 W The difference between the two exergy destructions is 412 W and represents the exergy destroyed in the air layers on both sides of the wall The exergy destruction in this case is entirely due to irreversible heat transfer through a finite temperature difference Discussion This problem was solved in Chap 7 for entropy generation We could have determined the exergy destroyed by simply multiplying the entropy generation by the environment temperature of T0 273 K FIGURE 835 Schematic for Example 810 27C 0C Q 30 cm Brick wall 20C 5C Final PDF to printer 439 CHAPTER 8 cen22672ch08413474indd 439 110817 1156 AM EXAMPLE 811 Exergy Destruction During Expansion of Steam A pistoncylinder device contains 005 kg of steam at 1 MPa and 300C Steam now expands to a final state of 200 kPa and 150C doing work Heat losses from the system to the surroundings are estimated to be 2 kJ during this process Assuming the surroundings to be at T0 25C and P0 100 kPa determine a the exergy of the steam at the initial and the final states b the exergy change of the steam c the exergy destroyed and d the secondlaw efficiency for the process SOLUTION Steam in a pistoncylinder device expands to a specified state The exergies of steam at the initial and final states the exergy change the exergy destroyed and the secondlaw efficiency for this process are to be determined Assumptions The kinetic and potential energies are negligible Analysis We take the steam contained within the pistoncylinder device as the system Fig 836 This is a closed system since no mass crosses the system boundary during the process We note that boundary work is done by the system and heat is lost from the system during the process a First we determine the properties of the steam at the initial and final states as well as the state of the surroundings State 1 P 1 1 MPa T 1 300C u 1 27937 kJ kg v 1 025799 m 3 kg s 1 71246 kJ kgK Table A6 State 2 P 2 200 kPa T 2 150C u 2 25771 kJ kg v 2 095986 m 3 kg s 2 72810 kJ kgK Table A6 Dead state P 0 100 kPa T 0 25C u 0 u f 25C 10483 kJ kg v 0 v f 25C 000103 m 3 kg s 0 s f 25C 03672 kJ kgK Table A4 The exergies of the system at the initial state X1 and the final state X2 are determined from Eq 815 to be X 1 m u 1 u 0 T 0 s 1 s 0 P 0 v 1 v 0 005 kg27937 10483 kJ kg 298 K71246 03672 kJ kgK 100 kPa025799 000103 m 3 kg kJ kPa m 3 350 kJ and X 2 m u 2 u 0 T 0 s 2 s 0 P 0 v 2 v 0 005 kg25771 10483 kJ kg 298 K72810 03672 kJ kgK 100 kPa095986 000103 m 3 kgkJ kPa m 3 254 kJ That is steam initially has an exergy content of 35 kJ which drops to 254 kJ at the end of the process In other words if the steam were allowed to undergo a reversible FIGURE 836 Schematic for Example 811 2 kJ P1 1 MPa T1 300C P2 200 kPa T2 150C Steam P0 100 kPa T0 25C State 1 State 2 Final PDF to printer 440 EXERGY cen22672ch08413474indd 440 110817 1156 AM process from the initial state to the state of the environment it would produce 35 kJ of useful work b The exergy change for a process is simply the difference between the exergy at the initial and final states of the process ΔX X 2 X 1 254 350 96 kJ That is if the process between states 1 and 2 were executed in a reversible manner the system would deliver 96 kJ of useful work c The total exergy destroyed during this process can be determined from the exergy balance applied on the extended system system immediate surroundings whose boundary is at the environment temperature of T0 so that there is no exergy transfer accompanying heat transfer to or from the environment X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy X workout X heatout X destroyed X 2 X 1 X destroyed X 2 X 1 W uout where Wuout is the useful boundary work delivered as the system expands By writing an energy balance on the system the total boundary work done during the process is determined to be E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies Q out W bout ΔU W bout Q out ΔU Q out m u 2 u 1 2 kJ 005 kg25771 27937 kJ kg 88 kJ This is the total boundary work done by the system including the work done against the atmosphere to push the atmospheric air out of the way during the expansion process The useful work is the difference between the two W u W W surr W bout P 0 V 2 V 1 W bout P 0 m v 2 v 1 88 kJ 100 kPa005 kg 09599 025799 m 3 kg 1 kJ 1 kPa m 3 53 kJ Substituting the exergy destroyed is determined to be X destroyed X 1 X 2 W uout 350 254 53 43 kJ That is 43 kJ of work potential is wasted during this process In other words an additional 43 kJ of energy could have been converted to work during this process but was not 0 Final PDF to printer 441 CHAPTER 8 cen22672ch08413474indd 441 110817 1156 AM The exergy destroyed could also be determined from X destroyed T 0 S gen T 0 m s 2 s 1 Q surr T 0 298 K 005 kg 72810 71246 kJ kgK 2 kJ 298 K 43 kJ which is the same result obtained before d Noting that the decrease in the exergy of the steam is the exergy expended and the useful work output is the exergy recovered the secondlaw efficiency for this process can be determined from η II Exergy recovered Exergy expended W u X 1 X 2 53 350 254 0552 or 552 That is 448 percent of the work potential of the steam is wasted during this process EXAMPLE 812 Exergy Destroyed During Stirring of a Gas An insulated rigid tank contains 2 lbm of air at 20 psia and 70F A paddle wheel inside the tank is now rotated by an external power source until the temperature in the tank rises to 130F Fig 837 If the surrounding air is at T0 70F determine a the exergy destroyed and b the reversible work for this process SOLUTION The air in an adiabatic rigid tank is heated by stirring it with a pad dle wheel The exergy destroyed and the reversible work for this process are to be determined Assumptions 1 Air at about atmospheric conditions can be treated as an ideal gas with constant specific heats at room temperature 2 The kinetic and potential energies are negligible 3 The volume of a rigid tank is constant and thus there is no boundary work 4 The tank is well insulated and thus there is no heat transfer Analysis We take the air contained within the tank as the system This is a closed system since no mass crosses the system boundary during the process We note that shaft work is done on the system a The exergy destroyed during a process can be determined from an exergy balance or directly from Xdestroyed T0Sgen We will use the second approach since it is usually easier But first we determine the entropy generated from an entropy balance S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy 0 S gen Δ S system m c v ln T 2 T 1 R ln V 2 V 1 S gen m c v ln T 2 T 1 0 FIGURE 837 Schematic for Example 812 Air m 2 lbm P1 20 psia T1 70F T0 70F Wpw Final PDF to printer 442 EXERGY cen22672ch08413474indd 442 110817 1156 AM Taking cv 0172 BtulbmF and substituting the exergy destroyed becomes X destroyed T 0 S gen T 0 m c v ln T 2 T 1 530 R 2 lbm 0172 Btu lbmF ln 590 R 530 R 196 Btu b The reversible work which represents the minimum work input Wrevin in this case can be determined from the exergy balance by setting the exergy destruction equal to zero since ΔKE ΔPE 0 and V2 V1 Noting that T0S2 S1 T0 ΔSsystem 196 Btu the reversible work becomes W revin m c v T 2 T 1 T 0 S 2 S 1 2 lbm 0172 Btu lbmF 130 70 F 196 Btu 206 196 Btu 10 Btu Therefore a work input of just 10 Btu would be sufficient to accomplish this process raise the temperature of air in the tank from 70 to 130F if all the irreversibilities were eliminated Discussion The solution is complete at this point However to gain some physical insight we will set the stage for a discussion First let us determine the actual work the paddlewheel work Wpw done during this process Applying the energy balance on the system E in E out Net energy transfer by heat work and mass Δ E system Change in internal kinetic potential etc energies W pwin ΔU 206 Btu from part b since the system is adiabatic Q 0 and involves no moving boundaries Wb 0 To put the information into perspective 206 Btu of work is consumed during the process 196 Btu of exergy is destroyed and the reversible work input for the process is 10 Btu What does all this mean It simply means that we could have created the same effect on the closed system raising its temperature to 130F at constant volume by consuming 10 Btu of work only instead of 206 Btu and thus saving 196 Btu of work from going to waste This would have been accomplished by a reversible heat pump To prove what we have just said consider a Carnot heat pump that absorbs heat from the surroundings at T0 530 R and transfers it to the air in the rigid tank until X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy W revin X 2 X 1 E 2 E 1 P 0 V 2 V 1 T 0 S 2 S 1 U 2 U 1 T 0 S 2 S 1 0 reversible 0 Final PDF to printer 443 CHAPTER 8 cen22672ch08413474indd 443 110817 1156 AM the air temperature T rises from 530 to 590 R as shown in Fig 838 The system involves no direct work interactions in this case and the heat supplied to the system can be expressed in differential form as δ Q H dU m c v dT The coefficient of performance of a reversible heat pump is given by COP HP δ Q H δ Q netin 1 1 T 0 T Thus δ W netin δ Q H COP HP 1 T 0 T m c v dT Integrating we get W netin 1 2 1 T 0 T m c v dT m c vavg T 2 T 1 T 0 m c vavg ln T 2 T 1 206 196 Btu 10 Btu The first term on the righthand side of the final expression is recognized as ΔU and the second term as the exergy destroyed whose values were determined earlier By substituting those values the total work input to the heat pump is determined to be 10 Btu proving our claim Notice that the system is still supplied with 206 Btu of energy all we did in the latter case was replace the 196 Btu of valuable work with an equal amount of useless energy captured from the surroundings Discussion It is also worth mentioning that the exergy of the system as a result of 206 Btu of paddlewheel work done on it has increased by 10 Btu only that is by the amount of the reversible work In other words if the system were returned to its initial state it would produce at most 10 Btu of work EXAMPLE 813 Exergy Analysis of Heating a Room with a Radiator A 50L electrical radiator containing heating oil is placed in a wellsealed 75m3 room Fig 839 Both the air in the room and the oil in the radiator are initially at the environment temperature of 6C Electricity with a rating of 24 kW is now turned on Heat is also lost from the room at an average rate of 075 kW The heater is turned off after some time when the temperatures of the room air and oil are measured to be 20C and 60C respectively Taking the density and the specific heat of oil to be 950 kgm3 and 22 kJkgC determine a how long the heater is kept on b the exergy destruction and c the secondlaw efficiency for this process SOLUTION An electrical radiator is placed in a room and it is turned on for a period of time The time period for which the heater was on the exergy destruction and the secondlaw efficiency are to be determined FIGURE 838 The same effect on the system can be accomplished by a reversible heat pump that consumes only 1 Btu of work Air 70F 130F Ambient air 70F Wnetin 1 Btu Reversible heat pump 196 Btu 206 Btu FIGURE 839 Schematic for Example 813 Room 6C Q Radiator Final PDF to printer 444 EXERGY cen22672ch08413474indd 444 110817 1156 AM Assumptions 1 Kinetic and potential energy changes are negligible 2 Air is an ideal gas with constant specific heats 3 The room is well sealed 4 Standard atmospheric pres sure of 1013 kPa is assumed Properties The properties of air at room temperature are R 0287 kPa m 3 kgK c p 1005 kJ kgK c v 0718 kJ kgK Table A2a The properties of oil are given to be ρ 950 kg m 3 c 22 kJ kgK Analysis a The masses of air and oil are m a P 1 V R T 1 1013 kPa 75 m 3 0287 kPa m 3 kgK 6 273 K 9488 kg m oil ρ V oil 950 kg m 3 0050 m 3 4750 kg An energy balance on the system can be used to determine time period for which the heater was kept on W in Q out Δt m c v T 2 T 1 a mc T 2 T 1 oil 24 075 kW Δt 9488 kg 0718 kJ kg C 20 6 C 4750 kg 22 kJ kg C 60 6 C Δt 3988 s 666 min b The pressure of the air at the final state is P a2 m a R T a2 V 9488 kg 0287 kPa m 3 kgK 20 273 K 75 m 3 1064 kPa The amount of heat transfer to the surroundings is Q out Q out Δt 075 kJ s 3988 s 2999 kJ The entropy generation is the sum of the entropy changes of air oil and the surroundings Δ S a m c p ln T 2 T 1 R ln P 2 P 1 9488 kg 1005 kJ kgK ln 20 273 K 6 273 K 0287 kJ kgK ln 1064 kPa 1013 kPa 3335 kJ K Δ S oil mc ln T 2 T 1 4750 kg 22 kJ kgK ln 60 273 K 6 273 K 1849 kJ K Δ S surr Q out T surr 2999 kJ 6 273 K 1075 kJ K S gen Δ S a Δ S oil Δ S surr 3335 1849 1075 3257 kJ K The exergy destruction is determined from X dest T 0 S gen 6 273 K 3257 kJ K 9088 kJ 909 MJ Final PDF to printer 445 CHAPTER 8 cen22672ch08413474indd 445 110817 1156 AM EXAMPLE 814 Work Potential of Heat Transfer Between Two Tanks Two constantvolume tanks each filled with 30 kg of air have temperatures of 900 K and 300 K Fig 840 A heat engine placed between the two tanks extracts heat from the hightemperature tank produces work and rejects heat to the lowtemperature tank Determine the maximum work that can be produced by the heat engine and the final temperatures of the tanks Assume constant specific heats at room temperature SOLUTION A heat engine operates between two tanks filled with air at different temperatures The maximum work that can be produced and the final temperature of the tanks are to be determined Assumptions Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is 0287 kPam3kgK Table A1 The constant volume specific heat of air at room temperature is cv 0718 kJkgK Table A2a Analysis For maximum work production the process must be reversible and thus the entropy generation must be zero We take the two tanks the heat source and heat sink and the heat engine as the system Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero the entropy balance can be expressed as S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy 0 S gen Δ S tanksource Δ S tanksink Δ S heat engine 0 Δ S tanksource Δ S tanksink or m c v ln T 2 T 1 mR ln V 2 V 1 source m c v ln T 2 T 1 mR ln V 2 V 1 sink 0 ln T 2 T 2 T 1A T 1B 0 T 2 2 T 1A T 1B 0 c The secondlaw efficiency may be defined in this case as the ratio of the exergy recovered to the exergy expended That is Δ X a m c v T 2 T 1 T 0 Δ S a 9488 kg 0718 kJ kg C 20 6 C 6 273 K 3335 kJ K 2316 kJ Δ X oil m c T 2 T 1 T 0 Δ S oil 4750 kg 22 kJ kg C 60 6 C 6 273 K 1849 kJK 4845 kJ η II X recovered X expended Δ X a Δ X oil W in Δt 2316 4845 kJ 24 kJ s 3998 s 00529 or 53 Discussion This is a highly irreversible process since the most valuable form of energy work is used to heat the room air Out of 9571 kJ of exergy expended by electrical work 9088 kJ is destroyed with a corresponding secondlaw efficiency of 53 percent FIGURE 840 Schematic for Example 814 Air 30 kg 300 K QH W QL Air 30 kg 900 K HE 0 0 0 0 Final PDF to printer 446 EXERGY cen22672ch08413474indd 446 110817 1156 AM where T1A and T1B are the initial temperatures of the source and the sink respectively and T2 is the common final temperature Therefore the final temperature of the tanks for maximum power production is T 2 T 1 A T 1 B 900 K 300 K 5196 K The energy balance Ein Eout ΔEsystem for the source and sink can be expressed as follows Source Q sourceout ΔU m c v T 2 T 1A Q sourceout m c v T 1A T 2 30 kg 0718 kJ kgK 900 5196 K 8193 kJ Sink Q sinkin m c v T 2 T 1B 30 kg 0718 kJ kgK 5196 300 K 4731 kJ Then the work produced in this case becomes W maxout Q H Q L Q sourceout Q sinkin 8193 4731 3462 kJ Discussion Note that 3462 kJ of the 8193 kJ heat transferred from the source can be converted to work and this is the best that can be done This corresponds to a first law efficiency of 34628193 0423 or 423 percent but to a secondlaw efficiency of 100 percent since the process involves no entropy generation and thus no exergy destruction FIGURE 841 Exergy is transferred into or out of a control volume by mass as well as heat and work transfer Surroundings Q Control volume XCV T mi ψi me ψe W Xwork Xheat 88 EXERGY BALANCE CONTROL VOLUMES The exergy balance relations for control volumes differ from those for closed systems in that they involve one more mechanism of exergy transfer mass flow across the boundaries As mentioned earlier mass possesses exergy as well as energy and entropy and the amounts of these three extensive prop erties are proportional to the amount of mass Fig 841 Again taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system the general exergy balance relations Eqs 836 and 837 can be expressed for a control volume more explicitly as X heat X work X massin X massout X destroyed X 2 X 1 CV 844 or 1 T 0 T k Q k W P 0 V 2 V 1 in mψ out mψ X destroyed X 2 X 1 CV It can also be expressed in the rate form as 1 T 0 T k Q k W P 0 d V CV dt in m ψ out m ψ X destroyed d X CV dt 846 845 Final PDF to printer 447 CHAPTER 8 cen22672ch08413474indd 447 110817 1156 AM The exergy balance relation above can be stated as the rate of exergy change within the control volume during a process is equal to the rate of net exergy transfer through the control volume boundary by heat work and mass flow minus the rate of exergy destruction within the boundaries of the control volume When the initial and final states of the control volume are specified the exergy change of the control volume is X2 X1 m2ϕ2 m1ϕ1 Exergy Balance for SteadyFlow Systems Most control volumes encountered in practice such as turbines compressors nozzles diffusers heat exchangers pipes and ducts operate steadily and thus they experience no changes in their mass energy entropy and exergy contents as well as their volumes Therefore dVCVdt 0 and dXCVdt 0 for such systems and the amount of exergy entering a steadyflow system in all forms heat work mass transfer must be equal to the amount of exergy leaving plus the exergy destroyed Then the rate form of the general exergy balance Eq 846 reduces for a steadyflow process to Fig 842 Steadyflow 1 T 0 T k Q k W in m ψ out m ψ X destroyed 0 847 For a singlestream oneinlet oneexit steadyflow device the relation above further reduces to Singlestream 1 T 0 T k Q k W m ψ 1 ψ 2 X destroyed 0 848 where the subscripts 1 and 2 represent inlet and exit states m is the mass flow rate and the change in the flow exergy is given by Eq 823 as ψ 1 ψ 2 h 1 h 2 T 0 s 1 s 2 V 1 2 V 2 2 2 g z 1 z 2 Dividing Eq 848 by m gives the exergy balance on a unitmass basis as 1 T 0 T k q k w ψ 1 ψ 2 x destroyed 0 kJkg 849 where q Q m and w W m are the heat transfer and work done per unit mass of the working fluid respectively For the case of an adiabatic singlestream device with no work interactions the exergy balance relation further simplifies to X destroyed m ψ 1 ψ 2 which indicates that the specific exergy of the fluid must decrease as it flows through a workfree adiabatic device or remain the same ψ2 ψ1 in the limiting case of a reversible process regardless of the changes in other properties of the fluid Reversible Work The exergy balance relations presented above can be used to determine the reversible work Wrev by setting the exergy destroyed equal to zero The work W in that case becomes the reversible work That is General W W rev when X destroyed 0 850 FIGURE 842 The exergy transfer to a steadyflow system is equal to the exergy transfer from it plus the exergy destruction within the system Steady flow system Xin Xout Heat Work Mass Heat Work Mass Xdestroyed Final PDF to printer 448 EXERGY cen22672ch08413474indd 448 110817 1156 AM For example the reversible power for a singlestream steadyflow device is from Eq 848 Singlestream W rev m ψ 1 ψ 2 1 T 0 T k Q k kW 851 which reduces for an adiabatic device to Adiabatic singlestream W rev m ψ 1 ψ 2 852 Note that the exergy destroyed is zero only for a reversible process and reversible work represents the maximum work output for workproducing devices such as turbines and the minimum work input for workconsuming devices such as compressors SecondLaw Efficiency of SteadyFlow Devices The secondlaw efficiency of various steadyflow devices can be determined from its general definition ηII Exergy recoveredExergy expended When the changes in kinetic and potential energies are negligible the secondlaw efficiency of an adiabatic turbine can be determined from η IIturb w out ψ 1 ψ 2 h 1 h 2 ψ 1 ψ 2 w out w revout or η IIturb 1 T 0 s gen ψ 1 ψ 2 853 where sgen s2 s1 For an adiabatic compressor with negligible kinetic and potential energies the secondlaw efficiency becomes η IIcomp ψ 2 ψ 1 w in ψ 2 ψ 1 h 2 h 1 w revin w in or η IIcomp 1 T 0 s gen h 2 h 1 854 where again sgen s2 s1 Note that in the case of turbine the exergy resource utilized is steam and the expended exergy is simply the decrease in the exergy of the steam The recovered exergy is the turbine shaft work In the case of compressor the exergy resource is mechanical work and the expended exergy is the work consumed by the compressor The recovered exergy in this case is the increase in the exergy of the compressed fluid For an adiabatic heat exchanger with two unmixed fluid streams Fig 843 the exergy expended is the decrease in the exergy of the hot stream and the exergy recovered is the increase in the exergy of the cold stream provided that the cold stream is not at a lower temperature than the surroundings Then the secondlaw efficiency of the heat exchanger becomes η IIHX m cold ψ 4 ψ 3 m hot ψ 1 ψ 2 or η IIHX 1 T 0 S gen m hot ψ 1 ψ 2 855 where S gen m hot s 2 s 1 m cold s 4 s 3 Perhaps you are wondering what happens if the heat exchanger is not adiabatic that is it is losing some heat to its surroundings at T0 If the temperature of the boundary the outer surface of the heat exchanger Tb is equal to T0 the definition above still holds except the entropy generation term needs to be modified if the second definition is used However if Tb T0 then the exergy of the lost heat at the boundary should be included in the recovered exergy η IIHX m cold ψ 4 ψ 3 Q loss 1 T 0 T b m hot ψ 1 ψ 2 1 T 0 S gen m hot ψ 1 ψ 2 856 FIGURE 843 A heat exchanger with two unmixed fluid streams T0 Hot stream Cold stream 1 2 3 4 Final PDF to printer 449 CHAPTER 8 cen22672ch08413474indd 449 110817 1156 AM where Tb is the temperature of the system boundary through which the lost heat crosses at a rate of Q loss Also S gen m hot s 2 s 1 m cold s 4 s 3 Q loss T b in this case Although no attempt is made in practice to utilize this exergy associated with lost heat and it is allowed to be destroyed the heat exchanger should not be held responsible for this destruction which occurs outside its boundaries If we are interested in the exergy destroyed during the process not just within the boundaries of the device then it makes sense to consider an extended system that includes the immediate surroundings of the device such that the boundaries of the new enlarged system are at T0 The secondlaw efficiency of the extended system reflects the effects of the irreversibilities that occur within and just outside the device An interesting situation arises when the temperature of the cold stream remains below the temperature of the surroundings at all times In that case the exergy of the cold stream actually decreases instead of increasing In such cases it is better to define the secondlaw efficiency as the ratio of the sum of the exergies of the outgoing streams to the sum of the exergies of the incom ing streams For an adiabatic mixing chamber where a hot stream 1 is mixed with a cold stream 2 forming a mixture 3 the exergy resource is the hot fluid Then the exergy expended is the exergy decrease of the hot fluid and the exergy recov ered is the exegy increase of the cold fluid Noting that state 3 is the common state of the mixture the secondlaw efficiency can be expressed as η IImix m cold ψ 3 ψ 2 m hot ψ 1 ψ 3 or η IImix 1 T 0 S gen m hot ψ 1 ψ 3 857 where S gen m hot m cold s 3 m hot s 1 m cold s 2 EXAMPLE 815 SecondLaw Analysis of a Steam Turbine Steam enters a turbine steadily at 3 MPa and 450C at a rate of 8 kgs and exits at 02 MPa and 150C Fig 844 The steam is losing heat to the surrounding air at 100 kPa and 25C at a rate of 300 kW and the kinetic and potential energy changes are negligible Determine a the actual power output b the maximum possible power output c the secondlaw efficiency d the exergy destroyed and e the exergy of the steam at the inlet conditions SOLUTION A steam turbine operating steadily between specified inlet and exit states is considered The actual and maximum power outputs the secondlaw effi ciency the exergy destroyed and the inlet exergy are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 ΔECV 0 and ΔXCV 0 2 The kinetic and potential energies are negligible Analysis We take the turbine as the system This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one exit and thus m 1 m 2 m Also heat is lost to the surrounding air and work is done by the system FIGURE 844 Schematic for Example 815 W T0 25C P0 100 kPa 3 MPa 450C 300 kW 02 MPa 150C Steam turbine Final PDF to printer 450 EXERGY cen22672ch08413474indd 450 110817 1156 AM The properties of the steam at the inlet and exit states and the state of the environment are Inlet state P 1 3 MPa T 1 450C h 1 33449 kJkg s 1 70856 kJkgK Table A6 Exit state P 2 02 MPa T 2 150C h 2 27691 kJkg s 2 72810 kJkgK Table A6 Dead state P 0 100 kPa T 0 25C h 0 h f 25C 10483 kJkg s 0 s f 25C 03672 kJkgK Table A4 a The actual power output of the turbine is determined from the rate form of the energy balance E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 steady 0 E in E out m h 1 W out Q out m h 2 since ke pe 0 W out m h 1 h 2 Q out 8 kgs33449 27691 kJkg 300 kW 4306 kW b The maximum power output reversible power is determined from the rate form of the exergy balance applied on the extended system system immediate surround ings whose boundary is at the environment temperature of T0 and by setting the exergy destruction term equal to zero Note that exergy transfer with heat is zero when the temperature at the point of trans fer is the environment temperature T0 Substituting W revout 8 kgs 33449 27691 kJ kg 298 K 70856 72810 kJ kgK 5072 kW c The secondlaw efficiency of a turbine is the ratio of the actual work delivered to the reversible work η II W out W in 4306 kW 5072 kW 0849 or 849 X in X out Rate of net exergy transfer by heat work and mass X destroyed Rate of exergy destruction d X system dt Rate of change in exergy 0 X in X out m ψ 1 W revout X heat m ψ 2 W revout m ψ 1 ψ 2 m h 1 h 2 T 0 s 1 s 2 Δ ke Δ pe 0reversible 0steady 0 0 0 Final PDF to printer 451 CHAPTER 8 cen22672ch08413474indd 451 110817 1156 AM EXAMPLE 816 Exergy Destroyed During Mixing of Fluid Streams Water at 20 psia and 50F enters a mixing chamber at a rate of 300 lbmmin where it is mixed steadily with steam entering at 20 psia and 240F The mixture leaves the chamber at 20 psia and 130F and heat is being lost to the surrounding air at T0 70F at a rate of 180 Btumin Fig 845 Neglecting the changes in kinetic and potential energies determine the reversible power and the rate of exergy destruction for this process SOLUTION Liquid water and steam are mixed in a chamber that is losing heat at a specified rate The reversible power and the rate of exergy destruction are to be determined Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus Δmcv 0 ΔEcv 0 and ΔScv 0 2 There are no work interactions involved 3 The kinetic and potential energies are negligible ke pe 0 Analysis We take the mixing chamber as the system Fig 845 This is a control volume since mass crosses the system boundary during the process We note that there are two inlets and one exit Under the stated assumptions and observations the mass and energy balances for this steadyflow system can be expressed in the rate form as follows Mass balance m in m out d m system dt 0 steady 0 m 1 m 2 m 3 Energy balance E in E out Rate of net energy transfer by heat work and mass d E system dt Rate of change in internal kinetic potential etc energies 0 E in E out m 1 h 1 m 2 h 2 m 3 h 3 Q out since W 0 ke pe 0 0 steady FIGURE 845 Schematic for Example 816 1 2 3 130F 240F 50F T0 70F Mixing chamber 20 psia 180 Btumin That is 151 percent of the work potential is wasted during this process d The difference between the reversible work and the actual useful work is the exergy destroyed which is determined to be X destroyed W revout W out 5072 4306 776 kW That is the potential to produce useful work is wasted at a rate of 776 kW during this process The exergy destroyed could also be determined by first calculating the rate of entropy generation S gen during the process e The exergy maximum work potential of the steam at the inlet conditions is simply the stream exergy and is determined from ψ 1 h 1 h 0 T 0 s 1 s 0 V 1 2 2 g z 1 h 1 h 0 T 0 s 1 s 0 33449 10483 kJ kg 298 K 70856 03672 kJkgK 1238 kJ kg That is not counting the kinetic and potential energies every kilogram of the steam entering the turbine has a work potential of 1238 kJ This corresponds to a power potential of 8 kgs1238 kJkg 9904 kW Obviously the turbine is converting 43069904 435 percent of the available work potential of the steam to work 0 0 Final PDF to printer 452 EXERGY cen22672ch08413474indd 452 110817 1156 AM Combining the mass and energy balances gives Q out m 1 h 1 m 2 h 2 m 1 m 2 h 3 The desired properties at the specified states are determined from the steam tables to be State 1 P 1 20 psia T 1 50F h 1 s 1 h f 50F s f 50F 1807 Btu lbm 003609 Btu lbmR State 2 P 2 20 psia T 2 240F h 2 s 2 11623 Btulbm 17406 BtulbmR State 3 P 3 20 psia T 3 130C h 3 s 3 h f 130C s f 130C 9799 Btulbm 008174 BtulbmR Substituting 180 Btumin 300 1807 m 2 11623 300 m 2 9799 Btumin which gives m 2 227 lbmmin The maximum power output reversible power is determined from the rate form of the exergy balance applied on the extended system system immediate surround ings whose boundary is at the environment temperature of T0 and by setting the exergy destruction term equal to zero X in X out Rate of net exergy transfer by heat work and mass X destroyed Rate of exergy destruction d X system dt Rate of change in exergy 0 steady 0 X in X out m 1 ψ 1 m 2 ψ 2 W revout X heat m 3 ψ 3 W revout m 1 ψ 1 m 2 ψ 2 m 3 ψ 3 Note that exergy transfer by heat is zero when the temperature at the point of transfer is the environment temperature T0 and the kinetic and potential energies are negli gible Therefore W revout m 1 h 1 T 0 s 1 m 2 h 2 T 0 s 2 m 3 h 3 T 0 s 3 300 lbm min1807 Btu lbm 530 R003609 Btu lbmR 227 lbm min11623 Btu lbm 530 R17406 Btu lbmR 3227 lbm min9799 Btu lbm 530 R018174 Btu lbmR 4588 Btu min That is we could have produced work at a rate of 4588 Btumin if we ran a heat engine between the hot and the cold fluid streams instead of allowing them to mix directly The exergy destroyed is determined from X destroyed W revout W u T 0 S gen 0 0 reversible 0 Final PDF to printer 453 CHAPTER 8 cen22672ch08413474indd 453 110817 1156 AM Thus X destroyed W revout 4588 Btu min since there is no actual work produced during the process Discussion The entropy generation rate for this process can be shown to be S gen 865 BtuminR Thus the exergy destroyed could also be determined from the sec ond part of the above equation X destroyed T 0 S gen 530 R865 Btu minR 4585 Btu min The slight difference between the two results is due to roundoff error EXAMPLE 817 Charging a Compressed Air Storage System A 200m3 rigid tank initially contains atmospheric air at 100 kPa and 300 K and is to be used as a storage vessel for compressed air at 1 MPa and 300 K Fig 846 Compressed air is to be supplied by a compressor that takes in atmospheric air at P0 100 kPa and T0 300 K Determine the minimum work requirement for this process SOLUTION Air is to be compressed and stored at high pressure in a large tank The minimum work required is to be determined Assumptions 1 Air is an ideal gas 2 The kinetic and potential energies are negligible 3 The properties of air at the inlet remain constant during the entire charging process Analysis We take the rigid tank combined with the compressor as the system This is a control volume since mass crosses the system boundary during the process We note that this is an unsteadyflow process since the mass content of the system changes as the tank is charged Also there is only one inlet and no exit The minimum work required for a process is the reversible work which can be determined from the exergy balance applied on the extended system system imme diate surroundings whose boundary is at the environment temperature of T0 so that there is no exergy transfer accompanying heat transfer to or from the environment and by setting the exergy destruction term equal to zero X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy X in X out X 2 X 1 W revin m 1 ψ 1 m 2 ϕ 2 m 1 ϕ 1 W revin m 2 ϕ 2 Note that ϕ1 ψ1 0 since the initial air in the tank and the air entering are at the state of the environment and the exergy of a substance at the state of the environment is zero The final mass of air and the exergy of the pressurized air in the tank at the end of the process are m 2 P 2 V R T 2 1000 kPa200 m 3 0287 kPa m 3 kgK300 K 2323 kg ϕ 2 u 2 u 0 0since T 2 T 0 P 0 v 2 v 0 T 0 s 2 s 0 V 2 2 2 g z 2 P 0 v 2 v 0 T 0 s 2 s 0 0 0 0 reversible 0 0 FIGURE 846 Schematic for Example 817 100 kPa 300 K Compressor Air V 200 m3 300 K 100 kPa 1 MPa Final PDF to printer 454 EXERGY cen22672ch08413474indd 454 110817 1156 AM Thermodynamics is a fundamental natural science that deals with various aspects of energy and even nontechnical people have a basic understanding of energy and the first law of thermodynamics since there is hardly any aspect of life that does not involve the transfer or transformation of energy in differ ent forms All the dieters for example base their lifestyle on the conservation of energy principle Although most people readily understand and accept the firstlaw aspects of thermodynamics they know less about the second law of thermodynamics and the secondlaw aspects are not fully appreciated even by people with technical backgrounds This causes some students to view the second law as something that is of theoretical interest rather than an impor tant and practical engineering tool As a result students show little interest in a detailed study of the second law of thermodynamics This is unfortunate because they end up with a onesided view of thermodynamics and miss the complete picture Many ordinary events can serve as excellent examples of thermodynamics concepts Next we will try to demonstrate the relevance of secondlaw con cepts such as exergy reversible work irreversibility and secondlaw efficiency to various aspects of daily life using examples with which even nontechnical people can identify Hopefully this will enhance our understanding and appre ciation of the second law of thermodynamics and encourage us to use it more TOPIC OF SPECIAL INTEREST SecondLaw Aspects of Daily Life This section can be skipped without a loss in continuity We note that P 0 v 2 v 0 P 0 R T 2 P 2 R T 0 P 0 R T 0 P 0 P 2 1 since T 2 T 0 T 0 s 2 s 0 T 0 c p ln T 2 T 0 R ln P 2 P 0 R T 0 ln P 2 P 0 since T 2 T 0 Therefore ϕ 2 R T 0 P 0 P 2 1 R T 0 ln P 2 P 0 R T 0 ln P 2 P 0 P 0 P 2 1 0287 kJ kgK 300 K ln 1000 kPa 100 kPa 100 kPa 1000 kPa 1 12076 kJ kg and W revin m 2 ϕ 2 2323 kg12076 kJ kg 280525 kJ 281 MJ Discussion Note that a minimum of 281 MJ of work input is required to fill the tank with compressed air at 300 K and 1 MPa In reality the required work input will be greater by an amount equal to the exergy destruction during the process Compare this to the result of Example 87 What can you conclude 0 Final PDF to printer 455 CHAPTER 8 cen22672ch08413474indd 455 110817 1156 AM often in technical and even nontechnical areas The critical reader is reminded that the concepts presented below are soft and difficult to quantize and that they are offered here to stimulate interest in the study of the second law of thermodynamics and to enhance our understanding and appreciation of it Secondlaw concepts are implicit in daily life Many successful people seem to make extensive use of them without even realizing it There is a growing awareness that quality plays as important a role as quantity in even ordinary activities The following appeared in an article in the Reno GazetteJournal on March 3 1991 Dr Held considers himself a survivor of the ticktock conspiracy About four years ago right around his 40th birthday he was putting in 21hour days working late working out taking care of his three children and getting involved in sports He got about four or five hours of sleep a night Now Im in bed by 930 and Im up by 6 he says I get twice as much done as I used to I dont have to do things twice or read things three times before I understand them This example illustrates that the problem is not how much time we have the first law but rather how effectively we use it the second law For a person to get more done in less time is no different than for a car to go more miles on less fuel In thermodynamics reversible work for a process is defined as the maximum useful work output or minimum work input for that process It is the useful work that a system would deliver or consume during a process between two specified states if that process were executed in a reversible perfect manner The difference between the reversible work and the actual useful work is due to imperfections and is called irreversibility the wasted work potential For the special case of the final state being the dead state or the state of the sur roundings the reversible work becomes a maximum and is called the exergy of the system at the initial state The irreversibility for a reversible or perfect process is zero The exergy of a person in daily life can be viewed as the best job that person can do under the most favorable conditions The reversible work in daily life on the other hand can be viewed as the best job a person can do under some speci fied conditions Then the difference between the reversible work and the actual work done under those conditions can be viewed as the irreversibility or the exergy destroyed In engineering systems we try to identify the major sources of irreversibilities and minimize them in order to maximize performance In daily life a person should do just that to maximize his or her performance The exergy of a person at a given time and place can be viewed as the maxi mum amount of work he or she can do at that time and place Exergy is cer tainly difficult to quantify because of the interdependence of our physical and intellectual capabilities The ability to perform physical and intellectual tasks simultaneously complicates things even further Schooling and training obvi ously increase ones exergy Aging decreases our physical exergy Unlike most mechanical things the exergy of human beings is a function of time and the physical andor intellectual exergy of a person goes to waste if it is not utilized at the time A barrel of oil loses nothing from its exergy if left unattended for 40 years However a person will lose much of his or her exergy during that period if he or she just sits back Final PDF to printer 456 EXERGY cen22672ch08413474indd 456 110817 1156 AM A hardworking laborer for example may make full use of his physical exergy but little use of his intellectual exergy That laborer for example could learn a foreign language or a science by listening to some educational CDs at the same time he is doing his physical work This is also true for people who spend considerable time in the car commuting to work It is hoped that some day we will be able to do exergy analysis for people and their activities Such an analysis will point the way for people to minimize their exergy destruction and get more done in less time Computers can perform several tasks at once Why shouldnt human beings be able to do the same Children are born with different levels of exergies talents in different areas Giving aptitude tests to children at an early age is simply an attempt to uncover the extent of their hidden exergies or talents The children are then directed to areas in which they have the greatest exergy As adults they are more likely to perform at high levels without stretching the limits if they are naturally fit to be in that area We can view the level of alertness of a person as his or her exergy for intel lectual affairs When a person is wellrested the degree of alertness and thus intellectual exergy is at a maximum and this exergy decreases with time as the person gets tired as illustrated in Fig 847 Different tasks in daily life require different levels of intellectual exergy and the difference between available and required alertness can be viewed as the wasted alertness or exergy destruction To minimize exergy destruction there should be a close match between avail able alertness and required alertness Consider a wellrested student who is planning to spend her next four hours studying and watching a twohourlong movie From the firstlaw point of view it makes no difference in what order these tasks are performed But from the secondlaw point of view it makes a lot of difference Of these two tasks studying requires more intellectual alertness than watching a movie does and thus it makes thermodynamic sense to study first when the alertness is high and to watch the movie later when the alertness is lower as shown in the fig ure A student who does it backwards wastes a lot of alertness while watching the movie as illustrated in Fig 847 and she has to keep going back and forth while studying because of insufficient alertness thus getting less done in the same time period In thermodynamics the firstlaw efficiency or thermal efficiency of a heat engine is defined as the ratio of net work output to total heat input That is it is the fraction of the heat supplied that is converted to net work In general the firstlaw efficiency can be viewed as the ratio of the desired output to the required input The firstlaw efficiency makes no reference to the best possible performance and thus the firstlaw efficiency alone is not a realistic measure of performance To overcome this deficiency we defined the secondlaw effi ciency which is a measure of actual performance relative to the best possible performance under the same conditions For heat engines the secondlaw effi ciency is defined as the ratio of the actual thermal efficiency to the maximum possible reversible thermal efficiency under the same conditions In daily life the firstlaw efficiency or performance of a person can be viewed as the accomplishment of that person relative to the effort he or she puts in The secondlaw efficiency of a person on the other hand can be FIGURE 847 The irreversibility associated with a student studying and watching a movie on television each for two hours Mental alertness Time h Wasted alertness irreversibility 0 2 4 Alertness required for studying Alertness required for watching TV Alertness required for watching TV Mental alertness Time h Wasted alertness irreversibility 0 2 4 Alertness required for studying a Studying first b Watching a movie first Final PDF to printer 457 CHAPTER 8 cen22672ch08413474indd 457 110817 1156 AM viewed as the performance of that person relative to the best possible perfor mance under the circumstances Happiness is closely related to secondlaw efficiency Small children are probably the happiest human beings because there is so little they can do but they do it so well considering their limited capabilities That is children have very high secondlaw efficiencies in their daily lives The term full life also refers to secondlaw efficiency A person is considered to have a full life and thus a very high secondlaw efficiency if he or she has used all of his or her abilities to the limit during a lifetime Even a person with some disabilities has to put in considerably more effort to do what a physically fit person does Yet despite doing less with more effort the person with disabilities who gives an impressive performance often gets more praise Thus we can say that this person with disabilities had a low first law efficiency achieving little with a lot of effort but a very high secondlaw efficiency achieving as much as possible under the circumstances In daily life exergy can also be viewed as the opportunities that we have and the exergy destruction as the opportunities wasted Time is the biggest asset that we have and the time wasted is the wasted opportunity to do something useful Fig 848 The second law of thermodynamics also has interesting philosophical rami fications Mass and energy are conserved quantities and are associated with the first law of thermodynamics while entropy and exergy are nonconserved quantities and are associated with the second law The universe we perceive through our five senses consists of conserved quantities and thus we tend to view the nonconserved quantities as being nonreal and even out of this uni verse The widely accepted big bang theory about the origin of the universe gave rise to the notion that this is an allmaterial universe and everything is made of matter more correctly massenergy only As conserved quantities mass and energy fit into the description of truly physical quantities but entropy and exergy do not since entropy can be created and exergy can be destroyed Thus entropy and exergy are not truly physical quantities although they are closely related to the physical quantities of mass and energy Therefore the second law deals with quantities that are of a different kind of existencea universe in which things come into existence out of nothing and go out of existence into nothingand opens up a universe that is beyond the conserved allmaterial universe we know of A similar argument can be given for the laws of nature that rule over matter There is no question that both the first and the second laws of thermodynamics exist and these and other laws like Newtons laws of motion govern the physi cal universe behind the scenes As Alfred Montapert put it Natures laws are the invisible government of the earth Albert Einstein expressed this phenom enon as A spirit is manifest in the laws of the universe Yet these laws that constitute the core of the sciences cannot be detected by our five senses they do not have a material existence and thus they are not subject to the limita tions of time and space As such the laws that seem to have infused all matter like a spirit rule everywhere but they are not anywhere It appears that quan tities like entropy and exergy which come into existence out of nothing and go out of existence into nothing together with the laws of nature like the first FIGURE 848 A poetic expression of exergy and exergy destruction anonymous I have only just a minute Only 60 seconds in it Forced upon mecant refuse it Didnt seek it didnt choose it But it is up to me to use it I must suffer if I lose it Give account if I abuse it Just a tiny little minute But eternity is in it Final PDF to printer 458 EXERGY cen22672ch08413474indd 458 110817 1156 AM and second laws that govern the bigbang universe with an invisible powerful hand are pointing the way to a broadened definition of existence that is more in line with perceived and observed phenomena The arguments presented here are exploratory they are intended to promote discussions and research that may lead to a better understanding of perfor mance in daily life The second law may eventually be used to determine quan titatively the most effective way to improve our performance and quality of life just as it is now used to improve the performance of engineering systems SUMMARY The energy content of the universe is constant just as its mass content is Yet at times of crisis we are bombarded with speeches and articles on how to conserve energy As engi neers we know that energy is already conserved What is not conserved is exergy which is the useful work potential of the energy Once the exergy is wasted it can never be recovered When we use energy to heat our homes for example we are not destroying any energy we are merely converting it to a less useful form a form of less exergy The useful work potential of a system at the specified state is called exergy Exergy is a property and is associated with the state of the system and the environment A system that is in equilibrium with its surroundings has zero exergy and is said to be at the dead state The exergy of heat supplied by thermal energy reservoirs is equivalent to the work output of a Carnot heat engine operating between the reservoir and the environment Reversible work Wrev is defined as the maximum amount of useful work that can be produced or the minimum work that needs to be supplied as a system undergoes a process between the specified initial and final states This is the use ful work output or input obtained when the process between the initial and final states is executed in a totally reversible manner The difference between the reversible work Wrev and the useful work Wu is due to the irreversibilities present during the process and is called the irreversibility I It is equivalent to the exergy destroyed and is expressed as I X destroyed T 0 S gen W revout W uout W uin W revin where Sgen is the entropy generated during the process For a totally reversible process the useful and reversible work terms are identical and thus exergy destruction is zero Exergy destroyed represents the lost work potential and is also called the wasted work or lost work The secondlaw efficiency is a measure of the performance of a device relative to the performance under reversible condi tions for the same end states and is given by η II η th η threv W u W rev for heat engines and other workproducing devices and η II COP COP rev W rev W u for refrigerators heat pumps and other workconsuming devices In general the secondlaw efficiency is expressed as η II Exergy recovered Exergy expended 1 Exergy destroyed exergy expended The exergies of a fixed mass nonflow exergy and of a flow stream are expressed as Nonflow exergy ϕ u u 0 P 0 v v 0 T 0 s s 0 V 2 2 gz e e 0 P 0 v v 0 T 0 s s 0 Flow exergy ψ h h 0 T 0 s s 0 V 2 2 gz Then the exergy change of a fixed mass or fluid stream as it undergoes a process from state 1 to state 2 is given by ΔX X 2 X 1 m ϕ 2 ϕ 1 E 2 E 1 P 0 V 2 V 1 T 0 S 2 S 1 U 2 U 1 P 0 V 2 V 1 T 0 S 2 S 1 m V 2 2 V 1 2 2 mg z 2 z 1 Δψ ψ 2 ψ 1 h 2 h 1 T 0 s 2 s 1 V 2 2 V 1 2 2 g z 2 z 1 Final PDF to printer 459 CHAPTER 8 cen22672ch08413474indd 459 110817 1156 AM Exergy can be transferred by heat work and mass flow and exergy transfer accompanied by heat work and mass transfer are given by X heat 1 T 0 T Q X work W W surr for boundary work W for other forms of work X mass mψ The exergy of an isolated system during a process always decreases or in the limiting case of a reversible process remains constant This is known as the decrease of exergy principle and is expressed as Δ X isolated X 2 X 1 isolated 0 Exergy balance for any system undergoing any process can be expressed as General X in X out Net exergy transfer by heat work and mass X destroyed Exergy destruction Δ X system Change in exergy X in X out Rate of net exergy transfer by heat work and mass X destroyed Rate of exergy destruction d X system dt Rate of change in exergy Exergy transfer by heat Exergy transfer by mass Exergy transfer by work x in x out x destroyed Δ x system where X heat 1 T 0 T Q X work W useful X mass m ψ For a reversible process the exergy destruction term Xdestroyed drops out Taking the positive direction of heat transfer to be to the system and the positive direction of work transfer to be from the system the general exergy balance relations can be expressed more explicitly as 1 T 0 T k Q k W P 0 V 2 V 1 in mψ out mψ X destroyed X 2 X 1 1 T 0 T k Q k W P 0 d V CV dt in mψ out mψ X destroyed d X CV dt General unitmass basis General rate form REFERENCES AND SUGGESTED READINGS 1 J E Ahern The Exergy Method of Energy Systems Analysis New York John Wiley Sons 1980 2 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley Interscience 2006 3 A Bejan Entropy Generation through Heat and Fluid Flow New York John Wiley Sons 1982 4 Y A Çengel A Unified and Intuitive Approach to Teaching Thermodynamics ASME International Congress and Exposition Atlanta Georgia November 1722 1996 Final PDF to printer cen22672ch08413474indd 460 110817 1156 AM 460 EXERGY PROBLEMS Exergy Irreversibility Reversible Work and SecondLaw Efficiency 81C What final state will maximize the work output of a device 82C Is the exergy of a system different in different environments 83C Under what conditions does the reversible work equal irreversibility for a process 84C How does useful work differ from actual work For what kinds of systems are these two identical 85C How does reversible work differ from useful work 86C Is a process during which no entropy is generated Sgen 0 necessarily reversible 87C Consider an environment of zero absolute pressure such as outer space How will the actual work and the useful work compare in that environment 88C It is well known that the actual work between the two specified states depends on the path followed during the pro cess Can we say the same for the reversible work 89C Consider two geothermal wells whose energy contents are estimated to be the same Will the exergies of these wells necessarily be the same Explain 810C Consider two systems that are at the same pressure as the environment The first system is at the same temperature as the environment whereas the second system is at a lower temperature than the environment How would you compare the exergies of these two systems 811C What is the secondlaw efficiency How does it dif fer from the firstlaw efficiency 812C Does a power plant that has a higher thermal effi ciency necessarily have a higher secondlaw efficiency than one with a lower thermal efficiency Explain 813C Does a refrigerator that has a higher COP necessar ily have a higher secondlaw efficiency than one with a lower COP Explain 814E Saturated steam is generated in a boiler by converting a saturated liquid to a saturated vapor at 200 psia This is done by transferring heat from the combustion gases which are at 700F to the water in the boiler tubes Calculate the wasted work potential associated with this heat transfer process How does increasing the temperature of the combustion gases affect the work potential of the steam stream Take T0 80F and P0 147 psia Answer 149 Btulbm FIGURE P814E Water 200 psia sat liq 200 psia sat vap q 815 One method of meeting the extra electric power demand at peak periods is to pump some water from a large body of water such as a lake to a reservoir at a higher eleva tion at times of low demand and to generate electricity at times of high demand by letting this water run down and rotate a turbine ie convert the electric energy to potential energy and then back to electric energy For an energy storage capac ity of 5 106 kWh determine the minimum amount of water that needs to be stored at an average elevation relative to the ground level of 75 m Answer 245 1010 kg FIGURE P815 h 75 m 816 A heat engine that receives heat from a furnace at 1200C and rejects waste heat to a river at 20C has a thermal efficiency of 40 percent Determine the secondlaw efficiency of this power plant 817 Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150000 kJh Determine the exergy of this supplied energy assuming an environment temperature of 25C 818 A heat engine receives heat from a source at 1100 K at a rate of 400 kJs and it rejects the waste heat to a medium at 320 K The measured power output of the heat engine is 120 kW and the environment temperature is 25C Determine Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch08413474indd 461 110817 1156 AM 461 CHAPTER 8 a the reversible power b the rate of irreversibility and c the secondlaw efficiency of this heat engine Answers a 284 kW b 164 kW c 423 percent 819 Reconsider Prob 818 Using appropriate soft ware study the effect of reducing the temperature at which the waste heat is rejected on the reversible power the rate of irreversibility and the secondlaw efficiency as the rejection temperature is varied from 500 to 298 K and plot the results 820E A heat engine that rejects waste heat to a sink at 510 R has a thermal efficiency of 25 percent and a secondlaw effi ciency of 50 percent Determine the temperature of the source that supplies heat to this engine Answer 1020 R FIGURE P820E 510 R Heat engine ηth 25 ηII 50 TH 821 A geothermal power plant uses geothermal liquid water at 150C at a rate of 210 kgs as the heat source and it produces 51 MW of net power in an environment at 25C If 75 MW of exergy entering the plant with the geothermal water is destroyed within the plant determine a the exergy of the geothermal water entering the plant b the secondlaw efficiency and c the exergy of the heat rejected from the plant 822 A house that is losing heat at a rate of 35000 kJh when the outside temperature drops to 4C is to be heated by electric resistance heaters If the house is to be maintained at 25C at all times determine the reversible work input for this process and the irreversibility Answers 0685 kW 904 kW 823E A freezer is maintained at 20F by removing heat from it at a rate of 75 Btumin The power input to the freezer is 070 hp and the surrounding air is at 75F Determine a the reversible power b the irreversibility and c the secondlaw efficiency of this freezer Answers a 020 hp b 050 hp c 289 percent 824 The electric power needs of a community are to be met by windmills with 40mdiameter rotors The windmills are to be located where the wind is blowing steadily at an average velocity of 6 ms Determine the minimum number of wind mills that need to be installed if the required power output is 1500 kW 825 Show that the power produced by a wind turbine is pro portional to the cube of the wind velocity and to the square of the blade span diameter 826 Two constantpressure devices each filled with 30 kg of air have temperatures of 900 K and 300 K A heat engine placed between the two devices extracts heat from the high temperature device produces work and rejects heat to the lowtemperature device Determine the maximum work that can be produced by the heat engine and the final tempera tures of the devices Assume constant specific heats at room temperature Exergy Analysis of Closed Systems 827C Can a system have a higher secondlaw efficiency than the firstlaw efficiency during a process Give examples 828 A mass of 8 kg of helium undergoes a process from an initial state of 3 m3kg and 15C to a final state of 05 m3 kg and 80C Assuming the surroundings to be at 25C and 100 kPa determine the increase in the useful work potential of the helium during this process 829E Which is a more valuable resource for work produc tion in a closed system 15 ft3 of air at 100 psia and 250F or 20 ft3 of helium at 60 psia and 200F Take T0 77F and P0 147 psia 830 Which has the capability to produce the most work in a closed system l kg of steam at 800 kPa and 180C or 1 kg of R134a at 800 kPa and 180C Take T0 25C and P0 100 kPa Answers 623 kJ steam 475 kJ R134a FIGURE P830 Steam 1 kg 800 kPa 180C R134a 1 kg 800 kPa 180C 831 The radiator of a steam heating system has a volume of 20 L and is filled with superheated water vapor at 200 kPa and 200C At this moment both the inlet and the exit valves to the radiator are closed After a while it is observed that the tem perature of the steam drops to 80C as a result of heat transfer to the room air which is at 21C Assuming the surround ings to be at 0C determine a the amount of heat transfer to the room and b the maximum amount of heat that can be Final PDF to printer 462 EXERGY cen22672ch08413474indd 462 110817 1156 AM supplied to the room if this heat from the radiator is supplied to a heat engine that is driving a heat pump Assume the heat engine operates between the radiator and the surroundings Answers a 303 kJ b 116 kJ FIGURE P831 Steam 20 L P1 200 kPa T1 200C Q 832 Reconsider Prob 831 Using appropriate soft ware investigate the effect of the final steam tem perature in the radiator on the amount of actual heat transfer and the maximum amount of heat that can be transferred Vary the final steam temperature from 80 to 21C and plot the actual and maximum heat transferred to the room as functions of final steam temperature 833E A wellinsulated rigid tank contains 6 lbm of a saturated liquidvapor mixture of water at 35 psia Initially threequarters of the mass is in the liquid phase An electric resistance heater placed in the tank is turned on and kept on until all the liquid in the tank is vaporized Assuming the surroundings to be at 75F and 147 psia determine a the exergy destruction and b the secondlaw efficiency for this process 834 A pistoncylinder device contains 8 kg of refrigerant 134a at 07 MPa and 60C The refrigerant is now cooled at constant pressure until it exists as a liquid at 20C If the sur roundings are at 100 kPa and 20C determine a the exergy of the refrigerant at the initial and the final states and b the exergy destroyed during this process 835 An insulated pistoncylinder device contains 0018 m3 of saturated refrigerant134a vapor at 06 MPa pressure The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 016 MPa Determine the change in the exergy of the refrigerant during this process and the reversible work Assume the surroundings to be at 25C and 100 kPa 836E A 12ft3 rigid tank contains refrigerant134a at 30 psia and 55 percent quality Heat is transferred now to the refrigerant from a source at 120F until the pressure rises to 50 psia Assuming the surroundings to be at 75F deter mine a the amount of heat transfer between the source and the refrigerant and b the exergy destroyed during this pro cess Answers a 447 Btu b 778 Btu 837E Oxygen gas is compressed in a pistoncylinder device from an initial state of 12 ft3lbm and 75F to a final state of 15 ft3lbm and 525F Determine the reversible work input and the increase in the exergy of the oxygen during this process Assume the surroundings to be at 147 psia and 75F 838 A pistoncylinder device initially contains 2 L of air at 100 kPa and 25C Air is now compressed to a final state of 600 kPa and 150C The useful work input is 12 kJ Assum ing the surroundings are at 100 kPa and 25C determine a the exergy of the air at the initial and the final states b the minimum work that must be supplied to accomplish this compression process and c the secondlaw efficiency of this process Answers a 0 0171 kJ b 0171 kJ c 143 percent FIGURE P838 V1 2 L P1 100 kPa T1 25C Air 839 A 08m3 insulated rigid tank contains 154 kg of carbon dioxide at 100 kPa Now paddlewheel work is done on the system until the pressure in the tank rises to 135 kPa Determine a the actual paddlewheel work done during this process and b the minimum paddlewheel work with which this process between the same end states could be accom plished Take T0 298 K Answers a 101 kJ b 718 kJ FIGURE P839 CO2 08 m3 154 kg 100 kPa 840 An insulated pistoncylinder device initially con tains 20 L of air at 140 kPa and 27C Air is now heated for 10 min by a 100W resistance heater placed inside the cylinder The pressure of air is kept constant during this process and the surroundings are at 27C and 100 kPa Determine the exergy destroyed during this process Answer 199 kJ 841 A rigid tank is divided into two equal parts by a parti tion One part of the tank contains 4 kg of compressed liquid Final PDF to printer 463 CHAPTER 8 cen22672ch08413474indd 463 110817 1156 AM water at 200 kPa and 80C and the other side is evacuated Now the partition is removed and the water expands to fill the entire tank If the final pressure in the tank is 40 kPa determine the exergy destroyed during this process Assume the surroundings to be at 25C and 100 kPa Answer 103 kJ 842 Reconsider Prob 841 Using appropriate soft ware study the effect of final pressure in the tank on the exergy destroyed during the process Plot the exergy destroyed as a function of the final pressure for final pressures between 45 and 5 kPa and discuss the results 843 An insulated rigid tank is divided into two equal parts by a partition Initially one part contains 3 kg of argon gas at 300 kPa and 70C and the other side is evacuated The parti tion is now removed and the gas fills the entire tank Assuming the surroundings to be at 25C determine the exergy destroyed during this process Answer 129 kJ 844 A 50kg iron block and a 20kg copper block both initially at 80C are dropped into a large lake at 15C Ther mal equilibrium is established after a while as a result of heat transfer between the blocks and the lake water Assuming the surroundings to be at 20C determine the amount of work that could have been produced if the entire process were executed in a reversible manner 845 Carbon steel balls ρ 7833 kgm3 and cp 0465 kJ kgC 8 mm in diameter are annealed by heating them first to 900C in a furnace and then allowing them to cool slowly to 100C in ambient air at 35C If 1200 balls are to be annealed per hour determine a the rate of heat transfer from the balls to the air and b the rate of exergy destruction due to heat loss from the balls to the air Answers a 260 W b 146 W FIGURE P845 Furnace 900C 100C Steel ball Air 35C 846E A 70lbm copper block initially at 220F is dropped into an insulated tank that contains 12 ft3 of water at 65F Determine a the final equilibrium temperature and b the work potential wasted during this process Assume the sur roundings to be at 65F 847 An ordinary egg can be approximated as a 55cm diameter sphere The egg is initially at a uniform temperature of 8C and is dropped into boiling water at 97C Taking the properties of egg to be ρ 1020 kgm3 and cp 332 kJkgC determine how much heat is transferred to the egg by the time the average temperature of the egg rises to 85C and the amount of exergy destruction associated with this heat transfer process Take T0 25C FIGURE P847 97C Boiling water Ti 8C Egg 848 A pistoncylinder device initially contains 14 kg of refrigerant134a at 100 kPa and 20C Heat is now transferred to the refrigerant from a source at 150C and the piston which is resting on a set of stops starts moving when the pressure inside reaches 120 kPa Heat transfer continues until the temperature reaches 80C Assuming the surroundings to be at 25C and 100 kPa determine a the work done b the heat transfer c the exergy destroyed and d the secondlaw efficiency of this process Answers a 0497 kJ b 679 kJ c 148 kJ d 262 percent FIGURE P848 Q R134a 14 kg 100 kPa 20C 150C 849 A 004m3 tank initially contains air at ambient condi tions of 100 kPa and 22C Now a 15L tank containing liquid water at 85C is placed into the tank without causing any air to escape After some heat transfer from the water to the air and the surroundings both the air and water are measured to be at 44C Determine a the amount of heat lost to the surround ings and b the exergy destruction during this process FIGURE P849 Q Air 22C Water 85C 15 L Final PDF to printer 464 EXERGY cen22672ch08413474indd 464 110817 1156 AM Exergy Analysis of Control Volumes 850 Steam is throttled from 8 MPa and 450C to 6 MPa Determine the wasted work potential during this throttling process Assume the surroundings to be at 25C Answer 366 kJkg 851 Refrigerant134a enters an expansion valve at 1200 kPa as a saturated liquid and leaves at 200 kPa Determine a the temperature of R134a at the outlet of the expansion valve and b the entropy generation and the exergy destruction during this process Take T0 25C 852 Air enters a nozzle steadily at 200 kPa and 65C with a velocity of 35 ms and exits at 95 kPa and 240 ms The heat loss from the nozzle to the surrounding medium at 17C is estimated to be 3 kJkg Determine a the exit temperature and b the exergy destroyed during this process Answers a 340C b 369 kJkg 853 Reconsider Prob 852 Using appropriate soft ware study the effect of varying the nozzle exit velocity from 100 to 300 ms on both the exit temperature and exergy destroyed and plot the results 854 An adiabatic steam nozzle has steam entering at 500 kPa 200C and 30 ms and leaving as a saturated vapor at 200 kPa Calculate the secondlaw efficiency of the nozzle Take T0 25C Answer 884 percent 855 Steam enters a diffuser at 10 kPa and 60C with a velocity of 375 ms and exits as saturated vapor at 50C and 70 ms The exit area of the diffuser is 3 m2 Determine a the mass flow rate of the steam and b the wasted work potential during this process Assume the surroundings to be at 25C 856 Argon gas enters an adiabatic compressor at 120 kPa and 30C with a velocity of 20 ms and exits at 12 MPa 530C and 80 ms The inlet area of the compressor is 130 cm2 Assuming the surroundings to be at 25C determine the reversible power input and exergy destroyed Answers 126 kW 412 kW 857E Air enters a compressor at 147 psia and 77F and is compressed to 140 psia and 200F Determine the minimum work required for this compression in Btulbm with the same inlet and outlet states Does the minimum work require an adiabatic compressor 858 Air is compressed by a compressor from 101 kPa and 27C to 400 kPa and 220C at a rate of 015 kgs Neglect ing the changes in kinetic and potential energies and assuming the surroundings to be at 25C determine the reversible power input for this process Answer 245 kW 859 Reconsider Prob 858 Using appropriate soft ware investigate the effect of compressor exit pressure on reversible power Vary the compressor exit pres sure from 200 to 600 kPa while keeping the exit temperature at 220C Plot the reversible power input for this process as a function of the compressor exit pressure 860 The adiabatic compressor of a refrigeration system compresses R134a from a saturated vapor at 160 kPa to 800 kPa and 50C What is the minimum power required by this compressor when its mass flow rate is 01 kgs Take T0 25C FIGURE P860 R134a 01 kgs 800 kPa 50 C 160 kPa sat vapor 861 Refrigerant134a at 140 kPa and 10C is compressed by an adiabatic 05kW compressor to an exit state of 700 kPa and 60C Neglecting the changes in kinetic and potential ener gies and assuming the surroundings to be at 27C determine a the isentropic efficiency and b the secondlaw efficiency of the compressor FIGURE P861 R134a 700 kPa 60C 140 kPa 10C 05 kW 862 Air enters a compressor at ambient conditions of 100 kPa and 20C at a rate of 62 m3s with a low velocity and exits at 900 kPa 60C and 80 ms The compressor is cooled by cooling water that experiences a temperature rise of 10C The isothermal efficiency of the compressor is 70 percent Determine a the actual and reversible power inputs b the secondlaw efficiency and c the mass flow rate of the cooling water 863 Combustion gases enter a gas turbine at 900C 800 kPa and 100 ms and leave at 650C 400 kPa and 220 ms Taking cp 115 kJkgC and k 13 for the combustion gases determine a the exergy of the combustion gases at the turbine inlet and b the work output of the turbine under Final PDF to printer 465 CHAPTER 8 cen22672ch08413474indd 465 110817 1156 AM reversible conditions Assume the surroundings to be at 25C and 100 kPa Can this turbine be adiabatic 864 Steam enters a turbine at 9 MPa 600C and 60 ms and leaves at 20 kPa and 90 ms with a moisture content of 5 percent The turbine is not adequately insulated and it estimated that heat is lost from the turbine at a rate of 220 kW The power output of the turbine is 45 MW Assuming the surroundings to be at 25C determine a the reversible power output of the turbine b the exergy destroyed within the turbine and c the secondlaw efficiency of the turbine d Also estimate the possible increase in the power output of the turbine if the tur bine were perfectly insulated FIGURE P864 Steam 9 MPa 600C 60 ms 20 kPa 90 ms x 095 Q Turbine 865 Refrigerant134a is condensed in a refrigeration sys tem by rejecting heat to ambient air at 25C R134a enters the condenser at 700 kPa and 50C at a rate of 005 kgs and leaves at the same pressure as a saturated liquid Determine a the rate of heat rejected in the condenser b the COP of this refrigeration cycle if the cooling load at these conditions is 6 kW and c the rate of exergy destruction in the condenser FIGURE P865 25C Condenser 700 kPa sat liq R134a 700 kPa 50C 005 kgs QH 866 Air enters the evaporator section of a window air conditioner at 100 kPa and 27C with a volume flow rate of 6 m3min Refrigerant134a at 120 kPa with a quality of 03 enters the evaporator at a rate of 2 kgmin and leaves as saturated vapor at the same pressure Determine the exit temperature of the air and the exergy destruction for this pro cess assuming a the outer surfaces of the air conditioner are insulated and b heat is transferred to the evaporator of the air conditioner from the surrounding medium at 32C at a rate of 30 kJmin 867E Refrigerant22 absorbs heat from a cooled space at 50F as it flows through an evaporator of a refrigeration sys tem R22 enters the evaporator at 10F at a rate of 008 lbms with a quality of 03 and leaves as a saturated vapor at the same pressure Determine a the rate of cooling provided in Btuh b the rate of exergy destruction in the evaporator and c the secondlaw efficiency of the evaporator Take T0 77F The properties of R22 at the inlet and exit of the evaporator are h1 1075 Btulbm s1 02851 BtulbmR h2 1721 Btu lbm s2 04225 BtulbmR FIGURE P867E 50F R22 T1 10F x1 03 P2 P1 x2 1 QL Evaporator 868 Steam expands in a turbine steadily at a rate of 18000 kgh entering at 7 MPa and 600C and leaving at 50 kPa as saturated vapor Assuming the surroundings to be at 100 kPa and 25C determine a the power potential of the steam at the inlet conditions and b the power output of the turbine if there were no irreversibilities present Answers a 7710 kW b 5775 kW 869 An adiabatic turbine operates with air entering at 550 kPa 425 K and 150 ms and leaving at 110 kPa 325 K and 50 ms Determine the actual and maximum work pro duction for this turbine in kJkg Why are the maximum and actual works not the same Take T0 25C 870E Air enters a compressor at ambient conditions of 15 psia and 60F with a low velocity and exits at 150 psia 620F and 350 fts The compressor is cooled by the ambient air at 60F at a rate of 1500 Btumin The power input to the compressor is 400 hp Determine a the mass flow rate of air and b the portion of the power input that is used just to over come the irreversibilities 871 Hot combustion gases enter the nozzle of a turbojet engine at 230 kPa 627C and 60 ms and exit at 70 kPa and 450C Assuming the nozzle to be adiabatic and the surround ings to be at 20C determine a the exit velocity and b the decrease in the exergy of the gases Take k 13 and cp 115 kJkgC for the combustion gases Final PDF to printer 466 EXERGY cen22672ch08413474indd 466 110817 1156 AM FIGURE P871 230 kPa 627C 60 ms Combustion gases 70 kPa 450C 872 Ambient air at 100 kPa and 300 K is compressed isen tropically in a steadyflow device to 08 MPa Determine a the work input to the compressor b the exergy of the air at the compressor exit and c the exergy of compressed air after it is cooled to 300 K at 08 MPa pressure 873 A 06m3 rigid tank is filled with saturated liquid water at 135C A valve at the bottom of the tank is now opened and onehalf of the total mass is withdrawn from the tank in liquid form Heat is transferred to water from a source of 210C so that the temperature in the tank remains constant Determine a the amount of heat transfer and b the reversible work and exergy destruction for this process Assume the surroundings to be at 25C and 100 kPa Answers a 1115 kJ b 126 kJ 126 kJ 874 How much exergy is lost in a rigid vessel filled with 1 kg of liquid R134a whose temperature remains constant at 30C as R134a vapor is released from the vessel This vessel may exchange heat with the surrounding atmosphere which is at 100 kPa and 30C The vapor is released until the last of the liquid inside the vessel disappears 875 A vertical pistoncylinder device initially contains 012 m3 of helium at 20C The mass of the piston is such that it maintains a constant pressure of 200 kPa inside A valve is now opened and helium is allowed to escape until the volume inside the cylinder is decreased by onehalf Heat transfer takes place between the helium and its surroundings at 20C and 95 kPa so that the temperature of helium in the cylinder remains constant Determine a the maximum work potential of the helium at the initial state and b the exergy destroyed during this process FIGURE P875 Helium 012 m3 20C 200 kPa Surroundings 20C 95 kPa Q 876 An insulated vertical pistoncylinder device initially contains 15 kg of water 13 kg of which is in the vapor phase The mass of the piston is such that it maintains a constant pressure of 300 kPa inside the cylinder Now steam at 2 MPa and 400C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder is vaporized Assuming the surroundings to be at 25C and 100 kPa determine a the amount of steam that has entered and b the exergy destroyed during this process Answers a 827 kg b 2832 kJ 877 Liquid water at 200 kPa and 15C is heated in a cham ber by mixing it with superheated steam at 200 kPa and 200C Liquid water enters the mixing chamber at a rate of 4 kgs and the chamber is estimated to lose heat to the surrounding air at 25C at a rate of 600 kJmin If the mixture leaves the mixing chamber at 200 kPa and 80C determine a the mass flow rate of the superheated steam and b the wasted work potential during this mixing process FIGURE P877 80C 200C 4 kgs 600 kJmin 15C 200 kPa Mixing chamber 878 Consider a family of four with each person taking a 6min shower every morning The average flow rate through the shower head is 10 Lmin City water at 15C is heated to 55C in an electric water heater and tempered to 42C by cold water at the Telbow of the shower before being routed to the shower head Determine the amount of exergy destroyed by this family per year as a result of taking daily showers Take T0 25C 879 Outdoor air cp 1005 kJkgC is to be preheated by hot exhaust gases in a crossflow heat exchanger before it enters the furnace Air enters the heat exchanger at 101 kPa and 30C at a rate of 05 m3s The combustion gases cp 110 kJkgC enter at 350C at a rate of 085 kgs and leave at 260C Determine the rate of heat transfer to the air and the rate of exergy destruction in the heat exchanger FIGURE P879 Air 101 kPa 30C 05 m3s Exhaust gases 085 kgs 260C Final PDF to printer 467 CHAPTER 8 cen22672ch08413474indd 467 110817 1156 AM 880 A wellinsulated shellandtube heat exchanger is used to heat water cp 418 kJkgC in the tubes from 20 to 70C at a rate of 45 kgs Heat is supplied by hot oil cp 230 kJ kgC that enters the shell side at 170C at a rate of 10 kgs Disregarding any heat loss from the heat exchanger deter mine a the exit temperature of oil and b the rate of exergy destruction in the heat exchanger Take T0 25C 881E Steam is to be condensed on the shell side of a heat exchanger at 120F Cooling water enters the tubes at 60F at a rate of 1153 lbms and leaves at 73F Assuming the heat exchanger to be well insulated determine a the rate of heat transfer in the heat exchanger and b the rate of exergy destruction in the heat exchanger Take T0 77F 882 A 01m3 rigid tank initially contains refrigerant134a at 12 MPa and 100 percent quality The tank is connected by a valve to a supply line that carries refrigerant134a at 16 MPa and 30C The valve is now opened allowing the refrigerant to enter the tank and it is closed when the tank contains only sat urated vapor at 14 MPa The refrigerant exchanges heat with a source at 200C during this process The surroundings are at 15C and 100 kPa Determine a the mass of the refrigerant that entered the tank and b the exergy destroyed during this process 883 A 02m3 rigid tank initially contains saturated refrig erant134a vapor at 1 MPa The tank is connected by a valve to a supply line that carries refrigerant134a at 14 MPa and 60C The valve is now opened and the refrigerant is allowed to enter the tank The valve is closed when onehalf of the vol ume of the tank is filled with liquid and the rest with vapor at 12 MPa The refrigerant exchanges heat during this process with the surroundings at 25C Determine a the amount of heat transfer and b the exergy destruction associated with this process 884 Derive an expression for the work potential of the sin glephase contents of a rigid adiabatic container when the ini tially empty container is filled through a single opening from a source of working fluid whose properties remain fixed Review Problems 885E A refrigerator has a secondlaw efficiency of 28 percent and heat is removed from the refrigerated space at a rate of 800 Btumin If the space is maintained at 25F while the surrounding air temperature is 90F determine the power input to the refrigerator 886 The inner and outer surfaces of a 05cmthick 2m 2m window glass in winter are 10C and 3C respectively If the rate of heat loss through the window is 44 kJs determine the amount of heat loss in kJ through the glass over a period of 5 h Also determine the exergy destruction associated with this process Take T0 5C 887 An aluminum pan has a flat bottom whose diameter is 30 cm Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 1100 W If the temperatures of the inner and outer surfaces of the bottom of the pan are 104C and 105C respectively determine the rate of exergy destruc tion within the bottom of the pan during this process in W Take T0 25C 888 A 5cmexternaldiameter 10mlong hot water pipe at 80C is losing heat to the surrounding air at 5C by natural convection at a rate of 1175 W Determine the rate at which the work potential is wasted during this process as a result of this heat loss 889 Steam is condensed in a closed system at a constant pressure of 75 kPa from a saturated vapor to a saturated liq uid by rejecting heat to a thermal energy reservoir at 37C Determine the secondlaw efficiency of this process Take T0 25C and P0 100 kPa 890 Refrigerant134a is converted from a saturated liq uid to a saturated vapor in a closed system using a reversible constantpressure process by transferring heat from a heat reservoir at 6C From a secondlaw point of view is it more effective to do this phase change at 100 kPa or 180 kPa Take T0 25C and P0 100 kPa FIGURE P890 R134a 100 kPa sat liquid q 891 A wellinsulated thinwalled counterflow heat exchanger is to be used to cool oil cp 220 kJkgC from 150 to 40C at a rate of 2 kgs with water cp 418 kJkgC that enters at 22C at a rate of 15 kgs The diameter of the tube is 25 cm and its length is 6 m Determine a the rate of heat transfer and b the rate of exergy destruction in the heat exchanger FIGURE P891 Hot oil 40C Cold water 15 kgs 22C 150C 2 kgs 892 A wellinsulated heat exchanger is to heat water cp 418 kJkgC from 25C to 60C at a rate of 04 kgs The heating is to be accomplished by geothermal water cp 431 kJkgC available at 140C at a mass flow rate of 03 kgs The inner tube is thinwalled and has a diameter of Final PDF to printer 468 EXERGY cen22672ch08413474indd 468 110817 1156 AM 06 cm Determine a the rate of heat transfer and b the rate of exergy destruction in the heat exchanger FIGURE P892 893 Hot exhaust gases leaving an internal combustion engine at 400C and 150 kPa at a rate of 08 kgs are to be used to produce saturated steam at 200C in an insulated heat exchanger Water enters the heat exchanger at the ambi ent temperature of 20C and the exhaust gases leave the heat exchanger at 350C Determine a the rate of steam production b the rate of exergy destruction in the heat exchanger and c the secondlaw efficiency of the heat exchanger FIGURE P893 Water 20C sat vap 200C Exhaust gases 400C 150 kPa 350C Heat exchanger 894 A crater lake has a base area of 20000 m2 and the water it contains is 12 m deep The ground surrounding the crater is nearly flat and is 105 m below the base of the lake Determine the maximum amount of electrical work in kWh that can be generated by feeding this water to a hydroelectric power plant Answer 72600 kWh 895 A 30cmlong 1500W electric resistance heating ele ment whose diameter is 12 cm is immersed in 70 kg of water initially at 20C Assuming the water container is well insu lated determine how long it will take for this heater to raise the water temperature to 80C Also determine the minimum work input required and the exergy destruction for this process in kJ Take T0 20C FIGURE P895 Heater Water 70 kg 896 Nitrogen gas enters a diffuser at 100 kPa and 110C with a velocity of 205 ms and leaves at 110 kPa and 45 ms It is estimated that 25 kJkg of heat is lost from the diffuser to the surroundings at 100 kPa and 27C The exit area of the diffuser is 004 m2 Accounting for the variation of the specific heats with temperature determine a the exit temperature b the rate of exergy destruction and c the secondlaw efficiency of the diffuser Answers a 127C b 124 kW c 761 percent 897 An adiabatic steam nozzle has steam entering at 300 kPa 150C and 45 ms and leaving as a saturated vapor at 150 kPa Calculate the actual and maximum outlet velocity Take T0 25C Answers 372 ms 473 ms 898 Steam enters an adiabatic nozzle at 35 MPa and 300C with a low velocity and leaves at 16 MPa and 250C at a rate of 04 kgs If the ambient state is 100 kPa and 18C deter mine a the exit velocity b the rate of exergy destruction and c the secondlaw efficiency 899 Two rigid tanks are connected by a valve Tank A is insulated and contains 02 m3 of steam at 400 kPa and 80 per cent quality Tank B is uninsulated and contains 3 kg of steam at 200 kPa and 250C The valve is now opened and steam flows from tank A to tank B until the pressure in tank A drops to 300 kPa During this process 900 kJ of heat is transferred from tank B to the surroundings at 0C Assuming the steam remaining inside tank A to have undergone a reversible adia batic process determine a the final temperature in each tank and b the work potential wasted during this process FIGURE P899 A 02 m3 Steam 400 kPa x 08 B 3 kg Steam 200 kPa 250C 8100E A pistoncylinder device initially contains 8 ft3 of helium gas at 40 psia and 70F Helium is now compressed in Final PDF to printer 469 CHAPTER 8 cen22672ch08413474indd 469 110817 1156 AM a polytropic process Pvn constant to 140 psia and 320F Assuming the surroundings to be at 147 psia and 70F deter mine a the actual useful work consumed and b the mini mum useful work input needed for this process Answers a 500 Btu b 463 Btu 8101 An adiabatic turbine operates with air entering at 550 kPa and 425 K and leaving at 110 kPa and 325 K Calcu late the secondlaw efficiency of this turbine Take T0 25C Answer 640 percent 8102 Steam at 7 MPa and 400C enters a twostage adia batic turbine at a rate of 15 kgs Ten percent of the steam is extracted at the end of the first stage at a pressure of 18 MPa for other use The remainder of the steam is further expanded in the second stage and leaves the turbine at 10 kPa If the tur bine has an isentropic efficiency of 88 percent determine the wasted power potential during this process as a result of irre versibilities Assume the surroundings to be at 25C 8103E Argon gas enters an adiabatic turbine at 1350F and 200 psia at a rate of 40 lbmmin and exhausts at 20 psia If the power output of the turbine is 105 hp determine a the isentropic efficiency and b the secondlaw efficiency of the turbine Assume the surroundings to be at 77F 8104 Steam enters a twostage adiabatic turbine at 8 MPa and 500C It expands in the first stage to a state of 2 MPa and 350C Steam is then reheated at constant pressure to a temper ature of 500C before it is routed to the second stage where it exits at 30 kPa and a quality of 97 percent The work output of the turbine is 5 MW Assuming the surroundings to be at 25C determine the reversible power output and the rate of exergy destruction within this turbine Answers 5457 kW 457 kW FIGURE P8104 8 MPa 500C 30 kPa x 97 Stage I Stage II Heat 2 MPa 500C 2 MPa 350C 5 MW 8105 To control an isentropic steam turbine a throttle valve is placed in the steam line leading to the turbine inlet Steam at 6 MPa and 600C is supplied to the throttle inlet and the turbine exhaust pressure is set at 40 kPa What is the effect on the stream exergy at the turbine inlet when the throttle valve is partially closed such that the pressure at the turbine inlet is 2 MPa Compare the secondlaw efficiency of this system when the valve is partially open to when it is fully open Take T0 25C FIGURE P8105 1 2 3 Turbine 8106 Consider a wellinsulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side Initially one side of the piston contains 1 m3 of N2 gas at 500 kPa and 80C while the other side contains 1 m3 of He gas at 500 kPa and 25C Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston Using constant specific heats at room temperature determine a the final equilibrium temperature in the cylinder and b the wasted work potential during this process What would your answer be if the piston were not free to move Take T0 25C FIGURE P8106 He 1 m3 500 kPa 25C N2 1 m3 500 kPa 80C 8107 Repeat Prob 8106 by assuming the piston is made of 5 kg of copper initially at the average temperature of the two gases on both sides 8108 One ton of liquid water at 65C is brought into a well insulated and wellsealed 3m 4m 7m room initially at 16C and 100 kPa Assuming constant specific heats for both the air and water at room temperature determine a the final Final PDF to printer 470 EXERGY cen22672ch08413474indd 470 110817 1156 AM equilibrium temperature in the room b the exergy destruc tion c the maximum amount of work that can be produced during this process in kJ Take T0 10C 8109 In large steam power plants the feedwater is often heated in closed feedwater heaters which are basically heat exchangers by steam extracted from the turbine at some stage Steam enters the feedwater heater at 16 MPa and 250C and leaves as saturated liquid at the same pressure Feedwater enters the heater at 4 MPa and 30C and leaves 10C below the exit temperature of the steam Neglecting any heat losses from the outer surfaces of the heater determine a the ratio of the mass flow rates of the extracted steam and the feedwater heater and b the reversible work for this process per unit mass of the feedwater Assume the surroundings to be at 25C Answers a 0333 b 110 kJkg FIGURE P8109 sat liquid Steam from turbine Feedwater 4 MPa 30C 16 MPa 250C 8110 Reconsider Prob 8109 Using appropriate software investigate the effect of the state of the steam at the inlet of the feedwater heater on the ratio of mass flow rates and the reversible power Vary the extracted steam pressure between 200 and 2000 kPa Plot both the ratio of the mass flow rates of the extracted steam and the feedwater heater and the reversible work for this process per unit mass of feed water as functions of the extraction pressure 8111 One method of passive solar heating is to stack gal lons of liquid water inside the buildings and expose them to the sun The solar energy stored in the water during the day is released at night to the room air providing some heating Con sider a house that is maintained at 22C and whose heating is assisted by a 270L water storage system If the water is heated to 45C during the day determine the amount of heating this water will provide to the house at night Assuming an outside temperature of 5C determine the exergy destruction associ ated with this process Answers 25900 kJ 904 kJ 8112 A passive solar house that was losing heat to the out doors at 5C at an average rate of 50000 kJh was maintained at 22C at all times during a winter night for 10 h The house was heated by 50 glass containers each containing 20 L of water that was heated to 80C during the day by absorbing solar energy A thermostatcontrolled 15kW backup electric resistance heater turned on whenever necessary to keep the house at 22C Determine a how long the electric heating system was on that night b the exergy destruction and c the minimum work input required for that night in kJ 8113 A 100L wellinsulated rigid tank is initially filled with nitrogen at 1000 kPa and 20C Now a valve is opened and onehalf of nitrogens mass is allowed to escape Deter mine the change in the exergy content of the tank 8114 A 4L pressure cooker has an operating pressure of 175 kPa Initially onehalf of the volume is filled with liquid water and the other half by water vapor The cooker is now placed on top of a 750W electrical heating unit that is kept on for 20 min Assuming the surroundings to be at 25C and 100 kPa determine a the amount of water that remained in the cooker and b the exergy destruction associated with the entire process Answers a 151 kg b 689 kJ FIGURE P8114 4 L 175 kPa 750 W 8115 Repeat Prob 8114 if heat were supplied to the pres sure cooker from a heat source at 180C instead of the electri cal heating unit 8116 Consider a 20L evacuated rigid bottle that is sur rounded by the atmosphere at 100 kPa and 25C A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle The air trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle The valve remains open during the process so that the trapped air Final PDF to printer 471 CHAPTER 8 cen22672ch08413474indd 471 110817 1156 AM also reaches mechanical equilibrium with the atmosphere Determine the net heat transfer through the wall of the bottle and the exergy destroyed during this filling process FIGURE P8116 20 L Evacuated 100 kPa 25C 8117 A rigid 50L nitrogen cylinder is equipped with a safety relief valve set at 1200 kPa Initially this cylinder con tains nitrogen at 1200 kPa and 20C Heat is now transferred to the nitrogen from a thermal energy reservoir at 500C and nitrogen is allowed to escape until the mass of nitrogen becomes onehalf of its initial mass Determine the change in the nitrogens work potential as a result of this heating Take T0 20C 8118 A frictionless pistoncylinder device shown in Fig P8118 initially contains 001 m3 of argon gas at 400 K and 350 kPa Heat is now transferred to the argon from a fur nace at 1200 K and the argon expands isothermally until its volume is doubled No heat transfer takes place between the argon and the surrounding atmospheric air which is at 300 K and 100 kPa Determine a the useful work output b the exergy destroyed and c the maximum work that can be pro duced during this process FIGURE P8118 Argon 400 K 350 kPa T0 300 K P0 100 kPa QR Furnace TR 1200 K 8119 A constantvolume tank contains 30 kg of nitrogen at 900 K and a constantpressure device contains 15 kg of argon at 300 K A heat engine placed between the tank and device extracts heat from the hightemperature tank produces work and rejects heat to the lowtemperature device Determine the maximum work that can be produced by the heat engine and the final temperatures of the nitrogen and argon Assume con stant specific heats at room temperature FIGURE P8119 Ar 15 kg 300 K QH W QL HE N2 30 kg 900 K 8120 The compressedair storage tank shown in Fig P8120 has a volume of 500000 m3 and it initially contains air at 100 kPa and 20C The isentropic compressor proceeds to compress air that enters the compressor at 100 kPa and 20C until the tank is filled at 600 kPa and 20C All heat exchanges are with the surrounding air at 20C Calculate the change in the work potential of the air stored in the tank How does this compare to the work required to compress the air as the tank is being filled FIGURE P8120 4 1 Storage tank Compressor Turbine 2 3 8121 Reconsider Prob 8120 The air stored in the tank is now released through the isentropic turbine until the tank contents are at 100 kPa and 20C The pressure is always 100 kPa at the turbine outlet and all heat exchanges are with Final PDF to printer 472 EXERGY cen22672ch08413474indd 472 110817 1156 AM the surrounding air which is at 20C How does the total work produced by the turbine compare to the change in the work potential of the air in the storage tank 8122E In a production facility 15inthick 1ft 3ft square brass plates ρ 5325 lbmft3 and cp 0091 Btu lbmF that are initially at a uniform temperature of 75F are heated by passing them through an oven at 1300F at a rate of 175 per minute If the plates remain in the oven until their average temperature rises to 1000F determine the rate of heat transfer to the plates in the furnace and the rate of exergy destruction associated with this heat transfer process FIGURE P8122E Oven 1300F Brass plate 75F 15 in 8123 Long cylindrical steel rods ρ 7833 kgm3 and cp 0465 kJkgC of 10cm diameter are heattreated by drawing them at a velocity of 3 mmin through a 6mlong oven maintained at 900C If the rods enter the oven at 30C and leave at 700C determine a the rate of heat transfer to the rods in the oven and b the rate of exergy destruction asso ciated with this heat transfer process Take T0 25C 8124 In a dairy plant milk at 4C is pasteurized continuously at 72C at a rate of 12 Ls for 24 hday and 365 daysyr The milk is heated to the pasteurizing temperature by hot water heated in a natural gasfired boiler having an efficiency of 82 percent The pasteurized milk is then cooled by cold water at 18C before it is finally refrigerated back to 4C To save energy and money the plant installs a regenerator that has an effectiveness of 82 percent If the cost of natural gas is 130therm 1 therm 105500 kJ deter mine how much energy and money the regenerator will save this company per year and the annual reduction in exergy destruction 8125E Refrigerant134a enters an adiabatic compressor as saturated vapor at 30 psia at a rate of 20 ft3min and exits at 70 psia pressure If the isentropic efficiency of the compres sor is 80 percent determine a the actual power input and b the secondlaw efficiency of the compressor Assume the surroundings to be at 75F Answers a 285 hp b 798 percent 8126 Refrigerant134a at 1600 kPa and 80C is expanded adiabatically in a closed system to 100 kPa with an isentropic expansion efficiency of 85 percent Determine the secondlaw efficiency of this expansion Take T0 25C and P0 100 kPa 8127 Combustion gases enter a gas turbine at 627C and 12 MPa at a rate of 25 kgs and leave at 527C and 500 kPa It is estimated that heat is lost from the turbine at a rate of 20 kW Using air properties for the combustion gases and assuming the surroundings to be at 25C and 100 kPa determine a the actual and reversible power outputs of the turbine b the exergy destroyed within the turbine and c the secondlaw efficiency of the turbine FIGURE P8127 Turbine 527C 500 kPa Q Exhaust gases 627C 12 MPa 8128 Water enters a pump at 100 kPa and 30C at a rate of 135 kgs and leaves at 4 MPa If the pump has an isentropic efficiency of 70 percent determine a the actual power input b the rate of frictional heating c the exergy destruction and d the secondlaw efficiency for an environment temperature of 20C 8129 Argon gas expands from 35 MPa and 100C to 500 kPa in an adiabatic expansion valve For environment con ditions of 100 kPa and 25C determine a the exergy of argon at the inlet b the exergy destruction during the process and c the secondlaw efficiency FIGURE P8129 Argon 35 MPa 100C 500 kPa 8130 Can closedsystem exergy be negative How about flow exergy Explain using an incompressible substance as an example 8131 Obtain a relation for the secondlaw efficiency of a heat engine that receives heat QH from a source at temperature Final PDF to printer 473 CHAPTER 8 cen22672ch08413474indd 473 110817 1156 AM TH and rejects heat QL to a sink at TL which is higher than T0 the temperature of the surroundings while producing work in the amount of W 8132 Writing the first and secondlaw relations and sim plifying obtain the reversible work relation for a closed system that exchanges heat with the surrounding medium at T0 in the amount of Q0 as well as a heat reservoir at TR in the amount of QR Hint Eliminate Q0 between the two equations 8133 Writing the first and secondlaw relations and sim plifying obtain the reversible work relation for a steadyflow system that exchanges heat with the surrounding medium at T0 at a rate of Q 0 as well as a thermal reservoir at TR at a rate of Q R Hint Eliminate Q 0 between the two equations Fundamentals of Engineering FE Exam Problems 8134 Keeping the limitations imposed by the second law of thermodynamics in mind choose the wrong statement below a A heat engine cannot have a thermal efficiency of 100 percent b For all reversible processes the secondlaw efficiency is 100 percent c The secondlaw efficiency of a heat engine cannot be greater than its thermal efficiency d The secondlaw efficiency of a process is 100 percent if no entropy is generated during that process e The coefficient of performance of a refrigerator can be greater than 1 8135 Heat is lost through a plane wall steadily at a rate of 800 W If the inner and outer surface temperatures of the wall are 20C and 9C respectively and the environment temperature is 0C the rate of exergy destruction within the wall is a 0 W b 11 W c 15 W d 29 W e 76 W 8136 Liquid water enters an adiabatic piping system at 15C at a rate of 3 kgs It is observed that the water tempera ture rises by 03C in the pipe due to friction If the environ ment temperature is also 15C the rate of exergy destruction in the pipe is a 38 kW b 24 kW c 72 kW d 98 kW e 124 kW 8137 A water reservoir contains 100 tons of water at an average elevation of 60 m The maximum amount of electric power that can be generated from this water is a 8 kWh b 16 kWh c 1630 kWh d 16300 kWh e 58800 kWh 8138 A house is maintained at 21C in winter by electric resistance heaters If the outdoor temperature is 3C the sec ondlaw efficiency of the resistance heaters is a 0 b 41 c 61 d 86 e 163 8139 A furnace can supply heat steadily at 1300 K at a rate of 500 kJs The maximum amount of power that can be produced by using the heat supplied by this furnace in an envi ronment at 300 K is a 115 kW b 192 kW c 385 kW d 500 kW e 650 kW 8140 A heat engine receives heat from a source at 1500 K at a rate of 600 kJs and rejects the waste heat to a sink at 300 K If the power output of the engine is 400 kW the secondlaw efficiency of this heat engine is a 42 b 53 c 83 d 67 e 80 8141 Air is throttled from 50C and 800 kPa to a pressure of 200 kPa at a rate of 05 kgs in an environment at 25C The change in kinetic energy is negligible and no heat transfer occurs during the process The power potential wasted during this process is a 0 b 020 kW c 47 kW d 59 kW e 119 kW 8142 Steam enters a turbine steadily at 4 MPa and 600C and exits at 02 MPa and 150C in an environment at 25C The decrease in the exergy of the steam as it flows through the turbine is a 879 kJkg b 1123 kJkg c 1645 kJkg d 1910 kJkg e 4260 kJkg 8143 A 12kg solid whose specific heat is 28 kJkgC is at a uniform temperature of 10C For an environment tempera ture of 20C the exergy content of this solid is a Less than zero b 0 kJ c 46 kJ d 55 kJ e 1008 kJ Design and Essay Problems 8144 Obtain the following information about a power plant that is closest to your town the net power output the type and amount of fuel used the power consumed by the pumps fans and other auxiliary equipment stack gas losses temperatures at several locations and the rate of heat rejection at the con denser Using these and other relevant data determine the rate of irreversibility in that power plant 8145 Human beings are probably the most capable crea tures and they have a high level of physical intellectual emotional and spiritual potentials or exergies Unfortunately people make little use of their exergies letting most of their exergies go to waste Draw four exergyversustime charts and plot your physical intellectual emotional and spiritual exer gies on each of these charts for a 24h period using your best judgment based on your experience On these four charts plot your respective exergies that you have utilized during the last 24 h Compare the two plots on each chart and determine if you are living a full life or if you are wasting your life away Can you think of any ways to reduce the mismatch between your exergies and your utilization of them 8146 Domestic hotwater systems involve a high level of irreversibility and thus they have low secondlaw efficiencies The water in these systems is heated from about 15C to about 60C and most of the hot water is mixed with cold water to Final PDF to printer 474 EXERGY cen22672ch08413474indd 474 110817 1156 AM reduce its temperature to 45C or even lower before it is used for any useful purpose such as taking a shower or washing clothes at a warm setting The water is discarded at about the same temperature at which it was used and replaced by fresh cold water at 15C Redesign a typical residential hotwater system such that the irreversibility is greatly reduced Draw a sketch of your proposed design 8147 Consider natural gas electric resistance and heat pump heating systems For a specified heating load which one of these systems will do the job with the least irreversibility Explain 8148 The temperature of the air in a building can be maintained at a desirable level during winter by using dif ferent methods of heating Compare heating this air in a heat exchanger unit with condensing steam to heating it with an electricresistance heater Perform a secondlaw analysis to determine the heating method that generates the least entropy and thus causes the least exergy destruction 8149 A steam boiler may be thought of as a heat exchanger The combustion gases may be modeled as a stream of air because their thermodynamic properties are close to those of air Using this model consider a boiler that is to convert saturated liquid water at 500 psia to a saturated vapor while keeping the water pressure constant Determine the tempera ture at which the air ie combustion gases must enter this unit so that the transfer of exergy from the air to the boiling water is done at the minimum loss 8150 An adiabatic nozzle is designed to accelerate an ideal gas from nearly 0 ms P1 and T1 to V ms As the efficiency of this noz zle decreases the pressure at the nozzle exit must also be decreased to maintain the speed at V Plot the change in the flow exergy as a function of the nozzle efficiency for an ideal gas say air Final PDF to printer cen22672ch09475542indd 475 110617 0921 AM 475 CHAPTER9 GAS P OW E R CYCL E S T wo important areas of application for thermodynamics are power gen eration and refrigeration Both are usually accomplished by systems that operate on a thermodynamic cycle Thermodynamic cycles can be divided into two general categories power cycles which are discussed in this chapter and Chap 10 and refrigeration cycles which are discussed in Chap 11 The devices or systems used to produce a net power output are often called engines and the thermodynamic cycles they operate on are called power cycles The devices or systems used to produce a refrigeration effect are called refrigerators air conditioners or heat pumps and the cycles they operate on are called refrigeration cycles Thermodynamic cycles can also be categorized as gas cycles and vapor cycles depending on the phase of the working fluid In gas cycles the work ing fluid remains in the gaseous phase throughout the entire cycle whereas in vapor cycles the working fluid exists in the vapor phase during one part of the cycle and in the liquid phase during another part Thermodynamic cycles can be categorized yet another way closed and open cycles In closed cycles the working fluid is returned to the initial state at the end of the cycle and is recirculated In open cycles the working fluid is renewed at the end of each cycle instead of being recirculated In automobile engines the combustion gases are exhausted and replaced by fresh airfuel mixture at the end of each cycle The engine operates on a mechanical cycle but the working fluid does not go through a complete thermodynamic cycle Heat engines are categorized as internal combustion and external combus tion engines depending on how the heat is supplied to the working fluid In external combustion engines such as steam power plants heat is supplied to the working fluid from an external source such as a furnace a geothermal well a nuclear reactor or even the sun In internal combustion engines such as automobile engines this is done by burning the fuel within the system boundaries In this chapter various gas power cycles are analyzed under some simplifying assumptions OBJECTIVES The objectives of Chapter 9 are to Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle Develop simplifying assump tions applicable to gas power cycles Review the operation of recip rocating engines Analyze both closed and open gas power cycles Solve problems based on the Otto Diesel Stirling and Erics son cycles Solve problems based on the Brayton cycle the Bray ton cycle with regeneration and the Brayton cycle with intercooling reheating and regeneration Analyze jetpropulsion cycles Perform secondlaw analysis of gas power cycles Final PDF to printer 476 GAS POWER CYCLES cen22672ch09475542indd 476 110617 0921 AM 91 BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES Most powerproducing devices operate on cycles and the study of power cycles is an exciting and important part of thermodynamics The cycles encountered in actual devices are difficult to analyze because of the pres ence of complicating effects such as friction and the absence of sufficient time for establishment of the equilibrium conditions during the cycle To make an analytical study of a cycle feasible we have to keep the complexi ties at a manageable level and utilize some idealizations Fig 91 When the actual cycle is stripped of all the internal irreversibilities and complexi ties we end up with a cycle that resembles the actual cycle closely but is made up totally of internally reversible processes Such a cycle is called an ideal cycle Fig 92 A simple idealized model enables engineers to study the effects of the major parameters that dominate the cycle without getting bogged down in the details The cycles discussed in this chapter are somewhat idealized but they still retain the general characteristics of the actual cycles they represent The conclusions reached from the analysis of ideal cycles are also applicable to actual cycles The thermal efficiency of the Otto cycle the ideal cycle for sparkignition automobile engines for example increases with the compres sion ratio This is also the case for actual automobile engines The numerical values obtained from the analysis of an ideal cycle however are not neces sarily representative of the actual cycles and care should be exercised in their interpretation The simplified analysis presented in this chapter for various power cycles of practical interest may also serve as the starting point for a more indepth study Heat engines are designed for the purpose of converting thermal energy to work and their performance is expressed in terms of the thermal efficiency ηth which is the ratio of the net work produced by the engine to the total heat input η th W net Q in or η th w net q in 91 Recall that heat engines that operate on a totally reversible cycle such as the Carnot cycle have the highest thermal efficiency of all heat engines oper ating between the same temperature levels That is nobody can develop a cycle more efficient than the Carnot cycle Then the following question arises naturally If the Carnot cycle is the best possible cycle why do we not use it as the model cycle for all the heat engines instead of bothering with several socalled ideal cycles The answer to this question is hardwarerelated Most cycles encountered in practice differ significantly from the Carnot cycle which makes it unsuitable as a realistic model Each ideal cycle discussed in this chapter is related to a specific workproducing device and is an idealized version of the actual cycle The ideal cycles are internally reversible but unlike the Carnot cycle they are not necessarily externally reversible That is they may involve irrevers ibilities external to the system such as heat transfer through a finite tempera ture difference Therefore the thermal efficiency of an ideal cycle in general is less than that of a totally reversible cycle operating between the same FIGURE 91 Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy Oven Ideal Actual 175C Water Potato FIGURE 92 The analysis of many complex pro cesses can be reduced to a manageable level by utilizing some idealizations P Actual cycle Ideal cycle v Final PDF to printer 477 CHAPTER 9 cen22672ch09475542indd 477 110617 0921 AM temperature limits However it is still considerably higher than the thermal efficiency of an actual cycle because of the idealizations utilized Fig 93 The idealizations and simplifications commonly employed in the analysis of power cycles can be summarized as follows 1 The cycle does not involve any friction Therefore the working fluid does not experience any pressure drop as it flows in pipes or devices such as heat exchangers 2 All expansion and compression processes take place in a quasi equilibrium manner 3 The pipes connecting the various components of a system are well insu lated and heat transfer through them is negligible Neglecting the changes in kinetic and potential energies of the working fluid is another commonly utilized simplification in the analysis of power cycles This is a reasonable assumption since in devices that involve shaft work such as turbines compressors and pumps the kinetic and potential energy terms are usually very small relative to the other terms in the energy equation Fluid velocities encountered in devices such as condensers boilers and mixing chambers are typically low and the fluid streams experience little change in their velocities again making kinetic energy changes negligible The only devices where the changes in kinetic energy are significant are the nozzles and diffusers which are specifically designed to create large changes in velocity In the preceding chapters property diagrams such as the Pv and Ts diagrams have served as valuable aids in the analysis of thermodynamic pro cesses On both the Pv and Ts diagrams the area enclosed by the process curves of a cycle represents the net work produced during the cycle Fig 94 which is also equivalent to the net heat transfer for that cycle The Ts diagram is particularly useful as a visual aid in the analysis of ideal power cycles An ideal power cycle does not involve any internal irreversibilities and so the only effect that can change the entropy of the working fluid during a process is heat transfer On a Ts diagram a heataddition process proceeds in the direction of increas ing entropy a heatrejection process proceeds in the direction of decreasing FIGURE 93 An automotive engine with the combustion chamber exposed Idealink PhotographyAlamy RF FIGURE 94 On both Pv and Ts diagrams the area enclosed by the process curve represents the net work of the cycle P T s v 1 2 3 4 1 2 3 4 wnet wnet Final PDF to printer 478 GAS POWER CYCLES cen22672ch09475542indd 478 110617 0921 AM entropy and an isentropic internally reversible adiabatic process proceeds at constant entropy The area under the process curve on a Ts diagram represents the heat transfer for that process The area under the heat addition process on a Ts diagram is a geometric measure of the total heat supplied during the cycle qin and the area under the heat rejection process is a mea sure of the total heat rejected qout The difference between these two the area enclosed by the cyclic curve is the net heat transfer which is also the net work produced during the cycle Therefore on a Ts diagram the ratio of the area enclosed by the cyclic curve to the area under the heataddition process curve represents the thermal efficiency of the cycle Any modification that increases the ratio of these two areas will also increase the thermal efficiency of the cycle Although the working fluid in an ideal power cycle operates on a closed loop the type of individual process that comprises the cycle depends on the individual devices used to execute the cycle In the Rankine cycle which is the ideal cycle for steam power plants the working fluid flows through a series of steadyflow devices such as the turbine and condenser whereas in the Otto cycle which is the ideal cycle for the sparkignition automobile engine the working fluid is alternately expanded and compressed in a pistoncylinder device Therefore equations pertaining to steadyflow systems should be used in the analysis of the Rankine cycle and equations pertaining to closed systems should be used in the analysis of the Otto cycle 92 THE CARNOT CYCLE AND ITS VALUE IN ENGINEERING The Carnot cycle is composed of four totally reversible processes isothermal heat addition isentropic expansion isothermal heat rejection and isentropic compression The Pv and Ts diagrams of a Carnot cycle are replotted in Fig 95 The Carnot cycle can be executed in a closed system a piston cylinder device or a steadyflow system utilizing two turbines and two com pressors as shown in Fig 96 and either a gas or a vapor can be utilized as the working fluid The Carnot cycle is the most efficient cycle that can be executed between a heat source at temperature TH and a sink at temperature TL and its thermal efficiency is expressed as η thCarnot 1 T L T H 92 Reversible isothermal heat transfer is very difficult to achieve in reality because it would require very large heat exchangers and it would take a very long time a power cycle in a typical engine is completed in a fraction of a second Therefore it is not practical to build an engine that would operate on a cycle that closely approximates the Carnot cycle The real value of the Carnot cycle comes from its being a standard against which the actual or the ideal cycles can be compared The thermal efficiency of the Carnot cycle is a function of the sink and source temperatures only and the thermal efficiency relation for the Carnot cycle Eq 92 conveys an important message that is equally applicable to both ideal and actual cycles Thermal efficiency increases with an increase in the average temperature FIGURE 95 Pv and Ts diagrams of a Carnot cycle P T s v 1 2 3 4 1 2 3 4 qout qin Isentropic TH TL qin qout TH const Isentropic Isentropic Isentropic TL const Final PDF to printer 479 CHAPTER 9 cen22672ch09475542indd 479 110617 0921 AM at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system The source and sink temperatures that can be used in practice are not with out limits however The highest temperature in the cycle is limited by the maximum temperature that the components of the heat engine such as the piston or the turbine blades can withstand The lowest temperature is limited by the temperature of the cooling medium utilized in the cycle such as a lake a river or the atmospheric air FIGURE 96 A steadyflow Carnot engine qin qout Isothermal compressor Isentropic compressor wnet Isentropic turbine 1 2 3 4 Isothermal turbine EXAMPLE 91 Derivation of the Efficiency of the Carnot Cycle Show that the thermal efficiency of a Carnot cycle operating between the temperature limits of TH and TL is solely a function of these two temperatures and is given by Eq 92 SOLUTION It is to be shown that the efficiency of a Carnot cycle depends on the source and sink temperatures alone Analysis The Ts diagram of a Carnot cycle is redrawn in Fig 97 All four processes that comprise the Carnot cycle are reversible and thus the area under each process curve represents the heat transfer for that process Heat is transferred to the system during pro cess 12 and rejected during process 34 Therefore the amount of heat input and heat output for the cycle can be expressed as q in T H s 2 s 1 and q out T L s 3 s 4 T L s 2 s 1 since processes 23 and 41 are isentropic and thus s2 s3 and s4 s1 Substituting these into Eq 91 we see that the thermal efficiency of a Carnot cycle is η th w net q in 1 q out q in 1 T L s 2 s 1 T H s 2 s 1 1 T L T H Discussion Notice that the thermal efficiency of a Carnot cycle is independent of the type of the working fluid used an ideal gas steam etc or whether the cycle is executed in a closed or steadyflow system FIGURE 97 Ts diagram for Example 91 T s 1 2 4 3 qin qout TH TL s1 s4 s2 s3 Final PDF to printer 480 GAS POWER CYCLES cen22672ch09475542indd 480 110617 0921 AM 93 AIRSTANDARD ASSUMPTIONS In gas power cycles the working fluid remains a gas throughout the entire cycle Sparkignition engines diesel engines and conventional gas turbines are familiar examples of devices that operate on gas cycles In all these engines energy is provided by burning a fuel within the system boundaries That is they are internal combustion engines Because of this combustion process the composition of the working fluid changes from air and fuel to combustion products during the course of the cycle However considering that air is predominantly nitrogen that undergoes hardly any chemical reac tions in the combustion chamber the working fluid closely resembles air at all times Even though internal combustion engines operate on a mechanical cycle the piston returns to its starting position at the end of each revolution the working fluid does not undergo a complete thermodynamic cycle It is thrown out of the engine at some point in the cycle as exhaust gases instead of being returned to the initial state Working on an open cycle is the characteristic of all internal combustion engines The actual gas power cycles are rather complex To reduce the analysis to a manageable level we utilize the following approximations commonly known as the airstandard assumptions 1 The working fluid is air which continuously circulates in a closed loop and always behaves as an ideal gas 2 All the processes that make up the cycle are internally reversible 3 The combustion process is replaced by a heataddition process from an external source Fig 98 4 The exhaust process is replaced by a heatrejection process that restores the working fluid to its initial state Another assumption that is often used to simplify the analysis even more is that air has constant specific heats whose values are determined at room temperature 25C or 77F When this assumption is used the airstandard assumptions are called the coldairstandard assumptions A cycle for which the airstandard assumptions are applicable is frequently referred to as an airstandard cycle The airstandard assumptions previously stated provide considerable sim plification in the analysis without significantly deviating from the actual cycles This simplified model enables us to study qualitatively the influence of major parameters on the performance of the actual engines FIGURE 98 The combustion process is replaced by a heataddition process in ideal cycles Combustion products Air Fuel a Actual Combustion chamber Air Air b Ideal Heating section Heat EXAMPLE 92 An AirStandard Cycle An airstandard cycle is executed in a closed system and is composed of the following four processes 12 Isentropic compression from 100 kPa and 27C to 1 MPa 23 P constant heat addition in amount of 2800 kJkg 34 v constant heat rejection to 100 kPa 41 P constant heat rejection to initial state Final PDF to printer 481 CHAPTER 9 cen22672ch09475542indd 481 110617 0921 AM FIGURE 99 Schematic for Example 92 P v T s 1 2 4 3 qin q41 q34 1 2 4 3 q34 q41 qin a Show the cycle on Pv and Ts diagrams b Calculate the maximum temperature in the cycle c Determine the thermal efficiency Assume constant specific heats at room temperature SOLUTION The four processes of an airstandard cycle are described The cycle is to be shown on Pv and Ts diagrams and the maximum temperature in the cycle and the thermal efficiency are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with constant specific heats Properties The properties of air at room temperature are cp 1005 kJkgK cv 0718 kJkgK and k 14 Table A2a Analysis a The cycle is shown on Pv and Ts diagrams in Fig 99 b From the ideal gas isentropic relations and energy balance T 2 T 1 P 2 P 1 k1 k 300 K 1000 kPa 100 kPa 0414 5792 K q in h 3 h 2 c p T 3 T 1 2800 kJ kg 1005 kJ kgK T 3 5792 T max T 3 3360 K c The temperature at state 4 is determined from the ideal gas relation for a fixed mass P 3 v 3 T 3 P 4 v 4 T 4 T 4 P 4 P 3 T 3 100 kPa 1000 kPa 3360 K 336 K The total amount of heat rejected from the cycle is q out q 34out q 41out u 3 u 4 h 4 h 1 c v T 3 T 4 c p T 4 T 1 0718 kJ kgK 3360 336 K 1005 kJ kgK 336 300 K 2212 kJ kg Then the thermal efficiency is determined from its definition to be η th 1 q out q in 1 2212 kJ kg 2800 kJ kg 0210 or 210 Discussion The assumption of constant specific heats at room temperature is not real istic in this case since the temperature changes involved are too large 94 AN OVERVIEW OF RECIPROCATING ENGINES Despite its simplicity the reciprocating engine basically a pistoncylinder device is one of the rare inventions that has proved to be very versatile and to have a wide range of applications It is the powerhouse of the vast majority of automobiles trucks light aircraft ships and electric power generators as well as many other devices The basic components of a reciprocating engine are shown in Fig 910 The piston reciprocates in the cylinder between two fixed positions called the FIGURE 910 Nomenclature for reciprocating engines Intake valve Exhaust valve Bore TDC BDC Stroke Final PDF to printer 482 GAS POWER CYCLES cen22672ch09475542indd 482 110617 0921 AM top dead center TDCthe position of the piston when it forms the smallest volume in the cylinderand the bottom dead center BDCthe position of the piston when it forms the largest volume in the cylinder The distance between the TDC and the BDC is the largest distance that the piston can travel in one direction and it is called the stroke of the engine The diameter of the piston is called the bore The air or airfuel mixture is drawn into the cylinder through the intake valve and the combustion products are expelled from the cylinder through the exhaust valve The minimum volume formed in the cylinder when the piston is at TDC is called the clearance volume Fig 911 The volume displaced by the piston as it moves between TDC and BDC is called the displacement volume The ratio of the maximum volume formed in the cylinder to the minimum clear ance volume is called the compression ratio r of the engine r V max V min V BDC V TDC 93 Notice that the compression ratio is a volume ratio and should not be con fused with the pressure ratio Another term often used in conjunction with reciprocating engines is the mean effective pressure MEP It is a fictitious pressure that if it acted on the piston during the entire power stroke would produce the same amount of net work as that produced during the actual cycle Fig 912 That is W net MEP Piston area Stroke MEP Displacement volume or MEP W net V max V min w net v max v min kPa 94 The mean effective pressure can be used as a parameter to compare the per formances of reciprocating engines of equal size The engine with a larger value of MEP delivers more net work per cycle and thus performs better Reciprocating engines are classified as sparkignition SI engines or compressionignition CI engines depending on how the combustion pro cess in the cylinder is initiated In SI engines the combustion of the airfuel mixture is initiated by a spark plug In CI engines the airfuel mixture is self ignited as a result of compressing the mixture above its selfignition tempera ture In the next two sections we discuss the Otto and Diesel cycles which are the ideal cycles for the SI and CI reciprocating engines respectively 95 OTTO CYCLE THE IDEAL CYCLE FOR SPARKIGNITION ENGINES The Otto cycle is the ideal cycle for sparkignition reciprocating engines It is named after Nikolaus A Otto who built a successful fourstroke engine in 1876 in Germany using the cycle proposed by Frenchman Beau de Rochas in 1862 In most sparkignition engines the piston executes four complete strokes two mechanical cycles within the cylinder and the crankshaft com pletes two revolutions for each thermodynamic cycle These engines are called fourstroke internal combustion engines A schematic of each stroke FIGURE 911 Displacement and clearance volumes of a reciprocating engine TDC BDC a Displacement volume b Clearance volume FIGURE 912 The net work output of a cycle is equivalent to the product of the mean effective pressure and the displace ment volume Wnet MEPVmax Vmin Vmin Vmax V MEP P TDC BDC Wnet Final PDF to printer 483 CHAPTER 9 cen22672ch09475542indd 483 110617 0921 AM as well as a Pv diagram for an actual fourstroke sparkignition engine is given in Fig 913a Initially both the intake and the exhaust valves are closed and the piston is at its lowest position BDC During the compression stroke the piston moves upward compressing the airfuel mixture Shortly before the pis ton reaches its highest position TDC the spark plug fires and the mixture ignites increasing the pressure and temperature of the system The highpres sure gases force the piston down which in turn forces the crankshaft to rotate producing a useful work output during the expansion or power stroke Toward the end of expansion stroke the exhaust valve opens and the combustion gases that are above the atmospheric pressure rush out of the cylinder through the open exhaust valve This process is called exhaust blowdown and most combustion gases leave the cylinder by the time the piston reaches BDC The cylinder is still filled by the exhaust gases at a lower pressure at BDC Now the piston moves upward one more time purging the exhaust gases through the exhaust valve the exhaust stroke and down a second time drawing in fresh airfuel mixture through the intake valve the intake stroke Notice that the pressure in the cylinder is slightly above the atmospheric value during the exhaust stroke and slightly below during the intake stroke In twostroke engines all four functions described above are executed in just two strokes the power stroke and the compression stroke In these FIGURE 913 Actual and ideal cycles in sparkignition engines and their Pv diagrams a Actual fourstroke sparkignition engine b Ideal Otto cycle qin qout 4 Isentropic Isentropic 3 2 1 P Isentropic compression Air 2 1 Air 41 Air 23 Air 3 4 Isentropic expansion v const heat addition v const heat rejection qout TDC BDC v qin Patm P Compression stroke Power expansion stroke Airfuel mixture Exhaust valve opens Expansion Compression Ignition TDC BDC Intake Exhaust Intake valve opens Airfuel mixture Exhaust stroke Intake stroke Exhaust gases v Final PDF to printer 484 GAS POWER CYCLES cen22672ch09475542indd 484 110617 0921 AM engines the crankcase is sealed and the outward motion of the piston is used to slightly pressurize the airfuel mixture in the crankcase as shown in Fig 914 Also the intake and exhaust valves are replaced by openings in the lower portion of the cylinder wall During the latter part of the power stroke the piston uncovers first the exhaust port allowing the exhaust gases to be partially expelled and then the intake port allowing the fresh airfuel mixture to rush in and drive most of the remaining exhaust gases out of the cylinder This mixture is then compressed as the piston moves upward during the compression stroke and is subsequently ignited by a spark plug The twostroke engines are generally less efficient than their fourstroke counterparts because of the incomplete expulsion of the exhaust gases and the partial expulsion of the fresh airfuel mixture with the exhaust gases However they are relatively simple and inexpensive and they have high powertoweight and powertovolume ratios which make them suitable for applications requiring small size and weight such as for motorcycles chain saws and lawn mowers Fig 915 Advances in several technologiessuch as direct fuel injection stratified charge combustion and electronic controlsbrought about a renewed inter est in twostroke engines that can offer high performance and fuel economy while satisfying stringent emission requirements For a given weight and displacement a welldesigned twostroke engine can provide significantly more power than its fourstroke counterpart because twostroke engines pro duce power on every engine revolution instead of every other one In the new twostroke engines the highly atomized fuel spray that is injected into the combustion chamber toward the end of the compression stroke burns much more completely The fuel is sprayed after the exhaust valve is closed which prevents unburned fuel from being ejected into the atmosphere With strat ified combustion the flame that is initiated by igniting a small amount of the rich fuelair mixture near the spark plug propagates through the com bustion chamber filled with a much leaner mixture and this results in much cleaner combustion Also the advances in electronics have made it possible to ensure the optimum operation under varying engine load and speed condi tions Major car companies have research programs underway on twostroke engines which are expected to make a comeback in the future The thermodynamic analysis of the actual fourstroke or twostroke cycles described is not a simple task However the analysis can be simplified signifi cantly if the airstandard assumptions are utilized The resulting cycle which closely resembles the actual operating conditions is the ideal Otto cycle It consists of four internally reversible processes 12 Isentropic compression 23 Constantvolume heat addition 34 Isentropic expansion 41 Constantvolume heat rejection The execution of the Otto cycle in a pistoncylinder device together with a Pv diagram is illustrated in Fig 913b The Ts diagram of the Otto cycle is given in Fig 916 The ideal Otto cycle shown in Fig 913b has one shortcoming This ideal cycle consists of two strokes equivalent to one mechanical cycle or one crank shaft rotation The actual engine operation shown in Fig 913a on the other FIGURE 914 Schematic of a twostroke reciprocating engine Exhaust port Intake port Crankcase Spark plug Fuelair mixture FIGURE 915 Twostroke engines are commonly used in motorcycles and lawn mowers Fineart1Shutterstock RF Final PDF to printer 485 CHAPTER 9 cen22672ch09475542indd 485 110617 0921 AM hand involves four strokes equivalent to two mechanical cycles or two crank shaft rotations This can be corrected by including intake and exhaust strokes in the ideal Otto cycle as shown in Fig 917 In this modified cycle airfuel mixture approximated as air due to airstandard assumptions enters the cyl inder through the open intake valve at atmospheric pressure P0 during process 01 as the piston moves from TDC to BDC The intake valve is closed at state 1 and air is compressed isentropically to state 2 Heat is transferred at con stant volume process 23 it is expanded isentropically to state 4 and heat is rejected at constant volume process 41 Exhaust gases again approximated as air are expelled through the open exhaust valve process 10 as the pres sure remains constant at P0 The modified Otto cycle shown in Fig 917 is executed in an open system during the intake and exhaust processes and in a closed system during the remaining four processes We should point out that the constantvolume heat addition process 23 in the ideal Otto cycle replaces the combustion process of the actual engine operation while the constantvolume heat rejection pro cess 41 replaces the exhaust blowdown The work interactions during the constantpressure intake 01 and con stantpressure exhaust 10 processes can be expressed as w out01 P 0 v 1 v 0 w in10 P 0 v 1 v 0 These two processes cancel each other as the work output during the intake is equal to work input during the exhaust Then the cycle reduces to the one in Fig 913b Therefore inclusion of the intake and exhaust processes has no effect on the net work output from the cycle However when calculating power output from the cycle during an ideal Otto cycle analysis we must consider the fact that the ideal Otto cycle has four strokes just like actual four stroke sparkignition engine This is illustrated in the last part of Example 93 The Otto cycle is executed in a closed system and disregarding the changes in kinetic and potential energies the energy balance for any of the processes is expressed on a unitmass basis as q in q out w in w out Δu kJkg 95 No work is involved during the two heat transfer processes since both take place at constant volume Therefore heat transfer to and from the working fluid can be expressed as q in u 3 u 2 c v T 3 T 2 96a and q out u 4 u 1 c v T 4 T 1 96b Then the thermal efficiency of the ideal Otto cycle under the cold air standard assumptions becomes η thOtto w net q in 1 q out q in 1 T 4 T 1 T 3 T 2 1 T 1 T 4 T 1 1 T 2 T 3 T 2 1 Processes 12 and 34 are isentropic and v2 v3 and v4 v1 Thus FIGURE 916 Ts diagram of the ideal Otto cycle T s 1 2 3 4 v const v const qout qin FIGURE 917 Pv diagram of the ideal Otto cycle that includes intake and exhaust strokes P v 2 3 4 1 0 TDC BDC Isentropic Isentropic qout P0 qin Final PDF to printer 486 GAS POWER CYCLES cen22672ch09475542indd 486 110617 0921 AM T 1 T 2 v 2 v 1 k 1 v 3 v 4 k 1 T 4 T 3 97 Substituting these equations into the thermal efficiency relation and simplify ing give η thOtto 1 1 r k1 98 where r V max V min V 1 V 2 v 1 v 2 99 is the compression ratio and k is the specific heat ratio cpcv Equation 98 shows that under the coldairstandard assumptions the ther mal efficiency of an ideal Otto cycle depends on the compression ratio of the engine and the specific heat ratio of the working fluid The thermal effi ciency of the ideal Otto cycle increases with both the compression ratio and the specific heat ratio This is also true for actual sparkignition internal com bustion engines A plot of thermal efficiency versus the compression ratio is given in Fig 918 for k 14 which is the specific heat ratio value of air at room temperature For a given compression ratio the thermal efficiency of an actual sparkignition engine is less than that of an ideal Otto cycle because of the irreversibilities such as friction and other factors such as incomplete combustion We can observe from Fig 918 that the thermal efficiency curve is rather steep at low compression ratios but flattens out starting with a compression ratio value of about 8 Therefore the increase in thermal efficiency with the compression ratio is not as pronounced at high compression ratios Also when high compression ratios are used the temperature in some regions of the airfuel mixture rises above the autoignition temperature of the fuel the temperature at which the fuel ignites without the help of a spark during the combustion process causing an early and rapid burn of the fuel at some point or points ahead of the flame front followed by almost instantaneous inflam mation of the end gas This premature ignition of the fuel called autoignition produces an audible noise which is called engine knock Autoignition in sparkignition engines cannot be tolerated because it hurts performance and can cause engine damage The requirement that autoignition not be allowed places an upper limit on the compression ratios that can be used in spark ignition internal combustion engines Improvement of the thermal efficiency of gasoline engines by utilizing higher compression ratios up to about 12 without facing the autoignition problem has been made possible by using gasoline blends that have good anti knock characteristics such as gasoline mixed with tetraethyl lead Tetraethyl lead had been added to gasoline since the 1920s because it is an inexpensive method of raising the octane rating which is a measure of the engine knock resistance of a fuel Leaded gasoline however has a very undesirable side effect it forms compounds during the combustion process that are hazard ous to health and pollute the environment In an effort to combat air pollu tion the government adopted a policy in the mid1970s that resulted in the eventual phaseout of leaded gasoline Unable to use lead the refiners devel oped other techniques to improve the antiknock characteristics of gasoline FIGURE 918 Thermal efficiency of the ideal Otto cycle as a function of compression ratio k 14 2 4 6 8 10 12 14 Compression ratio r 07 06 05 04 03 02 01 Typical compression ratios for gasoline engines ηthOtto Final PDF to printer 487 CHAPTER 9 cen22672ch09475542indd 487 110617 0921 AM Most cars made since 1975 have been designed to use unleaded gasoline and the compression ratios had to be lowered to avoid engine knock The ready availability of highoctane fuels made it possible to raise the compression ratios again in recent years Also owing to the improvements in other areas reduction in overall automobile weight improved aerodynamic design using variable compression ratios by the use of a multilink system hybrid engines that recover power lost during braking individually controlled intake and exhaust valves etc todays cars have better fuel economy and consequently get more miles per gallon of fuel This is an example of how engineering deci sions involve compromises and efficiency is only one of the considerations in final design The second parameter affecting the thermal efficiency of an ideal Otto cycle is the specific heat ratio k For a given compression ratio an ideal Otto cycle using a monatomic gas such as argon or helium k 1667 as the working fluid will have the highest thermal efficiency The specific heat ratio k and thus the thermal efficiency of the ideal Otto cycle decreases as the molecules of the working fluid get larger Fig 919 At room temperature it is 14 for air 13 for carbon dioxide and 12 for ethane The working fluid in actual engines contains larger molecules such as carbon dioxide and the spe cific heat ratio decreases with temperature which is one of the reasons that the actual cycles have lower thermal efficiencies than the ideal Otto cycle The thermal efficiencies of actual sparkignition engines range from about 25 to 30 percent FIGURE 919 The thermal efficiency of the Otto cycle increases with the specific heat ratio k of the working fluid 08 06 04 02 2 4 6 8 10 12 k 1667 k 14 k 13 Compression ratio r ηthOtto EXAMPLE 93 The Ideal Otto Cycle An ideal Otto cycle has a compression ratio of 8 At the beginning of the compression process air is at 100 kPa and 17C and 800 kJkg of heat is transferred to air during the constantvolume heataddition process Accounting for the variation of specific heats of air with temperature determine a the maximum temperature and pressure that occur during the cycle b the net work output c the thermal efficiency and d the mean effective pressure for the cycle e Also determine the power output from the cycle in kW for an engine speed of 4000 rpm revmin Assume that this cycle is operated on an engine that has four cylinders with a total displacement volume of 16 L SOLUTION An ideal Otto cycle is considered The maximum temperature and pressure the net work output the thermal efficiency the mean effective pressure and the power output for a given engine speed are to be determined Assumptions 1 The airstandard assumptions are applicable 2 Kinetic and potential energy changes are negligible 3 The variation of specific heats with temperature is to be accounted for Analysis The Pv diagram of the ideal Otto cycle described is shown in Fig 920 We note that the air contained in the cylinder forms a closed system a The maximum temperature and pressure in an Otto cycle occur at the end of the constantvolume heataddition process state 3 But first we need to determine the temperature and pressure of air at the end of the isentropic compression process state 2 using data from Table A17 T 1 290 K u 1 20691 kJkg v r1 6761 FIGURE 920 Pv diagram for the Otto cycle discussed in Example 93 1 2 3 4 P kPa 100 Isentropic Isentropic qin qout v2 v3 v1 v1 v4 v 18 Final PDF to printer 488 GAS POWER CYCLES cen22672ch09475542indd 488 110617 0921 AM Process 12 isentropic compression of an ideal gas v r2 v r1 v 2 v 1 1 r v r2 v r1 r 6761 8 8451 T 2 6524 K u 2 47511 kJkg P 2 v 2 T 2 P 1 v 1 T 1 P 2 P 1 T 2 T 1 v 1 v 2 100 kPa 6524 K 290 K 8 17997 kPa Process 23 constantvolume heat addition q in u 3 u 2 800 kJkg u 3 47511 kJkg u 3 127511 kJkg T 3 15751 K v r3 6108 P 3 v 3 T 3 P 2 v 2 T 2 P 3 P 2 T 3 T 2 v 2 v 3 17997 MPa 15751 K 6524 K 1 4345 MPa b The net work output for the cycle is determined either by finding the boundary P dV work involved in each process by integration and adding them or by finding the net heat transfer that is equivalent to the net work done during the cycle We take the latter approach However first we need to find the internal energy of the air at state 4 Process 34 isentropic expansion of an ideal gas v r4 v r3 v 4 v 3 r v r4 r v r3 8 6108 48864 T 4 7956 K u 4 58874 kJkg Process 41 constantvolume heat rejection q out u 1 u 4 q out u 4 u 1 q out 58874 20691 38183 kJkg Thus w net q net q in q out 800 38183 41817 kJkg c The thermal efficiency of the cycle is determined from its definition η th w net q in 41817 kJkg 800 kJkg 0523 or 523 Under the coldairstandard assumptions constant specific heat values at room tem perature the thermal efficiency would be Eq 98 η thOtto 1 1 r k1 1 r 1k 1 8 1 14 0565 or 565 which is considerably different from the value obtained above Therefore care should be exercised in utilizing the coldairstandard assumptions Final PDF to printer 489 CHAPTER 9 cen22672ch09475542indd 489 110617 0921 AM d The mean effective pressure is determined from its definition Eq 94 MEP w net v 1 v 2 w net v 1 v 1 r w net v 1 1 1 r where v 1 R T 1 P 1 0287 kPa m 3 kgK 290 K 100 kPa 08323 m 3 kg Thus MEP 41817 kJkg 08323 m 3 kg 1 1 8 1 kPa m 3 1 kJ 574 kPa e The total air mass taken by all four cylinders when they are charged is m V d v 1 00016 m 3 08323 m 3 kg 0001922 kg The net work produced by the cycle is W net m w net 0001922 kg 41817 kJkg 08037 kJ That is the net work produced per thermodynamic cycle is 08037 kJcycle Noting that there are two revolutions per thermodynamic cycle nrev 2 revcycle in a four stroke engine or in the ideal Otto cycle including intake and exhaust strokes the power produced by the engine is determined from W net W net n n rev 08037 kJcycle 4000 revmin 2 revcycle 1 min 60 s 268 kW Discussion If we analyzed a twostroke engine operating on an ideal Otto cycle with the same values the power output would be calculated as W net W net n n rev 08037 kJcycle 4000 revmin 1 revcycle 1 min 60 s 536 kW Note that there is one revolution in one thermodynamic cycle in twostroke engines 96 DIESEL CYCLE THE IDEAL CYCLE FOR COMPRESSIONIGNITION ENGINES The Diesel cycle is the ideal cycle for CI reciprocating engines The CI engine first proposed by Rudolph Diesel in the 1890s is very similar to the SI engine discussed in the last section differing mainly in the method of initiat ing combustion In sparkignition engines also known as gasoline engines the airfuel mixture is compressed to a temperature that is below the autoigni tion temperature of the fuel and the combustion process is initiated by firing a spark plug In CI engines also known as diesel engines the air is compressed to a temperature that is above the autoignition temperature of the fuel and combustion starts on contact as the fuel is injected into this hot air Therefore the spark plug is replaced by a fuel injector in diesel engines Fig 921 FIGURE 921 In diesel engines the spark plug is replaced by a fuel injector and only air is compressed during the com pression process Gasoline engine Diesel engine Spark plug Fuel injector Air Airfuel mixture Fuel spray Spark Final PDF to printer 490 GAS POWER CYCLES cen22672ch09475542indd 490 110617 0921 AM In gasoline engines a mixture of air and fuel is compressed during the compression stroke and the compression ratios are limited by the onset of autoignition or engine knock In diesel engines only air is compressed during the compression stroke eliminating the possibility of autoignition Therefore diesel engines can be designed to operate at much higher compression ratios typically between 12 and 24 Not having to deal with the problem of autoig nition has another benefit many of the stringent requirements placed on the gasoline can now be removed and fuels that are less refined thus less expen sive can be used in diesel engines The fuel injection process in diesel engines starts when the piston approaches TDC and continues during the first part of the power stroke Therefore the combustion process in these engines takes place over a longer interval Because of this longer duration the combustion process in the ideal Diesel cycle is approximated as a constantpressure heataddition process In fact this is the only process where the Otto and the Diesel cycles dif fer The remaining three processes are the same for both ideal cycles That is process 12 is isentropic compression 23 is constantpressure heat addition 34 is isentropic expansion and 41 is constantvolume heat rejection The similarity between the two cycles is also apparent from the Pv and Ts dia grams of the Diesel cycle shown in Fig 922 Noting that the Diesel cycle is executed in a pistoncylinder device which forms a closed system the amount of heat transferred to the working fluid at constant pressure and rejected from it at constant volume can be expressed as q in w bout u 3 u 2 q in P 2 v 3 v 2 u 3 u 2 h 3 h 2 c p T 3 T 2 910a and q out u 1 u 4 q out u 4 u 1 c v T 4 T 1 910b Then the thermal efficiency of the ideal Diesel cycle under the coldair standard assumptions becomes η thDiesel w net q in 1 q out q in 1 T 4 T 1 k T 3 T 2 1 T 1 T 4 T 1 1 k T 2 T 3 T 2 1 We now define a new quantity the cutoff ratio rc as the ratio of the cylinder volumes after and before the combustion process r c V 3 V 2 v 3 v 2 911 Utilizing this definition and the isentropic idealgas relations for processes 12 and 34 we see that the thermal efficiency relation reduces to η thDiesel 1 1 r k1 r c k 1 k r c 1 912 where r is the compression ratio defined by Eq 99 Looking at Eq 912 carefully one would notice that under the coldairstandard assumptions the efficiency of a Diesel cycle differs from the efficiency of an Otto cycle by the quantity in the brackets This quantity is always greater than 1 Therefore FIGURE 922 Ts and Pv diagrams for the ideal Diesel cycle 1 2 3 4 P Isentropic Isentropic s v 1 2 3 4 T P constant v constant a Pv diagram b Ts diagram qin qout qout qin Final PDF to printer 491 CHAPTER 9 cen22672ch09475542indd 491 110617 0921 AM η thOtto η thDiesel 913 when both cycles operate on the same compression ratio Also as the cutoff ratio decreases the efficiency of the Diesel cycle increases Fig 923 For the limiting case of rc 1 the quantity in the brackets becomes unity can you prove it and the efficiencies of the Otto and Diesel cycles become identical Remember though that diesel engines operate at much higher compression ratios and thus are usually more efficient than the sparkignition gasoline engines The diesel engines also burn the fuel more completely since they usually operate at lower revolutions per minute and the airfuel mass ratio is much higher than in sparkignition engines Thermal efficiencies of diesel automotive engines range from about 35 to 40 percent Some very large low speed CI engines have thermal efficiencies over 50 percent The higher efficiency and lower fuel costs of diesel engines make them attractive in applications requiring relatively large amounts of power such as in locomotive engines emergency power generation units large ships and heavy trucks As an example of how large a diesel engine can be a 12 cylinder diesel engine built in 1964 by the Fiat Corporation of Italy had a normal power output of 25200 hp 188 MW at 122 rpm a cylinder bore of 90 cm and a stroke of 91 cm In modern highspeed compression ignition engines fuel is injected into the combustion chamber much sooner than in the early diesel engines Fuel starts to ignite late in the compression stroke and consequently part of the combus tion occurs almost at constant volume Fuel injection continues until the piston reaches the top dead center and combustion of the fuel keeps the pressure high well into the expansion stroke Thus the entire combustion process can be better modeled as the combination of constantvolume and constantpressure processes The ideal cycle based on this concept is called the dual cycle and the Pv diagram for it is given in Fig 924 The relative amounts of heat trans ferred during each process can be adjusted to approximate the actual cycle more closely Note that both the Otto and the Diesel cycles can be obtained as special cases of the dual cycle Dual cycle is a more realistic model than Diesel cycle for representing modern highspeed compression ignition engines FIGURE 923 Thermal efficiency of the ideal Diesel cycle as a function of compression and cutoff ratios k 14 07 Compression ratio r 06 05 04 03 02 01 2 4 6 8 10 12 14 16 18 20 22 24 Typical compression ratios for diesel engines rc 1 Otto 2 3 4 ηthDiesel FIGURE 924 Pv diagram of an ideal dual cycle 1 2 3 4 P Isentropic Isentropic X qin qout v EXAMPLE 94 The Ideal Diesel Cycle An ideal Diesel cycle with air as the working fluid has a compression ratio of 18 and a cutoff ratio of 2 At the beginning of the compression process the working fluid is at 147 psia 80F and 117 in3 Utilizing the coldairstandard assumptions determine a the temperature and pressure of air at the end of each process b the net work output and the thermal efficiency and c the mean effective pressure SOLUTION An ideal Diesel cycle is considered The temperature and pressure at the end of each process the net work output the thermal efficiency and the mean effective pressure are to be determined Assumptions 1 The coldairstandard assumptions are applicable and thus air can be assumed to have constant specific heats at room temperature 2 Kinetic and potential energy changes are negligible Properties The gas constant of air is R 03704 psiaft3lbmR and its other proper ties at room temperature are cp 0240 BtulbmR cv 0171 BtulbmR and k 14 Table A2Ea Final PDF to printer 492 GAS POWER CYCLES cen22672ch09475542indd 492 110617 0921 AM Analysis The PV diagram of the ideal Diesel cycle described is shown in Fig 925 We note that the air contained in the cylinder forms a closed system a The temperature and pressure values at the end of each process can be determined by utilizing the idealgas isentropic relations for processes 12 and 34 But first we determine the volumes at the end of each process from the definitions of the compres sion ratio and the cutoff ratio V 2 V 1 r 117 in 3 18 65 in 3 V 3 r c V 2 265 in 3 13 in 3 V 4 V 1 117 in 3 Process 12 isentropic compression of an ideal gas constant specific heats T 2 T 1 V 1 V 2 k 1 540 R 18 14 1 1716 R P 2 P 1 V 1 V 2 k 147 psia 18 14 841 psia Process 23 constantpressure heat addition to an ideal gas P 3 P 2 841 psia P 2 V 2 T 2 P 3 V 3 T 3 T 3 T 2 V 3 V 2 1716 R 2 3432 R Process 34 isentropic expansion of an ideal gas constant specific heats T 4 T 3 V 3 V 4 k 1 3432 R 13 in 3 117 in 3 14 1 1425 R P 4 P 3 V 3 V 4 k 841 psia 13 in 3 117 in 3 14 388 psia b The net work for a cycle is equivalent to the net heat transfer But first we find the mass of air m P 1 V 1 R T 1 147 psia117 in 3 03704 psia ft 3 lbmR540 R 1 ft 3 1728 in 3 000498 lbm Process 23 is a constantpressure heataddition process for which the boundary work and Δu terms can be combined into Δh Thus Q in m h 3 h 2 m c p T 3 T 2 000498 lbm 0240 BtulbmR 3432 1716 R 2051 Btu Process 41 is a constantvolume heatrejection process it involves no work interac tions and the amount of heat rejected is Q out m u 4 u 1 m c v T 4 T 1 000498 lbm 0171 BtulbmR 1425 540 R 0754 Btu Thus W net Q in Q out 2051 0754 1297 Btu FIGURE 925 PV diagram for the ideal Diesel cycle discussed in Example 94 1 2 3 4 P psia Isentropic Isentropic 147 V2 V118 V3 2V2 V1 V4 V qin qout Final PDF to printer 493 CHAPTER 9 cen22672ch09475542indd 493 110617 0921 AM Then the thermal efficiency becomes η th W net Q in 1297 Btu 2051 Btu 0632 or 632 The thermal efficiency of this Diesel cycle under the coldairstandard assumptions could also be determined from Eq 912 c The mean effective pressure is determined from its definition Eq 94 MEP W net V max V min W net V 1 V 2 1297 Btu 117 65 in 3 77817 lbfft 1 Btu 12 in 1 ft 110 psia Discussion Note that a constant pressure of 110 psia during the power stroke would produce the same net work output as the entire Diesel cycle 97 STIRLING AND ERICSSON CYCLES The ideal Otto and Diesel cycles discussed in the preceding sections are com posed entirely of internally reversible processes and thus are internally revers ible cycles These cycles are not totally reversible however since they involve heat transfer through a finite temperature difference during the nonisothermal heataddition and heatrejection processes which are irreversible Therefore the thermal efficiency of an Otto or Diesel engine will be less than that of a Carnot engine operating between the same temperature limits Consider a heat engine operating between a heat source at TH and a heat sink at TL For the heatengine cycle to be totally reversible the temperature difference between the working fluid and the heat source or sink should never exceed a differential amount dT during any heattransfer process That is both the heataddition and heatrejection processes during the cycle must take place isothermally one at a temperature of TH and the other at a tempera ture of TL This is precisely what happens in a Carnot cycle There are two other cycles that involve an isothermal heataddition process at TH and an isothermal heatrejection process at TL the Stirling cycle and the Ericsson cycle They differ from the Carnot cycle in that the two isentro pic processes are replaced by two constantvolume regeneration processes in the Stirling cycle and by two constantpressure regeneration processes in the Ericsson cycle Both cycles utilize regeneration a process during which heat is transferred to a thermal energy storage device called a regenerator dur ing one part of the cycle and is transferred back to the working fluid during another part of the cycle Fig 926 Figure 927b shows the Ts and Pv diagrams of the Stirling cycle which is made up of four totally reversible processes 12 T constant expansion heat addition from the external source 23 v constant regeneration internal heat transfer from the working fluid to the regenerator 34 T constant compression heat rejection to the external sink 41 v constant regeneration internal heat transfer from the regenerator back to the working fluid FIGURE 926 A regenerator is a device that borrows energy from the working fluid during one part of the cycle and pays it back without interest during another part Energy Energy Regenerator Working fluid Final PDF to printer 494 GAS POWER CYCLES cen22672ch09475542indd 494 110617 0921 AM The execution of the Stirling cycle requires rather innovative hardware The actual Stirling engines including the original one patented by Robert Stirling are heavy and complicated To spare the reader the complexities the execu tion of the Stirling cycle in a closed system is explained with the help of the hypothetical engine shown in Fig 928 This system consists of a cylinder with two pistons on each side and a regenerator in the middle The regenerator can be a wire or a ceramic mesh or any kind of porous plug with a high thermal mass mass times specific heat It is used for the temporary storage of thermal energy The mass of the working fluid contained within the regenerator at any instant is considered negligible Initially the left chamber houses the entire working fluid a gas which is at a high temperature and pressure During process 12 heat is transferred to the gas at TH from a source at TH As the gas expands isothermally the left piston moves outward doing work and the gas pressure drops During process 23 both pistons are moved to the right at the same rate to keep the volume constant until the entire gas is forced into the right chamber As the gas passes through the regenerator heat is transferred to the regenerator and the gas temperature drops from TH to TL For this heat transfer process to be reversible the temperature difference between the gas and the regenerator should not exceed a differential amount dT at any point Thus the tempera ture of the regenerator will be TH at the left end and TL at the right end of the regenerator when state 3 is reached During process 34 the right piston is moved inward compressing the gas Heat is transferred from the gas to a sink at temperature TL so that the gas temperature remains constant at TL while the FIGURE 927 Ts and Pv diagrams of Carnot Stirling and Ericsson cycles s 1 2 3 4 T s const s const TH TL 1 2 3 4 P TH const TH const TH const TL const TL const TL const 1 2 3 4 P Regeneration Regeneration 1 2 3 4 P s 1 2 3 4 T v const v const Regeneration s 1 2 3 4 P const P const Regeneration a Carnot cycle b Stirling cycle c Ericsson cycle qin qout TH TL T TH TL qin qout qin qout qin qout qin qin qout qout v v v FIGURE 928 The execution of the Stirling cycle State 1 State 2 State 3 State 4 Regenerator TH TH TL TL qin qout Final PDF to printer 495 CHAPTER 9 cen22672ch09475542indd 495 110617 0921 AM pressure rises Finally during process 41 both pistons are moved to the left at the same rate to keep the volume constant forcing the entire gas into the left chamber The gas temperature rises from TL to TH as it passes through the regenerator and picks up the thermal energy stored there during process 23 This completes the cycle Notice that the second constantvolume process takes place at a smaller volume than the first one and the net heat transfer to the regenerator dur ing a cycle is zero That is the amount of energy stored in the regenera tor during process 23 is equal to the amount picked up by the gas during process 41 The Ts and Pv diagrams of the Ericsson cycle are shown in Fig 927c The Ericsson cycle is very much like the Stirling cycle except that the two constantvolume processes are replaced by two constantpressure processes A steadyflow system operating on an Ericsson cycle is shown in Fig 929 Here the isothermal compression and expansion processes are executed in a compressor and a turbine respectively and a counterflow heat exchanger serves as a regenerator Hot and cold fluid streams enter the heat exchanger from opposite ends and heat transfer takes place between the two streams In the ideal case the temperature difference between the two fluid streams does not exceed a differential amount at any point and the cold fluid stream leaves the heat exchanger at the inlet temperature of the hot stream Both the Stirling and Ericsson cycles are totally reversible as is the Carnot cycle and thus according to the Carnot principle all three cycles must have the same thermal efficiency when operating between the same temperature limits η thStirling η thEricsson η thCarnot 1 T L T H 914 This is proved for the Carnot cycle in Example 91 and can be proved in a similar manner for both the Stirling and Ericsson cycles EXAMPLE 95 Thermal Efficiency of the Ericsson Cycle Using an ideal gas as the working fluid show that the thermal efficiency of an Erics son cycle is identical to the efficiency of a Carnot cycle operating between the same temperature limits SOLUTION It is to be shown that the thermal efficiencies of Carnot and Ericsson cycles are identical Analysis Heat is transferred to the working fluid isothermally from an external source at temperature TH during process 12 and it is rejected isothermally to an external sink at temperature TL during process 34 For a reversible isothermal process heat transfer is related to the entropy change by q T Δs The entropy change of an ideal gas during an isothermal process is Δs c p ln T e T i 0 R ln P e P i R ln P e P i FIGURE 929 A steadyflow Ericsson engine Regenerator TH const turbine wnet qin TL const compressor qout Heat Final PDF to printer 496 GAS POWER CYCLES cen22672ch09475542indd 496 110617 0921 AM Stirling and Ericsson cycles are difficult to achieve in practice because they involve heat transfer through a differential temperature difference in all com ponents including the regenerator This would require providing infinitely large surface areas for heat transfer or allowing an infinitely long time for the process Neither is practical In reality all heat transfer processes take place through a finite temperature difference the regenerator does not have an efficiency of 100 percent and the pressure losses in the regenerator are considerable Because of these limitations both Stirling and Ericsson cycles have long been of only theoretical interest However there is renewed interest in engines that operate on these cycles because of their potential for higher efficiency and better emission control The Ford Motor Company General Motors Corporation and the Phillips Research Laboratories of the Nether lands have successfully developed Stirling engines suitable for trucks buses and even automobiles More research and development are needed before these engines can compete with the gasoline or diesel engines Both the Stirling and the Ericsson engines are external combustion engines That is the fuel in these engines is burned outside the cylinder as opposed to gasoline or diesel engines where the fuel is burned inside the cylinder External combustion offers several advantages First a variety of fuels can be used as a source of thermal energy Second there is more time for combus tion and thus the combustion process is more complete which means less air pollution and more energy extraction from the fuel Third these engines operate on closed cycles and thus a working fluid that has the most desir able characteristics stable chemically inert high thermal conductivity can be utilized Hydrogen and helium are two gases commonly employed in these engines Despite the physical limitations and impracticalities associated with them both the Stirling and Ericsson cycles give a strong message to design engi neers Regeneration can increase efficiency It is no coincidence that modern gasturbine and steam power plants make extensive use of regeneration In fact the Brayton cycle with intercooling reheating and regeneration which is used in large gasturbine power plants and discussed later in this chapter closely resembles the Ericsson cycle The heat input and heat output can be expressed as q in T H s 2 s 1 T H R ln P 2 P 1 R T H ln P 1 P 2 and q out T L s 4 s 3 T L R ln P 4 P 3 R T L ln P 4 P 3 Then the thermal efficiency of the Ericsson cycle becomes η thEricsson 1 q out q in 1 R T L ln P 4 P 3 R T H ln P 1 P 2 1 T L T H since P1 P4 and P3 P2 Notice that this result is independent of whether the cycle is executed in a closed or steadyflow system Final PDF to printer 497 CHAPTER 9 cen22672ch09475542indd 497 110617 0921 AM 98 BRAYTON CYCLE THE IDEAL CYCLE FOR GASTURBINE ENGINES The Brayton cycle was first proposed by George Brayton for use in the recip rocating oilburning engine that he developed around 1870 Today it is used for gas turbines only where both the compression and expansion processes take place in rotating machinery Gas turbines usually operate on an open cycle as shown in Fig 930 Fresh air at ambient conditions is drawn into the compressor where its temperature and pressure are raised The highpressure air proceeds into the combustion chamber where the fuel is burned at con stant pressure The resulting hightemperature gases then enter the turbine where they expand to the atmospheric pressure while producing power The exhaust gases leaving the turbine are thrown out not recirculated causing the cycle to be classified as an open cycle The open gasturbine cycle just described can be modeled as a closed cycle as shown in Fig 931 by using the airstandard assumptions Here the compression and expansion processes remain the same but the combus tion process is replaced by a constantpressure heataddition process from an external source and the exhaust process is replaced by a constantpressure heatrejection process to the ambient air The ideal cycle that the working fluid undergoes in this closed loop is the Brayton cycle which is made up of four internally reversible processes 12 Isentropic compression in a compressor 23 Constantpressure heat addition 34 Isentropic expansion in a turbine 41 Constantpressure heat rejection The Ts and Pv diagrams of an ideal Brayton cycle are shown in Fig 932 Notice that all four processes of the Brayton cycle are executed in steadyflow devices thus they should be analyzed as steadyflow processes When the changes in kinetic and potential energies are neglected the energy balance for a steadyflow process can be expressed on a unitmass basis as q in q out w in w out h exit h inlet 915 Therefore heat transfers to and from the working fluid are q in h 3 h 2 c p T 3 T 2 916a and q out h 4 h 1 c p T 4 T 1 916b Then the thermal efficiency of the ideal Brayton cycle under the coldair standard assumptions becomes η thBrayton w net q in 1 q out q in 1 c p T 4 T 1 c p T 3 T 1 1 T 1 T 4 T 1 1 T 2 T 3 T 2 1 Processes 12 and 34 are isentropic and P2 P3 and P4 P1 Thus T 2 T 1 P 2 P 1 k1 k P 3 P 4 k1 k T 3 T 4 FIGURE 930 An opencycle gasturbine engine Compressor wnet Combustion chamber Turbine Fresh air Exhaust gases 1 2 3 4 Fuel FIGURE 931 A closedcycle gasturbine engine qin qout Compressor wnet Heat exchanger Heat exchanger Turbine 1 2 3 4 Final PDF to printer 498 GAS POWER CYCLES cen22672ch09475542indd 498 110617 0921 AM Substituting these equations into the thermal efficiency relation and simplify ing give η thBrayton 1 1 r p k 1 k 917 where r p P 2 P 1 918 is the pressure ratio and k is the specific heat ratio Equation 917 shows that under the coldairstandard assumptions the thermal efficiency of an ideal Brayton cycle depends on the pressure ratio of the gas turbine and the specific heat ratio of the working fluid The thermal efficiency increases with both of these parameters which is also the case for actual gas turbines A plot of thermal efficiency versus the pressure ratio is given in Fig 933 for k 14 which is the specificheatratio value of air at room temperature The highest temperature in the cycle occurs at the end of the combustion process state 3 and it is limited by the maximum temperature that the tur bine blades can withstand This also limits the pressure ratios that can be used in the cycle For a fixed turbine inlet temperature T3 the net work output per cycle increases with the pressure ratio reaches a maximum and then starts to decrease as shown in Fig 934 Therefore there should be a compromise between the pressure ratio thus the thermal efficiency and the net work out put With less work output per cycle a larger mass flow rate thus a larger system is needed to maintain the same power output which may not be eco nomical In most common designs the pressure ratio of gas turbines ranges from about 11 to 16 The air in gas turbines performs two important functions It supplies the necessary oxidant for the combustion of the fuel and it serves as a coolant to keep the temperature of various components within safe limits The second function is accomplished by drawing in more air than is needed for the com plete combustion of the fuel In gas turbines an airfuel mass ratio of 50 or above is not uncommon Therefore in a cycle analysis treating the combus tion gases as air does not cause any appreciable error Also the mass flow rate through the turbine is greater than that through the compressor the difference being equal to the mass flow rate of the fuel Thus assuming a constant mass flow rate throughout the cycle yields conservative results for openloop gas turbine engines The two major application areas of gasturbine engines are aircraft propul sion and electric power generation When it is used for aircraft propulsion the gas turbine produces just enough power to drive the compressor and a small generator to power the auxiliary equipment The highvelocity exhaust gases are responsible for producing the necessary thrust to propel the aircraft Gas turbines are also used as stationary power plants to generate electricity as standalone units or in conjunction with steam power plants on the high temperature side In these plants the exhaust gases of the gas turbine serve as the heat source for the steam The gasturbine cycle can also be executed as a closed cycle for use in nuclear power plants This time the working fluid is not limited to air and a gas with more desirable characteristics such as helium can be used FIGURE 932 Ts and Pv diagrams for the ideal Brayton cycle P s const s const 2 1 4 3 s T 2 3 4 1 P const P const a Ts diagram b Pv diagram qout qin qout qin v FIGURE 933 Thermal efficiency of the ideal Bray ton cycle as a function of the pressure ratio 5 Pressure ratio rp 07 06 05 04 03 02 01 ηthBrayton Typical pressure ratios for gas turbine engines 10 15 20 25 Final PDF to printer 499 CHAPTER 9 cen22672ch09475542indd 499 110617 0921 AM The majority of the Western worlds naval fleets already use gasturbine engines for propulsion and electric power generation The General Electric LM2500 gas turbines used to power ships have a simplecycle thermal effi ciency of 37 percent The General Electric WR21 gas turbines equipped with intercooling and regeneration have a thermal efficiency of 43 per cent and produce 216 MW 29040 hp The regeneration also reduces the exhaust temperature from 600C 1100F to 350C 650F Air is com pressed to 3 atm before it enters the intercooler Compared to steamturbine and dieselpropulsion systems the gas turbine offers greater power for a given size and weight high reliability long life and more convenient opera tion The engine startup time has been reduced from 4 h required for a typical steam propulsion system to less than 2 min for a gas turbine Many modern marine propulsion systems use gas turbines together with diesel engines because of the high fuel consumption of simplecycle gasturbine engines In combined diesel and gasturbine systems diesel is used to pro vide for efficient lowpower and cruise operation and gas turbine is used when high speeds are needed In gasturbine power plants the ratio of the compressor work to the turbine work called the back work ratio is very high Fig 935 Usually more than onehalf of the turbine work output is used to drive the compressor The situ ation is even worse when the isentropic efficiencies of the compressor and the turbine are low This is quite in contrast to steam power plants where the back work ratio is only a few percent This is not surprising however since a liquid is compressed in steam power plants instead of a gas and the steadyflow work is proportional to the specific volume of the working fluid A power plant with a high back work ratio requires a larger turbine to pro vide the additional power requirements of the compressor Therefore the tur bines used in gasturbine power plants are larger than those used in steam power plants of the same net power output Development of Gas Turbines The gas turbine has experienced phenomenal progress and growth since its first successful development in the 1930s The early gas turbines built in the 1940s and even 1950s had simplecycle efficiencies of about 17 percent because of the low compressor and turbine efficiencies and low turbine inlet temperatures due to metallurgical limitations of those times Therefore gas turbines found only limited use despite their versatility and their ability to burn a variety of fuels The efforts to improve the cycle efficiency concen trated in three areas 1 Increasing the turbine inlet or firing temperatures This has been the primary approach taken to improve gasturbine efficiency The tur bine inlet temperatures have increased steadily from about 540C 1000F in the 1940s to 1425C 2600F and even higher today These increases were made possible by the development of new materials and innovative cooling techniques for the critical components such as coating the turbine blades with ceramic layers and cooling the blades with the discharge air from the com pressor Maintaining high turbine inlet temperatures with an aircooling tech nique requires the combustion temperature to be higher to compensate for the cooling effect of the cooling air However higher combustion temperatures FIGURE 934 For fixed values of Tmin and Tmax the net work of the Brayton cycle first increases with the pressure ratio then reaches a maximum at rp TmaxTmink2k 1 and finally decreases s T 2 3 wnetmax Tmax 1000 K rp 15 rp 82 rp 2 Tmin 300 K 1 4 FIGURE 935 The fraction of the turbine work used to drive the compressor is called the back work ratio wnet Back work wturbine wcompressor Final PDF to printer 500 GAS POWER CYCLES cen22672ch09475542indd 500 110617 0921 AM increase the production of nitrogen oxides NOx which are responsible for the formation of ozone at ground level and smog Using steam as the cool ant allowed an increase in the turbine inlet temperatures by 200F without an increase in the combustion temperature Steam is also a much more effective heat transfer medium than air 2 Increasing the efficiencies of turbomachinery components The performance of early turbines suffered greatly from the inefficiencies of turbines and compressors However the advent of computers and advanced techniques for computeraided design made it possible to design these com ponents aerodynamically with minimal losses The increased efficiencies of the turbines and compressors resulted in a significant increase in the cycle efficiency 3 Adding modifications to the basic cycle The simplecycle efficien cies of early gas turbines were practically doubled by incorporating intercool ing regeneration or recuperation and reheating discussed in the next two sections These improvements of course come at the expense of increased initial and operation costs and they cannot be justified unless the decrease in fuel costs offsets the increase in other costs Relatively low fuel prices the general desire in the industry to minimize installation costs and the tre mendous increase in the simplecycle efficiency to about 40 percent left little desire to make these modifications The first gas turbine for an electric utility was installed in 1949 in Okla homa as part of a combinedcycle power plant It was built by General Electric and produced 35 MW of power Gas turbines installed until the mid 1970s suffered from low efficiency and poor reliability In the past baseload electric power generation was dominated by large coal and nuclear power plants However there has been an historic shift toward natural gasfired gas turbines because of their higher efficiencies lower capital costs shorter installation times and better emission characteristics and the abundance of natural gas supplies and more and more electric utilities are using gas tur bines for baseload power production as well as for peaking The construction costs for gasturbine power plants are roughly half that of comparable con ventional fossilfuel steam power plants which were the primary baseload power plants until the early 1980s More than half of all power plants to be installed in the foreseeable future are forecast to be gasturbine or combined gassteam turbine types A gas turbine manufactured by General Electric in the early 1990s had a pressure ratio of 135 and generated 1357 MW of net power at a thermal efficiency of 33 percent in simplecycle operation A more recent gas turbine manufactured by General Electric uses a turbine inlet temperature of 1425C 2600F and produces up to 282 MW while achieving a thermal efficiency of 395 percent in the simplecycle mode A 13ton smallscale gas turbine labeled OP16 built by the Dutch firm Opra Optimal Radial Turbine can run on gas or liquid fuel and can replace a 16ton diesel engine It has a pressure ratio of 65 and produces up to 2 MW of power Its efficiency is 26 percent in the simplecycle operation which rises to 37 percent when equipped with a regenerator The most recent simplecycle gas turbine power plants have a thermal efficiency of up to 44 percent with a single unit producing over 500 MW of net power Final PDF to printer 501 CHAPTER 9 cen22672ch09475542indd 501 110617 0921 AM EXAMPLE 96 The Simple Ideal Brayton Cycle A gasturbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8 The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet Using the airstandard assumptions determine a the gas temperature at the exits of the compressor and the turbine b the back work ratio and c the thermal efficiency SOLUTION A power plant operating on the ideal Brayton cycle is considered The compressor and turbine exit temperatures back work ratio and thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible 4 The variation of spe cific heats with temperature is to be considered Analysis The Ts diagram of the ideal Brayton cycle described is shown in Fig 936 We note that the components involved in the Brayton cycle are steadyflow devices a The air temperatures at the compressor and turbine exits are determined from isentropic relations Process 12 isentropic compression of an ideal gas T 1 300 K h 1 30019 kJkg P r1 1386 P r2 P 2 P 1 P r1 8 1386 1109 T 2 540 K at compressor exit h 2 54435 kJkg Process 34 isentropic expansion of an ideal gas T 3 1300 K h 3 139597 kJkg P r3 3309 P r4 P 4 P 3 P r3 1 8 3309 4136 T 4 770 K at turbine exit h 4 78937 kJkg b To find the back work ratio we need to find the work input to the compressor and the work output of the turbine w compin h 2 h 1 54435 30019 24416 kJkg w turbout h 3 h 4 139597 78937 60660 kJkg Thus r bw w compin w turbout 24416 kJkg 60660 kJkg 0403 That is 403 percent of the turbine work output is used just to drive the compressor c The thermal efficiency of the cycle is the ratio of the net power output to the total heat input q in h 3 h 2 139597 54435 85162 kJkg w net w out w in 60660 24416 3624 kJkg Thus η th w net q in 3624 kJkg 85162 kJkg 0426 or 426 FIGURE 936 Ts diagram for the Brayton cycle dis cussed in Example 96 s T K 2 3 4 1 P const P const wturb wcomp rp 8 1300 300 qout qin Final PDF to printer 502 GAS POWER CYCLES cen22672ch09475542indd 502 110617 0921 AM The thermal efficiency could also be determined from η th 1 q out q in where q out h 4 h 1 78937 30019 4892 kJkg Discussion Under the coldairstandard assumptions constant specific heat values at room temperature the thermal efficiency would be from Eq 917 η thBrayton 1 1 r p k1 k 1 1 8 14 1 14 0448 or 448 which is sufficiently close to the value obtained by accounting for the variation of specific heats with temperature Deviation of Actual GasTurbine Cycles from Idealized Ones The actual gasturbine cycle differs from the ideal Brayton cycle on several accounts For one thing some pressure drop during the heataddition and heatrejection processes is inevitable More importantly the actual work input to the compressor is more and the actual work output from the turbine is less because of irreversibilities The deviation of actual compressor and turbine behavior from the idealized isentropic behavior can be accurately accounted for by utilizing the isentropic efficiencies of the turbine and compressor as η C w s w a h 2s h 1 h 2a h 1 919 and η T w a w s h 3 h 4a h 3 h 4s 920 where states 2a and 4a are the actual exit states of the compressor and the turbine respectively and 2s and 4s are the corresponding states for the isen tropic case as illustrated in Fig 937 The effect of the turbine and compres sor efficiencies on the thermal efficiency of gasturbine engines is illustrated in Example 97 FIGURE 937 The deviation of an actual gas turbine cycle from the ideal Brayton cycle as a result of irreversibilities s T 2s 3 1 2a Pressure drop during heat addition Pressure drop during heat rejection 4s 4a EXAMPLE 97 An Actual GasTurbine Cycle Assuming a compressor efficiency of 80 percent and a turbine efficiency of 85 per cent determine a the back work ratio b the thermal efficiency and c the turbine exit temperature of the gasturbine cycle discussed in Example 96 SOLUTION The Brayton cycle discussed in Example 96 is reconsidered For specified turbine and compressor efficiencies the back work ratio the thermal effi ciency and the turbine exit temperature are to be determined Final PDF to printer 503 CHAPTER 9 cen22672ch09475542indd 503 110617 0921 AM Analysis a The Ts diagram of the cycle is shown in Fig 938 The actual compres sor work and turbine work are determined by using the definitions of compressor and turbine efficiencies Eqs 919 and 920 Compressor w compin w s η C 24416 kJkg 080 30520 kJkg Turbine w turbout η T w s 085 60660 kJkg 51561 kJkg Thus r bw w compin w turbout 30520 kJkg 51561 kJkg 0592 That is the compressor is now consuming 592 percent of the work produced by the turbine up from 403 percent This increase is due to the irreversibilities that occur within the compressor and the turbine b In this case air leaves the compressor at a higher temperature and enthalpy which are determined to be w compin h 2a h 1 h 2a h 1 w compin 30019 30520 60539 kJkg and T 2a 598 K Thus q in h 3 h 2a 139597 60539 79058 kJkg w net w out w in 51561 30520 21041 kJkg and η th w net q in 21041 kJkg 79058 kJkg 0266 or 266 That is the irreversibilities occurring within the turbine and compressor caused the thermal efficiency of the gas turbine cycle to drop from 426 to 266 percent This example shows how sensitive the performance of a gasturbine power plant is to the efficiencies of the compressor and the turbine In fact gasturbine efficiencies did not reach competitive values until significant improvements were made in the design of gas turbines and compressors c The air temperature at the turbine exit is determined from an energy balance on the turbine w turbout h 3 h 4a h 4a h 3 w turbout 139597 51561 88036 kJkg Then from Table A17 T 4a 853 K Discussion The temperature at turbine exit is considerably higher than that at the com pressor exit T2a 598 K which suggests the use of regeneration to reduce fuel cost FIGURE 938 Ts diagram of the gasturbine cycle discussed in Example 97 s T K 2s 3 4a 1 qin qout 1300 300 2a 4s Final PDF to printer 504 GAS POWER CYCLES cen22672ch09475542indd 504 110617 0921 AM 99 THE BRAYTON CYCLE WITH REGENERATION In gasturbine engines the temperature of the exhaust gas leaving the turbine is often considerably higher than the temperature of the air leaving the com pressor Therefore the highpressure air leaving the compressor can be heated by transferring heat to it from the hot exhaust gases in a counterflow heat exchanger which is also known as a regenerator or a recuperator A sketch of the gasturbine engine utilizing a regenerator and the Ts diagram of the new cycle are shown in Figs 939 and 940 respectively The thermal efficiency of the Brayton cycle increases as a result of regener ation since the portion of energy of the exhaust gases that is normally rejected to the surroundings is now used to preheat the air entering the combustion chamber This in turn decreases the heat input thus fuel requirements for the same net work output Note however that the use of a regenerator is rec ommended only when the turbine exhaust temperature is higher than the com pressor exit temperature Otherwise heat will flow in the reverse direction to the exhaust gases decreasing the efficiency This situation is encountered in gasturbine engines operating at very high pressure ratios The highest temperature occurring within the regenerator is T4 the tem perature of the exhaust gases leaving the turbine and entering the regener ator Under no conditions can the air be preheated in the regenerator to a temperature above this value Air normally leaves the regenerator at a lower temperature T5 In the limiting ideal case the air exits the regenerator at the inlet temperature of the exhaust gases T4 Assuming the regenerator to be well insulated and any changes in kinetic and potential energies to be negligible the actual and maximum heat transfers from the exhaust gases to the air can be expressed as q regenact h 5 h 2 921 and q regenmax h 5 h 2 h 4 h 2 922 The extent to which a regenerator approaches an ideal regenerator is called the effectiveness and is defined as q regenact q regenmax h 5 h 2 h 4 h 2 923 FIGURE 939 A gasturbine engine with regenerator Heat Regenerator wnet 2 5 6 1 3 4 Turbine Compressor Combustion chamber FIGURE 940 Ts diagram of a Brayton cycle with regeneration Regeneration s T 4 qin qout 1 3 2 5 5 6 qsaved qregen qregen Final PDF to printer 505 CHAPTER 9 cen22672ch09475542indd 505 110617 0921 AM When the coldairstandard assumptions are utilized it reduces to T 5 T 2 T 4 T 2 924 A regenerator with a higher effectiveness obviously saves a greater amount of fuel since it preheats the air to a higher temperature prior to combustion However achieving a higher effectiveness requires the use of a larger regen erator which carries a higher price tag and causes a larger pressure drop Therefore the use of a regenerator with a very high effectiveness cannot be justified economically unless the savings from the fuel costs exceed the addi tional expenses involved The effectiveness of most regenerators used in prac tice is below 085 Under the coldairstandard assumptions the thermal efficiency of an ideal Brayton cycle with regeneration is η thregen 1 T 1 T 3 r p k 1 k 925 Therefore the thermal efficiency of an ideal Brayton cycle with regeneration depends on the ratio of the minimum to maximum temperatures as well as the pressure ratio The thermal efficiency is plotted in Fig 941 for various pres sure ratios and minimumtomaximum temperature ratios This figure shows that regeneration is most effective at lower pressure ratios and low minimum tomaximum temperature ratios EXAMPLE 98 Actual GasTurbine Cycle with Regeneration Determine the thermal efficiency of the gas turbine described in Example 97 if a regenerator having an effectiveness of 80 percent is installed SOLUTION The gas turbine discussed in Example 97 is equipped with a regen erator For a specified effectiveness the thermal efficiency is to be determined Analysis The Ts diagram of the cycle is shown in Fig 942 We first determine the enthalpy of the air at the exit of the regenerator using the definition of effectiveness h 5 h 2a h 4a h 2a 080 h 5 60539 kJkg 88036 60539 kJkg h 5 82537 kJkg Thus q in h 3 h 5 139597 82537 kJkg 57060 kJkg This represents a savings of 2200 kJkg from the heat input requirements The addi tion of a regenerator assumed to be frictionless does not affect the net work output Thus η th w net q in 21041 kJkg 57060 kJkg 0369 or 369 Discussion Note that the thermal efficiency of the gas turbine has gone up from 266 to 369 percent as a result of installing a regenerator that helps to recover some of the ther mal energy of the exhaust gases FIGURE 941 Thermal efficiency of the ideal Brayton cycle with and without regeneration 5 10 15 20 25 07 06 05 04 03 02 01 ηthBrayton Pressure ratio rp With regeneration Without regeneration T1T3 02 T1T3 025 T1T3 033 FIGURE 942 Ts diagram of the regenerative Bray ton cycle described in Example 98 3 5 s T K 4a 1 qregen qsaved 1300 300 2a qin Final PDF to printer 506 GAS POWER CYCLES cen22672ch09475542indd 506 110617 0921 AM 910 THE BRAYTON CYCLE WITH INTERCOOLING REHEATING AND REGENERATION The net work of a gasturbine cycle is the difference between the turbine work output and the compressor work input and it can be increased by either decreasing the compressor work or increasing the turbine work or both It was shown in Chap 7 that the work required to compress a gas between two specified pressures can be decreased by carrying out the compression process in stages and cooling the gas in between Fig 943that is using multi stage compression with intercooling As the number of stages is increased the compression process becomes nearly isothermal at the compressor inlet temperature and the compression work decreases Likewise the work output of a turbine operating between two pressure levels can be increased by expanding the gas in stages and reheating it in betweenthat is utilizing multistage expansion with reheating This is accomplished without raising the maximum temperature in the cycle As the number of stages is increased the expansion process becomes nearly iso thermal The foregoing argument is based on a simple principle The steady flow compression or expansion work is proportional to the specific volume of the fluid Therefore the specific volume of the working fluid should be as low as possible during a compression process and as high as possible dur ing an expansion process This is precisely what intercooling and reheating accomplish Combustion in gas turbines typically occurs at four times the amount of air needed for complete combustion to avoid excessive temperatures There fore the exhaust gases are rich in oxygen and reheating can be accomplished by simply spraying additional fuel into the exhaust gases between two expan sion states The working fluid leaves the compressor at a lower temperature and the turbine at a higher temperature when intercooling and reheating are utilized This makes regeneration more attractive since a greater potential for regenera tion exists Also the gases leaving the compressor can be heated to a higher temperature before they enter the combustion chamber because of the higher temperature of the turbine exhaust A schematic of the physical arrangement and the Ts diagram of an ideal twostage gasturbine cycle with intercooling reheating and regeneration are shown in Figs 944 and 945 The gas enters the first stage of the com pressor at state 1 is compressed isentropically to an intermediate pressure P2 is cooled at constant pressure to state 3 T3 T1 and is compressed in the second stage isentropically to the final pressure P4 At state 4 the gas enters the regenerator where it is heated to T5 at constant pressure In an ideal regenerator the gas leaves the regenerator at the temperature of the turbine exhaust that is T5 T9 The primary heat addition or combustion process takes place between states 5 and 6 The gas enters the first stage of the turbine at state 6 and expands isentropically to state 7 where it enters the reheater It is reheated at constant pressure to state 8 T8 T6 where it enters the second stage of the turbine The gas exits the turbine at state 9 and enters the regener ator where it is cooled to state 10 at constant pressure The cycle is completed by cooling the gas to the initial state or purging the exhaust gases FIGURE 943 Comparison of work inputs to a singlestage compressor 1AC and a twostage compressor with intercool ing 1ABD P P2 P1 D C A B Polytropic process paths Work saved as a result of intercooling Isothermal process paths Intercooling 1 v Final PDF to printer 507 CHAPTER 9 cen22672ch09475542indd 507 110617 0921 AM It was shown in Chap 7 that the work input to a twostage compressor is minimized when equal pressure ratios are maintained across each stage It can be shown that this procedure also maximizes the turbine work output Thus for best performance we have P 2 P 1 P 4 P 3 and P 6 P 7 P 8 P 9 926 In the analysis of the actual gasturbine cycles the irreversibilities that are present within the compressor the turbine and the regenerator as well as the pressure drops in the heat exchangers should be taken into consideration The back work ratio of a gasturbine cycle improves as a result of intercool ing and reheating However this does not mean that the thermal efficiency also improves The fact is intercooling and reheating always decrease the thermal efficiency unless they are accompanied by regeneration This is because inter cooling decreases the average temperature at which heat is added and reheat ing increases the average temperature at which heat is rejected This is also apparent from Fig 945 Therefore in gasturbine power plants intercooling and reheating are always used in conjunction with regeneration If the number of compression and expansion stages is increased the ideal gasturbine cycle with intercooling reheating and regeneration approaches the Ericsson cycle as illustrated in Fig 946 and the thermal efficiency approaches the theoretical limit the Carnot efficiency However the contri bution of each additional stage to the thermal efficiency is less and less and the use of more than two or three stages cannot be justified economically FIGURE 944 A gasturbine engine with twostage compression with intercooling twostage expansion with reheating and regeneration Regenerator wnet Compressor II Compressor I Turbine I Turbine II Intercooler 10 1 2 3 4 5 6 7 8 9 Reheater Combustion chamber FIGURE 945 Ts diagram of an ideal gasturbine cycle with intercooling reheating and regeneration s T 3 4 6 qin qout 1 2 10 8 9 7 5 qregen qsaved qregen Final PDF to printer 508 GAS POWER CYCLES cen22672ch09475542indd 508 110617 0921 AM FIGURE 946 As the number of compression and expansion stages increases the gasturbine cycle with intercool ing reheating and regeneration approaches the Ericsson cycle s T THavg TLavg P const P const FIGURE 947 Ts diagram of the gasturbine cycle discussed in Example 99 s T K qprimary qout 6 1 10 8 9 7 5 4 3 2 qreheat 1300 300 EXAMPLE 99 A Gas Turbine with Reheating and Intercooling An ideal gasturbine cycle with two stages of compression and two stages of expan sion has an overall pressure ratio of 8 Air enters each stage of the compressor at 300 K and each stage of the turbine at 1300 K Determine the back work ratio and the thermal efficiency of this gasturbine cycle assuming a no regenerators and b an ideal regenerator with 100 percent effectiveness Compare the results with those obtained in Example 96 SOLUTION An ideal gasturbine cycle with two stages of compression and two stages of expansion is considered The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of no regeneration and maximum regeneration Assumptions 1 Steady operating conditions exist 2 The airstandard assumptions are applicable 3 Kinetic and potential energy changes are negligible Analysis The Ts diagram of the ideal gasturbine cycle described is shown in Fig 947 We note that the cycle involves two stages of expansion two stages of compres sion and regeneration For twostage compression and expansion the work input is minimized and the work output is maximized when both stages of the compressor and the turbine have the same pressure ratio Thus P 2 P 1 P 4 P 3 8 283 and P 6 P 7 P 8 P 9 8 283 Air enters each stage of the compressor at the same temperature and each stage has the same isentropic efficiency 100 percent in this case Therefore the tempera ture and enthalpy of the air at the exit of each compression stage will be the same A similar argument can be given for the turbine Thus At inlets T 1 T 3 h 1 h 3 and T 6 T 8 h 6 h 8 At exits T 2 T 4 h 2 h 4 and T 7 T 9 h 7 h 9 Under these conditions the work input to each stage of the compressor will be the same and so will the work output from each stage of the turbine a In the absence of any regeneration the back work ratio and the thermal efficiency are determined by using data from Table A17 as follows T 1 300 K h 1 30019 kJ kg P r1 1386 P r2 P 2 P 1 P r1 8 1386 392 T 2 4033 K h 2 40431 kJ kg T 6 1300 K h 6 139597 kJ kg P r6 3309 P r7 P 7 P 6 P r6 1 8 3309 1170 T 7 10064 K h 7 105333 kJ kg Final PDF to printer 509 CHAPTER 9 cen22672ch09475542indd 509 110617 0921 AM Then w compin 2 w compinI 2 h 2 h 1 2 40431 30019 20824 kJ kg w turbout 2 w turboutI 2 h 6 h 7 2 139597 105333 68528 kJ kg w net w turbout w compin 68528 20824 47704 kJ kg q in q primary q reheat h 6 h 4 h 8 h 7 139597 40431 139597 105333 133430 kJ kg Thus r bw w compin w turbout 20824 kJ kg 68528 kJ kg 0304 and η th w net q in 47704 kJ kg 133430 kJ kg 0358 or 358 A comparison of these results with those obtained in Example 96 singlestage compression and expansion reveals that multistage compression with intercooling and multistage expansion with reheating improve the back work ratio it drops from 0403 to 0304 but hurt the thermal efficiency it drops from 426 to 358 percent Therefore intercooling and reheating are not recommended in gasturbine power plants unless they are accompanied by regeneration b The addition of an ideal regenerator no pressure drops 100 percent effectiveness does not affect the compressor work and the turbine work Therefore the net work output and the back work ratio of an ideal gasturbine cycle are identical whether there is a regenerator or not A regenerator however reduces the heat input require ments by preheating the air leaving the compressor using the hot exhaust gases In an ideal regenerator the compressed air is heated to the turbine exit temperature T9 before it enters the combustion chamber Thus under the airstandard assumptions h5 h7 h9 The heat input and the thermal efficiency in this case are q in q primary q reheat h 6 h 5 h 8 h 7 139597 105333 139597 105333 68528 kJ kg and η th w net q in 47704 kJ kg 68528 kJ kg 0696 or 696 Discussion Note that the thermal efficiency almost doubles as a result of regeneration compared to the noregeneration case The overall effect of twostage compression and expansion with intercooling reheating and regeneration on the thermal efficiency is an increase of 63 percent As the number of compression and expansion stages is increased the cycle will approach the Ericsson cycle and the thermal efficiency will approach η thEricsson η thCarnot 1 T L T H 1 300 K 1300 K 0769 Adding a second stage increases the thermal efficiency from 426 to 696 percent an increase of 27 percentage points This is a significant increase in efficiency and usually it is well worth the extra cost associated with the second stage Adding more stages however no matter how many can increase the efficiency an additional 73 percentage points at most and usually cannot be justified economically Final PDF to printer 510 GAS POWER CYCLES cen22672ch09475542indd 510 110617 0921 AM 911 IDEAL JETPROPULSION CYCLES Gasturbine engines are widely used to power aircraft because they are light and compact and have a high powertoweight ratio Aircraft gas turbines oper ate on an open cycle called a jetpropulsion cycle The ideal jet propulsion cycle differs from the simple ideal Brayton cycle in that the gases are not expanded to the ambient pressure in the turbine Instead they are expanded to a pressure such that the power produced by the turbine is just sufficient to drive the compressor and the auxiliary equipment such as a small generator and hydraulic pumps That is the net work output of a jetpropulsion cycle is zero The gases that exit the turbine at a relatively high pressure are sub sequently accelerated in a nozzle to provide the thrust to propel the aircraft Fig 948 Also aircraft gas turbines operate at higher pressure ratios typi cally between 10 and 25 and the fluid passes through a diffuser first where it is decelerated and its pressure is increased before it enters the compressor Aircraft are propelled by accelerating a fluid in the opposite direction to motion This is accomplished by either slightly accelerating a large mass of fluid propellerdriven engine or greatly accelerating a small mass of fluid jet or turbojet engine or both turboprop engine A schematic of a turbojet engine and the Ts diagram of the ideal turbojet cycle are shown in Fig 949 The pressure of air rises slightly as it is deceler ated in the diffuser Air is compressed by the compressor It is mixed with fuel in the combustion chamber where the mixture is burned at constant pressure The highpressure and hightemperature combustion gases partially expand in the turbine producing enough power to drive the compressor and other equip ment Finally the gases expand in a nozzle to the ambient pressure and leave the engine at a high velocity In the ideal case the turbine work is assumed to equal the compressor work Also the processes in the diffuser the compressor the turbine and the nozzle are assumed to be isentropic In the analysis of actual cycles however the irre versibilities associated with these devices should be considered The effect of the irreversibilities is to reduce the thrust that can be obtained from a turbojet engine The thrust developed in a turbojet engine is the unbalanced force that is caused by the difference in the momentum of the lowvelocity air entering FIGURE 948 In jet engines the hightemperature and highpressure gases leaving the turbine are accelerated in a nozzle to provide thrust Yunus Çengel FIGURE 949 Basic components of a turbojet engine and the Ts diagram for the ideal turbojet cycle P const P const s T qin qout 6 5 4 3 2 1 Diffuser Compressor Combustion chamber Turbine Nozzle 6 5 4 3 2 1 Final PDF to printer 511 CHAPTER 9 cen22672ch09475542indd 511 110617 0921 AM the engine and the highvelocity exhaust gases leaving the engine and it is determined from Newtons second law The pressures at the inlet and the exit of a turbojet engine are identical the ambient pressure thus the net thrust developed by the engine is F m V exit m V inlet m V exit V inlet N 927 where Vexit is the exit velocity of the exhaust gases and Vinlet is the inlet veloc ity of the air both relative to the aircraft Thus for an aircraft cruising in still air Vinlet is the aircraft velocity In reality the mass flow rates of the gases at the engine exit and the inlet are different the difference being equal to the combustion rate of the fuel However the airfuel mass ratio used in jet propulsion engines is usually very high making this difference very small Thus m in Eq 927 is taken as the mass flow rate of air through the engine For an aircraft cruising at a constant speed the thrust is used to overcome air drag and the net force acting on the body of the aircraft is zero Commercial airplanes save fuel by flying at higher altitudes during long trips since air at higher altitudes is thinner and exerts a smaller drag force on aircraft The power developed from the thrust of the engine is called the propulsive power W P which is the propulsive force thrust times the dis tance this force acts on the aircraft per unit time that is the thrust times the aircraft velocity Fig 950 W P F V aircraft m V exit V inlet V aircraft kW 928 The net work developed by a turbojet engine is zero Thus we cannot define the efficiency of a turbojet engine in the same way as stationary gasturbine engines Instead we should use the general definition of efficiency which is the ratio of the desired output to the required input The desired output in a turbojet engine is the power produced to propel the aircraft W P and the required input is the heating value of the fuel Q in The ratio of these two quan tities is called the propulsive efficiency and is given by η P Propulsive power Energy input rate W P Q in 929 Propulsive efficiency is a measure of how efficiently the thermal energy released during the combustion process is converted to propulsive energy The remaining part of the energy released shows up as the kinetic energy of the exhaust gases relative to a fixed point on the ground and as an increase in the enthalpy of the gases leaving the engine EXAMPLE 910 The Ideal JetPropulsion Cycle A turbojet aircraft flies with a velocity of 850 fts at an altitude where the air is at 5 psia and 40F The compressor has a pressure ratio of 10 and the temperature of the gases at the turbine inlet is 2000F Air enters the compressor at a rate of 100 lbms Utilizing the coldairstandard assumptions determine a the temperature and pres sure of the gases at the turbine exit b the velocity of the gases at the nozzle exit and c the propulsive efficiency of the cycle FIGURE 950 Propulsive power is the thrust acting on the aircraft through a distance per unit time V ms WP FV F F Final PDF to printer 512 GAS POWER CYCLES cen22672ch09475542indd 512 110617 0921 AM FIGURE 951 Ts diagram for the turbojet cycle described in Example 910 P const P const s T F qin qout 6 5 4 3 2 2000 40 1 SOLUTION The operating conditions of a turbojet aircraft are specified The tem perature and pressure at the turbine exit the velocity of gases at the nozzle exit and the propulsive efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 The coldairstandard assump tions are applicable and thus air can be assumed to have constant specific heats at room temperature cp 0240 BtulbmF and k 14 3 Kinetic and potential energies are negligible except at the diffuser inlet and the nozzle exit 4 The turbine work output is equal to the compressor work input Analysis The Ts diagram of the ideal jet propulsion cycle described is shown in Fig 951 We note that the components involved in the jet propulsion cycle are steady flow devices a Before we can determine the temperature and pressure at the turbine exit we need to find the temperatures and pressures at other states Process 12 isentropic compression of an ideal gas in a diffuser For conve nience we can assume that the aircraft is stationary and the air is moving toward the aircraft at a velocity of V1 850 fts Ideally the air exits the diffuser with a negli gible velocity V2 0 h 2 V 2 2 2 h 1 V 1 2 2 0 c p T 2 T 1 V 1 2 2 T 2 T 1 V 1 2 2 c p 420 R 850 ft s 2 2 0240 Btu lbmR 1 Btu lbm 25037 ft 2 s 2 480 R P 2 P 1 T 2 T 1 k k 1 5 psia 480 R 420 R 14 14 1 80 psia Process 23 isentropic compression of an ideal gas in a compressor P 3 r p P 2 10 80 psia 80 psia P 4 T 3 T 2 P 3 P 2 k1 k 480 R 10 141 14 927 R Process 45 isentropic expansion of an ideal gas in a turbine Neglecting the kinetic energy changes across the compressor and the turbine and assuming the tur bine work to be equal to the compressor work we find the temperature and pressure at the turbine exit to be w compin w turbout h 3 h 2 h 4 h 5 c p T 3 T 2 c p T 4 T 5 T 5 T 4 T 3 T 2 2460 927 480 2013 R P 5 P 4 T 5 T 4 k k 1 80 psia 2013 R 2460 R 14 14 1 397 psia 0 Final PDF to printer 513 CHAPTER 9 cen22672ch09475542indd 513 110617 0921 AM FIGURE 952 Energy supplied to an aircraft from the burning of a fuel manifests itself in various forms Aircraft Qin WP propulsive power Qout excess thermal energy KEout excess kinetic energy b To find the air velocity at the nozzle exit we need to first determine the nozzle exit temperature and then apply the steadyflow energy equation Process 56 isentropic expansion of an ideal gas in a nozzle T 6 T 5 P 6 P 5 k 1 k 2013 R 5 psia 397 psia 14 1 14 1114 R h 6 V 6 2 2 h 5 V 5 2 2 0 c p T 6 T 5 V 6 2 2 V 6 2 c p T 5 T 6 2 0240 Btu lbmR 2013 1114 R 25037 ft 2 s 2 1 Btu lbm 3288 ft s c The propulsive efficiency of a turbojet engine is the ratio of the propulsive power developed W P to the total heat transfer rate to the working fluid W P m V exit V inlet V aircraft 100 lbm s 3288 850 ft s 850 ft s 1 Btu lbm 25037 ft 2 s 2 8276 Btu s or 11707 hp Q in m h 4 h 3 m c p T 4 T 3 100 lbm s 0240 Btu lbmR 2460 927 R 36794 Btu s η P W P Q in 8276 Btu s 36794 Btu s 0255 or 225 That is 225 percent of the energy input is used to propel the aircraft and to overcome the drag force exerted by the atmospheric air Discussion For those who are wondering what happened to the rest of the energy here is a brief account KE out m V g 2 2 100 lbm s 3288 850 ft s 2 2 1 Btu lbm 25037 ft 2 s 2 11867 Btu s 322 Q out m h 6 h 1 m c p T 6 T 1 100 lbm s 024 Btu lbmR 1114 420 R 16651 Btu s 453 Thus 322 percent of the energy shows up as excess kinetic energy kinetic energy of the gases relative to a fixed point on the ground Notice that for the highest propul sion efficiency the velocity of the exhaust gases relative to the ground Vg should be zero That is the exhaust gases should leave the nozzle at the velocity of the aircraft The remaining 453 percent of the energy shows up as an increase in enthalpy of the gases leaving the engine These last two forms of energy eventually become part of the internal energy of the atmospheric air Fig 952 0 Final PDF to printer 514 GAS POWER CYCLES cen22672ch09475542indd 514 110617 0921 AM Modifications to Turbojet Engines The first airplanes built were all propellerdriven with propellers powered by engines essentially identical to automobile engines The major breakthrough in commercial aviation occurred with the introduction of the turbojet engine in 1952 Both propellerdriven engines and jetpropulsiondriven engines have their own strengths and limitations and several attempts have been made to combine the desirable characteristics of both in one engine Two such mod ifications are the propjet engine and the turbofan engine The most widely used engine in aircraft propulsion is the turbofan or fanjet engine wherein a large fan driven by the turbine forces a consider able amount of air through a duct cowl surrounding the engine as shown in Figs 953 and 954 The fan exhaust leaves the duct at a higher velocity enhancing the total thrust of the engine significantly A turbofan engine is based on the principle that for the same power a large volume of slower moving air produces more thrust than a small volume of fastmoving air The first commercial turbofan engine was successfully tested in 1955 FIGURE 953 A turbofan engine Lowpressure compressor Fan Duct Combustion chamber Fan Highpressure compressor Lowpressure turbine Highpressure turbine Fan exhaust Turbine exhaust FIGURE 954 A modern jet engine used to power Boeing 777 aircraft This is a Pratt Whitney PW4084 turbofan capable of producing 84000 pounds of thrust It is 487 m 192 in long has a 284 m 112 in diameter fan and it weighs 6800 kg 15000 lbm Reproduced by permission of United Technologies Corporation Pratt Whitney Fan Air inlet Low pressure compressor Fan air bypassing the jet engine High pressure compressor Low pressure turbine to turn inner shaft 2stage high pressure turbine to turn outer shaft Combustors Thrust Thrust Twin spool shaft to turn the fan and the compressors Final PDF to printer 515 CHAPTER 9 cen22672ch09475542indd 515 110617 0921 AM The turbofan engine on an airplane can be distinguished from the less effi cient turbojet engine by its fat cowling covering the large fan All the thrust of a turbojet engine is due to the exhaust gases leaving the engine at about twice the speed of sound In a turbofan engine the highspeed exhaust gases are mixed with the lowerspeed air which results in a considerable reduction in noise New cooling techniques have resulted in considerable increases in efficien cies by allowing gas temperatures at the burner exit to reach over 1500C which is more than 100C above the melting point of the turbine blade mate rials Turbofan engines deserve most of the credit for the success of jumbo jets that weigh almost 400000 kg and are capable of carrying over 400 pas sengers for up to a distance of 10000 km at speeds over 950 kmh with less fuel per passenger mile The ratio of the mass flow rate of air bypassing the combustion chamber to that of air flowing through it is called the bypass ratio The first commercial highbypassratio engines had a bypass ratio of 5 Increasing the bypass ratio of a turbofan engine increases thrust Thus it makes sense to remove the cowl from the fan The result is a propjet engine as shown in Fig 955 Turbofan and propjet engines differ primarily in their bypass ratios 5 or 6 for turbo fans and as high as 100 for propjets As a general rule propellers are more efficient than jet engines but they are limited to lowspeed and lowaltitude operation since their efficiency decreases at high speeds and altitudes The old propjet engines turboprops were limited to speeds of about Mach 062 and to altitudes of around 9100 m The new propjet engines propfans are expected to achieve speeds of about Mach 082 and altitudes of about 12200 m Commercial airplanes of medium size and range propelled by propfans are expected to fly as high and as fast as the planes propelled by turbofans and to do so on less fuel Another modification that is popular in military aircraft is the addition of an afterburner section between the turbine and the nozzle Whenever a need for extra thrust arises such as for short takeoffs or combat conditions additional fuel is injected into the oxygenrich combustion gases leaving the turbine As a result of this added energy the exhaust gases leave at a higher velocity providing a greater thrust A ramjet engine is a properly shaped duct with no compressor or turbine as shown in Fig 956 and is sometimes used for highspeed propulsion of mis siles and aircraft The pressure rise in the engine is provided by the ram effect of the incoming highspeed air being rammed against a barrier Therefore a FIGURE 955 A turboprop engine Propeller Compressor Combustion chamber Turbine Gear reduction Final PDF to printer 516 GAS POWER CYCLES cen22672ch09475542indd 516 110617 0921 AM ramjet engine needs to be brought to a sufficiently high speed by an external source before it can be fired The ramjet performs best in aircraft flying above Mach 2 or 3 two or three times the speed of sound In a ramjet the air is slowed down to about Mach 02 fuel is added to the air and burned at this low velocity and the combus tion gases are expanded and accelerated in a nozzle A scramjet engine is essentially a ramjet in which air flows through at supersonic speeds above the speed of sound Ramjets that convert to scram jet configurations at speeds above Mach 6 are successfully tested at speeds of about Mach 8 Finally a rocket is a device where a solid or liquid fuel and an oxidizer react in the combustion chamber The highpressure combustion gases are then expanded in a nozzle The gases leave the rocket at very high velocities producing the thrust to propel the rocket 912 SECONDLAW ANALYSIS OF GAS POWER CYCLES The ideal Carnot Ericsson and Stirling cycles are totally reversible thus they do not involve any irreversibilities The ideal Otto Diesel and Brayton cycles however are only internally reversible and they may involve irrevers ibilities external to the system A secondlaw analysis of these cycles reveals where the largest irreversibilities occur and where to start improvements Relations for exergy and exergy destruction for both closed and steadyflow systems are developed in Chap 8 The exergy destruction for a closed system can be expressed as X dest T 0 S gen T 0 Δ S sys S in S out T 0 S 2 S 1 sys Q in T bin Q out T bout kJ 930 where Tbin and Tbout are the temperatures of the system boundary where heat is transferred into and out of the system respectively A similar relation for steadyflow systems can be expressed in rate form as X dest T 0 S gen T 0 S out S in T 0 out m s in m s Q in T bin Q out T bout kW 931 FIGURE 956 A ramjet engine Air inlet Fuel sprayers Exhaust Flame holders Final PDF to printer 517 CHAPTER 9 cen22672ch09475542indd 517 110617 0921 AM or on a unitmass basis for a oneinlet oneexit steadyflow device as x dest T 0 s gen T 0 s e s i q in T bin q out T bout kJ kg 932 where subscripts i and e denote the inlet and exit states respectively The exergy destruction of a cycle is the sum of the exergy destructions of the processes that compose that cycle The exergy destruction of a cycle can also be determined without tracing the individual processes by consider ing the entire cycle as a single process and using one of the relations above Entropy is a property and its value depends on the state only For a cycle reversible or actual the initial and the final states are identical thus se si Therefore the exergy destruction of a cycle depends on the magnitude of the heat transfer with the high and lowtemperature reservoirs involved and on their temperatures It can be expressed on a unitmass basis as x dest T 0 q out T bout q in T bin kJ kg 933 For a cycle that involves heat transfer only with a source at TH and a sink at TL the exergy destruction becomes x dest T 0 q out T L q in T H kJ kg 934 The exergies of a closed system ϕ and a fluid stream ψ at any state can be determined from ϕ u u 0 T 0 s s 0 P 0 v v 0 V 2 2 gz kJ kg 935 and ψ h h 0 T 0 s s 0 V 2 2 gz kJ kg 936 where subscript 0 denotes the state of the surroundings EXAMPLE 911 SecondLaw Analysis of an Otto Cycle Consider an engine operating on the ideal Otto cycle with a compression ratio of 8 Fig 957 At the beginning of the compression process air is at 100 kPa and 17C During the constantvolume heataddition process 800 kJkg of heat is transferred to air from a source at 1700 K and waste heat is rejected to the surroundings at 300 K Accounting for the variation of specific heats of air with temperature determine a the exergy destruction associated with each of the four processes and the cycle and b the secondlaw efficiency of this cycle SOLUTION An engine operating on the ideal Otto cycle is considered For specific source and sink temperatures the exergy destruction associated with this cycle and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible FIGURE 957 Schematic for Example 911 1 2 3 4 P kPa 100 Isentropic Isentropic qin qout v2 v3 v1 v1 v4 v 18 Final PDF to printer 518 GAS POWER CYCLES cen22672ch09475542indd 518 110617 0921 AM Analysis a We take the engine bordering the heat source at temperature TH and the environment at temperature T0 as the system This cycle was analyzed in Example 93 and various quantities were given or determined to be r 8 P 2 17997 MPa T 0 290 K P 3 4345 MPa T 1 290 K q in 800 kJ kg T 2 6524 K q out 38183 kJ kg T 3 15751 K w net 41817 kJ kg Processes 12 and 34 are isentropic s1 s2 s3 s4 and therefore do not involve any internal or external irreversibilities that is Xdest12 0 and Xdest34 0 Processes 23 and 41 are constantvolume heataddition and heatrejection pro cesses respectively and are internally reversible However the heat transfer between the working fluid and the source or the sink takes place through a finite temperature difference rendering both processes irreversible The exergy destruction associated with each process is determined from Eq 932 However first we need to determine the entropy change of air during these processes s 3 s 2 s 3 s 2 R ln P 3 P 2 35045 24975 kJ kgK 0287 kJ kgK ln 4345 MPa 17997 MPa 07540 kJ kgK Also q in 800 kJ kg and T source 1700 K Thus x dest23 T 0 s 3 s 2 sys q in T source 290 K 07540 kJ kgK 800 kJ kg 1700 K 822 kJ kg For process 41 s1 s4 s2 s3 07540 kJkgK q41 qout 38183 kJkg and Tsink 290 K Thus x dest41 T 0 s 1 s 4 sys q out T sink 290 K 07540 kJ kgK 38183 kJ kg 290 K 1632 kJ kg Therefore the irreversibility of the cycle is x destcycle x dest12 x dest23 x dest34 x dest41 0 822 kJ kg 0 1632 kJ kg 2454 kJ kg Final PDF to printer 519 CHAPTER 9 cen22672ch09475542indd 519 110617 0921 AM Twothirds of the oil used in the United States is used for transportation Half of this oil is consumed by passenger cars and light trucks that are used to com mute to and from work 38 percent to run a family business 35 percent and for recreational social and religious activities 27 percent The overall fuel efficiency of the vehicles has increased considerably over the years due to improvements primarily in aerodynamics materials and electronic controls However the average fuel consumption of new vehicles has not changed much from about 20 miles per gallon mpg because of the increasing consumer trend toward purchasing larger and less fuelefficient cars trucks and sport utility vehicles Motorists also continue to drive more each year 13476 miles in 2010 compared to 10277 miles in 1990 Also the annual gasoline use per vehicle in the United States has increased to 580 gallons in 2010 worth 2320 at 400gal from 506 gallons in 1990 Fig 958 The exergy destruction of the cycle could also be determined from Eq 934 Notice that the largest exergy destruction in the cycle occurs during the heatrejection pro cess Therefore any attempt to reduce the exergy destruction should start with this process b The secondlaw efficiency is defined as η II Exergy recovered Exergy expended x recovered x expended 1 x destroyed x expended Here the expended energy is the energy content of the heat supplied to the air in the engine which is its work potential and the energy recovered is the net work output x expended x heatin 1 T 0 T H q in 1 2900 K 1700 K 800 kJ kg 6635 kJ kg x recovered w netout 41817 kJ kg Substituting the secondlaw efficiency of this cycle is determined to be η II x recovered x expended 41817 kJ kg 6635 kJ kg 0630 or 630 Discussion The secondlaw efficiency can also be determined using the exergy destruction data η II 1 x destroyed x expended 1 2454 kJ kg 6635 kJ kg 0630 or 630 Note that the exergy destruction associated with heat transfer involving both the heat source and the environment are accounted for in the results FIGURE 958 The average car in the United States is driven about 13500 miles a year and uses about 580 gallons of gasoline worth 2300 at 400gal TOPIC OF SPECIAL INTEREST Saving Fuel and Money by Driving Sensibly This section can be skipped without a loss in continuity Information in this section is based largely on the publications of the US Department of Energy Environmental Pro tection Agency and the American Automotive Association Final PDF to printer 520 GAS POWER CYCLES cen22672ch09475542indd 520 110617 0921 AM Saving fuel is not limited to good driving habits It also involves purchasing the right car using it responsibly and maintaining it properly A car does not burn any fuel when it is not running and thus a sure way to save fuel is not to drive the car at allbut this is not the reason we buy a car We can reduce driv ing and thus fuel consumption by considering viable alternatives such as living close to work and shopping areas working at home working longer hours in fewer days joining a car pool or starting one using public transportation combining errands into a single trip and planning ahead avoiding rush hours and roads with heavy traffic and many traffic lights and simply walking or bicycling instead of driving to nearby places with the added benefit of good health and physical fitness Driving only when necessary is the best way to save fuel money and the environment too Driving efficiently starts before buying a car just like raising good children starts before getting married The buying decision made now will affect the fuel consumption for many years Under average driving conditions the owner of a 30mpg vehicle will spend 900 less each year on fuel than the owner of a 20mpg vehicle assuming a fuel cost of 400 per gallon and 13500 miles of driving per year If the vehicle is owned for 5 years the 30mpg vehicle will save 4500 during this period Fig 959 The fuel consumption of a car depends on many factors such as the type of the vehicle the weight the trans mission type the size and efficiency of the engine and the accessories and the options installed The most fuelefficient cars are aerodynamically designed compact cars with a small engine manual transmission low frontal area the height times the width of the car and bare essentials At highway speeds most fuel is used to overcome aerodynamic drag or air resistance to motion which is the force needed to move the vehicle through the air This resistance force is proportional to the drag coefficient and the frontal area Therefore for a given frontal area a sleeklooking aerodynami cally designed vehicle with contoured lines that coincide with the streamlines of airflow has a smaller drag coefficient and thus better fuel economy than a boxlike vehicle with sharp corners Fig 960 For the same overall shape a compact car has a smaller frontal area and thus better fuel economy compared to a large car Moving around the extra weight requires more fuel and thus it hurts fuel economy Therefore the lighter the vehicle the more fuelefficient it is Also as a general rule the larger the engine is the greater its rate of fuel consump tion is So you can expect a car with a 18 L engine to be more fuel efficient than one with a 30 L engine For a given engine size diesel engines operate on much higher compression ratios than the gasoline engines and thus they are inherently more fuel efficient Manual transmissions are usually more effi cient than the automatic ones but this is not always the case A car with auto matic transmission generally uses 10 percent more fuel than a car with manual transmission because of the losses associated with the hydraulic connection between the engine and the transmission and the added weight Transmis sions with an overdrive gear found in fourspeed automatic transmissions and fivespeed manual transmissions save fuel and reduce noise and engine wear during highway driving by decreasing the engine rpm while maintaining the same vehicle speed FIGURE 959 Under average driving conditions the owner of a 30mpg vehicle spends 900 less each year on gasoline than the owner of a 20mpg vehicle assum ing 400gal and 13500 milesyr 30 MPG 20 MPG 1800yr 2700yr FIGURE 960 Aerodynamically designed vehicles have a smaller drag coefficient and thus better fuel economy than boxlike vehicles with sharp corners Final PDF to printer 521 CHAPTER 9 cen22672ch09475542indd 521 110617 0921 AM Front wheel drive offers better traction because of the engine weight on top of the front wheels reduced vehicle weight and thus better fuel econ omy with an added benefit of increased space in the passenger compartment Fourwheeldrive mechanisms provide better traction and braking and thus safer driving on slippery roads and loose gravel by transmitting torque to all four wheels However the added safety comes with increased weight noise and cost and decreased fuel economy Radial tires usually reduce the fuel consumption by 5 to 10 percent by reducing the rolling resistance but their pressure should be checked regularly since they can look normal and still be underinflated Cruise control saves fuel during long trips on open roads by maintaining steady speed Tinted windows and light interior and exterior col ors reduce solar heat gain and thus the need for air conditioning BEFORE DRIVING Certain things done before driving can make a significant difference on the fuel cost of the vehicle while driving Next we discuss some measures such as using the right kind of fuel minimizing idling removing extra weight and keeping the tires properly inflated Use Fuel with the Minimum Octane Number Recommended by the Vehicle Manufacturer Many motorists buy higherpriced premium fuel thinking that it is better for the engine Most of todays cars are designed to operate on regular unleaded fuel If the owners manual does not call for premium fuel using anything other than regular gas is simply a waste of money Octane number is not a measure of the power or quality of the fuel it is simply a measure of fuels resistance to engine knock caused by premature ignition Despite the implica tions of flashy names like premium super or power plus a fuel with a higher octane number is not a better fuel it is simply more expensive because of the extra processing involved to raise the octane number Fig 961 Older cars may need to go up one grade level from the recommended new car octane number if they start knocking Do Not Overfill the Gas Tank Topping off the gas tank may cause the fuel to backflow during pumping In hot weather an overfilled tank may also cause the fuel to overflow due to ther mal expansion This wastes fuel pollutes the environment and may damage the cars paint Also fuel tank caps that do not close tightly allow some gaso line to be lost by evaporation Buying fuel in cool weather such as early in the mornings minimizes evaporative losses Each gallon of spilled or evaporated fuel emits as much hydrocarbon to the air as 7500 miles of driving Park in the Garage The engine of a car parked in a garage overnight is warmer the next morning This reduces the problems associated with the warmingup period such as start ing excessive fuel consumption and environmental pollution In hot weather a garage blocks the direct sunlight and reduces the need for air conditioning FIGURE 961 Despite the implications of flashy names a fuel with a higher octane number is not a better fuel it is simply more expensive Shutterstock RF Final PDF to printer 522 GAS POWER CYCLES cen22672ch09475542indd 522 110617 0921 AM Start the Car Properly and Avoid Extended Idling With todays cars it is not necessary to prime the engine first by pumping the accelerator pedal repeatedly before starting This only wastes fuel Warming up the engine isnt necessary either Keep in mind that an idling engine wastes fuel and pollutes the environment Dont race a cold engine to warm it up An engine warms up faster on the road under a light load and the catalytic con verter begins to function sooner Start driving as soon as the engine is started but avoid rapid acceleration and highway driving before the engine and thus the oil fully warms up to prevent engine wear In cold weather the warmup period is much longer the fuel consumption during warmup is much higher and the exhaust emissions are much greater At 20C for example a car needs to be driven at least three miles to warm up fully A gasoline engine uses up to 50 percent more fuel during warmup than it does after it is warmed up Exhaust emissions from a cold engine dur ing warmup are much higher since the catalytic converters do not function properly before reaching their normal operating temperature of about 390C Dont Carry Unnecessary Weight In or On the Vehicle Remove any snow or ice from the vehicle and avoid carrying unneeded items especially heavy ones such as snow chains old tires books in the passenger compartment trunk or the cargo area of the vehicle Fig 962 This wastes fuel since it requires extra fuel to carry around the extra weight An extra 100 lbm decreases fuel economy of a car by about 12 percent Some people find it convenient to use a roof rack or carrier for additional cargo space However if you must carry some extra items place them inside the vehicle rather than on roof racks to reduce drag Any snow that accumu lates on a vehicle and distorts its shape must be removed for the same reason A loaded roof rack can increase fuel consumption by up to 5 percent in high way driving Even the most streamlined empty rack increases aerodynamic drag and thus fuel consumption Therefore the roof rack should be removed when it is no longer needed Keep Tires Inflated to the Recommended Maximum Pressure Keeping the tires inflated properly is one of the easiest and most important things one can do to improve fuel economy If a range is recommended by the manufacturer the higher pressure should be used to maximize fuel efficiency Tire pressure should be checked when the tire is cold since tire pressure changes with temperature it increases by 1 psi for every 10F rise in temperature due to a rise in ambient temperature or just road friction Underinflated tires run hot and jeopardize safety cause the tires to wear prematurely affect the vehicles handling adversely and hurt the fuel economy by increasing the rolling resis tance Overinflated tires cause unpleasant bumpy rides and cause the tires to wear unevenly Tires lose about 1 psi pressure per month due to air loss caused FIGURE 962 A loaded roof rack can increase fuel consumption by up to 5 percent in highway driving Final PDF to printer 523 CHAPTER 9 cen22672ch09475542indd 523 110617 0921 AM FIGURE 963 Tire pressure should be checked at least once a month since underinflated tires often cause fuel consumption of vehicles to increase by 5 or 6 percent ShutterstockMinerva Studio by the tire hitting holes bumps and curbs Therefore the tire pressure should be checked at least once a month Just one tire underinflated by 2 psi results in a 1 percent increase in fuel consumption Underinflated tires often cause fuel consumption of vehicles to increase by 5 or 6 percent Fig 963 It is also important to keep the wheels aligned Driving a vehicle with the front wheels out of alignment increases rolling resistance and thus fuel con sumption while causing handling problems and uneven tire wear Therefore the wheels should be aligned properly whenever necessary Recently devel oped low rolling resistance tires can provide up to 9 percent fuel savings WHILE DRIVING The driving habits can make a significant difference in the amount of fuel used Driving sensibly and practicing some fuelefficient driving techniques such as those discussed below can improve fuel economy easily by more than 10 percent Avoid Quick Starts and Sudden Stops Despite the attention they may attract abrupt aggressive jackrabbit starts waste fuel wear the tires jeopardize safety and are harder on vehicle com ponents and connectors Sudden squealing stops wear the brake pads prema turely and may cause the driver to lose control of the vehicle Easy starts and stops save fuel reduce wear and tear reduce pollution and are safer and more courteous to other drivers Drive at Moderate Speeds Avoiding high speeds on open roads results in safer driving and better fuel economy In highway driving over 50 percent of the power produced by the engine is used to overcome aerodynamic drag ie to push air out of the way Aerodynamic drag and thus fuel consumption increase rapidly at speeds above 55 mph as shown in Fig 964 On average a car uses about 15 percent more fuel at 65 mph and 25 percent more fuel at 70 mph than it does at 55 mph A car uses about 10 percent more fuel at 100 kmh and 20 percent more fuel at 110 kmh than it does at 90 kmh The preceding discussion should not lead one to conclude that the lower the speed the better the fuel economybecause this is not the case The number of miles that can be driven per gallon of fuel drops sharply at speeds below 30 mph or 50 kmh as shown in Fig 964 Besides speeds slower than the flow of traffic can create a traffic hazard Therefore a car should be driven at moderate speeds for safety and best fuel economy Maintain a Constant Speed The fuel consumption remains at a minimum during steady driving at a moder ate speed Keep in mind that every time the accelerator is pressed hard more fuel is pumped into the engine The vehicle should be accelerated gradually and smoothly since extra fuel is squirted into the engine during quick accelera tion Using cruise control on highway trips can help maintain a constant speed and reduce fuel consumption Steady driving is also safer easier on the nerves and better for the heart FIGURE 964 Aerodynamic drag increases and thus fuel economy decreases rapidly at speeds above 55 mph Source EPA and US Dept of Energy 15 15 20 25 MPG 30 35 25 35 45 Speed mph 55 65 75 Final PDF to printer 524 GAS POWER CYCLES cen22672ch09475542indd 524 110617 0921 AM Anticipate Traffic Ahead and Avoid Tailgating A driver can reduce fuel consumption by up to 10 percent by anticipating traf fic conditions ahead and adjusting the speed accordingly and by avoiding tailgating and thus unnecessary braking and acceleration Fig 965 Accel erations and decelerations waste fuel Braking and abrupt stops can be mini mized for example by not following too closely and slowing down gradually by releasing the gas pedal when approaching a red light a stop sign or slow traffic This relaxed driving style is safer saves fuel and money reduces pollu tion reduces wear on the tires and brakes and is appreciated by other drivers Allowing sufficient time to reach the destination makes it easier to resist the urge to tailgate Avoid Sudden Acceleration and Sudden Braking Except in Emergencies Accelerate gradually and smoothly when passing other vehicles or merging with faster traffic Pumping or stomping on the accelerator pedal while driv ing causes the engine to switch to a fuel enrichment mode of operation that wastes fuel In city driving nearly half of the engine power is used for accel eration When accelerating with stickshifts the RPM of the engine should be kept to a minimum Braking wastes the mechanical energy produced by the engine and wears the brake pads Avoid Resting Feet on the Clutch or Brake Pedal While Driving Resting the left foot on the brake pedal increases the temperature of the brake components and thus reduces their effectiveness and service life while wasting fuel Similarly resting the left foot on the clutch pedal lessens the pressure on the clutch pads causing them to slip and wear prematurely wasting fuel Use Highest Gear Overdrive During Highway Driving Overdrive improves fuel economy during highway driving by decreasing the vehicles engine speed or RPM The lower engine speed reduces fuel con sumption per unit time as well as engine wear Therefore overdrive the fifth gear in cars with overdrive manual transmission should be used as soon as the vehicles speed is high enough Turn the Engine Off Rather Than Letting It Idle Unnecessary idling during lengthy waits such as waiting for someone or for service at a driveup window being stuck in traffic etc wastes fuel pollutes the air and causes engine wear more wear than driving Fig 966 There fore the engine should be turned off rather than letting it idle Idling for more than a minute consumes much more fuel than restarting the engine Fuel con sumption in the lines of driveup windows and the pollution emitted can be avoided altogether by simply parking the car and going inside FIGURE 965 Fuel consumption can be decreased by up to 10 percent by anticipating traffic conditions ahead and adjusting accordingly PhotoDiscGetty Images RF FIGURE 966 Unnecessary idling during lengthy waits wastes fuel costs money and pollutes the air Final PDF to printer 525 CHAPTER 9 cen22672ch09475542indd 525 110617 0921 AM Use the Air Conditioner Sparingly Air conditioning consumes considerable power and thus increases fuel con sumption by 3 to 4 percent during highway driving and by as much as 10 per cent during city driving Fig 967 The best alternative to air conditioning is to supply fresh outdoor air to the car through the vents by turning on the flowthrough ventilation system usually by running the air conditioner in the economy mode while keeping the windows and the sunroof closed This measure is adequate to achieve comfort in pleasant weather and it saves the most fuel since the compressor of the air conditioner is off In warmer weather however ventilation cannot provide adequate cooling effect In that case we can try to achieve comfort by rolling down the windows or opening the sun roof This is certainly a viable alternative for city driving but not so on high ways since the aerodynamic drag caused by wideopen windows or sunroof at highway speeds consumes more fuel than does the air conditioner Therefore at highway speeds the windows or the sunroof should be closed and the air conditioner should be turned on instead to save fuel This is especially the case for the newer aerodynamically designed cars Most air conditioners have a maximum or recirculation setting that reduces the amount of hot outside air that must be cooled and thus the fuel consumption for air conditioning A passive measure to reduce the need for air conditioning is to park the vehicle in the shade and to leave the windows slightly open to allow for air circulation AFTER DRIVING You cannot be an efficient person and accomplish much unless you take good care of yourself eating right maintaining physical fitness having checkups etc and the cars are no exception Regular maintenance improves perfor mance increases gas mileage reduces pollution lowers repair costs and extends engine life A little time and money saved now may cost a lot later in increased fuel repair and replacement costs Proper maintenance such as checking the levels of fluids engine oil coolant transmission brake power steering windshield washer etc the tightness of all belts and formation of cracks or frays on hoses belts and wires keep ing tires properly inflated lubricating the moving components and replacing clogged air fuel or oil filters maximizes fuel efficiency Fig 968 Clogged air filters increase fuel consumption by up to 10 percent and pollution by restricting airflow to the engine and thus they should be replaced The car should be tuned up regularly unless it has electronic controls and a fuel injection system High temperatures which may be due to a malfunction of the cooling fan should be avoided as they may cause the breakdown of the engine oil and thus excessive wear of the engine and low temperatures which may be due to a malfunction of the thermostat may extend the engines warmup period and may prevent the engine from reaching the optimum operating con ditions Both effects reduce fuel economy Clean oil extends engine life by reducing engine wear caused by fric tion removes acids sludge and other harmful substances from the engine improves performance reduces fuel consumption and decreases air pollution Oil also helps to cool the engine provides a seal between the cylinder walls FIGURE 967 Air conditioning increases fuel con sumption by 3 to 4 percent during highway driving and by as much as 10 percent during city driving FIGURE 968 Proper maintenance maximizes fuel efficiency and extends engine life Final PDF to printer 526 GAS POWER CYCLES cen22672ch09475542indd 526 110617 0921 AM and the pistons and prevents the engine from rusting Therefore oil and oil filter should be changed as recommended by the vehicle manufacturer Fuel efficient oils indicated by Energy Efficient API label contain certain addi tives that reduce friction and increase a vehicles fuel economy by 3 percent or more In summary a person can save fuel money and the environment by purchas ing an energyefficient vehicle minimizing the amount of driving being fuel conscious while driving and maintaining the car properly These measures have the added benefits of enhanced safety reduced maintenance costs and extended vehicle life SUMMARY A cycle during which a net amount of work is produced is called a power cycle and a power cycle during which the working fluid remains a gas throughout is called a gas power cycle The most efficient cycle operating between a heat source at temperature TH and a sink at temperature TL is the Carnot cycle and its thermal efficiency is given by η thCarnot 1 T L T H The actual gas cycles are rather complex The approximations used to simplify the analysis are known as the airstandard assumptions Under these assumptions all the processes are assumed to be internally reversible the working fluid is assumed to be air which behaves as an ideal gas and the com bustion and exhaust processes are replaced by heat addition and heatrejection processes respectively The airstandard assumptions are called coldairstandard assumptions if air is also assumed to have constant specific heats at room temperature In reciprocating engines the compression ratio r and the mean effective pressure MEP are defined as r V max V min V BDC V TDC MEP w net v max v min The Otto cycle is the ideal cycle for sparkignition recipro cating engines and it consists of four internally reversible processes isentropic compression constantvolume heat addition isentropic expansion and constantvolume heat rejection Under coldairstandard assumptions the thermal efficiency of the ideal Otto cycle is η thOtto 1 1 r k1 where r is the compression ratio and k is the specific heat ratio cpcv The Diesel cycle is the ideal cycle for compression ignition reciprocating engines It is very similar to the Otto cycle except that the constantvolume heataddition process is replaced by a constantpressure heataddition process Its thermal efficiency under coldairstandard assumptions is η thDiesel 1 1 r k1 r c k 1 k r c 1 where rc is the cutoff ratio defined as the ratio of the cylinder volumes after and before the combustion process Stirling and Ericsson cycles are two totally reversible cycles that involve an isothermal heataddition process at TH and an isothermal heatrejection process at TL They differ from the Carnot cycle in that the two isentropic processes are replaced by two constantvolume regeneration processes in the Stirling cycle and by two constantpressure regeneration processes in the Ericsson cycle Both cycles utilize regeneration a process during which heat is transferred to a thermal energy storage device called a regenerator during one part of the cycle then transferred back to the working fluid during another part of the cycle The ideal cycle for modern gasturbine engines is the Brayton cycle which is made up of four internally reversible processes isentropic compression constantpressure heat addition isentropic expansion and constantpressure heat rejection Under coldairstandard assumptions its thermal efficiency is η thBrayton 1 1 r p k1 k Final PDF to printer 527 CHAPTER 9 cen22672ch09475542indd 527 110617 0921 AM where rp PmaxPmin is the pressure ratio and k is the specific heat ratio The thermal efficiency of the simple Brayton cycle increases with the pressure ratio The deviation of the actual compressor and the turbine from the idealized isentropic ones can be accurately accounted for by utilizing their isentropic efficiencies defined as η C w s w a h 2s h 1 h 2a h 1 and η T w a w s h 3 h 4a h 3 h 4s where states 1 and 3 are the inlet states 2a and 4a are the actual exit states and 2s and 4s are the isentropic exit states In gasturbine engines the temperature of the exhaust gas leaving the turbine is often considerably higher than the tem perature of the air leaving the compressor Therefore the highpressure air leaving the compressor can be heated by transferring heat to it from the hot exhaust gases in a coun terflow heat exchanger which is also known as a regenerator The extent to which a regenerator approaches an ideal regen erator is called the effectiveness and is defined as q regenact q regenmax Under coldairstandard assumptions the thermal efficiency of an ideal Brayton cycle with regeneration becomes η thregen 1 T 1 T 3 r p k1 k where T1 and T3 are the minimum and maximum tempera tures respectively in the cycle The thermal efficiency of the Brayton cycle can also be increased by utilizing multistage compression with intercool ing regeneration and multistage expansion with reheating The work input to the compressor is minimized when equal pressure ratios are maintained across each stage This proce dure also maximizes the turbine work output Gasturbine engines are widely used to power aircraft because they are light and compact and have a high power toweight ratio The ideal jetpropulsion cycle differs from the simple ideal Brayton cycle in that the gases are partially expanded in the turbine The gases that exit the turbine at a relatively high pressure are subsequently accelerated in a noz zle to provide the thrust needed to propel the aircraft The net thrust developed by the engine is F m V exit V inlet where m is the mass flow rate of gases Vexit is the exit velocity of the exhaust gases and Vinlet is the inlet velocity of the air both relative to the aircraft The power developed from the thrust of the engine is called the propulsive power W P and is given by W P m V exit V inlet V aircraft Propulsive efficiency is a measure of how efficiently the energy released during the combustion process is converted to propulsive energy and it is defined as η P Propulsive power Energy input rate W P Q in For an ideal cycle that involves heat transfer only with a source at TH and a sink at TL the exergy destruction is x dest T 0 q out T L q in T H REFERENCES AND SUGGESTED READINGS 1 V D Chase Propfans A New Twist for the Propeller Mechanical Engineering November 1986 pp 4750 2 C R Ferguson and A T Kirkpatrick Internal Combustion Engines Applied Thermosciences 2nd ed New York Wiley 2000 3 R A Harmon The Keys to Cogeneration and Combined Cycles Mechanical Engineering February 1988 pp 6473 4 J Heywood Internal Combustion Engine Fundamen tals New York McGrawHill 1988 5 L C Lichty Combustion Engine Processes New York McGrawHill 1967 6 H McIntosh Jumbo Jet 10 Outstanding Achievements 19641989 Washington DC National Academy of Engineering 1989 pp 3033 7 W Pulkrabek Engineering Fundamentals of the Internal Combustion Engine 2nd ed Upper Saddle River NJ PrenticeHall 2004 8 W Siuru TwoStroke Engines Cleaner and Meaner Mechanical Engineering June 1990 pp 6669 9 C F Taylor The Internal Combustion Engine in Theory and Practice Cambridge MA MIT Press 1968 Final PDF to printer cen22672ch09475542indd 528 110617 0921 AM 528 GAS POWER CYCLES PROBLEMS Actual and Ideal Cycles Carnot Cycle AirStandard Assumptions Reciprocating Engines 91C What are the airstandard assumptions 92C What is the difference between airstandard assump tions and the coldairstandard assumptions 93C Why is the Carnot cycle not suitable as an ideal cycle for all powerproducing cyclic devices 94C How does the thermal efficiency of an ideal cycle in general compare to that of a Carnot cycle operating between the same temperature limits 95C How are the combustion and exhaust processes mod eled under the airstandard assumptions 96C What does the area enclosed by the cycle represent on a Pv diagram How about on a Ts diagram 97C Define the compression ratio for reciprocating engines 98C Can the mean effective pressure of an automobile engine in operation be less than the atmospheric pressure 99C What is the difference between sparkignition and compressionignition engines 910C Define the following terms related to reciprocating engines stroke bore top dead center and clearance volume 911C What is the difference between the clearance volume and the displacement volume of reciprocating engines 912 Can any ideal gas power cycle have a thermal effi ciency greater than 55 percent when using thermal energy res ervoirs at 627C and 17C 913 An airstandard cycle is executed within a closed piston cylinder system and consists of three processes as follows 12 V constant heat addition from 100 kPa and 27C to 850 kPa 23 Isothermal expansion until V3 7V2 31 P constant heat rejection to the initial state Assume air has constant properties with cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK and k 14 a Sketch the Pv and Ts diagrams for the cycle b Determine the ratio of the compression work to the expansion work the back work ratio c Determine the cycle thermal efficiency Answers b 0453 c 256 percent 914 An airstandard cycle with variable specific heats is executed in a closed system with 0003 kg of air and consists of the following three processes 12 v constant heat addition from 95 kPa and 17C to 380 kPa 23 Isentropic expansion to 95 kPa 31 P constant heat rejection to initial state a Show the cycle on Pv and Ts diagrams b Calculate the net work per cycle in kJ c Determine the thermal efficiency 915 Repeat Prob 914 using constant specific heats at room temperature 916E An airstandard cycle with variable specific heats is executed in a closed system and is composed of the following four processes 12 v constant heat addition from 147 psia and 80F in the amount of 300 Btulbm 23 P constant heat addition to 3200 R 34 Isentropic expansion to 147 psia 41 P constant heat rejection to initial state a Show the cycle on Pv and Ts diagrams b Calculate the total heat input per unit mass c Determine the thermal efficiency Answers b 612 Btulbm c 242 percent 917E Repeat Prob 916E using constant specific heats at room temperature 918 An ideal gas is contained in a pistoncylinder device and undergoes a power cycle as follows 12 isentropic compression from an initial temperature T1 20C with a compression ratio r 5 23 constantpressure heat addition 31 constantvolume heat rejection The gas has constant specific heats with cv 07 kJkgK and R 03 kJkgK a Sketch the Pv and Ts diagrams for the cycle b Determine the heat and work interactions for each process in kJkg c Determine the cycle thermal efficiency d Obtain the expression for the cycle thermal efficiency as a function of the compression ratio r and ratio of specific heats k 919 An airstandard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K The pressures before and after the isothermal compression are 150 and 300 kPa respectively If the net work output per cycle is Problems designated by a C are concept questions and stu dents are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch09475542indd 529 110617 0921 AM 529 CHAPTER 9 05 kJ determine a the maximum pressure in the cycle b the heat transfer to air and c the mass of air Assume vari able specific heats for air Answers a 300 MPa b 0706 kJ c 000296 kg 920 Repeat Prob 919 using helium as the working fluid 921E The thermal energy reservoirs of an ideal gas Carnot cycle are at 1240F and 40F and the device executing this cycle rejects 100 Btu of heat each time the cycle is executed Determine the total heat supplied to and the total work pro duced by this cycle each time it is executed 922 Consider a Carnot cycle executed in a closed system with 06 kg of air The temperature limits of the cycle are 300 and 1100 K and the minimum and maximum pressures that occur during the cycle are 20 and 3000 kPa Assuming con stant specific heats determine the net work output per cycle 923 Consider a Carnot cycle executed in a closed system with air as the working fluid The maximum pressure in the cycle is 1300 kPa while the maximum temperature is 950 K If the entropy increase during the isothermal heat addition process is 025 kJkgK and the net work output is 110 kJkg determine a the minimum pressure in the cycle b the heat rejection from the cycle and c the thermal efficiency of the cycle d If an actual heat engine cycle operates between the same tempera ture limits and produces 5200 kW of power for an airflow rate of 95 kgs determine the secondlaw efficiency of this cycle Otto Cycle 924C What four processes make up the ideal Otto cycle 925C Are the processes that make up the Otto cycle analyzed as closedsystem or steadyflow processes Why 926C How do the efficiencies of the ideal Otto cycle and the Carnot cycle compare for the same temperature limits Explain 927C How does the thermal efficiency of an ideal Otto cycle change with the compression ratio of the engine and the specific heat ratio of the working fluid 928C Why are high compression ratios not used in spark ignition engines 929C An ideal Otto cycle with a specified compression ratio is executed using a air b argon and c ethane as the working fluid For which case will the thermal efficiency be the highest Why 930C How is the rpm revolutions per minute of an actual fourstroke gasoline engine related to the number of thermo dynamic cycles What would your answer be for a twostroke engine 931C What is the difference between fuelinjected gasoline engines and diesel engines 932E Determine the mean effective pressure of an ideal Otto cycle that uses air as the working fluid its state at the begin ning of the compression is 14 psia and 60F its temperature at the end of the combustion is 1500F and its compression ratio is 9 Use constant specific heats at room temperature 933E Reconsider Prob 932E Determine the rate of heat addition and rejection for this ideal Otto cycle when it pro duces 140 hp 934 An ideal Otto cycle has a compression ratio of 8 At the beginning of the compression process air is at 95 kPa and 27C and 750 kJkg of heat is transferred to air during the constantvolume heataddition process Taking into account the variation of specific heats with temperature determine a the pressure and temperature at the end of the heataddition process b the net work output c the thermal efficiency and d the mean effective pressure for the cycle Answers a 3898 kPa 1539 K b 392 kJkg c 523 percent d 495 kPa 935 Reconsider Prob 934 Using appropriate soft ware study the effect of varying the compression ratio from 5 to 10 Plot the net work output and thermal effi ciency as a function of the compression ratio Plot the Ts and Pv diagrams for the cycle when the compression ratio is 8 936 Repeat Prob 934 using constant specific heats at room temperature 937E A sparkignition engine has a compression ratio of 10 an isentropic compression efficiency of 85 percent and an isentropic expansion efficiency of 95 percent At the beginning of the compression the air in the cylinder is at 13 psia and 60F The maximum gas temperature is found to be 2300F by measurement Determine the heat supplied per unit mass the thermal efficiency and the mean effective pressure of this engine when modeled with the Otto cycle Use constant specific heats at room temperature Answers 225 Btulbm 497 percent 453 psia 938 An ideal Otto cycle has a compression ratio of 7 At the beginning of the compression process P1 90 kPa T1 27C and V1 0004 m3 The maximum cycle tempera ture is 1127C For each repetition of the cycle calculate the heat rejection and the net work production Also calculate the thermal efficiency and mean effective pressure for this cycle Use constant specific heats at room temperature Answers 103 kJ 121 kJ 541 percent 354 kPa 939 A sixcylinder 4L sparkignition engine operating on the ideal Otto cycle takes in air at 90 kPa and 20C The mini mum enclosed volume is 15 percent of the maximum enclosed volume When operated at 2500 rpm this engine produces 90 hp Determine the rate of heat addition to this engine Use constant specific heats at room temperature 940E An ideal Otto cycle with air as the working fluid has a compression ratio of 8 The minimum and maximum tempera tures in the cycle are 540 and 2400 R Accounting for the varia tion of specific heats with temperature determine a the amount of heat transferred to the air during the heataddition process b the thermal efficiency and c the thermal efficiency of a Carnot cycle operating between the same temperature limits Final PDF to printer cen22672ch09475542indd 530 110617 0921 AM 530 GAS POWER CYCLES 941E Repeat Prob 940E using argon as the working fluid 942 Someone has suggested that the airstandard Otto cycle is more accurate if the two isentropic processes are replaced with polytropic processes with a polytropic exponent n 13 Consider such a cycle when the compression ratio is 8 P1 95 kPa T1 15C and the maximum cycle temperature is 1200C Determine the heat transferred to and rejected from this cycle as well as the cycles thermal efficiency Use con stant specific heats at room temperature Answers 835 kJkg 420 kJkg 498 percent 943 Repeat Prob 942 when isentropic processes are used in place of the polytropic processes 944 When we double the compression ratio of an ideal Otto cycle what happens to the maximum gas temperature and pressure when the state of the air at the beginning of the compression and the amount of heat addition remain the same Use constant specific heats at room temperature Diesel Cycle 945C How does a diesel engine differ from a gasoline engine 946C How does the ideal Diesel cycle differ from the ideal Otto cycle 947C What is the cutoff ratio How does it affect the ther mal efficiency of a Diesel cycle 948C For a specified compression ratio is a diesel or gaso line engine more efficient 949C Do diesel or gasoline engines operate at higher com pression ratios Why 950 An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 15 Determine the maximum air temperature and the rate of heat addition to this cycle when it produces 200 hp of power the cycle is repeated 1200 times per minute and the state of the air at the beginning of the compression is 95 kPa and 17C Use constant specific heats at room temperature 951 Rework Prob 950 when the isentropic compression efficiency is 90 percent and the isentropic expansion effi ciency is 95 percent 952 An airstandard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2 At the beginning of the compression process air is at 95 kPa and 27C Accounting for the variation of specific heats with temperature determine a the tempera ture after the heataddition process b the thermal efficiency and c the mean effective pressure Answers a 1725 K b 563 percent c 6759 kPa 953 Repeat Prob 952 using constant specific heats at room temperature 954E An airstandard Diesel cycle has a compression ratio of 182 Air is at 120F and 147 psia at the beginning of the compression process and at 3200 R at the end of the heat addition process Accounting for the variation of specific heats with temperature determine a the cutoff ratio b the heat rejection per unit mass and c the thermal efficiency 955E Repeat Prob 954E using constant specific heats at room temperature 956 An ideal Diesel cycle has a maximum cycle tempera ture of 2000C The state of the air at the beginning of the compression is P1 95 kPa and T1 15C This cycle is exe cuted in a fourstroke eightcylinder engine with a cylinder bore of 10 cm and a piston stroke of 12 cm The minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume Determine the power produced by this engine when it is operated at 1600 rpm Use constant specific heats at room temperature Answer 965 kW 957 An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid The state of air at the begin ning of the compression process is 95 kPa and 20C If the maximum temperature in the cycle is not to exceed 2200 K determine a the thermal efficiency and b the mean effec tive pressure Assume constant specific heats for air at room temperature Answers a 635 percent b 933 kPa 958 Repeat Prob 957 but replace the isentropic expan sion process with a polytropic expansion process with the polytropic exponent n 135 Use variable specific heats 959 Reconsider Prob 958 Using appropriate soft ware study the effect of varying the compression ratio from 14 to 24 Plot the net work output mean effective pressure and thermal efficiency as a function of the compres sion ratio Plot the Ts and Pv diagrams for the cycle when the compression ratio is 20 960 A fourcylinder twostroke 24L diesel engine that operates on an ideal Diesel cycle has a compression ratio of 22 and a cutoff ratio of 18 Air is at 70C and 97 kPa at the beginning of the compression process Using the coldair standard assumptions determine how much power the engine will deliver at 4250 rpm 961 Repeat Prob 960 using nitrogen as the working fluid 962 The compression ratio of an ideal dual cycle is 14 Air is at 100 kPa and 300 K at the beginning of the compression process and at 2200 K at the end of the heataddition process Heat transfer to air takes place partly at constant volume and partly at constant pressure and it amounts to 15204 kJkg Assuming variable specific heats for air determine a the fraction of heat transferred at constant volume and b the ther mal efficiency of the cycle 963 Reconsider Prob 962 Using appropriate soft ware study the effect of varying the compression ratio from 10 to 18 For the compression ratio equal to 14 plot the Ts and Pv diagrams for the cycle 964 Repeat Prob 962 using constant specific heats at room temperature Is the constant specific heat assumption reasonable in this case Final PDF to printer cen22672ch09475542indd 531 110617 0921 AM 531 CHAPTER 9 965E An airstandard dual cycle has a compression ratio of 20 and a cutoff ratio of 13 The pressure ratio during the con stantvolume heat addition process is 12 Determine the ther mal efficiency amount of heat added and the maximum gas pressure and temperature when this cycle is operated at 14 psia and 70F at the beginning of the compression Use constant specific heats at room temperature 966E Repeat Prob 965E if the compression ratio were reduced to 12 967 Develop an expression for cutoff ratio rc which expresses it in terms of qincpT1rk1 for an airstandard Diesel cycle 968 An airstandard cycle called the dual cycle with con stant specific heats is executed in a closed pistoncylinder sys tem and is composed of the following five processes 12 Isentropic compression with a compression ratio r V1V2 23 Constantvolume heat addition with a pressure ratio rp P3P2 34 Constantpressure heat addition with a volume ratio rc V4V3 45 Isentropic expansion while work is done until V5 V1 51 Constantvolume heat rejection to the initial state a Sketch the Pv and Ts diagrams for this cycle b Obtain an expression for the cycle thermal efficiency as a function of k r rc and rp c Evaluate the limit of the efficiency as rp approaches unity and compare your answer with the expression for the Diesel cycle efficiency d Evaluate the limit of the efficiency as rc approaches unity and compare your answer with the expression for the Otto cycle efficiency Stirling and Ericsson Cycles 969C What cycle is composed of two isothermal and two constantvolume processes 970C How does the ideal Ericsson cycle differ from the Carnot cycle 971C Consider the ideal Otto Stirling and Carnot cycles operating between the same temperature limits How would you compare the thermal efficiencies of these three cycles 972C Consider the ideal Diesel Ericsson and Carnot cycles operating between the same temperature limits How would you compare the thermal efficiencies of these three cycles 973E An ideal Ericsson engine using helium as the work ing fluid operates between temperature limits of 550 and 3000 R and pressure limits of 25 and 200 psia Assuming a mass flow rate of 14 lbms determine a the thermal efficiency of the cycle b the heat transfer rate in the regenerator and c the power delivered 974 An ideal Stirling engine using helium as the working fluid operates between temperature limits of 300 and 2000 K and pressure limits of 150 kPa and 3 MPa Assuming the mass of the helium used in the cycle is 012 kg determine a the thermal efficiency of the cycle b the amount of heat transfer in the regenerator and c the work output per cycle 975E An airstandard Stirling cycle operates with a maxi mum pressure of 600 psia and a minimum pressure of 10 psia The maximum volume of the air is 10 times the minimum volume The temperature during the heat rejection process is 100F Calculate the specific heat added to and rejected by this cycle as well as the net specific work produced by the cycle Use constant specific heats at room temperature 976E Reconsider Prob 975E How much heat is stored and recovered in the regenerator 977 Consider an ideal Ericsson cycle with air as the work ing fluid executed in a steadyflow system Air is at 27C and 120 kPa at the beginning of the isothermal compression pro cess during which 150 kJkg of heat is rejected Heat transfer to air occurs at 950 K Determine a the maximum pressure in the cycle b the net work output per unit mass of air and c the thermal efficiency of the cycle Answers a 685 kPa b 325 kJkg c 684 percent 978E An ideal Stirling cycle uses energy reservoirs at 40F and 640F and uses hydrogen as the working gas It is designed such that its minimum volume is 01 ft3 maximum volume is 1 ft3 and maximum pressure is 400 psia Calculate the amount of external heat addition external heat rejection and heat trans fer between the working fluid and regenerator for each com plete cycle Use constant specific heats at room temperature Ideal and Actual GasTurbine Brayton Cycles 979C What four processes make up the simple ideal Brayton cycle 980C For fixed maximum and minimum temperatures what is the effect of the pressure ratio on a the thermal effi ciency and b the net work output of a simple ideal Brayton cycle 981C What is the back work ratio What are typical back work ratio values for gasturbine engines 982C Why are the back work ratios relatively high in gasturbine engines 983C How do the inefficiencies of the turbine and the compressor affect a the back work ratio and b the thermal efficiency of a gasturbine engine 984E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10 The air enters the compres sor at 520 R and the turbine at 2000 R Accounting for the Final PDF to printer cen22672ch09475542indd 532 110617 0921 AM 532 GAS POWER CYCLES FIGURE P995 Compressor Combustion chamber Turbine 16 MPa 100 kPa 40C 650C 1 2 3 4 Wnet variation of specific heats with temperature determine a the air temperature at the compressor exit b the back work ratio and c the thermal efficiency 985 A stationary gasturbine power plant operates on a simple ideal Brayton cycle with air as the working fluid The air enters the compressor at 95 kPa and 290 K and the turbine at 760 kPa and 1100 K Heat is transferred to air at a rate of 35000 kJs Determine the power delivered by this plant a assuming constant specific heats at room temperature and b accounting for the variation of specific heats with temperature 986 A gasturbine power plant operates on the simple Bray ton cycle with air as the working fluid and delivers 32 MW of power The minimum and maximum temperatures in the cycle are 310 and 900 K and the pressure of air at the compressor exit is 8 times the value at the compressor inlet Assuming an isentro pic efficiency of 80 percent for the compressor and 86 percent for the turbine determine the mass flow rate of air through the cycle Account for the variation of specific heats with temperature 987 Repeat Prob 986 using constant specific heats at room temperature 988 A simple ideal Brayton cycle operates with air with minimum and maximum temperatures of 27C and 727C It is designed so that the maximum cycle pressure is 2000 kPa and the minimum cycle pressure is 100 kPa Determine the net work produced per unit mass of air each time this cycle is executed and the cycles thermal efficiency Use constant spe cific heats at room temperature respectively and there is a 50kPa pressure drop across the combustion chamber Answers 73 kJ 38 percent 992 Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12 a compressor inlet temperature of 300 K and a turbine inlet temperature of 1000 K Determine the required mass flow rate of air for a net power output of 70 MW assuming both the compressor and the turbine have an isentropic efficiency of a 100 percent and b 85 percent Assume constant specific heats at room tem perature Answers a 352 kgs b 1037 kgs 993 An aircraft engine operates on a simple ideal Brayton cycle with a pressure ratio of 10 Heat is added to the cycle at a rate of 500 kW air passes through the engine at a rate of 1 kgs and the air at the beginning of the compression is at 70 kPa and 0C Determine the power produced by this engine and its thermal efficiency Use constant specific heats at room temperature 994 Repeat Prob 993 for a pressure ratio of 15 995 A gasturbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1600 kPa The working fluid is air which enters the compressor at 40C at a rate of 850 m3min and leaves the turbine at 650C Assuming a compressor isentropic efficiency of 85 percent and a turbine isentropic efficiency of 88 percent determine a the net power output b the back work ratio and c the ther mal efficiency Use constant specific heats with cv 0821 kJ kgK cp 1108 kJkgK and k 135 Answers a 6488 kW b 0511 c 378 percent FIGURE P988 Qout Compressor Wnet Turbine 1 4 2 3 Qin 989 Repeat Prob 988 when the isentropic efficiency of the turbine is 90 percent 990 Repeat Prob 988 when the isentropic efficiency of the turbine is 90 percent and that of the compressor is 80 percent 991 Repeat Prob 988 when the isentropic efficiencies of the turbine and compressor are 90 percent and 80 percent 996E A simple ideal Brayton cycle uses argon as the work ing fluid At the beginning of the compression P1 15 psia and T1 80F the maximum cycle temperature is 1200F and the pressure in the combustion chamber is 150 psia The argon enters the compressor through a 3 ft2 opening with a velocity of 200 fts Determine the rate of heat addition to this engine the power produced and the cycles thermal efficiency 997 A gasturbine power plant operates on a modified Brayton cycle shown in the figure with an overall pressure Final PDF to printer cen22672ch09475542indd 533 110617 0921 AM 533 CHAPTER 9 FIGURE P9105 4 3 1 2 6 5 Comb Heat exchanger Compressor Turbine ratio of 8 Air enters the compressor at 0C and 100 kPa The maximum cycle temperature is 1500 K The compres sor and the turbines are isentropic The highpressure turbine develops just enough power to run the compressor Assume constant properties for air at 300 K with cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK k 14 a Sketch the Ts diagram for the cycle Label the data states b Determine the temperature and pressure at state 4 the exit of the highpressure turbine c If the net power output is 200 MW determine the mass flow rate of the air into the compressor in kgs Answers b 1279 K 457 kPa c 442 kgs efficiency of a gasturbine engine Is there any truth in this claim Explain 9102C In an ideal regenerator is the air leaving the com pressor heated to the temperature at a the turbine inlet b the turbine exit c slightly above the turbine exit 9103C In 1903 Aegidius Elling of Norway designed and built an 11hp gas turbine that used steam injection between the combustion chamber and the turbine to cool the combus tion gases to a safe temperature for the materials available at the time Currently there are several gasturbine power plants that use steam injection to augment power and improve ther mal efficiency For example the thermal efficiency of the General Electric LM5000 gas turbine is reported to increase from 358 percent in simplecycle operation to 43 percent when steam injection is used Explain why steam injection increases the power output and the efficiency of gas turbines Also explain how you would obtain the steam 9104 Develop an expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator of effectiveness 100 percent Use constant specific heats at room temperature 9105 A gas turbine for an automobile is designed with a regenerator Air enters the compressor of this engine at 100 kPa and 30C The compressor pressure ratio is 8 the maximum cycle temperature is 800C and the cold airstream leaves the regenerator 10C cooler than the hot airstream at the inlet of the regenerator Assuming both the compressor and the tur bine to be isentropic determine the rates of heat addition and rejection for this cycle when it produces 115 kW Use constant specific heats at room temperature Answers 240 kW 125 kW FIGURE P997 Compressor Combustion chamber Highpressure turbine 1 2 3 4 5 Lowpressure turbine 998 A gasturbine power plant operating on the simple Bray ton cycle has a pressure ratio of 7 Air enters the compressor at 0C and 100 kPa The maximum cycle temperature is 1500 K The compressor has an isentropic efficiency of 80 percent and the turbine has an isentropic efficiency of 90 percent Assume constant properties for air at 300 K with cv 0718 kJkgK cp 1005 kJkgK R 0287 kJkgK k 14 a Sketch the Ts diagram for the cycle b If the net power output is 150 MW determine the volume flow rate of the air into the compressor in m3s c For a fixed compressor inlet velocity and flow area explain the effect of increasing compressor inlet tempera ture ie summertime operation versus wintertime opera tion on the inlet mass flow rate and the net power output with all other parameters of the problem being the same Brayton Cycle with Regeneration 999C How does regeneration affect the efficiency of a Brayton cycle and how does it accomplish it 9100C Define the effectiveness of a regenerator used in gasturbine cycles 9101C Somebody claims that at very high pressure ratios the use of regeneration actually decreases the thermal 9106 Rework Prob 9105 when the compressor isentropic efficiency is 87 percent and the turbine isentropic efficiency is 90 percent 9107 A gasturbine engine operates on the ideal Bray ton cycle with regeneration as shown in Fig P9105 Now the regenerator is rearranged so that the airstreams of states 2 and 5 enter at one end of the regenerator and streams 3 and 6 exit at the other end ie parallel flow arrangement of a heat exchanger Consider such a system when air enters the com pressor at 100 kPa and 20C the compressor pressure ratio is Final PDF to printer cen22672ch09475542indd 534 110617 0921 AM 534 GAS POWER CYCLES 7 the maximum cycle temperature is 727C and the differ ence between the hot and cold airstream temperatures is 6C at the end of the regenerator where the cold stream leaves the regenerator Is the cycle arrangement shown in the figure more or less efficient than this arrangement Assume both the com pressor and the turbine are isentropic and use constant spe cific heats at room temperature 9108E An ideal regenerator T3 T5 is added to a simple ideal Brayton cycle see Fig P9105 Air enters the compres sor of this cycle at 16 psia and 100F the pressure ratio is 11 and the maximum cycle temperature is 1940F What is the thermal efficiency of this cycle Use constant specific heats at room temperature What would the thermal efficiency of the cycle be without the regenerator 9109 The 7FA gas turbine manufactured by General Electric is reported to have an efficiency of 359 percent in the simple cycle mode and to produce 159 MW of net power The pressure ratio is 147 and the turbine inlet temperature is 1288C The mass flow rate through the turbine is 1536000 kgh Taking the ambient conditions to be 30C and 100 kPa determine the isentropic efficiencies of the turbine and the compressor Also determine the thermal efficiency of this gas turbine if a regen erator with an effectiveness of 65 percent is added 9110 Reconsider Prob 9109 Using appropriate software develop a solution that allows differ ent isentropic efficiencies for the compressor and turbine and study the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle Plot the Ts diagram for the cycle 9111E The idea of using gas turbines to power automobiles was conceived in the 1930s and considerable research was done in the 1940s and 1950s to develop automotive gas tur bines by major automobile manufacturers such as the Chrys ler and Ford corporations in the United States and Rover in the United Kingdom The worlds first gasturbinepowered automobile the 200hp Rover Jet 1 was built in 1950 in the United Kingdom This was followed by the production of the Plymouth Sport Coupe by Chrysler in 1954 under the leader ship of G J Huebner Several hundred gasturbinepowered Plymouth cars were built in the early 1960s for demonstration purposes and were loaned to a select group of people to gather field experience The users had no complaints other than slow acceleration But the cars were never massproduced because of the high production especially material costs and the fail ure to satisfy the provisions of the 1966 Clean Air Act A gasturbinepowered Plymouth car built in 1960 had a tur bine inlet temperature of 1700F a pressure ratio of 4 and a regenerator effectiveness of 09 Using isentropic efficiencies of 80 percent for both the compressor and the turbine deter mine the thermal efficiency of this car Also determine the mass flow rate of air for a net power output of 130 hp Assume the ambient air to be at 510 R and 145 psia 9112 A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7 The minimum and max imum temperatures in the cycle are 310 and 1150 K Assuming an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator determine a the air temperature at the turbine exit b the net work output and c the thermal efficiency Use variable specific heats Answers a 783 K b 108 kJkg c 225 percent 9113 A stationary gasturbine power plant operates on an ideal regenerative Brayton cycle 100 percent with air as the working fluid Air enters the compressor at 95 kPa and 290 K and the turbine at 880 kPa and 1100 K Heat is trans ferred to air from an external source at a rate of 30000 kJs Determine the power delivered by this plant a assum ing constant specific heats for air at room temperature and b accounting for the variation of specific heats with temperature 9114 Air enters the compressor of a regenerative gas turbine engine at 310 K and 100 kPa where it is compressed to 900 kPa and 650 K The regenerator has an effectiveness of 80 percent and the air enters the turbine at 1400 K For a turbine efficiency of 90 percent determine a the amount of heat transfer in the regenerator and b the thermal efficiency Assume variable specific heats for air Answers a 193 kJkg b 400 percent 9115 Repeat Prob 9114 using constant specific heats at room temperature 9116 Repeat Prob 9114 for a regenerator effectiveness of 70 percent Brayton Cycle with Intercooling Reheating and Regeneration 9117C For a specified pressure ratio why does multistage compression with intercooling decrease the compressor work and multistage expansion with reheating increase the turbine work 9118C In an ideal gasturbine cycle with intercooling reheating and regeneration as the number of compression and expansion stages is increased the cycle thermal efficiency approaches a 100 percent b the Otto cycle efficiency or c the Carnot cycle efficiency 9119C The singlestage compression process of an ideal Brayton cycle without regeneration is replaced by a multistage compression process with intercooling between the same pres sure limits As a result of this modification a Does the compressor work increase decrease or remain the same b Does the back work ratio increase decrease or remain the same c Does the thermal efficiency increase decrease or remain the same Final PDF to printer cen22672ch09475542indd 535 110617 0921 AM 535 CHAPTER 9 9120C The singlestage expansion process of an ideal Brayton cycle without regeneration is replaced by a multistage expansion process with reheating between the same pressure limits As a result of this modification a Does the turbine work increase decrease or remain the same b Does the back work ratio increase decrease or remain the same c Does the thermal efficiency increase decrease or remain the same 9121C A simple ideal Brayton cycle without regenera tion is modified to incorporate multistage compression with intercooling and multistage expansion with reheating without changing the pressure or temperature limits of the cycle As a result of these two modifications a Does the net work output increase decrease or remain the same b Does the back work ratio increase decrease or remain the same c Does the thermal efficiency increase decrease or remain the same d Does the heat rejected increase decrease or remain the same 9122C A simple ideal Brayton cycle is modified to incor porate multistage compression with intercooling multistage expansion with reheating and regeneration without chang ing the pressure limits of the cycle As a result of these modifications a Does the net work output increase decrease or remain the same b Does the back work ratio increase decrease or remain the same c Does the thermal efficiency increase decrease or remain the same d Does the heat rejected increase decrease or remain the same 9123 Consider a regenerative gasturbine power plant with two stages of compression and two stages of expansion The overall pressure ratio of the cycle is 9 The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K Accounting for the variation of specific heats with temperature determine the minimum mass flow rate of air needed to develop a net power output of 110 MW Answer 250 kgs 9124 Repeat Prob 9123 using argon as the working fluid 9125 Consider an ideal gasturbine cycle with two stages of compression and two stages of expansion The pressure ratio across each stage of the compressor and turbine is 3 The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K Determine the back work ratio and the thermal efficiency of the cycle assuming a no regenerator is used and b a regenerator with 75 percent effectiveness is used Use variable specific heats 9126 Repeat Prob 9125 assuming an efficiency of 86 per cent for each compressor stage and an efficiency of 90 percent for each turbine stage 9127E A gas turbine operates with a regenerator and two stages of reheating and intercooling Air enters this engine at 14 psia and 60F the pressure ratio for each stage of com pression is 3 the air temperature when entering a turbine is 940F and the regenerator operates perfectly Determine the mass flow rate of the air passing through this engine and the rates of heat addition and rejection when this engine produces 1000 hp Assume isentropic operations for all compressor and the turbine stages and use constant specific heats at room temperature FIGURE P9127E 8 7 qreheat qin 1 10 3 6 4 5 9 2 qintercool 9128E Reconsider Prob 9127E Determine the change in the rate of heat addition to the cycle when the isentropic efficiency of each compressor is 88 percent and that of each turbine is 93 percent JetPropulsion Cycles 9129C What is propulsive power How is it related to thrust 9130C What is propulsive efficiency How is it determined 9131C Is the effect of turbine and compressor irrevers ibilities of a turbojet engine to reduce a the net work b the thrust or c the fuel consumption rate 9132 Air at 7C enters a turbojet engine at a rate of 16 kgs and at a velocity of 220 ms relative to the engine Air is heated in the combustion chamber at a rate 15000 kJs and it leaves the engine at 427C Determine the thrust produced by this turbojet engine Hint Choose the entire engine as your control volume Final PDF to printer cen22672ch09475542indd 536 110617 0921 AM 536 GAS POWER CYCLES 9133E A turbojet is flying with a velocity of 900 fts at an altitude of 20000 ft where the ambient conditions are 7 psia and 10F The pressure ratio across the compressor is 13 and the temperature at the turbine inlet is 2400 R Assuming ideal operation for all components and constant specific heats for air at room temperature determine a the pressure at the turbine exit b the velocity of the exhaust gases and c the propul sive efficiency 9134E Repeat Prob 9133E accounting for the variation of specific heats with temperature 9135 A pure jet engine propels an aircraft at 240 ms through air at 45 kPa and 13C The inlet diameter of this engine is 16 m the compressor pressure ratio is 13 and the temperature at the turbine inlet is 557C Determine the veloc ity at the exit of this engines nozzle and the thrust produced Assume ideal operation for all components and constant spe cific heats at room temperature 9136 A turbojet aircraft is flying with a velocity of 280 ms at an altitude of 9150 m where the ambient conditions are 32 kPa and 32C The pressure ratio across the compressor is 12 and the temperature at the turbine inlet is 1100 K Air enters the compressor at a rate of 50 kgs and the jet fuel has a heating value of 42700 kJkg Assuming ideal operation for all components and constant specific heats for air at room temperature determine a the velocity of the exhaust gases b the propulsive power developed and c the rate of fuel consumption 9137 Repeat Prob 9136 using a compressor efficiency of 80 percent and a turbine efficiency of 85 percent 9138E A turboprop aircraft propulsion engine operates where the air is at 8 psia and 10F on an aircraft flying at a speed of 600 fts The Brayton cycle pressure ratio is 10 and the air temperature at the turbine inlet is 940F The propeller diameter is 10 ft and the mass flow rate through the propeller is 20 times that through the compressor Determine the thrust force generated by this propulsion system Assume ideal oper ation for all components and constant specific heats at room temperature 9139E Reconsider Prob 9138E How much change would result in the thrust if the propeller diameter were reduced to 8 ft while maintaining the same mass flow rate through the com pressor Note The mass flow rate ratio will no longer be 20 9140 Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9 The aircraft is stationary on the ground held in position by its brakes The ambient air is at 7C and 95 kPa and enters the engine at a rate of 20 kgs The jet fuel has a heating value of 42700 kJkg and it is burned com pletely at a rate of 05 kgs Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components determine the force that must be applied on the brakes to hold the plane stationary Answer 19370 N 9141 Reconsider Prob 9140 In the problem state ment replace the inlet mass flow rate with an inlet volume flow rate of 181 m3s Using appropriate soft ware investigate the effect of compressor inlet temperature in the range of 20 to 30C on the force that must be applied to the brakes to hold the plane stationary Plot this force as a func tion of compressor inlet temperature SecondLaw Analysis of Gas Power Cycles 9142 An ideal Otto cycle has a compression ratio of 8 At the beginning of the compression process air is at 95 kPa and 27C and 750 kJkg of heat is transferred to air during the con stantvolume heataddition process Determine the total exergy destruction associated with the cycle assuming a source tem perature of 2000 K and a sink temperature of 300 K Also determine the exergy at the end of the power stroke Account for the variation of specific heats with temperature Answers 245 kJkg 145 kJkg 9143 An airstandard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2 At the beginning of the com pression process air is at 95 kPa and 27C Determine the total exergy destruction associated with the cycle assuming a source temperature of 2000 K and a sink temperature of 300 K Also determine the exergy at the end of the isentropic com pression process Account for the variation of specific heats with temperature Answers 293 kJkg 349 kJkg 9144E An airstandard Diesel cycle has a compression ratio of 182 Air is at 120F and 147 psia at the beginning of the com pression process and at 3200 R at the end of the heat addition process Determine the exergy destruction associated with the heat rejection process of the cycle assuming a source tempera ture of 3200 R and a sink temperature of 540 R Also deter mine the exergy at the end of the isentropic expansion process Account for the variation of specific heats with temperature 9145E An airstandard dual cycle has a compression ratio of 20 and a cutoff ratio of 13 The pressure ratio during the con stantvolume heat addition process is 12 This cycle is oper ated at 14 psia and 70F at the beginning of the compression Calculate the exergy that is lost each time the cycle is repeated The surroundings are at 147 psia and 70F The source tem perature is the same as the maximum cycle temperature and the sink temperature is the same as the minimum cycle tem perature Use constant specific heats at room temperature 9146E A simple ideal Brayton cycle uses argon as the work ing fluid At the beginning of the compression P1 15 psia and T1 80F the maximum cycle temperature is 1200F and the pressure in the combustion chamber is 150 psia The argon enters the compressor through a 3 ft2 opening with a velocity of 200 fts Determine the rate at which entropy is generated by the cycle The temperature of the source is the same as the maxi mum cycle temperature and the temperature of the sink is the same as the minimum cycle temperature Answer 0320 BtusR Final PDF to printer cen22672ch09475542indd 537 110617 0921 AM 537 CHAPTER 9 9147 A gas turbine for an automobile is designed with a regenerator Air enters the compressor of this engine at 100 kPa and 20C The compressor pressure ratio is 8 the maximum cycle temperature is 800C and the cold airstream leaves the regenerator 10C cooler than the hot airstream at the inlet of the regenerator The cycle produces 150 kW The compressor isen tropic efficiency is 87 percent and the turbine isentropic effi ciency is 93 percent Determine the exergy destruction for each of the processes of the cycle The temperature of the hot reser voir is the same as the maximum cycle temperature and the tem perature of the cold reservoir is the same as the minimum cycle temperature Use constant specific heats at room temperature 9148 A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7 The minimum and maximum temperatures in the cycle are 310 and 1150 K Take an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator Determine the total exergy destruction associ ated with the cycle assuming a source temperature of 1500 K and a sink temperature of 290 K Also determine the exergy of the exhaust gases at the exit of the regenerator Use variable specific heats for air 9149 Reconsider Prob 9148 Using appropriate soft ware investigate the effect of varying the cycle pressure ratio from 6 to 14 on the total exergy destruction for the cycle and the exergy of the exhaust gas leaving the regenerator Plot these results as functions of pressure ratio Discuss the results 9150 Air enters the compressor of a regenerative gas turbine engine at 310 K and 100 kPa where it is compressed to 900 kPa and 650 K The regenerator has an effectiveness of 80 percent the air enters the turbine at 1400 K and the tur bine isentropic efficiency is 90 percent Determine the exergy destruction associated with each of the processes of the cycle assuming a source temperature of 1260 K and a sink tempera ture of 300 K Also determine the exergy of the exhaust gases at the exit of the regenerator Take Pexhaust P0 100 kPa and assume variable specific heats for air 9151E A gas turbine operates with a regenerator and two stages of reheating and intercooling Air enters this engine at 14 psia and 60F the pressure ratio for each stage of com pression is 3 the air temperature when entering a turbine is 940F the engine produces 1000 hp and the regenerator oper ates perfectly The isentropic efficiency of each compressor is 88 percent and that of each turbine is 93 percent Which pro cess of the cycle loses the greatest amount of work potential The temperature of the heat source is the same as the maxi mum cycle temperature and the temperature of the heat sink is the same as the minimum cycle temperature Use constant specific heats at room temperature 9152 A gasturbine power plant operates on the regenerative Brayton cycle between the pressure limits of 100 and 700 kPa Air enters the compressor at 30C at a rate of 126 kgs and leaves at 260C It is then heated in a regenerator to 400C by the hot combustion gases leaving the turbine A diesel fuel with a heating value of 42000 kJkg is burned in the combustion chamber with a combustion efficiency of 97 percent The com bustion gases leave the combustion chamber at 871C and enter the turbine whose isentropic efficiency is 85 percent Treating combustion gases as air and using constant specific heats at 500C determine a the isentropic efficiency of the compres sor b the effectiveness of the regenerator c the airfuel ratio in the combustion chamber d the net power output and the back work ratio e the thermal efficiency and f the second law efficiency of the plant Also determine g the secondlaw efficiencies of the compressor the turbine and the regenerator and h the rate of the exergy flow with the combustion gases at the regenerator exit Answers a 0881 b 0632 c 781 d 2267 kW 0583 e 0345 f 0469 g 0929 0932 0890 h 1351 kW FIGURE P9152 Compressor Regenerator 100 kPa 30C 700 kPa 260C 400C 1 2 5 3 4 6 Turbine Combustion chamber 871C 9153 A fourcylinder fourstroke 18L modern high speed compressionignition engine operates on the ideal dual cycle with a compression ratio of 16 The air is at 95 kPa and 70C at the beginning of the compression process and the engine speed is 2200 rpm Equal amounts of fuel are burned at constant volume and at constant pressure The maximum allowable pressure in the cycle is 75 MPa due to material strength limitations Using constant specific heats at 1000 K determine a the maximum temperature in the cycle b the net work output and the thermal efficiency c the mean effec tive pressure and d the net power output Also determine e the secondlaw efficiency of the cycle and the rate of exergy output with the exhaust gases when they are purged Answers a 2308 K b 836 kJkg 595 percent c 860 kPa d 284 kW e 683 percent 103 kW Review Problems 9154 An airstandard cycle with variable specific heats is executed in a closed system with 0003 kg of air and it consists of the following three processes 12 Isentropic compression from 100 kPa and 27C to 700 kPa 23 P constant heat addition to initial specific volume Final PDF to printer cen22672ch09475542indd 538 110617 0921 AM 538 GAS POWER CYCLES 31 v constant heat rejection to initial state a Show the cycle on Pv and Ts diagrams b Calculate the maximum temperature in the cycle c Determine the thermal efficiency Answers b 2100 K c 158 percent 9155 Repeat Prob 9154 using constant specific heats at room temperature 9156 A Carnot cycle is executed in a closed system and uses 00025 kg of air as the working fluid The cycle efficiency is 60 percent and the lowest temperature in the cycle is 300 K The pressure at the beginning of the isentropic expansion is 700 kPa and at the end of the isentropic compression it is 1 MPa Determine the net work output per cycle 9157 An ideal gas Carnot cycle uses helium as the working fluid and rejects heat to a lake at 15C Determine the pressure ratio compression ratio and minimum temperature of the heat source for this cycle to have a thermal efficiency of 50 percent Answers 565 283 576 K 9158E Repeat Prob 9157 when the lake is at 60F and the Carnot cycles thermal efficiency is to be 60 percent 9159E A fourstroke turbocharged V16 diesel engine built by GE Transportation Systems to power fast trains produces 4400 hp at 1500 rpm Determine the amount of work produced per cylin der per a mechanical cycle and b thermodynamic cycle 9160 An Otto cycle with a compression ratio of 8 begins its compression at 94 kPa and 10C The maximum cycle tem perature is 900C Utilizing airstandard assumptions deter mine the thermal efficiency of this cycle using a constant specific heats at room temperature and b variable specific heats Answers a 565 percent b 537 percent 9161 A Diesel cycle has a compression ratio of 22 and begins its compression at 85 kPa and 15C The maximum cycle temperature is 1200C Utilizing airstandard assump tions determine the thermal efficiency of this cycle using a constant specific heats at room temperature and b variable specific heats 9162 Consider an engine operating on the ideal Diesel cycle with air as the working fluid The volume of the cylin der is 1200 cm3 at the beginning of the compression process 75 cm3 at the end and 150 cm3 after the heataddition process Air is at 17C and 100 kPa at the beginning of the compres sion process Determine a the pressure at the beginning of the heatrejection process b the net work per cycle in kJ and c the mean effective pressure 9163 Repeat Prob 9162 using argon as the working fluid 9164 A fourcylinder fourstroke sparkignition engine operates on the ideal Otto cycle with a compression ratio of 11 and a total displacement volume of 18 L The air is at 90 kPa and 50C at the beginning of the compression pro cess The heat input is 05 kJ per cycle per cylinder Determine a the maximum temperature and pressure that occur during the cycle b the net work per cycle per cylinder and the ther mal efficiency of the cycle c the mean effective pressure and d the power output for an engine speed of 3000 rpm Use constant specific heats with cv 0821 kJkgK cp 1108 kJ kgK and k 135 9165 A typical hydrocarbon fuel produces 43000 kJkg of heat when used in a sparkignition engine Determine the com pression ratio required for an ideal Otto cycle to use 0039 g of fuel to produce 1 kJ of work Use constant specific heats at room temperature Answer 966 9166E An ideal dual cycle has a compression ratio of 14 and uses air as the working fluid At the beginning of the compression process air is at 147 psia and 120F and it occupies a volume of 98 in3 During the heataddition process 06 Btu of heat is trans ferred to air at constant volume and 11 Btu at constant pressure Using constant specific heats evaluated at room temperature determine the thermal efficiency of the cycle 9167 A fourcylinder fourstroke 16L gasoline engine operates on the Otto cycle with a compression ratio of 11 The air is at 100 kPa and 37C at the beginning of the compres sion process and the maximum pressure in the cycle is 8 MPa The compression and expansion processes may be modeled as polytropic with a polytropic constant of 13 Using constant specific heats at 850 K determine a the temperature at the end of the expansion process b the net work output and the thermal efficiency c the mean effective pressure d the engine speed for a net power output of 50 kW and e the spe cific fuel consumption in gkWh defined as the ratio of the mass of the fuel consumed to the net work produced The air fuel ratio defined as the amount of air divided by the amount of fuel intake is 16 9168 Consider an ideal Stirling cycle using air as the work ing fluid Air is at 400 K and 200 kPa at the beginning of the isothermal compression process and heat is supplied to air from a source at 1800 K in the amount of 900 kJkg Determine a the maximum pressure in the cycle and b the net work output per unit mass of air Answers a 5139 kPa b 700 kJkg 9169 Consider a simple ideal Brayton cycle operating between the temperature limits of 300 and 1250 K Using constant specific heats at room temperature determine the pressure ratio for which the compressor and the turbine exit temperatures of air are equal 9170 Consider a simple ideal Brayton cycle with air as the working fluid The pressure ratio of the cycle is 6 and the min imum and maximum temperatures are 300 and 1300 K respec tively Now the pressure ratio is doubled without changing the minimum and maximum temperatures in the cycle Determine the change in a the net work output per unit mass and b the thermal efficiency of the cycle as a result of this modification Assume variable specific heats for air Answers a 415 kJkg b 106 percent Final PDF to printer cen22672ch09475542indd 539 110617 0921 AM 539 CHAPTER 9 9171 Repeat Prob 9170 using constant specific heats at room temperature 9172 A Brayton cycle with a pressure ratio of 15 operates with air entering the compressor at 70 kPa and 0C and the turbine at 600C Calculate the net specific work produced by this cycle treating the air as an ideal gas with a constant spe cific heats and b variable specific heats variable specific heats determine a the back work ratio and the net work output b the thermal efficiency and c the sec ondlaw efficiency of the cycle Also determine d the exergies at the exits of the combustion chamber and the regenerator This problem is solved using appropriate software FIGURE P9175 Regenerator wnet Comp I Comp II Turb I Turb II Intercooler 10 1 2 3 4 5 6 7 8 9 Reheater Comb cham FIGURE P9172 Qout Compressor Wnet Turbine 1 4 2 3 Qin 9173 Helium is used as the working fluid in a Brayton cycle with regeneration The pressure ratio of the cycle is 8 the compressor inlet temperature is 300 K and the turbine inlet temperature is 1800 K The effectiveness of the regenerator is 75 percent Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW assuming both the compressor and the turbine have an isentro pic efficiency of a 100 percent and b 80 percent 9174 Consider an ideal gasturbine cycle with one stage of compression and two stages of expansion and regeneration The pressure ratio across each turbine stage is the same The highpressure turbine exhaust gas enters the regenerator and then enters the lowpressure turbine for expansion to the com pressor inlet pressure Determine the thermal efficiency of this cycle as a function of the compressor pressure ratio and the highpressureturbinetocompressor inlet temperature ratio Compare your result with the efficiency of the standard regen erative cycle 9175 A gasturbine plant operates on the regenerative Brayton cycle with two stages of reheating and two stages of intercooling between the pressure limits of 100 and 1200 kPa The working fluid is air The air enters the first and the second stages of the compressor at 300 K and 350 K respectively and the first and the second stages of the turbine at 1400 K and 1300 K respectively Assuming both the compressor and the turbine have an isentropic efficiency of 80 percent and the regenerator has an effectiveness of 75 percent and using 9176 Compare the thermal efficiency of a twostage gas turbine with regeneration reheating and intercooling to that of a threestage gas turbine with the same equipment when a all components operate ideally b air enters the first compres sor at 100 kPa and 20C c the total pressure ratio across all stages of compression is 16 and d the maximum cycle tem perature is 800C 9177E The specific impulse of an aircraftpropulsion sys tem is the force produced per unit of thrustproducing mass flow rate Consider a jet engine that operates in an environ ment at 10 psia and 30F and propels an aircraft cruising at 1200 fts Determine the specific impulse of this engine when the compressor pressure ratio is 9 and the temperature at the turbine inlet is 700F Assume ideal operations for all compo nents and constant specific heats at room temperature 9178 Electricity and process heat requirements of a manufacturing facility are to be met by a cogen eration plant consisting of a gas turbine and a heat exchanger for steam production The plant operates on the simple Brayton cycle between the pressure limits of 100 and 1000 kPa with air as the working fluid Air enters the compressor at 20C Combustion gases leave the turbine and enter the heat exchanger at 450C and leave the heat exchanger at 325C while the liquid water enters the heat exchanger at 15C and leaves at 200C as a saturated vapor The net power produced by the gasturbine cycle is 1500 kW Assuming a compressor isentropic efficiency of 86 percent and a turbine isentropic efficiency of 88 percent and using variable specific heats determine a the mass flow rate of air b the back work ratio and the thermal efficiency and c the rate at which steam is produced in the heat exchanger Also determine d the Final PDF to printer cen22672ch09475542indd 540 110617 0921 AM 540 GAS POWER CYCLES 9179 A turbojet aircraft flies with a velocity of 1100 kmh at an altitude where the air tempera ture and pressure are 35C and 40 kPa Air leaves the diffuser at 50 kPa with a velocity of 15 ms and combustion gases enter the turbine at 450 kPa and 950C The turbine produces 800 kW of power all of which is used to drive the compressor Assuming an isentropic efficiency of 83 percent for the com pressor turbine and nozzle and using variable specific heats determine a the pressure of combustion gases at the turbine exit b the mass flow rate of air through the compressor c the velocity of the gases at the nozzle exit and d the pro pulsive power and the propulsive efficiency for this engine This problem is solved using appropriate software 9180 An airstandard cycle with constant specific heats is executed in a closed pistoncylinder system and is composed of the following three processes 12 Isentropic compression with a compression ratio r V1V2 23 Constantpressure heat addition 31 Constantvolume heat rejection a Sketch the Pv and Ts diagrams for this cycle b Obtain an expression for the back work ratio as a func tion of k and r c Obtain an expression for the cycle thermal efficiency as a function of k and r c Determine the value of the back work ratio and thermal efficiency as r goes to unity What do your results imply about the net work done by the cycle 9181 Consider the ideal regenerative Brayton cycle Deter mine the pressure ratio that maximizes the thermal efficiency of the cycle and compare this value with the pressure ratio that max imizes the cycle net work For the same maximumtominimum temperature ratios explain why the pressure ratio for maximum efficiency is less than the pressure ratio for maximum work 9182 Using the cutoff ratio rc and the pressure ratio dur ing constantvolume heat addition process rp determine the amount of heat added to the dual cycle Develop an equation for qin cvT1r k1 in terms of k rc and rp Use constant specific heats at room temperature 9183 Using appropriate software study the effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid At the beginning of the compression process air is at 100 kPa and 300 K Determine the percentage of error involved in using constant specific heat values at room temperature for the fol lowing combinations of compression ratios and maximum cycle temperatures r 6 8 10 12 and Tmax 1000 1500 2000 2500 K 9184 Using appropriate software determine the effects of pressure ratio maximum cycle temperature and compressor and turbine isentropic efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid Air is at 100 kPa and 300 K at the compressor inlet Also assume constant spe cific heats for air at room temperature Determine the net work output and the thermal efficiency for all combinations of the fol lowing parameters and draw conclusions from the results Pressure ratio 5 8 14 Maximum cycle temperature 800 1200 1600 K Compressor isentropic efficiency 80 100 percent Turbine isentropic efficiency 80 100 percent 9185 Repeat Prob 9184 by considering the variation of specific heats of air with temperature 9186 Repeat Prob 9184 using helium as the working fluid 9187 Using appropriate software determine the effects of pressure ratio maximum cycle tem perature regenerator effectiveness and compressor and tur bine efficiencies on the net work output per unit mass and on the thermal efficiency of a regenerative Brayton cycle with air as the working fluid Air is at 100 kPa and 300 K at the compressor inlet Also assume constant specific heats for air at room temperature Determine the net work output and the thermal efficiency for all combinations of the following parameters Pressure ratio 6 10 Maximum cycle temperature 1500 2000 K Compressor isentropic efficiency 80 100 percent Turbine isentropic efficiency 80 100 percent Regenerator effectiveness 70 90 percent FIGURE P9178 450C 325C 15C Combustion chamber 100 kPa 20C 1 2 3 2 MPa sat vapor 200C 4 Turbine Compressor Heat exchanger utilization efficiency of the cogeneration plant defined as the ratio of the total energy utilized to the energy supplied to the plant This problem is solved using appropriate software Final PDF to printer cen22672ch09475542indd 541 110617 0921 AM 541 CHAPTER 9 9188 Repeat Prob 9187 by considering the variation of specific heats of air with temperature 9189 Repeat Prob 9187 using helium as the working fluid 9190 Using appropriate software determine the effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion Assume that the overall pressure ratio of the cycle is 18 and the air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K Using constant specific heats for air at room temperature determine the thermal efficiency of the cycle by varying the number of stages from 1 to 22 in increments of 3 Plot the thermal efficiency versus the number of stages Compare your results to the efficiency of an Erics son cycle operating between the same temperature limits 9191 Repeat Prob 9190 using helium as the working fluid Fundamentals of Engineering FE Exam Problems 9192 For specified limits for the maximum and minimum temperatures the ideal cycle with the lowest thermal efficiency is a Carnot b Stirling c Ericsson d Otto e All are the same 9193 A Carnot cycle operates between the temperature lim its of 300 and 2000 K and produces 400 kW of net power The rate of entropy change of the working fluid during the heat addition process is a 0 kWK b 0200 kWK c 0174 kWK d 0235 kWK e 133 kWK 9194 An Otto cycle with air as the working fluid has a com pression ratio of 104 Under coldairstandard conditions the thermal efficiency of this cycle is a 10 percent b 39 percent c 61 percent d 79 percent e 82 percent 9195 Air in an ideal Diesel cycle is compressed from 2 to 013 L and then it expands during the constantpressure heat addition process to 030 L Under coldairstandard conditions the thermal efficiency of this cycle is a 41 percent b 59 percent c 66 percent d 70 percent e 78 percent 9196 Helium gas in an ideal Otto cycle is compressed from 20C and 25 to 025 L and its temperature increases by an additional 700C during the heataddition process The tem perature of helium before the expansion process is a 1790C b 2060C c 1240C d 620C e 820C 9197 In an ideal Otto cycle air is compressed from 120 kgm3 and 22 to 026 L and the net work output of the cycle is 440 kJkg The mean effective pressure MEP for this cycle is a 612 kPa b 599 kPa c 528 kPa d 416 kPa e 367 kPa 9198 Air enters a turbojet engine at 320 ms at a rate of 30 kgs and exits at 570 ms relative to the aircraft The thrust developed by the engine is a 25 kN b 50 kN c 75 kN d 10 kN e 125 kN 9199 In an ideal Brayton cycle air is compressed from 95 kPa and 25C to 1400 kPa Under coldairstandard condi tions the thermal efficiency of this cycle is a 40 percent b 44 percent c 49 percent d 54 percent e 58 percent 9200 In an ideal Brayton cycle air is compressed from 100 kPa and 25C to 1 MPa and then heated to 927C before entering the turbine Under coldairstandard conditions the air temperature at the turbine exit is a 349C b 426C c 622C d 733C e 825C 9201 Consider an ideal Brayton cycle executed between the pressure limits of 1200 and 100 kPa and temperature limits of 20 and 1000C with argon as the working fluid The net work output of the cycle is a 68 kJkg b 93 kJkg c 158 kJkg d 186 kJkg e 310 kJkg 9202 An ideal Brayton cycle has a net work output of 150 kJkg and a back work ratio of 04 If both the turbine and the compressor had an isentropic efficiency of 85 percent the net work output of the cycle would be a 74 kJkg b 95 kJkg c 109 kJkg d 128 kJkg e 177 kJkg 9203 In an ideal Brayton cycle with regeneration argon gas is compressed from 100 kPa and 25C to 400 kPa and then heated to 1200C before entering the turbine The highest tem perature that argon can be heated in the regenerator is a 246C b 846C c 689C d 368C e 573C 9204 In an ideal Brayton cycle with regeneration air is compressed from 80 kPa and 10C to 400 kPa and 175C is heated to 450C in the regenerator and is then further heated to 1000C before entering the turbine Under coldairstandard conditions the effectiveness of the regenerator is a 33 percent b 44 percent c 62 percent d 77 percent e 89 percent 9205 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20 and 900C If the specific heat ratio of the working fluid is 13 the highest thermal efficiency this gas turbine can have is a 38 percent b 46 percent c 62 percent d 58 percent e 97 percent 9206 An ideal gas turbine cycle with many stages of com pression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10 Air enters every stage of the compressor at 290 K and every stage of the Final PDF to printer cen22672ch09475542indd 542 110617 0921 AM 542 GAS POWER CYCLES turbine at 1200 K The thermal efficiency of this gasturbine cycle is a 36 percent b 40 percent c 52 percent d 64 percent e 76 percent Design and Essay Problems 9207 The amount of fuel introduced into a sparkignition engine is used in part to control the power produced by the engine Gasoline produces approximately 42000 kJkg when burned with air in a sparkignition engine Develop a sched ule for gasoline consumption and maximum cycle temperature versus power production for an Otto cycle with a compression ratio of 8 9208 The weight of a diesel engine is directly proportional to the compression ratio W kr because extra metal must be used to strengthen the engine for the higher pressures Exam ine the net specific work produced by a diesel engine per unit of weight as the pressure ratio is varied and the specific heat input remains fixed Do this for several heat inputs and propor tionality constants k Are there any optimal combinations of k and specific heat inputs 9209 In response to concerns about the environment some major car manufacturers are currently marketing electric cars Write an essay on the advantages and disadvantages of electric cars and discuss when it is advisable to purchase an electric car instead of a traditional internal combustion car 9210 Intense research is underway to develop adiabatic engines that require no cooling of the engine block Such engines are based on ceramic materials because of the abil ity of such materials to withstand high temperatures Write an essay on the current status of adiabatic engine development Also determine the highest possible efficiencies with these engines and compare them to the highest possible efficiencies of current engines 9211 Write an essay on the most recent developments in twostroke engines and find out when we might be seeing cars on the market powered by twostroke engines Why do the major car manufacturers have a renewed interest in twostroke engines 9212 Exhaust gases from the turbine of a simple Brayton cycle are quite hot and may be used for other thermal purposes One proposed use is generating saturated steam at 110C from water at 30C in a boiler This steam will be distributed to several buildings on a college campus for space heating A Brayton cycle with a pressure ratio of 6 is to be used for this purpose Plot the power produced the flow rate of produced steam and the maximum cycle temperature as functions of the rate at which heat is added to the cycle The temperature at the turbine inlet is not to exceed 2000C 9213 A gas turbine operates with a regenerator and two stages of reheating and intercooling This system is designed so that when air enters the compressor at 100 kPa and 15C the pressure ratio for each stage of compression is 3 the air temperature when entering a turbine is 500C and the regen erator operates perfectly At full load this engine produces 800 kW For this engine to service a partial load the heat addi tion in both combustion chambers is reduced Develop an opti mal schedule of heat addition to the combustion chambers for partial loads ranging from 400 to 800 kW 9214 Since its introduction in 1903 by Aegidius Elling of Norway steam injection between the combustion chamber and the turbine is used even in some modern gas turbines currently in operation to cool the combustion gases to a metallurgically safe temperature while increasing the mass flow rate through the turbine Currently there are several gasturbine power plants that use steam injection to augment power and improve thermal efficiency Consider a gasturbine power plant whose pressure ratio is 8 The isentropic efficiencies of the compressor and the turbine are 80 percent and there is a regenerator with an effective ness of 70 percent When the mass flow rate of air through the compressor is 40 kgs the turbine inlet temperature becomes 1700 K But the turbine inlet temperature is limited to 1500 K and thus steam injection into the combustion gases is being considered However to avoid the complexities associated with steam injection it is proposed to use excess air that is to take in much more air than needed for complete combustion to lower the combustion and thus turbine inlet temperatures while increasing the mass flow rate and thus power output of the turbine Evaluate this proposal and compare the thermo dynamic performance of high air flow to that of a steam injection gasturbine power plant under the following design conditions the ambient air is at 100 kPa and 25C adequate water supply is available at 20C and the amount of fuel sup plied to the combustion chamber remains constant Final PDF to printer cen22672ch10543596indd 543 110917 1148 AM 543 OBJECTIVES The objectives of Chapter 10 are to Analyze vapor power cycles in which the working fluid is alternately vaporized and condensed Investigate ways to modify the basic Rankine vapor power cycle to increase the cycle thermal efficiency Analyze the reheat and regenerative vapor power cycles Perform secondlaw analysis of vapor power cycles Analyze power generation coupled with process heating called cogeneration Analyze power cycles that consist of two separate cycles known as combined cycles VA P O R AN D CO MBIN E D P OW E R CYC LE S I n Chap 9 we discussed gas power cycles for which the working fluid remains a gas throughout the entire cycle In this chapter we consider vapor power cycles in which the working fluid is alternately vaporized and condensed We also consider power generation coupled with process heating called cogeneration The continued quest for higher thermal efficiencies has resulted in some innovative modifications to the basic vapor power cycle Among these we discuss the reheat and regenerative cycles as well as combined gasvapor power cycles Steam is the most common working fluid used in vapor power cycles because of its many desirable characteristics such as low cost availability and high enthalpy of vaporization Therefore this chapter is mostly devoted to the discussion of steam power plants Steam power plants are commonly referred to as coal plants nuclear plants or natural gas plants depending on the type of fuel used to supply heat to the steam However the steam goes through the same basic cycle in all of them Therefore all can be analyzed in the same manner 10 CHAPTER Final PDF to printer 544 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 544 110917 1148 AM 101 THE CARNOT VAPOR CYCLE We have mentioned repeatedly that the Carnot cycle is the most efficient cycle operating between two specified temperature limits Thus it is natural to look at the Carnot cycle first as a prospective ideal cycle for vapor power plants If we could we would certainly adopt it as the ideal cycle As will be explained however the Carnot cycle is not a suitable model for power cycles Throughout the discussions we assume steam to be the working fluid since it is the working fluid predominantly used in vapor power cycles Consider a steadyflow Carnot cycle executed within the saturation dome of a pure substance as shown in Fig 101a The fluid is heated reversibly and isothermally in a boiler process 12 expanded isentropically in a tur bine process 23 condensed reversibly and isothermally in a condenser process 34 and compressed isentropically by a compressor to the initial state process 41 Several impracticalities are associated with this cycle 1 Isothermal heat transfer to or from a twophase system is not difficult to achieve in practice since maintaining a constant pressure in the device automatically fixes the temperature at the saturation value Therefore pro cesses 12 and 34 can be approached closely in actual boilers and condens ers Limiting the heat transfer processes to twophase systems however severely limits the maximum temperature that can be used in the cycle it has to remain under the criticalpoint value which is 374C for water Limiting the maximum temperature in the cycle also limits the thermal effi ciency Any attempt to raise the maximum temperature in the cycle involves heat transfer to the working fluid in a single phase which is not easy to accomplish isothermally 2 The isentropic expansion process process 23 can be approximated closely by a welldesigned turbine However the quality of the steam decreases during this process as shown on the Ts diagram in Fig 101a Thus the turbine has to handle steam with low quality that is steam with a high moisture content The impingement of liquid droplets on the turbine blades causes erosion and is a major source of wear Thus steam with quali ties less than about 90 percent cannot be tolerated in the operation of power plants This problem could be eliminated by using a working fluid with a very steep saturated vapor line 3 The isentropic compression process process 41 involves the compression of a liquidvapor mixture to a saturated liquid There are two difficulties associated with this process First it is not easy to control the condensation process so precisely as to end up with the desired quality at state 4 Second it is not practical to design a compressor that handles two phases Some of these problems could be eliminated by executing the Carnot cycle in a different way as shown in Fig 101b This cycle however presents other problems such as isentropic compression to extremely high pressures and iso thermal heat transfer at variable pressures Thus we conclude that the Carnot cycle cannot be approximated in actual devices and is not a realistic model for vapor power cycles FIGURE 101 Ts diagram of two Carnot vapor cycles s T 3 4 1 2 s T 3 4 1 2 a b Final PDF to printer 545 CHAPTER 10 cen22672ch10543596indd 545 110917 1148 AM 102 RANKINE CYCLE THE IDEAL CYCLE FOR VAPOR POWER CYCLES Many of the impracticalities associated with the Carnot cycle can be elimi nated by superheating the steam in the boiler and condensing it completely in the condenser as shown schematically on a Ts diagram in Fig 102 The cycle that results is the Rankine cycle which is the ideal cycle for vapor power plants The ideal Rankine cycle does not involve any internal irrevers ibilities and consists of the following four processes 12 Isentropic compression in a pump 23 Constantpressure heat addition in a boiler 34 Isentropic expansion in a turbine 41 Constantpressure heat rejection in a condenser Water enters the pump at state 1 as saturated liquid and is compressed isentro pically to the operating pressure of the boiler The water temperature increases somewhat during this isentropic compression process due to a slight decrease in the specific volume of water The vertical distance between states 1 and 2 on the Ts diagram is greatly exaggerated for clarity If water were truly incompress ible would there be a temperature change at all during this process Water enters the boiler as a compressed liquid at state 2 and leaves as a superheated vapor at state 3 The boiler is basically a large heat exchanger where the heat originating from combustion gases nuclear reactors or other sources is transferred to the water essentially at constant pressure The boiler together with the section where the steam is superheated the superheater is often called the steam generator The superheated vapor at state 3 enters the turbine where it expands isen tropically and produces work by rotating the shaft connected to an electric generator The pressure and the temperature of steam drop during this pro cess to the values at state 4 where steam enters the condenser At this state steam is usually a saturated liquidvapor mixture with a high quality Steam is condensed at constant pressure in the condenser which is basically a large heat exchanger by rejecting heat to a cooling medium such as a lake a river or the atmosphere Steam leaves the condenser as saturated liquid and enters the pump completing the cycle In areas where water is precious the power plants are cooled by air instead of water This method of cooling which is also used in car engines is called dry cooling Several power plants in the world including some in the United States use dry cooling to conserve water Remembering that the area under the process curve on a Ts diagram rep resents the heat transfer for internally reversible processes we see that the area under process curve 23 represents the heat transferred to the water in the boiler and the area under the process curve 41 represents the heat rejected in the condenser The difference between these two the area enclosed by the cycle curve is the net work produced during the cycle Energy Analysis of the Ideal Rankine Cycle All four components associated with the Rankine cycle the pump boiler turbine and condenser are steadyflow devices and thus all four processes FIGURE 102 The simple ideal Rankine cycle s T 3 2 4 1 wturbout wpumpin qout qin Boiler wturbout Condenser Pump 3 qin 4 2 wpumpin 1 qout Turbine Final PDF to printer 546 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 546 110917 1148 AM that make up the Rankine cycle can be analyzed as steadyflow processes The kinetic and potential energy changes of the steam are usually small relative to the work and heat transfer terms and are therefore usually neglected Then the steadyflow energy equation per unit mass of steam reduces to q in q out w in w out h e h i kJkg 101 The boiler and the condenser do not involve any work and the pump and the turbine are assumed to be isentropic Then the conservation of energy relation for each device can be expressed as follows Pump q 0 w pumpin h 2 h 1 102 or w pumpin v P 2 P 1 103 where h 1 h f P 1 and v v 1 v f P 1 104 Boiler w 0 q in h 3 h 2 105 Turbine q 0 w turbout h 3 h 4 106 Condenser w 0 q out h 4 h 1 107 The thermal efficiency of the Rankine cycle is determined from η th w net q in 1 q out q in 108 where w net q in q out w turbout w pumpin The conversion efficiency of power plants in the United States is often expressed in terms of heat rate which is the amount of heat supplied in Btus to generate 1 kWh of electricity The smaller the heat rate the greater the efficiency Considering that 1 kWh 3412 Btu and disregarding the losses associated with the conversion of shaft power to electric power the relation between the heat rate and the thermal efficiency can be expressed as η th 3412 BtukWh Heat rate BtukWh 109 For example a heat rate of 11363 BtukWh is equivalent to 30 percent efficiency The thermal efficiency can also be interpreted as the ratio of the area enclosed by the cycle on a Ts diagram to the area under the heataddition process The use of these relations is illustrated in the following example Final PDF to printer 547 CHAPTER 10 cen22672ch10543596indd 547 110917 1148 AM EXAMPLE 101 The Simple Ideal Rankine Cycle Consider a steam power plant operating on the simple ideal Rankine cycle Steam enters the turbine at 3 MPa and 350C and is condensed in the condenser at a pressure of 75 kPa Determine the thermal efficiency of this cycle SOLUTION A steam power plant operating on the simple ideal Rankine cycle is considered The thermal efficiency of the cycle is to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The schematic of the power plant and the Ts diagram of the cycle are shown in Fig 103 We note that the power plant operates on the ideal Rankine cycle Therefore the pump and the turbine are isentropic there are no pressure drops in the boiler and condenser and steam leaves the condenser and enters the pump as satu rated liquid at the condenser pressure First we determine the enthalpies at various points in the cycle using data from steam tables Tables A4 A5 and A6 State 1 P 1 75 kPa Sat liquid h 1 h f 75 kPa 38444 kJ kg v 1 v f 75 kPa 0001037 m 3 kg State 2 P 2 3 MPa s 2 s 1 w pumpin v 1 P 2 P 1 0001037 m 3 kg 3000 75 kPa 1 kJ 1 kPa m 3 303 kJkg h 2 h 1 w pumpin 38444 303 kJkg 38747 kJkg State 3 P 3 3 MPa T 3 350C h 3 31161 kJkg s 3 67450 kJkgK State 4 P 4 75 kPa sat mixture s 4 s 3 x 4 s 4 s f s fg 67450 12132 62426 08861 h 4 h f x 4 h fg 38444 08861 22780 24030 kJkg Thus q in h 3 h 2 31161 38747 kJkg 27286 kJkg q out h 4 h 1 24030 38444 kJkg 20186 kJkg and η th 1 q out q in 1 20186 kJkg 27286 kJkg 0260 or 2𝟔0 The thermal efficiency could also be determined from w turbout h 3 h 4 31161 24030 kJkg 7131 kJkg w net w turbout w pumpin 7131 303 kJkg 7101 kJkg FIGURE 103 Schematic and Ts diagram for Example 101 s T C 2 4 1 350 s3 s4 s1 s2 3 75 kPa 3 MPa 3 MPa 350C 3 MPa 75 kPa 75 kPa Boiler wturbout Condenser Pump 3 qin 4 2 wpumpin 1 qout Turbine Final PDF to printer 548 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 548 110917 1148 AM 103 DEVIATION OF ACTUAL VAPOR POWER CYCLES FROM IDEALIZED ONES The actual vapor power cycle differs from the ideal Rankine cycle as illus trated in Fig 104a as a result of irreversibilities in various components Fluid friction and heat loss to the surroundings are the two common sources of irreversibilities Fluid friction causes pressure drops in the boiler the condenser and the piping between various components As a result steam leaves the boiler at a somewhat lower pressure Also the pressure at the turbine inlet is somewhat lower than that at the boiler exit due to the pressure drop in the connect ing pipes The pressure drop in the condenser is usually very small To com pensate for these pressure drops the water must be pumped to a sufficiently higher pressure than the ideal cycle calls for This requires a larger pump and larger work input to the pump The other major source of irreversibility is the heat loss from the steam to the surroundings as the steam flows through various components To main tain the same level of net work output more heat needs to be transferred to the steam in the boiler to compensate for these undesired heat losses As a result cycle efficiency decreases or w net q in q out 27286 20186 kJkg 7100 kJkg and η th w net q in 7100 kJkg 27286 kJkg 0260 or 260 That is this power plant converts 26 percent of the heat it receives in the boiler to net work An actual power plant operating between the same temperature and pressure limits will have a lower efficiency because of the irreversibilities such as friction Discussion Notice that the back work ratio rbw winwout of this power plant is 0004 and thus only 04 percent of the turbine work output is required to operate the pump Having such low back work ratios is characteristic of vapor power cycles This is in contrast to the gas power cycles which typically involve very high back work ratios about 40 to 80 percent It is also interesting to note the thermal efficiency of a Carnot cycle operating between the same temperature limits η thCarnot 1 T min T max 1 9176 273 K 350 273 K 0415 Here Tmin is taken as the saturation temperature of water at 75 kPa The differ ence between the two efficiencies is due to the large external irreversibility in the Rankine cycle caused by the large temperature difference between steam and the heat source Final PDF to printer 549 CHAPTER 10 cen22672ch10543596indd 549 110917 1148 AM Of particular importance are the irreversibilities occurring within the pump and the turbine A pump requires a greater work input and a turbine produces a smaller work output as a result of irreversibilities Under ideal conditions the flow through these devices is isentropic The deviation of actual pumps and turbines from the isentropic ones can be accounted for by utilizing isen tropic efficiencies defined as η P w s w a h 2s h 1 h 2a h 1 1010 and η T w a w s h 3 h 4a h 3 h 4s 1011 where states 2a and 4a are the actual exit states of the pump and the turbine respectively and 2s and 4s are the corresponding states for the isentropic case Fig 104b Other factors also need to be considered in the analysis of actual vapor power cycles In actual condensers for example the liquid is usually subcooled to prevent the onset of cavitation the rapid vaporization and condensation of the fluid at the lowpressure side of the pump impeller which may damage it Additional losses occur at the bearings between the moving parts as a result of friction Steam that leaks out during the cycle and air that leaks into the con denser represent two other sources of loss Finally the power consumed by the auxiliary equipment such as fans that supply air to the furnace should also be considered in evaluating the overall performance of power plants The effect of irreversibilities on the thermal efficiency of a steam power cycle is illustrated next with an example FIGURE 104 a Deviation of actual vapor power cycle from the ideal Rankine cycle b The effect of pump and turbine irreversibilities on the ideal Rankine cycle 3 Ideal cycle Actual cycle Pressure drop in the condenser Irreversibility in the turbine Irreversibility in the pump Pressure drop in the boiler s T 2 4 1 a 3 s T 2s 4s 1 4a 2a b EXAMPLE 102 An Actual Steam Power Cycle A steam power plant operates on the cycle shown in Fig 105 If the isentropic efficiency of the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent determine a the thermal efficiency of the cycle and b the net power output of the plant for a mass flow rate of 15 kgs SOLUTION A steam power cycle with specified turbine and pump efficien cies is considered The thermal efficiency and the net power output are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The schematic of the power plant and the Ts diagram of the cycle are shown in Fig 105 The temperatures and pressures of steam at various points are also indicated on the figure We note that the power plant involves steadyflow com ponents and operates on the Rankine cycle but the imperfections at various compo nents are accounted for a The thermal efficiency of a cycle is the ratio of the net work output to the heat input and it is determined as follows Final PDF to printer 550 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 550 110917 1148 AM Pump work input w pumpin w spumpin η P v 1 P 2 P 1 η P 0001009 m 3 kg16000 9 kPa 085 1 kJ 1 kPa m 3 190 kJkg Turbine work output w turbout η T w sturbout η T h 5 h 6s 087 35831 21153 kJkg 12770 kJkg Boiler heat input q in h 4 h 3 36476 1601 kJkg 34875 kJkg Thus w net w turbout w pumpin 12770 190 kJkg 12580 kJkg η th w net q in 12580 kJkg 34875 kJkg 0361 or 361 b The power produced by this power plant is W net m w net 15 kgs 12580 kJkg 189 MW Discussion Without the irreversibilities the thermal efficiency of this cycle would be 430 percent see Example 103c FIGURE 105 Schematic and Ts diagram for Example 102 s T 4 5 6s 6 1 3 2s 2 Boiler wturbout Condenser Turbine T 087 Pump P 085 5 15 MPa 600C 152 MPa 625C 10 kPa 159 MPa 35C 4 3 6 2 1 9 kPa 38C 16 MPa wpumpin qin qout Final PDF to printer 551 CHAPTER 10 cen22672ch10543596indd 551 110917 1148 AM 104 HOW CAN WE INCREASE THE EFFICIENCY OF THE RANKINE CYCLE Steam power plants are responsible for the production of most electric power in the world and even small increases in thermal efficiency can mean large savings from the fuel requirements Therefore every effort is made to improve the efficiency of the cycle on which steam power plants operate The basic idea behind all the modifications to increase the thermal effi ciency of a power cycle is the same Increase the average temperature at which heat is transferred to the working fluid in the boiler or decrease the average temperature at which heat is rejected from the working fluid in the condenser That is the average fluid temperature should be as high as possible during heat addition and as low as possible during heat rejec tion Next we discuss three ways of accomplishing this for the simple ideal Rankine cycle Lowering the Condenser Pressure Lowers Tlowavg Steam exists as a saturated mixture in the condenser at the saturation tempera ture corresponding to the pressure inside the condenser Therefore lowering the operating pressure of the condenser automatically lowers the temperature of the steam and thus the temperature at which heat is rejected The effect of lowering the condenser pressure on the Rankine cycle effi ciency is illustrated on a Ts diagram in Fig 106 For comparison purposes the turbine inlet state is maintained the same The colored area on this dia gram represents the increase in net work output as a result of lowering the condenser pressure from P4 to P4 The heat input requirements also increase represented by the area under curve 22 but this increase is very small Thus the overall effect of lowering the condenser pressure is an increase in the thermal efficiency of the cycle To take advantage of the increased efficiencies at low pressures the con densers of steam power plants usually operate well below the atmospheric pressure This does not present a major problem since the vapor power cycles operate in a closed loop However there is a lower limit on the condenser pressure that can be used It cannot be lower than the saturation pressure corresponding to the temperature of the cooling medium Con sider for example a condenser that is to be cooled by a nearby river at 15C Allowing a temperature difference of 10C for effective heat trans fer the steam temperature in the condenser must be above 25C thus the condenser pressure must be above 32 kPa which is the saturation pressure at 25C Lowering the condenser pressure is not without any side effects however For one thing it creates the possibility of air leakage into the condenser More importantly it increases the moisture content of the steam at the final stages of the turbine as can be seen from Fig 106 The presence of large quantities of moisture is highly undesirable in turbines because it decreases the turbine efficiency and erodes the turbine blades Fortunately this problem can be cor rected as discussed next FIGURE 106 The effect of lowering the condenser pressure on the ideal Rankine cycle 3 s T 4 1 2 1 2 4 P4 P4 Increase in wnet Final PDF to printer 552 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 552 110917 1148 AM Superheating the Steam to High Temperatures Increases Thighavg The average temperature at which heat is transferred to steam can be increased without increasing the boiler pressure by superheating the steam to high temperatures The effect of superheating on the performance of vapor power cycles is illustrated on a Ts diagram in Fig 107 The colored area on this diagram represents the increase in the net work The total area under the process curve 33 represents the increase in the heat input Thus both the net work and heat input increase as a result of superheating the steam to a higher temperature The overall effect is an increase in thermal efficiency however since the average temperature at which heat is added increases Superheating the steam to higher temperatures has another very desirable effect It decreases the moisture content of the steam at the turbine exit as can be seen from the Ts diagram the quality at state 4 is higher than that at state 4 The temperature to which steam can be superheated is limited however by metallurgical considerations At present the highest steam temperature allowed at the turbine inlet is about 620C 1150F Any increase in this value depends on improving the present materials or finding new ones that can withstand higher temperatures Ceramics are very promising in this regard Increasing the Boiler Pressure Increases Thighavg Another way of increasing the average temperature during the heataddition process is to increase the operating pressure of the boiler which automatically raises the temperature at which boiling takes place This in turn raises the average temperature at which heat is transferred to the steam and thus raises the thermal efficiency of the cycle The effect of increasing the boiler pressure on the performance of vapor power cycles is illustrated on a Ts diagram in Fig 108 Notice that for a fixed turbine inlet temperature the cycle shifts to the left and the moisture content of steam at the turbine exit increases This undesirable side effect can be corrected however by reheating the steam as discussed in the next section Operating pressures of boilers have gradually increased over the years from about 27 MPa 400 psia in 1922 to over 30 MPa 4500 psia today generating enough steam to produce a net power output of 1000 MW or more in a large power plant Today many modern steam power plants oper ate at supercritical pressures P 2206 MPa and have thermal efficien cies of about 40 percent for fossilfuel plants and 34 percent for nuclear plants There are over 150 supercriticalpressure steam power plants in operation in the United States The lower efficiencies of nuclear power plants are due to the lower maximum temperatures used in those plants for safety reasons The Ts diagram of a supercritical Rankine cycle is shown in Fig 109 The effects of lowering the condenser pressure superheating to a higher temperature and increasing the boiler pressure on the thermal efficiency of the Rankine cycle are illustrated next with an example FIGURE 108 The effect of increasing the boiler pressure on the ideal Rankine cycle 3 s T 1 2 Increase in wnet 3 4 2 4 Decrease in wnet Tmax FIGURE 107 The effect of superheating the steam to higher temperatures on the ideal Rankine cycle 3 s T 4 1 2 Increase in wnet 3 4 Final PDF to printer 553 CHAPTER 10 cen22672ch10543596indd 553 110917 1148 AM FIGURE 109 A supercritical Rankine cycle 3 s T 1 2 4 Critical point EXAMPLE 103 Effect of Boiler Pressure and Temperature on Efficiency Consider a steam power plant operating on the ideal Rankine cycle Steam enters the turbine at 3 MPa and 350C and is condensed in the condenser at a pressure of 10 kPa Determine a the thermal efficiency of this power plant b the thermal efficiency if steam is superheated to 600C instead of 350C and c the thermal effi ciency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600C SOLUTION A steam power plant operating on the ideal Rankine cycle is consid ered The effects of superheating the steam to a higher temperature and raising the boiler pressure on thermal efficiency are to be investigated Analysis The Ts diagrams of the cycle for all three cases are given in Fig 1010 a This is the steam power plant discussed in Example 101 except that the con denser pressure is lowered to 10 kPa The thermal efficiency is determined in a simi lar manner State 1 P 1 10 kPa Sat liquid h 1 h f 10 kPa 19181 kJkg v 1 v f 10 kPa 000101 m 3 kg State 2 P 2 3 MPa s 2 s 1 w pumpin v 1 P 2 P 1 000101 m 3 kg 3000 10 kPa 1 kJ 1 kPa m 3 302 kJkg h 2 h 1 w pumpin 19181 302 kJkg 19483 kJkg State 3 P 3 3 MPa T 3 350C h 3 31161 kJkg s 3 67450 kJkgK FIGURE 1010 Ts diagrams of the three cycles discussed in Example 103 3 s T 1 2 4 T3 350C 3 MPa 10 kPa a 3 s T 1 2 4 T3 600C 3 MPa 10 kPa b 3 s T 1 2 4 15 MPa 10 kPa c T3 600C Final PDF to printer 554 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 554 110917 1148 AM State 4 P 4 10 kPa sat mixture s 4 s 3 x 4 s 4 s f s fg 67450 06492 74996 08128 Thus h 4 h f x 4 h fg 19181 08128 23921 21361 kJkg q in h 3 h 2 31161 19483 kJkg 29213 kJkg q out h 4 h 1 21361 19181 kJkg 19443 kJkg and η th 1 q out q in 1 19443 kJkg 29213 kJkg 0334 or 334 Therefore the thermal efficiency increases from 260 to 334 percent as a result of lowering the condenser pressure from 75 to 10 kPa At the same time however the quality of the steam decreases from 886 to 813 percent in other words the moisture content increases from 114 to 187 percent b States 1 and 2 remain the same in this case and the enthalpies at state 3 3 MPa and 600C and state 4 10 kPa and s4 s3 are determined to be h 3 36828 kJkg h 4 23803 kJkg x 4 0915 Thus q in h 3 h 2 36828 19483 34880 kJkg q out h 4 h 1 23803 19181 21885 kJkg and η th 1 q out q in 1 21885 kJkg 34880 kJkg 0373 or 373 Therefore the thermal efficiency increases from 334 to 373 percent as a result of superheating the steam from 350 to 600C At the same time the quality of the steam increases from 813 to 915 percent in other words the moisture content decreases from 187 to 85 percent c State 1 remains the same in this case but the other states change The enthalpies at state 2 15 MPa and s2 s1 state 3 15 MPa and 600C and state 4 10 kPa and s4 s3 are determined in a similar manner to be h 2 20695 kJkg h 3 35831 kJkg h 4 21153 kJkg x 4 0804 Final PDF to printer 555 CHAPTER 10 cen22672ch10543596indd 555 110917 1148 AM 105 THE IDEAL REHEAT RANKINE CYCLE We noted in Sec 104 that increasing the boiler pressure increases the thermal efficiency of the Rankine cycle but it also increases the moisture content of the steam to unacceptable levels Then it is natural to ask the fol lowing question How can we take advantage of the increased efficiencies at higher boiler pressures without facing the problem of excessive moisture at the final stages of the turbine Two possibilities come to mind 1 Superheat the steam to very high temperatures before it enters the tur bine This would be the desirable solution since the average temperature at which heat is added would also increase thus increasing the cycle efficiency This is not a viable solution however since it requires raising the steam tem perature to metallurgically unsafe levels 2 Expand the steam in the turbine in two stages and reheat it in between In other words modify the simple ideal Rankine cycle with a reheat process Reheating is a practical solution to the excessive moisture problem in tur bines and it is commonly used in modern steam power plants The Ts diagram of the ideal reheat Rankine cycle and the schematic of the power plant operating on this cycle are shown in Fig 1011 The ideal reheat Rankine cycle differs from the simple ideal Rankine cycle in that the expansion process takes place in two stages In the first stage the high pressure turbine steam is expanded isentropically to an intermediate pressure and sent back to the boiler where it is reheated at constant pres sure usually to the inlet temperature of the first turbine stage Steam then expands isentropically in the second stage lowpressure turbine to the con denser pressure Thus the total heat input and the total turbine work output for a reheat cycle become q in q primary q reheat h 3 h 2 h 5 h 4 1012 Thus q in h 3 h 2 35831 20695 33762 kJkg q out h 4 h 1 21153 19181 19235 kJkg and η th 1 q out q in 1 19235 kJkg 33762 kJkg 0430 or 430 Discussion The thermal efficiency increases from 373 to 430 percent as a result of raising the boiler pressure from 3 to 15 MPa while maintaining the turbine inlet temperature at 600C At the same time however the quality of the steam decreases from 915 to 804 percent in other words the moisture content increases from 85 to 196 percent Final PDF to printer 556 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 556 110917 1148 AM and w turbout w turbI w turbII h 3 h 4 h 5 h 6 1013 The incorporation of the single reheat in a modern power plant improves the cycle efficiency by 4 to 5 percent by increasing the average temperature at which heat is transferred to the steam The average temperature during the reheat process can be increased by increasing the number of expansion and reheat stages As the number of stages is increased the expansion and reheat processes approach an isothermal pro cess at the maximum temperature as shown in Fig 1012 The use of more than two reheat stages however is not practical The theoretical improve ment in efficiency from the second reheat is about half of that which results from a single reheat If the turbine inlet pressure is not high enough double reheat would result in superheated exhaust This is undesirable as it would cause the average temperature for heat rejection to increase and thus the cycle efficiency to decrease Therefore double reheat is used only on supercritical pressure P 2206 MPa power plants A third reheat stage would increase the cycle efficiency by about half of the improvement attained by the second reheat This gain is too small to justify the added cost and complexity The reheat cycle was introduced in the mid1920s but it was abandoned in the 1930s because of the operational difficulties The steady increase in boiler pressures over the years made it necessary to reintroduce single reheat in the late 1940s and double reheat in the early 1950s The reheat temperatures are very close or equal to the turbine inlet tempera ture The optimum reheat pressure is about onefourth of the maximum cycle pressure For example the optimum reheat pressure for a cycle with a boiler pressure of 12 MPa is about 3 MPa Remember that the sole purpose of the reheat cycle is to reduce the mois ture content of the steam at the final stages of the expansion process If we had materials that could withstand sufficiently high temperatures there would be no need for the reheat cycle FIGURE 1011 The ideal reheat Rankine cycle LowP turbine s T 4 5 6 1 3 2 Highpressure turbine Lowpressure turbine Reheating 6 Boiler HighP turbine Reheater P4 P5 Preheat Condenser Pump 3 2 5 4 1 FIGURE 1012 The average temperature at which heat is transferred during reheating increases as the number of reheat stages is increased Tavgreheat s T Final PDF to printer 557 CHAPTER 10 cen22672ch10543596indd 557 110917 1148 AM FIGURE 1013 Schematic and Ts diagram for Example 104 s T 4 3 5 6 1 2 200 psia 10 psia 600 psia 600 psia 600F 600 psia 10 psia 10 psia 200 psia 600F LowP turbine 6 Boiler HighP turbine Reheater Condenser Pump 3 2 5 4 1 EXAMPLE 104 The Ideal Reheat Rankine Cycle Consider a steam power plant that operates on the ideal reheat Rankine cycle The plant maintains the inlet of the highpressure turbine at 600 psia and 600F the inlet of the lowpressure turbine at 200 psia and 600F and the condenser at 10 psia The net power produced by this plant is 5000 kW Determine the rate of heat addition and rejection and the thermal efficiency of the cycle Is there any advantage to operating the reheat section of the boiler at 100 psia rather than 200 psia while maintaining the same lowpressure turbine inlet temperature SOLUTION An ideal reheat steam Rankine cycle produces 5000 kW of power The rates of heat addition and rejection and the thermal efficiency of the cycle are to be determined Also the effect of changing reheat pressure is to be investigated Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The schematic of the power plant and the Ts diagram of the cycle are shown in Fig 1013 The power plant operates on the ideal reheat Rankine cycle Therefore the pump and the turbines are isentropic there are no pressure drops in the boiler and condenser and steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure From the steam tables Tables A4E A5E and A6E h 1 h f 10 psia 16125 Btulbm v 1 v f 10 psia 001659 ft 3 lbm w pumpin v 1 P 2 P 1 001659 ft 3 lbm600 100 psia 1 Btu 5404 psia ft 3 181 Btulbm Final PDF to printer 558 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 558 110917 1148 AM h 2 h 1 w pumpin 16125 181 16306 Btulbm P 3 600 psia T 3 600 F h 3 12899 Btulbm s 3 15325 BtulbmR P 4 200 psia s 4 s 3 x 4 s 4 s f s fg 15325 054379 100219 09865 h 4 h f x 4 h fg 35546 0986584333 11875 Btu lbm P 5 200 psia T 5 600 F h 5 13223 Btulbm s 5 16771 BtulbmR P 6 10 psia s 6 s 5 x 6 s 6 s f s fg 16771 028362 150391 09266 h 6 h f x 6 h fg 16125 0926698182 10710 Btu lbm Thus q in h 3 h 2 h 5 h 4 12899 16306 13223 11875 12617 Btu lbm q out h 6 h 1 10710 16125 9097 Btu lbm w net q in q out 12617 9098 3520 Btu lbm The mass flow rate of steam in the cycle is determined from W net m w net m W net w net 5000 kJ s 3520 Btu lbm 094782 Btu 1 kJ 1347 lbm s The rates of heat addition and rejection are Q in m q in 1347 lbm s 12617 Btu lbm 16995 Btu s Q out m q out 1347 lbm s 9097 Btu lbm 12250 Btu s and the thermal efficiency of the cycle is η th W net Q in 5000 kJ s 16995 Btu s 094782 Btu 1 kJ 0279 or 279 If we repeat the analysis for a reheat pressure of 100 psia at the same reheat tempera ture we obtain a thermal efficiency of 273 percent Thus operating the reheater at 100 psia causes a slight decrease in the thermal efficiency Discussion Now we try to address this question At what reheat pressure will the thermal efficiency be maximum We repeat the analysis at various reheat pressures using appropriate software The results are plotted in Fig 1014 The thermal effi ciency reaches a maximum value of 281 percent at an optimum reheat pressure of about 325 psia Final PDF to printer 559 CHAPTER 10 cen22672ch10543596indd 559 110917 1148 AM 106 THE IDEAL REGENERATIVE RANKINE CYCLE A careful examination of the Ts diagram of the Rankine cycle redrawn in Fig 1015 reveals that heat is transferred to the working fluid during process 22 at a relatively low temperature This lowers the average heataddition temperature and thus the cycle efficiency To remedy this shortcoming we look for ways to raise the temperature of the liquid leaving the pump called the feedwater before it enters the boiler One such possibility is to transfer heat to the feedwater from the expanding steam in a counterflow heat exchanger built into the turbine that is to use regeneration This solution is also impractical because it is difficult to design such a heat exchanger and because it would increase the moisture content of the steam at the final stages of the turbine A practical regeneration process in steam power plants is accomplished by extracting or bleeding steam from the turbine at various points This steam which could have produced more work by expanding further in the turbine is used to heat the feedwater instead The device where the feedwater is heated by regeneration is called a regenerator or a feedwater heater FWH Regeneration not only improves cycle efficiency but also provides a conve nient means of deaerating the feedwater removing the air that leaks in at the condenser to prevent corrosion in the boiler It also helps control the large volume flow rate of the steam at the final stages of the turbine due to the large specific volumes at low pressures Therefore regeneration has been used in all modern steam power plants since its introduction in the early 1920s A feedwater heater is basically a heat exchanger where heat is transferred from the steam to the feedwater either by mixing the two fluid streams open feedwater heaters or without mixing them closed feedwater heaters Regen eration with both types of feedwater heaters is discussed next Open Feedwater Heaters An open or directcontact feedwater heater is basically a mixing chamber where the steam extracted from the turbine mixes with the feedwater exiting FIGURE 1015 The first part of the heataddition process in the boiler takes place at relatively low temperatures s T 4 1 3 2 2 Steam entering boiler Lowtemperature heat addition Steam exiting boiler FIGURE 1014 There is an optimum reheat pressure in the reheat Rankine cycle for which the thermal efficiency is maximum The values refer to Example 104 0285 028 0275 0265 026 50 100 150 200 250 Reheat pressure psia ηth 300 350 400 450 500 027 Final PDF to printer 560 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 560 110917 1148 AM the pump Ideally the mixture leaves the heater as a saturated liquid at the heater pressure The schematic of a steam power plant with one open feedwa ter heater also called singlestage regenerative cycle and the Ts diagram of the cycle are shown in Fig 1016 In an ideal regenerative Rankine cycle steam enters the turbine at the boiler pressure state 5 and expands isentropically to an intermediate pressure state 6 Some steam is extracted at this state and routed to the feedwater heater while the remaining steam continues to expand isentropically to the condenser pressure state 7 This steam leaves the condenser as a saturated liquid at the condenser pressure state 1 The condensed water which is also called the feedwater then enters an isentropic pump where it is compressed to the feedwater heater pressure state 2 and is routed to the feedwater heater where it mixes with the steam extracted from the turbine The fraction of the steam extracted is such that the mixture leaves the heater as a saturated liquid at the heater pressure state 3 A second pump raises the pressure of the water to the boiler pressure state 4 The cycle is completed by heating the water in the boiler to the turbine inlet state state 5 In the analysis of steam power plants it is more convenient to work with quantities expressed per unit mass of the steam flowing through the boiler For each 1 kg of steam leaving the boiler y kg expands partially in the turbine and is extracted at state 6 The remaining 1 y kg expands completely to the condenser pressure Therefore the mass flow rates are different in different components If the mass flow rate through the boiler is m for example it is 1 y m through the condenser This aspect of the regenerative Rankine cycle should be considered in the analysis of the cycle as well as in the interpreta tion of the areas on the Ts diagram In light of Fig 1016 the heat and work interactions of a regenerative Rankine cycle with one feedwater heater can be expressed per unit mass of steam flowing through the boiler as follows q in h 5 h 4 1014 q out 1 y h 7 h 1 1015 FIGURE 1016 The ideal regenerative Rankine cycle with an open feedwater heater 7 5 6 1 3 4 2 s T 1 y y 7 Boiler Turbine Condenser Pump II 5 Pump I 4 6 3 2 Open FWH 1 Final PDF to printer 561 CHAPTER 10 cen22672ch10543596indd 561 110917 1148 AM w turbout h 5 h 6 1 y h 6 h 7 1016 w pumpin 1 y w pump Iin w pump IIin 1017 where y m 6 m 5 fraction of steam extracted w pump Iin v 1 P 2 P 1 w pump IIin v 3 P 4 P 3 The thermal efficiency of the Rankine cycle increases as a result of regenera tion This is because regeneration raises the average temperature at which heat is transferred to the steam in the boiler by raising the temperature of the water before it enters the boiler The cycle efficiency increases further as the number of feedwater heaters is increased Many large plants in operation today use as many as eight feedwater heaters The optimum number of feedwater heaters is determined from economic considerations The use of an additional feedwater heater cannot be justified unless it saves more in fuel costs than its own cost Closed Feedwater Heaters Another type of feedwater heater often used in steam power plants is the closed feedwater heater in which heat is transferred from the extracted steam to the feedwater without any mixing taking place The two streams now can be at different pressures since they do not mix The schematic of a steam power plant with one closed feedwater heater and the Ts diagram of the cycle are shown in Fig 1017 In an ideal closed feedwater heater the feedwater is heated to the exit temperature of the extracted steam which ideally leaves the heater as a saturated liquid at the extraction pressure In actual power plants FIGURE 1017 The ideal regenerative Rankine cycle with a closed feedwater heater s T 6 8 1 2 7 3 4 5 9 4 3 Mixing chamber Turbine 8 7 5 9 6 Boiler Condenser Closed FWH Pump I Pump II 1 2 Boiler Final PDF to printer 562 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 562 110917 1148 AM the feedwater leaves the heater below the exit temperature of the extracted steam because a temperature difference of at least a few degrees is required for any effective heat transfer to take place The condensed steam is then either pumped to the feedwater line or routed to another heater or to the condenser through a device called a trap A trap allows the liquid to be throttled to a lowerpressure region but traps the vapor The enthalpy of steam remains constant during this throttling process The open and closed feedwater heaters can be compared as follows Open feed water heaters are simple and inexpensive and have good heat transfer charac teristics They also bring the feedwater to the saturation state For each heater however a pump is required to handle the feedwater The closed feedwater heaters are more complex because of the internal tubing network and thus they are more expensive Heat transfer in closed feedwater heaters is also less effective since the two streams are not allowed to be in direct contact However closed feedwater heaters do not require a separate pump for each heater since the extracted steam and the feedwater can be at different pressures Most steam power plants use a combination of open and closed feedwater heaters as shown in Fig 1018 FIGURE 1018 A steam power plant with one open and three closed feedwater heaters Trap Trap Condenser Pump Deaerating Trap Pump Turbine Closed FWH Closed FWH Closed FWH Boiler Open FWH EXAMPLE 105 The Ideal Regenerative Rankine Cycle Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open feedwater heater Steam enters the turbine at 15 MPa and 600C and is con densed in the condenser at a pressure of 10 kPa Some steam leaves the turbine at a pressure of 12 MPa and enters the open feedwater heater Determine the fraction of steam extracted from the turbine and the thermal efficiency of the cycle SOLUTION A steam power plant operates on the ideal regenerative Rankine cycle with one open feedwater heater The fraction of steam extracted from the turbine and the thermal efficiency are to be determined Final PDF to printer 563 CHAPTER 10 cen22672ch10543596indd 563 110917 1148 AM Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The schematic of the power plant and the Ts diagram of the cycle are shown in Fig 1019 We note that the power plant operates on the ideal regenerative Rankine cycle Therefore the pumps and the turbines are isentropic there are no pressure drops in the boiler condenser and feedwater heater and steam leaves the condenser and the feedwater heater as saturated liquid First we determine the enthalpies at various states State 1 P 1 10 kPa Sat liquid h 1 h f 10 kPa 19181 kJkg v 1 v f 10 kPa 000101 m 3 kg State 2 P 2 12 MPa s 2 s 1 w pumpin v 1 P 2 P 1 000101 m 3 kg 1200 10 kPa 1 kJ 1 kPa m 3 120 kJkg h 2 h 1 w pump Iin 19181 120 kJkg 19301 kJkg State 3 P 3 12 MPa Sat liquid v 3 v f 12 MPa 0001138 m 3 kg h 3 h f 12 MPa 79833 kJ kg State 4 P 4 15 MPa s 4 s 3 w pump IIin v 3 P 4 P 3 0001138 m 3 kg 15000 1200 kPa 1 kJ 1 kPa m 3 1570 kJkg h 4 h 3 w pump IIin 79833 1570 kJkg 81403 kJkg FIGURE 1019 Schematic and Ts diagram for Example 105 s T 5 7 1 2 6 3 4 7 qin Turbine 15 MPa 600C wturbout Condenser Pump II 15 MPa 5 4 6 3 12 MPa 12 MPa 12 MPa 10 kPa 2 Open FWH qout Pump I 1 10 kPa Boiler Final PDF to printer 564 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 564 110917 1148 AM State 5 P 5 15 MPa T 5 600C h 5 35831 kJkg s 5 66796 kJkgK State 6 P 6 12 MPa s 6 s 5 h 6 28602 kJkg T 6 2184C State 7 P 7 10 kPa s 7 s 5 x 7 s 7 s f s fg 66796 06492 74996 08041 h 7 h f x 7 h fg 19181 08041 23921 21153 kJkg The energy analysis of open feedwater heaters is identical to the energy analysis of mixing chambers The feedwater heaters are generally well insulated Q 0 and they do not involve any work interactions W 0 By neglecting the kinetic and potential energies of the streams the energy balance reduces for a feedwater heater to E in E out in m h out m h or y h 6 1 y h 2 1 h 3 where y is the fraction of steam extracted from the turbine m 6 m 5 Solving for y and substituting the enthalpy values we find y h 3 h 2 h 6 h 2 79833 19301 28602 19301 02270 Thus q in h 5 h 4 35831 81403 kJkg 27691 kJkg q out 1 y h 7 h 1 1 0227021153 19181 kJkg 14869 kJkg and η th 1 q out q in 1 14869 kJkg 27691 kJkg 0463 or 463 Discussion This problem was worked out in Example 103c for the same pressure and temperature limits but without the regeneration process A comparison of the two results reveals that the thermal efficiency of the cycle has increased from 430 to 463 percent as a result of regeneration The net work output decreased by 171 kJkg but the heat input decreased by 607 kJkg which results in a net increase in the ther mal efficiency Final PDF to printer 565 CHAPTER 10 cen22672ch10543596indd 565 110917 1148 AM EXAMPLE 106 The Ideal ReheatRegenerative Rankine Cycle Consider a steam power plant that operates on an ideal reheatregenerative Rankine cycle with one open feedwater heater one closed feedwater heater and one reheater Steam enters the turbine at 15 MPa and 600C and is condensed in the condenser at a pressure of 10 kPa Some steam is extracted from the turbine at 4 MPa for the closed feedwater heater and the remaining steam is reheated at the same pressure to 600C The extracted steam is completely condensed in the heater and is pumped to 15 MPa before it mixes with the feedwater at the same pressure Steam for the open feedwater heater is extracted from the lowpressure turbine at a pressure of 05 MPa Determine the fractions of steam extracted from the turbine as well as the thermal efficiency of the cycle SOLUTION A steam power plant operates on the ideal reheatregenerative Ran kine cycle with one open feedwater heater one closed feedwater heater and one reheater The fractions of steam extracted from the turbine and the thermal efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 In both open and closed feedwater heaters feedwater is heated to the saturation temperature at the feedwater heater pressure Note that this is a conservative assumption since extracted steam enters the closed feedwater heater at 376C and the saturation temperature at the closed feedwater pressure of 4 MPa is 250C Analysis The schematic of the power plant and the Ts diagram of the cycle are shown in Fig 1020 The power plant operates on the ideal reheatregenerative Ran kine cycle and thus the pumps and the turbines are isentropic there are no pressure drops in the boiler reheater condenser and feedwater heaters and steam leaves the condenser and the feedwater heaters as saturated liquid FIGURE 1020 Schematic and Ts diagram for Example 106 Closed FWH Mixing chamber LowP turbine Condenser s T 5 9 1 2 Pump II 11 13 9 12 3 8 7 6 Pump III Pump I 4 MPa 8 05 MPa 1 kg 15 MPa 600C y P10 P11 4 MPa z 600C 11 10 12 13 5 4 15 MPa 4 MPa 05 MPa z 10 kPa 1 y 1 y 1 kg y 1 y z 1 y 10 kPa 1 y z Boiler HighP turbine 10 Reheater 1 2 3 4 Open FWH 7 6 Final PDF to printer 566 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 566 110917 1148 AM The enthalpies at the various states and the pump work per unit mass of fluid flow ing through them are h 1 19181 kJkg h 9 35831 kJkg h 2 19230 kJkg h 10 31550 kJkg h 3 64009 kJkg h 11 36749 kJkg h 4 64392 kJkg h 12 30148 kJkg h 5 10874 kJkg h 13 23357 kJkg h 6 10874 kJkg w pump Iin 049 kJkg h 7 11012 kJkg w pump IIin 383 kJkg h 8 10898 kJkg w pump IIIin 1377 kJkg The fractions of steam extracted are determined from the mass and energy balances of the feedwater heaters Closed feedwater heater E in E out y h 10 1 y h 4 1 y h 5 y h 6 y h 5 h 4 h 10 h 6 h 5 h 4 10874 64392 31550 10874 10874 64392 01766 Open feedwater heater E in E out z h 12 1 y z h 2 1 y h 3 z 1 y h 3 h 2 h 12 h 2 1 0176664009 19230 30148 19230 01306 The enthalpy at state 8 is determined by applying the mass and energy equations to the mixing chamber which is assumed to be insulated E in E out 1 h 8 1 y h 5 y h 7 h 8 1 01766 10874 kJkg 01766 11012 kJkg 10898 kJkg Thus q in h 9 h 8 1 y h 11 h 10 35831 10898 kJkg 1 0176636749 31550 kJkg 29214 kJkg q out 1 y z h 13 h 1 1 01766 0130623357 19181 kJkg 14853 kJkg and η th 1 q out q in 1 14853 kJkg 29214 kJkg 0492 or 492 Final PDF to printer 567 CHAPTER 10 cen22672ch10543596indd 567 110917 1148 AM Discussion This problem was worked out in Example 104 for the same pressure and temperature limits with reheat but without the regeneration process A compari son of the two results reveals that the thermal efficiency of the cycle has increased from 450 to 492 percent as a result of regeneration The thermal efficiency of this cycle could also be determined from η th w net q in w turbout w pumpin q in where w turbout h 9 h 10 1 y h 11 h 12 1 y z h 12 h 13 w pumpin 1 y z w pump Iin 1 y w pump IIin y w pump IIIin Also if we assume that the feedwater leaves the closed FWH as a saturated liquid at 15 MPa and thus at T5 342C and h5 16103 kJkg it can be shown that the thermal efficiency would be 506 percent 107 SECONDLAW ANALYSIS OF VAPOR POWER CYCLES The ideal Carnot cycle is a totally reversible cycle and thus it does not involve any irreversibilities The ideal Rankine cycles simple reheat or regenerative however are only internally reversible and they may involve irreversibilities external to the system such as heat transfer through a finite temperature difference A secondlaw analysis of these cycles reveals where the largest irreversibilities occur and what their magnitudes are Relations for exergy and exergy destruction for steadyflow systems are developed in Chap 8 The exergy destruction for a steadyflow system can be expressed in the rate form as X dest T 0 S gen T 0 S out S in T 0 out m s Q out T bout in m s Q in T bin kW 1018 or on a unitmass basis for a oneinlet oneexit steadyflow device as x dest T 0 s gen T 0 s e s i q out T bout q in T bin kJkg 1019 where Tbin and Tbout are the temperatures of the system boundary where heat is transferred into and out of the system respectively The exergy destruction associated with a cycle depends on the magnitude of the heat transfer with the high and lowtemperature reservoirs involved and their temperatures It can be expressed on a unitmass basis as x dest T 0 q out T bout q in T bin kJkg 1020 Final PDF to printer 568 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 568 110917 1148 AM For a cycle that involves heat transfer only with a source at TH and a sink at TL the exergy destruction becomes x dest T 0 q out T L q in T H kJkg 1021 The exergy of a fluid stream ψ at any state can be determined from ψ h h 0 T 0 s s 0 V 2 2 gz kJkg 1022 where the subscript 0 denotes the state of the surroundings FIGURE 1021 Schematic for Example 107 3 MPa 350C Boiler Turbine wturbout Condenser Pump 3 MPa 75 kPa 3 qin 4 2 wpumpin 75 kPa 75 kPa 1 qout EXAMPLE 107 SecondLaw Analysis of an Ideal Rankine Cycle Consider a steam power plant operating on the simple ideal Rankine cycle Fig 1021 Steam enters the turbine at 3 MPa and 350C and is condensed in the condenser at a pressure of 75 kPa Heat is supplied to the steam in a furnace main tained at 800 K and waste heat is rejected to the surroundings at 300 K Determine a the exergy destruction associated with each of the four processes and the whole cycle and b the secondlaw efficiency of this cycle SOLUTION A steam power plant operating on the simple ideal Rankine cycle is considered For specified source and sink temperatures the exergy destruction associ ated with this cycle and the secondlaw efficiency are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis We take the power plant bordering the furnace at temperature TH and the environment at temperature T0 as the control volume This cycle was analyzed in Example 101 and various quantities were determined to be qin 2729 kJkg wpumpin 30 kJkg wturbout 713 kJkg qout 2019 kJkg and ηth 260 percent a Processes 12 and 34 are isentropic s1 s2 s3 s4 and therefore do not involve any internal or external irreversibilities that is x dest12 0 and x dest34 0 Processes 23 and 41 are constantpressure heataddition and heatrejection pro cesses respectively and they are internally reversible But the heat transfer between the working fluid and the source or the sink takes place through a finite temperature difference rendering both processes irreversible The irreversibility associated with each process is determined from Eq 1019 The entropy of the steam at each state is determined from the steam tables s 2 s 1 s f 75 kPa 12132 kJkgK s 4 s 3 67450 kJkgK at 3 MPa 350C Thus x dest23 T 0 s 3 s 2 q in23 T source 300 K 67450 12132 kJkgK 2729 kJkg 800 K 636 kJkg Final PDF to printer 569 CHAPTER 10 cen22672ch10543596indd 569 110917 1148 AM x dest41 T 0 s 1 s 4 q out41 T sink 300 K 12132 67450 kJ kgK 2019 kJ kg 300 K 360 kJ kg Therefore the irreversibility of the cycle is x destcycle x dest12 x dest23 x dest34 x dest41 0 636 kJ kg 0 360 kJ kg 996 kJ kg The total exergy destroyed during the cycle could also be determined from Eq 1021 Notice that the largest exergy destruction in the cycle occurs during the heataddition process Therefore any attempt to reduce the exergy destruction should start with this process Raising the turbine inlet temperature of the steam for exam ple would reduce the temperature difference and thus the exergy destruction b The secondlaw efficiency is defined as η II Exergy recovered Exergy expended x recovered x expended 1 x destroyed x expended Here the expended exergy is the exergy content of the heat supplied to steam in boiler which is its work potential and the pump input and the exergy recovered is the work output of the turbine x heatin 1 T 0 T H q in 1 300 K 800 K 2729 kJ kg 1706 kJ kg x expended x heatin x pumpin 1706 30 1709 kJ kg x recovered w turbineout 713 kJ kg Substituting the secondlaw efficiency of this power plant is determined to be η II x recovered x expended 713 kJ kg 1709 kJ kg 0417 or 417 Discussion The secondlaw efficiency can also be determined using the exergy destruction data η II 1 x destroyed x expended 1 996 kJ kg 1709 kJ kg 0417 or 417 Also the system considered contains both the furnace and the condenser and thus the exergy destruction associated with heat transfer involving both the furnace and the condenser are accounted for 108 COGENERATION In all the cycles discussed so far the sole purpose was to convert a portion of the heat transferred to the working fluid to work which is the most valuable form of energy The remaining portion of the heat is rejected to rivers lakes Final PDF to printer 570 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 570 110917 1148 AM oceans or the atmosphere as waste heat because its quality or grade is too low to be of any practical use Wasting a large amount of heat is a price we have to pay to produce work because electrical or mechanical work is the only form of energy on which many engineering devices such as a fan can operate Many systems or devices however require energy input in the form of heat called process heat Some industries that rely heavily on process heat are chemical pulp and paper oil production and refining steelmaking food processing and textile industries Process heat in these industries is usually supplied by steam at 5 to 7 atm and 150 to 200C 300 to 400F Energy is usually transferred to the steam by burning coal oil natural gas or another fuel in a furnace Now let us examine the operation of a processheating plant closely Disre garding any heat losses in the piping all the heat transferred to the steam in the boiler is used in the processheating units as shown in Fig 1022 Therefore process heating seems like a perfect operation with practically no waste of energy From the secondlaw point of view however things do not look so per fect The temperature in furnaces is typically very high around 1400C and thus the energy in the furnace is of very high quality This highquality energy is transferred to water to produce steam at about 200C or below a highly irreversible process Associated with this irreversibility is of course a loss in exergy or work potential It is simply not wise to use highquality energy to accomplish a task that could be accomplished with lowquality energy Industries that use large amounts of process heat also consume a large amount of electric power Therefore it makes economical as well as engineer ing sense to use the alreadyexisting work potential to produce power instead of letting it go to waste The result is a plant that produces electricity while meeting the processheat requirements of certain industrial processes Such a plant is called a cogeneration plant In general cogeneration is the produc tion of more than one useful form of energy such as process heat and electric power from the same energy source Either a steamturbine Rankine cycle or a gasturbine Brayton cycle or even a combined cycle discussed later can be used as the power cycle in a cogeneration plant The schematic of an ideal steamturbine cogeneration plant is shown in Fig 1023 Let us say this plant is to supply process heat Q p at 500 kPa at a rate of 100 kW To meet this demand steam is expanded in the turbine to a pressure of 500 kPa producing power at a rate of say 20 kW The flow rate of the steam can be adjusted such that steam leaves the process heating section as a saturated liquid at 500 kPa Steam is then pumped to the boiler pressure and is heated in the boiler to state 3 The pump work is usually very small and can be neglected Disregarding any heat losses the rate of heat input in the boiler is determined from an energy balance to be 120 kW Probably the most striking feature of the ideal steamturbine cogeneration plant shown in Fig 1023 is the absence of a condenser Thus no heat is rejected from this plant as waste heat In other words all the energy trans ferred to the steam in the boiler is utilized as either process heat or electric power Thus it is appropriate to define a utilization factor 𝜖u for a cogenera tion plant as 𝜖 u Net power output Process heat delivered Total heat input W net Q p Q in 1023 FIGURE 1022 A simple processheating plant Pump Process heater Qp Boiler Qin FIGURE 1023 An ideal cogeneration plant 1 Boiler 20 kW Pump 120 kW 3 2 4 Turbine Wpump 0 Process heater 100 kW Final PDF to printer 571 CHAPTER 10 cen22672ch10543596indd 571 110917 1148 AM or 𝜖 u 1 Q out Q in 1024 where Q out represents the heat rejected in the condenser Strictly speak ing Q out also includes all the undesirable heat losses from the piping and other components but they are usually small and thus neglected It also includes combustion inefficiencies such as incomplete combustion and stack losses when the utilization factor is defined on the basis of the heating value of the fuel The utilization factor of the ideal steamturbine cogeneration plant is obviously 100 percent Actual cogeneration plants have utilization factors as high as 80 percent Some recent cogeneration plants have even higher uti lization factors Notice that without the turbine we would need to supply heat to the steam in the boiler at a rate of only 100 kW instead of at 120 kW The additional 20 kW of heat supplied is converted to work Therefore a cogeneration power plant is equivalent to a processheating plant combined with a power plant that has a thermal efficiency of 100 percent The ideal steamturbine cogeneration plant just described is not practical because it cannot adjust to the variations in power and processheat loads The schematic of a more practical but more complex cogeneration plant is shown in Fig 1024 Under normal operation some steam is extracted from the tur bine at some predetermined intermediate pressure P6 The rest of the steam expands to the condenser pressure P7 and is then cooled at constant pressure The heat rejected from the condenser represents the waste heat for the cycle At times of high demand for process heat all the steam is routed to the process heating units and none to the condenser m 7 0 The waste heat is zero in this mode If this is not sufficient some steam leaving the boiler is throttled by an expansion or pressurereducing valve PRV to the extraction pressure P6 and is directed to the processheating unit Maximum process heating is realized when all the steam leaving the boiler passes through the expansion valve m 5 m 4 No power is produced in this mode When there is no demand for process heat all the steam passes through the turbine and the condenser m 5 m 6 0 and the cogeneration plant operates as an ordinary steam power plant The rates of heat input heat rejected and process heat supply as well as the power produced for this cogeneration plant can be expressed as follows Q in m 3 h 4 h 3 1025 Q out m 7 h 7 h 1 1026 Q p m 5 h 5 m 6 h 6 m 8 h 8 1027 W turb m 4 m 5 h 4 h 6 m 7 h 6 h 7 1028 Under optimum conditions a cogeneration plant simulates the ideal cogen eration plant discussed earlier That is all the steam expands in the turbine to the extraction pressure and continues to the processheating unit No steam passes through the expansion valve or the condenser thus no waste heat is rejected m 4 m 6 and m 5 m 7 0 This condition may be difficult to achieve in practice because of the constant variations in the processheat and power FIGURE 1024 A cogeneration plant with adjustable loads Pump I 4 3 7 6 Turbine Pump II Expansion valve Condenser 5 8 2 Process heater 1 Boiler Mixing chamber Final PDF to printer 572 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 572 110917 1148 AM loads But the plant should be designed so that the optimum operating condi tions are approximated most of the time The use of cogeneration dates to the beginning of last century when power plants were integrated into communities to provide district heating that is space hot water and process heating for residential and commercial buildings The district heating systems lost their popularity in the 1940s owing to low fuel prices However the rapid rise in fuel prices in the 1970s brought about renewed interest in district heating Cogeneration plants have proved to be economically very attractive Con sequently more and more such plants have been installed in recent years and more are being installed FIGURE 1025 Schematic and Ts diagram for Example 108 8 Pump I 5 kPa 9 Mixing chamber 6 500 kPa 5 kPa 500 kPa 7 MPa 500C 7 MPa 5 Turbine Pump II Expansion valve Condenser 4 2 3 s T 1 2 3 4 5 6 10 8 7 11 9 11 Boiler 10 7 Process heater 1 EXAMPLE 108 An Ideal Cogeneration Plant Consider the cogeneration plant shown in Fig 1025 Steam enters the turbine at 7 MPa and 500C Some steam is extracted from the turbine at 500 kPa for process heating The remaining steam continues to expand to 5 kPa Steam is then condensed at constant pressure and pumped to the boiler pressure of 7 MPa At times of high demand for process heat some steam leaving the boiler is throttled to 500 kPa and is routed to the process heater The extraction fractions are adjusted so that steam leaves the process heater as a saturated liquid at 500 kPa It is subsequently pumped to 7 MPa The mass flow rate of steam through the boiler is 15 kgs Disregarding any pressure drops and heat losses in the piping and assuming the turbine and the pump to be isentropic determine a the maximum rate at which process heat can be supplied b the power produced and the utilization factor when no process heat is supplied and c the rate of process heat supply when 10 percent of the steam is extracted before it enters the turbine and 70 percent of the steam is extracted from the turbine at 500 kPa for process heating Final PDF to printer 573 CHAPTER 10 cen22672ch10543596indd 573 110917 1148 AM SOLUTION A cogeneration plant is considered The maximum rate of process heat supply the power produced and the utilization factor when no process heat is supplied and the rate of process heat supply when steam is extracted from the steam line and turbine at specified ratios are to be determined Assumptions 1 Steady operating conditions exist 2 Pressure drops and heat losses in piping are negligible 3 Kinetic and potential energy changes are negligible Analysis The schematic of the cogeneration plant and the Ts diagram of the cycle are shown in Fig 1025 The power plant operates on an ideal cycle and thus the pumps and the turbines are isentropic there are no pressure drops in the boiler pro cess heater and condenser and steam leaves the condenser and the process heater as saturated liquid The work inputs to the pumps and the enthalpies at various states are as follows w pump Iin v 8 P 9 P 8 0001005 m 3 kg 7000 5 kPa 1 kJ 1 kPa m 3 703 kJkg w pump IIin v 7 P 10 P 7 0001093 m 3 kg 7000 500 kPa 1 kJ 1 kPa m 3 710 kJkg h 1 h 2 h 3 h 4 34114 kJkg h 5 27393 kJkg h 6 20730 kJkg h 7 h f 500 kPa 64009 kJkg h 8 h f 5 kPa 13775 kJkg h 9 h 8 w pump Iin 13775 703 kJkg 14478 kJkg h 10 h 7 w pump IIin 64009 710 kJkg 64719 kJkg a The maximum rate of process heat is achieved when all the steam leaving the boiler is throttled and sent to the process heater and none is sent to the turbine that is m 4 m 7 m 1 15 kgs and m 3 m 5 m 6 0 Thus Q pmax m 1 h 4 h 7 15 kg s 34114 64009 kJ kg 41570 kW The utilization factor is 100 percent in this case since no heat is rejected in the con denser heat losses from the piping and other components are assumed to be negli gible and combustion losses are not considered b When no process heat is supplied all the steam leaving the boiler passes through the turbine and expands to the condenser pressure of 5 kPa that is m 3 m 6 m 1 15 kgs and m 2 m 5 0 Maximum power is produced in this mode which is determined to be W turbout m h 3 h 6 15 kg s 34114 20730 kJ kg 20076 kW W pumpin 15 kg s703 kJ kg 105 kW W netout W turbout W pumpin 20076 105 kW 19971 kW 200 MW Q in m 1 h 1 h 11 15 kg s 34114 14478 kJ kg 48999 kW Final PDF to printer 574 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 574 110917 1148 AM 109 COMBINED GASVAPOR POWER CYCLES The continued quest for higher thermal efficiencies has resulted in rather innovative modifications to conventional power plants The binary vapor cycle discussed later is one such modification A more popular modification involves a gas power cycle topping a vapor power cycle which is called the combined gasvapor cycle or just the combined cycle The combined cycle of greatest interest is the gasturbine Brayton cycle topping a steamturbine Rankine cycle which has a higher thermal efficiency than either of the cycles executed individually Gasturbine cycles typically operate at considerably higher temperatures than steam cycles The maximum fluid temperature at the turbine inlet is about 620C 1150F for modern steam power plants but over 1425C 2600F for gasturbine power plants It is over 1500C at the burner exit of turbojet engines The use of higher temperatures in gas turbines is made pos sible by developments in cooling the turbine blades and coating the blades with hightemperatureresistant materials such as ceramics Because of the Thus 𝜖 u W net Q p Q in 19971 0 kW 48999 kW 0408 or 408 That is 408 percent of the energy is utilized for a useful purpose Notice that the utilization factor is equivalent to the thermal efficiency in this case c Neglecting any kinetic and potential energy changes an energy balance on the process heater yields E in E out m 4 h 4 m 5 h 5 Q pout m 7 h 7 or Q pout m 4 h 4 m 5 h 5 m 7 h 7 where m 4 01 15 kg s 15 kg s m 5 07 15 kg s 105 kg s m 7 m 4 m 5 15 105 12 kg s Thus Q pout 15 kg s 34114 kJ kg 105 kg s 27393 kJ kg 12 kg s 64009 kJ kg 262 MW Discussion Note that 262 MW of the heat transferred will be utilized in the process heater We could also show that 110 MW of power is produced in this case and the rate of heat input in the boiler is 430 MW Thus the utilization factor is 865 percent Final PDF to printer 575 CHAPTER 10 cen22672ch10543596indd 575 110917 1148 AM higher average temperature at which heat is supplied gasturbine cycles have a greater potential for higher thermal efficiencies However the gasturbine cycles have one inherent disadvantage The gas leaves the gas turbine at very high temperatures usually above 500C which erases any potential gains in the thermal efficiency The situation can be improved somewhat by using regeneration but the improvement is limited It makes engineering sense to take advantage of the very desirable characteris tics of the gasturbine cycle at high temperatures and to use the hightemperature exhaust gases as the energy source for the bottoming cycle such as a steam power cycle The result is a combined gassteam cycle as shown in Fig 1026 In this cycle energy is recovered from the exhaust gases by transferring it to the steam in a heat exchanger that serves as the boiler In general more than one gas tur bine is needed to supply sufficient heat to the steam Also the steam cycle may involve regeneration as well as reheating Energy for the reheating process can be supplied by burning some additional fuel in the oxygenrich exhaust gases Developments in gasturbine technology have made the combined gassteam cycle economically very attractive The combined cycle increases the efficiency without increasing the initial cost greatly Consequently many new power plants operate on combined cycles and many more existing steam or gas turbine plants are being converted to combinedcycle power plants Thermal efficiencies well over 50 percent are reported as a result of conversion FIGURE 1026 Combined gassteam power plant Qin 7 8 3 4 9 5 6 4 5 8 1 T s 6 7 3 9 Qin Qout Steam cycle Gas cycle Gas cycle Heat exchanger Steam cycle Exhaust gases Air in Condenser Pump Gas turbine 2 Steam turbine Combustion chamber 2 1 Qout Compressor Final PDF to printer 576 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 576 110917 1148 AM A 1090MW Tohoku combined plant that was put in commercial operation in 1985 in Niigata Japan is reported to operate at a thermal efficiency of 44 percent This plant has two 191MW steam turbines and six 118MW gas turbines Hot combustion gases enter the gas turbines at 1154C and steam enters the steam turbines at 500C Steam is cooled in the condenser by cool ing water at an average temperature of 15C The compressors have a pressure ratio of 14 and the mass flow rate of air through the compressors is 443 kgs A 1350MW combinedcycle power plant built in Ambarli Turkey in 1988 by Siemens is the first commercially operating thermal plant in the world to attain an efficiency level as high as 525 percent at design operating condi tions This plant has six 150MW gas turbines and three 173MW steam tur bines Another plant built by Siemens in Irsching Germany in 2011 reached a thermal efficiency of 608 percent with an electrical output of 578 MW In 2016 General Electric reported 622 percent efficiency for its combined cycle plant in Bouchain France with an output of 594 MW The new target for ther mal efficiency is 65 percent approaching the Carnot limit FIGURE 1027 Ts diagram of the gassteam combined cycle described in Example 109 4 1 4 1 T K s 2 3 5 500C 7 MPa 5 kPa 7 MPa 450 300 3 2 1300 EXAMPLE 109 A Combined GasSteam Power Cycle Consider the combined gassteam power cycle shown in Fig 1027 The topping cycle is a gasturbine cycle that has a pressure ratio of 8 Air enters the compressor at 300 K and the turbine at 1300 K The isentropic efficiency of the compressor is 80 percent and that of the gas turbine is 85 percent The bottoming cycle is a simple ideal Rankine cycle operating between the pressure limits of 7 MPa and 5 kPa Steam is heated in a heat exchanger by the exhaust gases to a temperature of 500C The exhaust gases leave the heat exchanger at 450 K Determine a the ratio of the mass flow rates of the steam and the combustion gases and b the thermal efficiency of the combined cycle SOLUTION A combined gassteam cycle is considered The ratio of the mass flow rates of the steam and the combustion gases and the thermal efficiency are to be determined Analysis The Ts diagrams of both cycles are given in Fig 1027 The gasturbine cycle alone was analyzed in Example 96 and the steam cycle in Example 108b with the following results Gas cycle h4 88036 kJ kg T4 853 K q in 79058 kJ kg w net 21041 kJ kg η th 266 h5 h 450 K 45180 kJ kg Steam cycle h 2 14478 kJ kg T 2 33C h 3 34114 kJ kg T 3 500C w net 13314 kJ kg η th 408 a The ratio of mass flow rates is determined from an energy balance on the heat exchanger E in E out m g h5 m s h 3 m g h4 m s h 2 m s h 3 h 2 m g h4 h5 m s 34114 14478 m g 88036 45180 Final PDF to printer 577 CHAPTER 10 cen22672ch10543596indd 577 110917 1148 AM Thus m s m g y 0131 That is 1 kg of exhaust gases can heat only 0131 kg of steam from 33 to 500C as they are cooled from 853 to 450 K Then the total net work output per kilogram of combustion gases becomes w net w netgas y w netsteam 21041 kJ kg gas 0131 kg steam kg gas 13314 kJ kg steam 3848 kJ kg gas Therefore for each kg of combustion gases produced the combined plant will deliver 3848 kJ of work The net power output of the plant is determined by multi plying this value by the mass flow rate of the working fluid in the gasturbine cycle b The thermal efficiency of the combined cycle is determined from η th w net q in 3848 kJ kg gas 7906 kJ kg gas 0487 or 487 Discussion Note that this combined cycle converts to useful work 487 percent of the energy supplied to the gas in the combustion chamber This value is consider ably higher than the thermal efficiency of the gasturbine cycle 266 percent or the steamturbine cycle 408 percent operating alone With the exception of a few specialized applications the working fluid predomi nantly used in vapor power cycles is water Water is the best working fluid cur rently available but it is far from being the ideal one The binary cycle is an attempt to overcome some of the shortcomings of water and to approach the ideal working fluid by using two fluids Before we discuss the binary cycle let us list the characteristics of a working fluid most suitable for vapor power cycles 1 A high critical temperature and a safe maximum pressure A critical temperature above the metallurgically allowed maximum temperature about 620C makes it possible to transfer a considerable portion of the heat isother mally at the maximum temperature as the fluid changes phase This makes the cycle approach the Carnot cycle Very high pressures at the maximum tem perature are undesirable because they create materialstrength problems 2 Low triplepoint temperature A triplepoint temperature below the tem perature of the cooling medium prevents any solidification problems 3 A condenser pressure that is not too low Condensers usually operate below atmospheric pressure Pressures well below the atmospheric pressure TOPIC OF SPECIAL INTEREST Binary Vapor Cycles This section can be skipped without a loss in continuity Final PDF to printer 578 VAPOR AND COMBINED POWER CYCLES cen22672ch10543596indd 578 110917 1148 AM create airleakage problems Therefore a substance whose saturation pressure at the ambient temperature is too low is not a good candidate 4 A high enthalpy of vaporization hfg so that heat transfer to the working fluid is nearly isothermal and large mass flow rates are not needed 5 A saturation dome that resembles an inverted U This eliminates the for mation of excessive moisture in the turbine and the need for reheating 6 Good heat transfer characteristics high thermal conductivity 7 Other properties such as being inert inexpensive readily available and nontoxic Not surprisingly no fluid possesses all these characteristics Water comes the closest although it does not fare well with respect to characteristics 1 3 and 5 We can cope with its subatmospheric condenser pressure by care ful sealing and with the inverted Vshaped saturation dome by reheating but there is not much we can do about item 1 Water has a low critical temperature 374C well below the metallurgical limit and very high saturation pressures at high temperatures 165 MPa at 350C Well we cannot change the way water behaves during the hightemperature part of the cycle but we certainly can replace it with a more suitable fluid The result is a power cycle that is actually a combination of two cycles one in the hightemperature region and the other in the lowtemperature region Such a cycle is called a binary vapor cycle In binary vapor cycles the condenser of the hightemperature cycle also called the topping cycle serves as the boiler of the lowtemperature cycle also called the bottoming cycle That is the heat output of the hightemperature cycle is used as the heat input to the low temperature one Some working fluids found suitable for the hightemperature cycle are mer cury sodium potassium and sodiumpotassium mixtures The schematic and Ts diagram for a mercurywater binary vapor cycle are shown in Fig 1028 The critical temperature of mercury is 898C well above the current metallur gical limit and its critical pressure is only about 18 MPa This makes mercury a very suitable working fluid for the topping cycle Mercury is not suitable as the sole working fluid for the entire cycle however since at a condenser temperature of 32C its saturation pressure is 007 Pa A power plant can not operate at this vacuum because of airleakage problems At an acceptable condenser pressure of 7 kPa the saturation temperature of mercury is 237C which is too high as the minimum temperature in the cycle Therefore the use of mercury as a working fluid is limited to the hightemperature cycles Other disadvantages of mercury are its toxicity and high cost The mass flow rate of mercury in binary vapor cycles is several times that of water because of its low enthalpy of vaporization It is evident from the Ts diagram in Fig 1028 that the binary vapor cycle approximates the Carnot cycle more closely than the steam cycle for the same temperature limits Therefore the thermal efficiency of a power plant can be increased by switching to binary cycles The use of mercurywater binary cycles in the United States dates back to 1928 Several such plants have been built since then in New England where fuel costs are typically higher A small 40MW mercurysteam power plant that was in service in New Hampshire FIGURE 1028 Mercurywater binary vapor cycle Superheater Mercury pump 4 7 8 Mercury cycle Heat exchanger Steam cycle Condenser Steam pump Steam turbine 6 5 1 Boiler 3 2 4 5 8 1 T s 6 3 Steam cycle 2 7 Saturation dome steam Saturation dome mercury Mercury cycle Final PDF to printer 579 CHAPTER 10 cen22672ch10543596indd 579 110917 1148 AM in 1950 had a higher thermal efficiency than most of the large modern power plants in use at that time Studies show that thermal efficiencies of 50 percent or higher are possible with binary vapor cycles However binary vapor cycles are not economically attractive because of their high initial cost and the competition offered by the combined gassteam power plants SUMMARY The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice The model cycle for vapor power cycles is the Rankine cycle which is composed of four internally reversible processes constantpressure heat addition in a boiler isentropic expan sion in a turbine constantpressure heat rejection in a con denser and isentropic compression in a pump Steam leaves the condenser as a saturated liquid at the condenser pressure The thermal efficiency of the Rankine cycle can be increased by increasing the average temperature at which heat is trans ferred to the working fluid andor by decreasing the average temperature at which heat is rejected to the cooling medium The average temperature during heat rejection can be decreased by lowering the turbine exit pressure Consequently the con denser pressure of most vapor power plants is well below the atmospheric pressure The average temperature during heat addition can be increased by raising the boiler pressure or by superheating the fluid to high temperatures There is a limit to the degree of superheating however since the fluid tempera ture is not allowed to exceed a metallurgically safe value Superheating has the added advantage of decreasing the moisture content of the steam at the turbine exit Lowering the exhaust pressure or raising the boiler pressure however increases the moisture content To take advantage of the improved efficiencies at higher boiler pressures and lower condenser pressures steam is usually reheated after expanding partially in the highpressure turbine This is done by extract ing the steam after partial expansion in the highpressure turbine sending it back to the boiler where it is reheated at constant pressure and returning it to the lowpressure turbine for complete expansion to the condenser pressure The average temperature during the reheat process and thus the thermal efficiency of the cycle can be increased by increasing the num ber of expansion and reheat stages As the number of stages is increased the expansion and reheat processes approach an isothermal process at maximum temperature Reheating also decreases the moisture content at the turbine exit Another way of increasing the thermal efficiency of the Rankine cycle is regeneration During a regeneration process liquid water feedwater leaving the pump is heated by steam bled off the turbine at some intermediate pressure in devices called feedwater heaters The two streams are mixed in open feedwater heaters and the mixture leaves as a saturated liq uid at the heater pressure In closed feedwater heaters heat is transferred from the steam to the feedwater without mixing The production of more than one useful form of energy such as process heat and electric power from the same energy source is called cogeneration Cogeneration plants produce electric power while meeting the process heat requirements of certain industrial processes This way more of the energy transferred to the fluid in the boiler is utilized for a useful purpose The fraction of energy that is used for either process heat or power generation is called the utilization factor of the cogeneration plant The overall thermal efficiency of a power plant can be increased by using a combined cycle The most common combined cycle is the gassteam combined cycle where a gasturbine cycle operates at the hightemperature range and a steamturbine cycle at the lowtemperature range Steam is heated by the hightemperature exhaust gases leaving the gas turbine Combined cycles have a higher thermal efficiency than the steam or gasturbine cycles operating alone REFERENCES AND SUGGESTED READINGS 1 R L Bannister and G J Silvestri The Evolution of Central Station Steam Turbines Mechanical Engineering February 1989 pp 7078 2 R L Bannister G J Silvestri A Hizume and T Fujikawa High Temperature Supercritical Steam Turbines Mechanical Engineering February 1987 pp 6065 Final PDF to printer cen22672ch10543596indd 580 110917 1148 AM 580 VAPOR AND COMBINED POWER CYCLES PROBLEMS Carnot Vapor Cycle 101C Why is the Carnot cycle not a realistic model for steam power plants 102C Why is excessive moisture in steam undesirable in steam turbines What is the highest moisture content allowed 103 A steadyflow Carnot cycle uses water as the working fluid Water changes from saturated liquid to saturated vapor as heat is transferred to it from a source at 250C Heat rejection takes place at a pressure of 20 kPa Show the cycle on a Ts diagram relative to the saturation lines and determine a the thermal efficiency b the amount of heat rejected and c the net work output 104 Repeat Prob 103 for a heat rejection pressure of 10 kPa 105 Consider a steadyflow Carnot cycle with water as the working fluid The maximum and minimum temperatures in the cycle are 350 and 60C The quality of water is 0891 at the beginning of the heatrejection process and 01 at the end Show the cycle on a Ts diagram relative to the saturation lines and determine a the thermal efficiency b the pres sure at the turbine inlet and c the net work output Answers a 465 percent b 140 MPa c 1623 kJkg 106E Water enters the boiler of a steadyflow Carnot engine as a saturated liquid at 300 psia and leaves with a qual ity of 095 Steam leaves the turbine at a pressure of 20 psia Show the cycle on a Ts diagram relative to the saturation lines and determine a the thermal efficiency b the quality at the end of the isothermal heatrejection process and c the net work output Answers a 216 percent b 0181 c 166 Btulbm The Simple Rankine Cycle 107C What four processes make up the simple ideal Rankine cycle 108C Consider a simple ideal Rankine cycle with fixed tur bine inlet conditions What is the effect of lowering the con denser pressure on 109C Consider a simple ideal Rankine cycle with fixed turbine inlet temperature and condenser pressure What is the effect of increasing the boiler pressure on Pump work input a increases b decreases c remains the same Turbine work output a increases b decreases c remains the same Heat supplied a increases b decreases c remains the same Heat rejected a increases b decreases c remains the same Cycle efficiency a increases b decreases c remains the same Moisture content at turbine exit a increases b decreases c remains the same Pump work input a increases b decreases c remains the same Turbine work output a increases b decreases c remains the same Heat supplied a increases b decreases c remains the same Heat rejected a increases b decreases c remains the same Cycle efficiency a increases b decreases c remains the same Moisture content at turbine exit a increases b decreases c remains the same Pump work input a increases b decreases c remains the same Turbine work output a increases b decreases c remains the same Heat supplied a increases b decreases c remains the same Heat rejected a increases b decreases c remains the same Cycle efficiency a increases b decreases c remains the same Moisture content at turbine exit a increases b decreases c remains the same Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software 3 M M ElWakil Powerplant Technology New York McGrawHill 1984 4 K W Li and A P Priddy Power Plant System Design New York John Wiley Sons 1985 5 H Sorensen Energy Conversion Systems New York John Wiley Sons 1983 6 Steam Its Generation and Use 39th ed New York Babcock and Wilcox Co 1978 7 Turbomachinery 28 no 2 MarchApril 1987 Norwalk CT Business Journals Inc 8 J Weisman and R Eckart Modern Power Plant Engineering Englewood Cliffs NJ PrenticeHall 1985 1010C Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures What is the effect of super heating the steam to a higher temperature on Final PDF to printer cen22672ch10543596indd 581 110917 1148 AM 581 CHAPTER 10 1011C How do actual vapor power cycles differ from ideal ized ones 1012C Compare the pressures at the inlet and the exit of the boiler for a actual and b ideal cycles 1013C The entropy of steam increases in actual steam turbines as a result of irreversibilities In an effort to control entropy increase it is proposed to cool the steam in the tur bine by running cooling water around the turbine casing It is argued that this will reduce the entropy and the enthalpy of the steam at the turbine exit and thus increase the work output How would you evaluate this proposal 1014C Is it possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20C 1015 A simple ideal Rankine cycle with water as the work ing fluid operates between the pressure limits of 3 MPa in the boiler and 30 kPa in the condenser If the quality at the exit of the turbine cannot be less than 85 percent what is the maximum thermal efficiency this cycle can have Answer 297 percent 1016 A simple ideal Rankine cycle with water as the work ing fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet tem perature of 700C The boiler is sized to provide a steam flow of 50 kgs Determine the power produced by the turbine and consumed by the pump 1017 A simple ideal Rankine cycle which uses water as the working fluid operates its condenser at 40C and its boiler at 250C Calculate the work produced by the turbine the heat supplied in the boiler and the thermal efficiency of this cycle when the steam enters the turbine without any superheating with respect to saturation lines and determine a the thermal efficiency of the cycle and b the power output of this plant 1019 Consider a 210MW steam power plant that operates on a simple ideal Rankine cycle Steam enters the turbine at 10 MPa and 500C and is cooled in the condenser at a pressure of 10 kPa Show the cycle on a Ts diagram with respect to sat uration lines and determine a the quality of the steam at the turbine exit b the thermal efficiency of the cycle and c the mass flow rate of the steam Answers a 0793 b 402 percent c 165 kgs 1020 Repeat Prob 1019 assuming an isentropic efficiency of 85 percent for both the turbine and the pump Answers a 0874 b 341 percent c 194 kgs 1021 A simple ideal Rankine cycle with water as the work ing fluid operates between the pressure limits of 15 MPa in the boiler and 100 kPa in the condenser Saturated steam enters the turbine Determine the work produced by the turbine the heat transferred in the boiler and thermal efficiency of the cycle Answer 699 kJkg 2178 kJkg 314 percent 1022 Reconsider Prob 1021 Irreversibilities in the tur bine cause the steam quality at the outlet of the turbine to be 70 percent Determine the isentropic efficiency of the turbine and the thermal efficiency of the cycle Answers 877 percent 274 percent 1023E A steam Rankine cycle operates between the pres sure limits of 1500 psia in the boiler and 2 psia in the con denser The turbine inlet temperature is 800F The turbine isentropic efficiency is 90 percent the pump losses are neg ligible and the cycle is sized to produce 2500 kW of power Calculate the mass flow rate through the boiler the power pro duced by the turbine the rate of heat supply in the boiler and the thermal efficiency 1024E Reconsider Prob 1023E How much error is caused in the thermal efficiency if the power required by the pump were completely neglected 1025 A simple Rankine cycle uses water as the work ing fluid The boiler operates at 6000 kPa and the condenser at 50 kPa At the entrance to the turbine the temperature is 450C The isentropic efficiency of the turbine is 94 percent pressure and pump losses are negligible and the water leav ing the condenser is subcooled by 63C The boiler is sized for a mass flow rate of 20 kgs Determine the rate at which heat is added in the boiler the power required to operate the pumps the net power produced by the cycle and the thermal efficiency Answers 59660 kW 122 kW 18050 kW 303 percent 1026 Reconsider Prob 1025 Using appropriate software determine how much the thermal effi ciency of the cycle would change if there were a 50 kPa pres sure drop across the boiler 1027 The net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared Steam enters FIGURE P1017 Boiler wturbout Condenser Pump 3 qin 4 2 wpumpin 1 qout Turbine 1018 Consider a solarpond power plant that operates on a simple ideal Rankine cycle with refrigerant134a as the work ing fluid The refrigerant enters the turbine as a saturated vapor at 14 MPa and leaves at 07 MPa The mass flow rate of the refrigerant is 3 kgs Show the cycle on a Ts diagram Final PDF to printer cen22672ch10543596indd 582 110917 1148 AM 582 VAPOR AND COMBINED POWER CYCLES the turbine in both cases at 5 MPa as a saturated vapor and the condenser pressure is 50 kPa In the Rankine cycle the condenser exit state is saturated liquid and in the Carnot cycle the boiler inlet state is saturated liquid Draw the Ts diagrams for both cycles 1028 A binary geothermal power plant uses geothermal water at 160C as the heat source The plant operates on the simple Rankine cycle with isobutane as the working fluid Heat is transferred to the cycle by a heat exchanger in which geothermal liquid water enters at 160C at a rate of 5559 kgs and leaves at 90C Isobutane enters the turbine at 325 MPa and 147C and leaves at 795C and 410 kPa Isobutane is condensed in an aircooled condenser and pumped to the heat exchanger pressure Assuming the pump to have an isentro pic efficiency of 90 percent determine a the isentropic effi ciency of the turbine b the net power output of the plant and c the thermal efficiency of the plant The properties of isobutane are h1 27301 kJkg v1 0001842 m3kg h3 76154 kJkg h4 68974 kJkg h4s 67040 kJkg Take the specific heat of geothermal water to be cp 4258 kJkgC temperature is the same for all stages How does the cycle effi ciency vary with the number of reheat stages 1031C Is there an optimal pressure for reheating the steam of a Rankine cycle Explain 1032C How do the following quantities change when a sim ple ideal Rankine cycle is modified with reheating Assume the mass flow rate is maintained the same FIGURE P1028 Heat exchanger Pump Aircooled condenser Geothermal water in Geothermal water out 4 1 2 5 6 3 Isobutane turbine 1029 Consider a coalfired steam power plant that produces 175 MW of electric power The power plant operates on a sim ple ideal Rankine cycle with turbine inlet conditions of 7 MPa and 550C and a condenser pressure of 15 kPa The coal has a heating value energy released when the fuel is burned of 29300 kJkg Assuming that 85 percent of this energy is trans ferred to the steam in the boiler and that the electric generator has an efficiency of 96 percent determine a the overall plant efficiency the ratio of net electric power output to the energy input as fuel and b the required rate of coal supply Answers a 315 percent b 683 th The Reheat Rankine Cycle 1030C Show the ideal Rankine cycle with three stages of reheating on a Ts diagram Assume the turbine inlet Pump work input a increases b decreases c remains the same Turbine work output a increases b decreases c remains the same Heat supplied a increases b decreases c remains the same Heat rejected a increases b decreases c remains the same Moisture content at turbine exit a increases b decreases c remains the same 1033C Consider a simple ideal Rankine cycle and an ideal Rankine cycle with three reheat stages Both cycles operate between the same pressure limits The maximum temperature is 700C in the simple cycle and 450C in the reheat cycle Which cycle do you think will have a higher thermal efficiency 1034 Consider a steam power plant that operates on the ideal reheat Rankine cycle The plant maintains the boiler at 175 MPa the reheater at 2 MPa and the condenser at 50 kPa The temperature is 550C at the entrance of the highpressure turbine and 300C at the entrance of the lowpressure turbine Determine the thermal efficiency of this system 1035 Reconsider Prob 1034 How much does the ther mal efficiency of the cycle change when the temperature at the entrance to the lowpressure turbine is increased to 550C 1036 An ideal reheat Rankine cycle with water as the work ing fluid operates the boiler at 15000 kPa the reheater at 2000 kPa and the condenser at 100 kPa The temperature is 450C at the entrance of the highpressure and lowpressure turbines The mass flow rate through the cycle is 174 kgs Determine the power used by pumps the power produced by the cycle the rate of heat transfer in the reheater and the ther mal efficiency of this system 1037E Steam enters the highpressure turbine of a steam power plant that operates on the ideal reheat Rankine cycle at 800 psia and 900F and leaves as saturated vapor Steam is then reheated to 800F before it expands to a pressure of 1 psia Heat is transferred to the steam in the boiler at a rate of 6 104 Btus Steam is cooled in the condenser by the cooling water from a nearby river which enters the condenser at 45F Show the cycle on a Ts diagram with respect to saturation lines and determine a the pressure at which reheating takes place b the net power output and thermal efficiency and c the minimum mass flow rate of the cooling water required Final PDF to printer cen22672ch10543596indd 583 110917 1148 AM 583 CHAPTER 10 FIGURE P1040 Turbine 6 Boiler Condenser Pump 3 2 5 4 1 1038 An ideal reheat Rankine cycle with water as the work ing fluid operates the inlet of the highpressure turbine at 8000 kPa and 450C the inlet of the lowpressure turbine at 500 kPa and 500C and the condenser at 10 kPa Determine the mass flow rate through the boiler needed for this system to produce a net 5000 kW of power and the thermal efficiency of the cycle 1039 A steam power plant operates on an ideal reheat Ran kine cycle between the pressure limits of 15 MPa and 10 kPa The mass flow rate of steam through the cycle is 12 kgs Steam enters both stages of the turbine at 500C If the moisture con tent of the steam at the exit of the lowpressure turbine is not to exceed 5 percent determine a the pressure at which reheating takes place b the total rate of heat input in the boiler and c the thermal efficiency of the cycle Also show the cycle on a Ts diagram with respect to saturation lines 1040 A steam power plant operates on the reheat Ran kine cycle Steam enters the highpressure tur bine at 125 MPa and 550C at a rate of 77 kgs and leaves at 2 MPa Steam is then reheated at constant pressure to 450C before it expands in the lowpressure turbine The isentropic efficiencies of the turbine and the pump are 85 percent and 90 percent respectively Steam leaves the condenser as a satu rated liquid If the moisture content of the steam at the exit of the turbine is not to exceed 5 percent determine a the con denser pressure b the net power output and c the thermal efficiency This problem is solved using appropriate soft ware Answers a 973 kPa b 102 MW c 369 percent Steam enters the highpressure turbine at 10 MPa and 500C and the lowpressure turbine at 1 MPa and 500C Steam leaves the condenser as a saturated liquid at a pressure of 10 kPa The isentropic efficiency of the turbine is 80 percent and that of the pump is 95 percent Show the cycle on a Ts diagram with respect to saturation lines and determine a the quality or temperature if superheated of the steam at the turbine exit b the thermal efficiency of the cycle and c the mass flow rate of the steam Answers a 881C b 341 percent c 627 kgs 1042 Repeat Prob 1041 assuming both the pump and the turbine are isentropic Answers a 0949 b 413 percent c 500 kgs Regenerative Rankine Cycle 1043C Devise an ideal regenerative Rankine cycle that has the same thermal efficiency as the Carnot cycle Show the cycle on a Ts diagram 1044C During a regeneration process some steam is extracted from the turbine and is used to heat the liquid water leaving the pump This does not seem like a smart thing to do since the extracted steam could produce some more work in the turbine How do you justify this action 1045C Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater The two cycles are very much alike except the feedwater in the regenerative cycle is heated by extracting some steam just before it enters the turbine How would you compare the effi ciencies of these two cycles 1046C How do open feedwater heaters differ from closed feedwater heaters 1047C How do the following quantities change when the simple ideal Rankine cycle is modified with regeneration Assume the mass flow rate through the boiler is the same Turbine work output a increases b decreases c remains the same Heat supplied a increases b decreases c remains the same Heat rejected a increases b decreases c remains the same Moisture content at turbine exit a increases b decreases c remains the same 1041 Consider a steam power plant that operates on a reheat Rankine cycle and has a net power output of 80 MW 1048 Cold feedwater enters a 200kPa open feedwater heater of a regenerative Rankine cycle at 70C with a flow rate of 10 kgs Bleed steam is available from the turbine at 200 kPa and 160C At what rate must bleed steam be supplied to the open feedwater heater so the feedwater leaves this unit as a saturated liquid Final PDF to printer cen22672ch10543596indd 584 110917 1148 AM 584 VAPOR AND COMBINED POWER CYCLES 1049E In a regenerative Rankine cycle the closed feed water heater with a pump as shown in the figure is arranged so that the water at state 5 is mixed with the water at state 2 to form a feedwater which is a saturated liquid at 200 psia Feedwater enters this heater at 350F and 200 psia with a flow rate of 2 lbms Bleed steam is taken from the turbine at 160 psia and 400F and enters the pump as a saturated liquid at 160 psia Determine the mass flow rate of bleed steam required to operate this unit Answer 00782 Ibms enters the turbine at 10 MPa and 600C and exhausts to the condenser at 10 kPa Steam is extracted from the turbine at 12 MPa for the closed feedwater heater and at 06 MPa for the open one The feedwater is heated to the condensation tem perature of the extracted steam in the closed feedwater heater The extracted steam leaves the closed feedwater heater as a saturated liquid which is subsequently throttled to the open feedwater heater Show the cycle on a Ts diagram with respect to saturation lines and determine a the mass flow rate of steam through the boiler for a net power output of 400 MW and b the thermal efficiency of the cycle FIGURE P1049E 2 Bleed steam from turbine Feedwater 1 m1 m3 3 5 4 1050 A steam power plant operates on an ideal regenera tive Rankine cycle Steam enters the turbine at 6 MPa and 450C and is condensed in the condenser at 20 kPa Steam is extracted from the turbine at 04 MPa to heat the feedwater in an open feedwater heater Water leaves the feedwater heater as a saturated liquid Show the cycle on a Ts diagram and deter mine a the net work output per kilogram of steam flowing through the boiler and b the thermal efficiency of the cycle Answers a 1017 kJkg b 378 percent 1051 Repeat Prob 1050 by replacing the open feedwater heater with a closed feedwater heater Assume that the feed water leaves the heater at the condensation temperature of the extracted steam and that the extracted steam leaves the heater as a saturated liquid and is pumped to the line carrying the feedwater 1052 A steam power plant operates on an ideal regenera tive Rankine cycle with two open feedwater heaters Steam enters the turbine at 8 MPa and 550C and exhausts to the con denser at 15 kPa Steam is extracted from the turbine at 06 and 02 MPa Water leaves both feedwater heaters as a satu rated liquid The mass flow rate of steam through the boiler is 24 kgs Show the cycle on a Ts diagram and determine a the net power output of the power plant and b the thermal efficiency of the cycle Answers a 288 MW b 422 percent 1053 Consider an ideal steam regenerative Rankine cycle with two feedwater heaters one closed and one open Steam FIGURE P1053 1 y z y z Closed FWH 7 6 Open FWH Turbine 11 1 5 3 10 9 4 8 Condenser Pump I Pump II 2 Boiler 1054 Reconsider Prob 1053 Using appropriate software investigate the effects of turbine and pump efficiencies as they are varied from 70 percent to 100 percent on the mass flow rate and thermal efficiency Plot the mass flow rate and the thermal efficiency as a function of turbine efficiency for pump efficiencies of 70 85 and 100 percent and discuss the results Also plot the Ts diagram for turbine and pump efficiencies of 85 percent 1055 Consider a steam power plant that operates on the ideal regenerative Rankine cycle with a closed feedwater heater as shown in the figure The plant maintains the turbine inlet at 3000 kPa and 350C and operates the condenser at 20 kPa Steam is extracted at 1000 kPa to serve the closed feedwater heater which discharges into the condenser after being throt tled to condenser pressure Calculate the work produced by the turbine the work consumed by the pump and the heat supply in the boiler for this cycle per unit of boiler flow rate Answers 741 kJkg 30 kJkg 2353 kJkg Final PDF to printer cen22672ch10543596indd 585 110917 1148 AM 585 CHAPTER 10 FIGURE P1055 Turbine 6 5 Condenser Closed FWH Pump 1 2 4 7 8 3 Boiler 1056 Reconsider Prob 1055 Using appropriate software determine the optimum bleed pressure for the closed feedwater heater that maximizes the thermal efficiency of the cycle Answer 220 kPa 1057 Reconsider Prob 1055 Determine the thermal effi ciency of the regenerative Rankine cycle when the isentropic efficiency of the turbine is 90 percent before and after the steam extraction point 1058 Reconsider Prob 1055 Determine the thermal effi ciency of the regenerative Rankine cycle when the isentropic efficiency of the turbine before and after the steam extraction point is 90 percent and the condenser condensate is subcooled by 10C 1059 Reconsider Prob 1055 Using appropriate software determine how much additional heat must be supplied to the boiler when the turbine isentropic effi ciency before and after the extraction point is 90 percent and there is a 10 kPa pressure drop across the boiler 1060 A steam power plant operates on an ideal reheat regenerative Rankine cycle and has a net power output of 80 MW Steam enters the highpressure turbine at 10 MPa and 550C and leaves at 08 MPa Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater The rest of the steam is reheated to 500C and is expanded in the lowpressure turbine to the condenser pressure of 10 kPa Show the cycle on a Ts diagram with respect to saturation lines and determine a the mass flow rate of steam through the boiler and b the thermal efficiency of the cycle Answers a 545 kgs b 444 percent 1061 Repeat Prob 1060 but replace the open feedwater heater with a closed feedwater heater Assume that the feedwater leaves the heater at the condensation temperature of the extracted steam and that the extracted steam leaves the heater as a saturated liquid and is pumped to the line carrying the feedwater FIGURE P1061 1 y y Mixing chamber LowP turbine Boiler HighP turbine Condenser 5 7 6 9 8 3 Pump I Pump II 1 Closed FWH 2 4 10 1062E A steam power plant operates on an ideal reheat regenerative Rankine cycle with one reheater and two open feed water heaters Steam enters the highpressure turbine at 1500 psia and 1100F and leaves the lowpressure turbine at 1 psia Steam is extracted from the turbine at 250 and 40 psia and it is reheated to 1000F at a pressure of 140 psia Water leaves both feedwater heaters as a saturated liquid Heat is transferred to the steam in the boiler at a rate of 4 105 Btus Show the cycle on a Ts dia gram with respect to saturation lines and determine a the mass flow rate of steam through the boiler b the net power output of the plant and c the thermal efficiency of the cycle 1 y y z z LowP turbine Boiler Reheater HighP turbine Condenser 7 9 8 12 1yz 11 10 3 Pump I Pump II Pump III 1 2 6 4 5 Open FWH I Open FWH II FIGURE P1062E Final PDF to printer cen22672ch10543596indd 586 110917 1148 AM 586 VAPOR AND COMBINED POWER CYCLES SecondLaw Analysis of Vapor Power Cycles 1063 A simple ideal Rankine cycle with water as the work ing fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet tempera ture of 700C Calculate the exergy destruction in each of the components of the cycle when heat is being rejected to the atmospheric air at 15C and heat is supplied from an energy res ervoir at 750C Answers 928 kJkg boiler 307 kJkg condenser 1064 Consider a steam power plant that operates on a sim ple ideal Rankine cycle Steam enters the turbine at 10 MPa and 500C and is cooled in the condenser at a pressure of 10 kPa Determine the exergy destruction associated with each of the processes of the cycle assuming a source temperature of 1500 K and a sink temperature of 290 K 1065 An ideal reheat Rankine cycle with water as the work ing fluid operates the inlet of the highpressure turbine at 8000 kPa and 450C the inlet of the lowpressure turbine at 500 kPa and 500C and the condenser at 10 kPa Which com ponent of the cycle offers the greatest opportunity to regain lost power potential The sink is at 10C and the source is at 600C 1066 Consider a steam power plant that operates on a reheat Rankine cycle Steam enters the highpressure turbine at 10 MPa and 500C and the lowpressure turbine at 1 MPa and 500C Steam leaves the condenser as a saturated liquid at a pressure of 10 kPa The isentropic efficiency of the turbine is 80 percent and that of the pump is 95 percent Determine the exergy destruction associated with the heat addition pro cess and the expansion process Assume a source temperature of 1600 K and a sink temperature of 285 K Also determine the exergy of the steam at the boiler exit Take P0 100 kPa Answers 1289 kJkg 2479 kJkg 1495 kJkg 1067 A steam power plant operates on an ideal regenera tive Rankine cycle Steam enters the turbine at 6 MPa and 450C and is condensed in the condenser at 20 kPa Steam is extracted from the turbine at 04 MPa to heat the feedwater in an open feedwater heater Water leaves the feedwater heater as a saturated liquid Determine the exergy destruction associated with the cycle Assume a source temperature of 1350 K and a sink temperature of 290 K Answer 1097 kJkg 1068 A steam power plant operates on an ideal reheat regenerative Rankine cycle Steam enters the highpressure tur bine at 10 MPa and 550C and leaves at 08 MPa Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater The rest of the steam is reheated to 500C and is expanded in the lowpressure turbine to the condenser pressure of 10 kPa Determine the exergy destruction associ ated with the reheating and regeneration processes Assume a source temperature of 1800 K and a sink temperature of 290 K 1069 The schematic of a singleflash geothermal power plant with state numbers is given in Fig P1069 Geothermal resource exists as saturated liquid at 230C The geothermal liq uid is withdrawn from the production well at a rate of 230 kgs and is flashed to a pressure of 500 kPa by an essentially isenthalpic flashing process where the resulting vapor is sepa rated from the liquid in a separator and is directed to the tur bine The steam leaves the turbine at 10 kPa with a moisture content of 5 percent and enters the condenser where it is con densed it is routed to a reinjection well along with the liquid coming off the separator Determine a the power output of the turbine and the thermal efficiency of the plant b the exergy of the geothermal liquid at the exit of the flash chamber and the exergy destructions and the secondlaw efficiencies for c the turbine and d the entire plant Answers a 108 MW 53 percent b 173 MW c 109 MW 500 percent d 390 MW 218 percent FIGURE P1069 Flash chamber Production well Reinjection well Separator Steam turbine 3 4 5 6 2 1 Condenser Cogeneration 1070C What is the difference between cogeneration and regeneration 1071C How is the utilization factor 𝜖u for cogeneration plants defined Could 𝜖u be unity for a cogeneration plant that does not produce any power 1072C Consider a cogeneration plant for which the utiliza tion factor is 1 Is the irreversibility associated with this cycle necessarily zero Explain 1073C Consider a cogeneration plant for which the utiliza tion factor is 05 Can the exergy destruction associated with this plant be zero If yes under what conditions 1074E Steam is generated in the boiler of a cogeneration plant at 600 psia and 650F at a rate of 32 lbms The plant is to produce power while meeting the process steam requirements Final PDF to printer cen22672ch10543596indd 587 110917 1148 AM 587 CHAPTER 10 for a certain industrial application Onethird of the steam leaving the boiler is throttled to a pressure of 120 psia and is routed to the process heater The rest of the steam is expanded in an isentropic turbine to a pressure of 120 psia and is also routed to the process heater Steam leaves the process heater at 240F Neglecting the pump work determine a the net power produced b the rate of process heat supply and c the utili zation factor of this plant 1075E A large foodprocessing plant requires 15 lbms of saturated or slightly superheated steam at 140 psia which is extracted from the turbine of a cogeneration plant The boiler generates steam at 800 psia and 1000F at a rate of 10 lbms and the condenser pressure is 2 psia Steam leaves the process heater as a saturated liquid It is then mixed with the feedwater at the same pressure and this mixture is pumped to the boiler pressure Assuming both the pumps and the turbine have isen tropic efficiencies of 91 percent determine a the rate of heat transfer to the boiler and b the power output of the cogenera tion plant Answers a 13810 Btus b 4698 kW 1076 An ideal cogeneration steam plant is to generate power and 8600 kJs of process heat Steam enters the turbine from the boiler at 7 MPa and 500C Onefourth of the steam is extracted from the turbine at 600kPa pressure for process heating The remainder of the steam continues to expand and exhausts to the condenser at 10 kPa The steam extracted for the process heater is condensed in the heater and mixed with the feedwater at 600 kPa The mixture is pumped to the boiler pressure of 7 MPa Show the cycle on a Ts diagram with respect to saturation lines and determine a the mass flow rate of steam that must be supplied by the boiler b the net power produced by the plant and c the utilization factor 1077 Steam is generated in the boiler of a cogeneration plant at 10 MPa and 450C at a steady rate of 5 kgs In normal operation steam expands in a turbine to a pressure of 05 MPa and is then routed to the process heater where it supplies the process heat Steam leaves the process heater as a saturated liquid and is pumped to the boiler pressure In this mode no steam passes through the condenser which operates at 20 kPa a Determine the power produced and the rate at which pro cess heat is supplied in this mode b Determine the power produced and the rate of process heat supplied if only 60 percent of the steam is routed to the pro cess heater and the remainder is expanded to the condenser pressure 1078 Consider a cogeneration power plant modified with regeneration Steam enters the turbine at 9 MPa and 400C and expands to a pressure of 16 MPa At this pressure 35 percent of the steam is extracted from the turbine and the remainder expands to 10 kPa Part of the extracted steam is used to heat the feedwater in an open feedwater heater The rest of the extracted steam is used for process heating and leaves the pro cess heater as a saturated liquid at 16 MPa It is subsequently mixed with the feedwater leaving the feedwater heater and the mixture is pumped to the boiler pressure Assuming the turbines and the pumps to be isentropic show the cycle on a Ts diagram with respect to saturation lines and determine the mass flow rate of steam through the boiler for a net power out put of 25 MW Answer 291 kgs FIGURE P1076 Qprocess Process heater Turbine Condenser Pum I p II p um P 5 1 8 3 7 6 4 2 Boiler FIGURE P1078 6 Pump II 5 4 9 3 Pump I 2 8 7 Condenser FWH 1 Boiler Process heater Turbine 1079 Reconsider Prob 1078 Using appropriate software investigate the effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate Plot the mass flow rate through the boiler as a func tion of the extraction pressure and discuss the results Final PDF to printer cen22672ch10543596indd 588 110917 1148 AM 588 VAPOR AND COMBINED POWER CYCLES Combined GasVapor Power Cycles 1080C In combined gassteam cycles what is the energy source for the steam 1081C Why is the combined gassteam cycle more effi cient than either of the cycles operated alone 1082 The gasturbine portion of a combined gassteam power plant has a pressure ratio of 16 Air enters the compres sor at 300 K at a rate of 14 kgs and is heated to 1500 K in the combustion chamber The combustion gases leaving the gas turbine are used to heat the steam to 400C at 10 MPa in a heat exchanger The combustion gases leave the heat exchanger at 420 K The steam leaving the turbine is condensed at 15 kPa Assuming all the compression and expansion processes to be isentropic determine a the mass flow rate of the steam b the net power output and c the thermal efficiency of the combined cycle For air assume constant specific heats at room temperature Answers a 1275 kgs b 7819 kW c 664 percent 1083 A combined gassteam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle Atmospheric air enters the gas turbine at 101 kPa and 20C and the maximum gas cycle temperature is 1100C The compressor pressure ratio is 8 the compres sor isentropic efficiency is 85 percent and the gas turbine isentropic efficiency is 90 percent The gas stream leaves the heat exchanger at the saturation temperature of the steam flow ing through the heat exchanger Steam flows through the heat exchanger with a pressure of 6000 kPa and leaves at 320C The steamcycle condenser operates at 20 kPa and the isen tropic efficiency of the steam turbine is 90 percent Determine the mass flow rate of air through the air compressor required for this system to produce 100 MW of power Use constant specific heats for air at room temperature Answer 279 kgs 1084 Reconsider Prob 1083 An ideal regenerator is added to the gas cycle portion of the combined cycle How much does this change the efficiency of this combined cycle 1085 Reconsider Prob 1083 Determine which compo nents of the combined cycle are the most wasteful of work potential 1086 Consider a combined gassteam power plant that has a net power output of 280 MW The pressure ratio of the gas turbine cycle is 11 Air enters the compressor at 300 K and the turbine at 1100 K The combustion gases leaving the gas turbine are used to heat the steam at 5 MPa to 350C in a heat exchanger The combustion gases leave the heat exchanger at 420 K An open feedwater heater incorporated with the steam cycle operates at a pressure of 08 MPa The condenser pres sure is 10 kPa Assuming isentropic efficiencies of 100 percent for the pump 82 percent for the compressor and 86 percent for the gas and steam turbines determine a the mass flow rate ratio of air to steam b the required rate of heat input in the combustion chamber and c the thermal efficiency of the combined cycle 1087 Reconsider Prob 1086 Using appropriate software study the effects of the gas cycle pres sure ratio as it is varied from 10 to 20 on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency Plot your results as functions of gas cycle pressure ratio and discuss the results 1088 Consider a combined gassteam power cycle The topping cycle is a simple Brayton cycle that has a pressure ratio of 7 Air enters the compressor at 15C at a rate of 40 kgs and the gas turbine at 950C The bottoming cycle is a reheat Rankine cycle between the pressure limits of 6 MPa and 10 kPa Steam is heated in a heat exchanger at a rate of 46 kgs by the exhaust gases leaving the gas turbine and the exhaust gases leave the heat exchanger at 200C Steam leaves the highpressure turbine at 10 MPa and is reheated to 400C in the heat exchanger before it expands in the lowpressure tur bine Assuming 80 percent isentropic efficiency for all pumps and turbines determine a the moisture content at the exit of the lowpressure turbine b the steam temperature at the inlet of the highpressure turbine c the net power output and the thermal efficiency of the combined plant This problem is solved using appropriate software FIGURE P1088 Heat exchanger 9 10 6 2 11 3 4 5 7 8 Condenser Pump Compressor Steam turbine Combustion chamber 1 Gas turbine Final PDF to printer cen22672ch10543596indd 589 110917 1148 AM 589 CHAPTER 10 Special Topic Binary Vapor Cycles 1089C What is a binary power cycle What is its purpose 1090C What is the difference between the binary vapor power cycle and the combined gassteam power cycle 1091C Why is mercury a suitable working fluid for the top ping portion of a binary vapor cycle but not for the bottoming cycle 1092C Why is steam not an ideal working fluid for vapor power cycles 1093 By writing an energy balance on the heat exchanger of a binary vapor power cycle obtain a relation for the ratio of mass flow rates of two fluids in terms of their enthalpies Review Problems 1094 Feedwater at 4000 kPa is heated at a rate of 6 kgs from 200C to 245C in a closed feedwater heater of a regen erative Rankine cycle Bleed steam enters this unit at 3000 kPa with a quality of 90 percent and leaves as a saturated liquid Calculate the rate at which bleed steam is required 1095 Steam enters the turbine of a steam power plant that operates on a simple ideal Rankine cycle at a pressure of 6 MPa and it leaves as a saturated vapor at 75 kPa Heat is transferred to the steam in the boiler at a rate of 40000 kJs Steam is cooled in the condenser by the cooling water from a nearby river which enters the condenser at 15C Show the cycle on a Ts diagram with respect to saturation lines and determine a the turbine inlet temperature b the net power output and thermal efficiency and c the minimum mass flow rate of the cooling water required 1096 Consider a steam power plant operating on the ideal Rankine cycle with reheat between the pressure limits of 30 MPa and 10 kPa with a maximum cycle temperature of 700C and a moisture content of 5 percent at the turbine exit For a reheat temperature of 700C determine the reheat pressures of the cycle for the cases of a single and b double reheat 1097 A steam power plant operates on an ideal Rankine cycle with two stages of reheat and has a net power output of 75 MW Steam enters all three stages of the turbine at 550C The maximum pressure in the cycle is 10 MPa and the mini mum pressure is 30 kPa Steam is reheated at 4 MPa the first time and at 2 MPa the second time Show the cycle on a Ts diagram with respect to saturation lines and determine a the thermal efficiency of the cycle and b the mass flow rate of the steam Answers a 405 percent b 485 kgs 1098 Consider a steam power plant that operates on a regenerative Rankine cycle and has a net power output of 150 MW Steam enters the turbine at 10 MPa and 500C and the condenser at 10 kPa The isentropic efficiency of the tur bine is 80 percent and that of the pumps is 95 percent Steam is extracted from the turbine at 05 MPa to heat the feedwater in an open feedwater heater Water leaves the feedwater heater as a saturated liquid Show the cycle on a Ts diagram and determine a the mass flow rate of steam through the boiler and b the thermal efficiency of the cycle Also determine the exergy destruction associated with the regeneration process Assume a source temperature of 1300 K and a sink tempera ture of 303 K FIGURE P1098 5 Pump II 4 Pump I 2 3 6 y 1 y 7 Condenser Open FWH 1 Turbine Boiler 1099 Repeat Prob 1098 assuming both the pump and the turbine are isentropic 10100 Consider an ideal reheatregenerative Rankine cycle with one open feedwater heater The boiler pressure is 10 MPa the condenser pressure is 15 kPa the reheater pressure is 1 MPa and the feedwater pressure is 06 MPa Steam enters both the high and lowpressure turbines at 500C Show the cycle on a Ts diagram with respect to saturation lines and determine a the fraction of steam extracted for regeneration and b the thermal efficiency of the cycle Answers a 0144 b 421 percent 10101 Repeat Prob 10100 assuming an isentropic effi ciency of 84 percent for the turbines and 89 percent for the pumps 10102 A textile plant requires 4 kgs of saturated steam at 2 MPa which is extracted from the turbine of a cogeneration plant Steam enters the turbine at 8 MPa and 500C at a rate of 11 kgs and leaves at 20 kPa The extracted steam leaves the process heater as a saturated liquid and mixes with the feedwa ter at constant pressure The mixture is pumped to the boiler pressure Assuming an isentropic efficiency of 88 percent for both the turbine and the pumps determine a the rate of process heat supply b the net power output and c the uti lization factor of the plant Answers a 856 MW b 860 MW c 538 percent Final PDF to printer cen22672ch10543596indd 590 110917 1148 AM 590 VAPOR AND COMBINED POWER CYCLES 10103 Consider a cogeneration power plant that is modi fied with reheat and that produces 3 MW of power and sup plies 7 MW of process heat Steam enters the highpressure turbine at 8 MPa and 500C and expands to a pressure of 1 MPa At this pressure part of the steam is extracted from the turbine and routed to the process heater while the remainder FIGURE P10103 LowP turbine 3 MW 7 MW Boiler HighP turbine Condenser 6 7 Mixing chamber 2 9 8 Pump I Pump II 5 1 4 Process heater 3 FIGURE P10102 Process heater Turbine Condenser Pump II Pump I 5 1 8 3 7 6 4 2 Boiler is reheated to 500C and expanded in the lowpressure turbine to the condenser pressure of 15 kPa The condensate from the condenser is pumped to 1 MPa and is mixed with the extracted steam which leaves the process heater as a compressed liquid at 120C The mixture is then pumped to the boiler pressure Assuming the turbine to be isentropic show the cycle on a Ts diagram with respect to saturation lines and disregarding pump work determine a the rate of heat input in the boiler and b the fraction of steam extracted for process heating 10104 Steam is to be supplied from a boiler to a high pressure turbine whose isentropic efficiency is 85 percent at conditions to be determined The steam is to leave the high pressure turbine as a saturated vapor at 14 MPa and the tur bine is to produce 55 MW of power Steam at the turbine exit is extracted at a rate of 1000 kgmin and routed to a process heater while the rest of the steam is supplied to a lowpressure turbine whose isentropic efficiency is 80 percent The low pressure turbine allows the steam to expand to 10 kPa pressure and produces 15 MW of power Determine the temperature pressure and the flow rate of steam at the inlet of the high pressure turbine 10105E Atmospheric air enters the air compressor of a sim ple combined gassteam power system at 147 psia and 80F The air compressors compression ratio is 10 the gas cycles maximum temperature is 2100F and the air compressor and turbine have an isentropic efficiency of 90 percent The gas leaves the heat exchanger 50F hotter than the saturation tem perature of the steam in the heat exchanger The steam pressure in the heat exchanger is 800 psia and the steam leaves the heat exchanger at 600F The steamcondenser pressure is 5 psia and the isentropic efficiency of the steam turbine is 95 percent Determine the overall thermal efficiency of this combined cycle For air use constant specific heats at room temperature Answer 464 percent 10106E Reconsider Prob 10105E It has been suggested that the steam passing through the condenser of the com bined cycle be routed to buildings during the winter to heat them When this is done the pressure in the heating system where the steam is now condensed will have to be increased to 10 psia How does this change the overall thermal efficiency of the combined cycle 10107E Reconsider Prob 10106E During winter the sys tem must supply 2 106 Btuh of heat to the buildings What is the mass flow rate of air through the air compressor and the systems total electrical power production in winter Answers 27340 lbmh 1286 kW 10108 The gasturbine cycle of a combined gassteam power plant has a pressure ratio of 12 Air enters the compres sor at 310 K and the turbine at 1400 K The combustion gases leaving the gas turbine are used to heat the steam at 125 MPa to 500C in a heat exchanger The combustion gases leave the heat exchanger at 247C Steam expands in a highpressure tur bine to a pressure of 25 MPa and is reheated in the combustion Final PDF to printer cen22672ch10543596indd 591 110917 1148 AM 591 CHAPTER 10 chamber to 550C before it expands in a lowpressure turbine to 10 kPa The mass flow rate of steam is 12 kgs Assuming all the compression and expansion processes to be isentropic determine a the mass flow rate of air in the gasturbine cycle b the rate of total heat input and c the thermal efficiency of the combined cycle Answers a 154 kgs b 144 105 kJs c 591 percent 10109 Repeat Prob 10108 assuming isentropic efficien cies of 100 percent for the pump 85 percent for the compres sor and 90 percent for the gas and steam turbines 10110 A steam power plant operates on an ideal reheat regenerative Rankine cycle with one reheater and two feed water heaters one open and one closed Steam enters the highpressure turbine at 15 MPa and 600C and the low pressure turbine at 1 MPa and 500C The condenser pressure is 5 kPa Steam is extracted from the turbine at 06 MPa for the closed feedwater heater and at 02 MPa for the open feedwater heater In the closed feedwater heater the feedwater is heated to the condensation temperature of the extracted steam The extracted steam leaves the closed feedwater heater as a satu rated liquid which is subsequently throttled to the open feed water heater Show the cycle on a Ts diagram with respect to saturation lines Determine a the fraction of steam extracted from the turbine for the open feedwater heater b the thermal efficiency of the cycle and c the net power output for a mass flow rate of 42 kgs through the boiler FIGURE P10110 Closed FWH z y LowP turbine Boiler HighP turbine Condenser 8 11 6 7 10 9 12 13 P I P II 1 Open FWH 2 5 3 4 1 y z 10111 A Rankine steam cycle modified for reheat a closed feedwater heater and an open feedwater heater is shown below The highpressure turbine receives 100 kgs of steam from the steam boiler The feedwater heater exit states for the boiler feedwater and the condensed steam are the nor mally assumed ideal states The following data tables give the saturation data for the pressures and data for h and s at selected states a Sketch the Ts diagram for the ideal cycle b Determine the net power output of the cycle in MW c If cooling water is available at 25C what is the minimum flow rate of the cooling water required for the ideal cycle in kgs Take cpwater 418 kJkgK Process states and selected data State P kPa T C h kJkg s kJkgK 1 20 2 1400 3 1400 4 1400 5 5000 6 5000 700 3894 7504 7 1400 3400 7504 8 1200 3349 7504 9 1200 600 3692 7938 10 245 3154 7938 11 20 2620 7938 FIGURE P10111 z y LowP turbine HighP turbine 5 Condenser 12 Steam trap Water in Water out 13 11 10 3 2 7 8 9 1 4 6 Boiler 10112 A Rankine steam cycle modified for reheat and three closed feedwater heaters is shown below The highpressure turbine receives 100 kgs of steam from the steam boiler The feedwater heater exit states for the boiler feedwater and the con densed steam are the normally assumed ideal states The fol lowing data tables give the saturation data for the pressures and data for h and s at selected states a Sketch the Ts diagram for the ideal cycle b Determine the net power output of the cycle in MW c If the cooling water is limited to a 10C temperature Final PDF to printer cen22672ch10543596indd 592 110917 1148 AM 592 VAPOR AND COMBINED POWER CYCLES Process states and selected data State P kPa T C h kJkg s kJkgK 1 10 2 5000 3 5000 4 5000 5 5000 6 5000 7 5000 700 3900 75136 8 2500 3615 75136 9 2500 600 3687 75979 10 925 3330 75979 11 300 3011 75979 12 75 2716 75979 13 10 2408 75979 rise what is the flow rate of the cooling water required for the ideal cycle in kgs Take cpwater 418 kJkgK cycle and plot it against the boiler pressure and discuss the results 10114 Using appropriate software investigate the effect of the condenser pressure on the perfor mance of a simple ideal Rankine cycle Turbine inlet condi tions of steam are maintained constant at 10 MPa and 550C while the condenser pressure is varied from 5 to 100 kPa Determine the thermal efficiency of the cycle and plot it against the condenser pressure and discuss the results 10115 Using appropriate software investigate the effect of superheating the steam on the perfor mance of a simple ideal Rankine cycle Steam enters the tur bine at 3 MPa and exits at 10 kPa The turbine inlet temperature is varied from 250 to 1100C Determine the thermal effi ciency of the cycle and plot it against the turbine inlet tempera ture and discuss the results 10116 Using appropriate software investigate the effect of reheat pressure on the performance of an ideal Rankine cycle The maximum and minimum pressures in the cycle are 15 MPa and 10 kPa respectively and steam enters both stages of the turbine at 500C The reheat pressure is varied from 125 to 05 MPa Determine the thermal effi ciency of the cycle and plot it against the reheat pressure and discuss the results 10117 Show that the thermal efficiency of a combined gas steam power plant ηcc can be expressed as η cc η g η s η g η s where ηg Wg Qin and ηs Ws Qgout are the thermal efficien cies of the gas and steam cycles respectively Using this rela tion determine the thermal efficiency of a combined power cycle that consists of a topping gasturbine cycle with an effi ciency of 40 percent and a bottoming steamturbine cycle with an efficiency of 30 percent 10118 It can be shown that the thermal efficiency of a com bined gassteam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steamturbine cycles as η cc η g η s η g η s Prove that the value of ηcc is greater than either ηg or ηs That is the combined cycle is more efficient than either the gas turbine or steamturbine cycle alone 10119 A solar collector system delivers heat to a power plant It is well known that the thermal collection efficiency ηsc of a solar collector diminishes with increasing solar collec tion output temperature TH or ηsc A BTH where A and B are known constants The thermal efficiency of the power plant ηth is a fixed fraction of the Carnot thermal efficiency such that ηth F1 TLTH where F is a known constant assumed here independent of temperatures and TL is the condenser tempera ture also constant for this problem Here the solar collection Saturation data P kPa T C vf m3kg hf kJkg sg kJkgK 10 458 0001010 1918 8149 75 918 0001037 3844 7456 300 1335 0001073 5614 6992 925 1765 0001123 7477 6612 2500 2240 0001197 9619 6256 5000 2639 0001286 11545 5974 FIGURE P10112 Steam trap 15 16 14 4 Mixing chamber y z w LowP turbine HighP turbine Condenser Steam trap Water in Water out 13 11 10 12 5 3 2 7 8 9 1 17 18 19 6 Boiler 10113 Using appropriate software investigate the effect of the boiler pressure on the perfor mance of a simple ideal Rankine cycle Steam enters the tur bine at 500C and exits at 10 kPa The boiler pressure is varied from 05 to 20 MPa Determine the thermal efficiency of the Final PDF to printer cen22672ch10543596indd 593 110917 1148 AM 593 CHAPTER 10 temperature TH is also taken to be the source temperature for the power plant a At what temperature TH should the solar collector be oper ated to obtain the maximum overall system efficiency b Develop an expression for the maximum overall system efficiency 10120 Starting with Eq 1020 show that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as xdest qinηthCarnot ηth where ηth is the effi ciency of the Rankine cycle and ηthCarnot is the efficiency of the Carnot cycle operating between the same temperature limits Fundamentals of Engineering FE Exam Problems 10121 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the steam is superheated to a higher temperature a the turbine work output will decrease b the amount of heat rejected will decrease c the cycle efficiency will decrease d the moisture content at turbine exit will decrease e the amount of heat input will decrease 10122 Consider a simple ideal Rankine cycle If the con denser pressure is lowered while keeping the turbine inlet state the same a the turbine work output will decrease b the amount of heat rejected will decrease c the cycle efficiency will decrease d the moisture content at turbine exit will decrease e the pump work input will decrease 10123 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the cycle is modified with reheating a the turbine work output will decrease b the amount of heat rejected will decrease c the pump work input will decrease d the moisture content at turbine exit will decrease e the amount of heat input will decrease 10124 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures If the cycle is modified with regenera tion that involves one open feedwater heater select the correct statement per unit mass of steam flowing through the boiler a the turbine work output will decrease b the amount of heat rejected will increase c the cycle thermal efficiency will decrease d the quality of steam at turbine exit will decrease e the amount of heat input will increase 10125 Consider a steadyflow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 1 MPa and 10 kPa Water changes from saturated liquid to saturated vapor during the heat addition pro cess The net work output of this cycle is a 596 kJkg b 666 kJkg c 708 kJkg d 822 kJkg e 1500 kJkg 10126 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa with a turbine inlet tem perature of 600C The mass fraction of steam that condenses at the turbine exit is a 6 percent b 9 percent c 12 percent d 15 percent e 18 percent 10127 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kPa and 5 MPa with a turbine inlet temperature of 600C The rate of heat transfer in the boiler is 450 kJs Disregarding the pump work the power output of this plant is a 118 kW b 140 kW c 177 kW d 286 kW e 450 kW 10128 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa with a turbine inlet tem perature of 600C Disregarding the pump work the cycle effi ciency is a 24 percent b 37 percent c 52 percent d 63 percent e 71 percent 10129 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa with reheat occurring at 4 MPa The temperature of steam at the inlets of both turbines is 500C and the enthalpy of steam is 3185 kJkg at the exit of the highpressure turbine and 2247 kJkg at the exit of the lowpressure turbine Disregarding the pump work the cycle efficiency is a 29 percent b 32 percent c 36 percent d 41 percent e 49 percent 10130 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 2 MPa with steam extracted from the turbine If the enthalpy of the feedwater is 252 kJkg and the enthalpy of the extracted steam is 2810 kJkg the mass fraction of steam extracted from the turbine is a 10 percent b 14 percent c 26 percent d 36 percent e 50 percent 10131 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater The enthalpy of the steam is 3374 kJkg at the turbine inlet 2797 kJkg at the location of bleeding and 2346 kJkg at the turbine exit The net power output of the plant is 120 MW and the fraction of steam bled off the turbine for regeneration is 0172 If the pump work is negligible the mass flow rate of steam at the turbine inlet is a 117 kgs b 126 kgs c 219 kgs d 268 kgs e 679 kgs 10132 Consider a combined gassteam power plant Water for the steam cycle is heated in a wellinsulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kgs and leave at 400 K Water enters the heat exchanger at 200C and 8 MPa and leaves at 350C and 8 MPa If the exhaust gases are treated as air with constant specific heats at room Final PDF to printer cen22672ch10543596indd 594 110917 1148 AM 594 VAPOR AND COMBINED POWER CYCLES temperature the mass flow rate of water through the heat exchanger becomes a 11 kgs b 24 kgs c 46 kgs d 53 kgs e 60 kgs 10133 Consider a cogeneration power plant modified with regeneration Steam enters the turbine at 6 MPa and 450C at a rate of 20 kgs and expands to a pressure of 04 MPa At this pressure 60 percent of the steam is extracted from the turbine and the remainder expands to a pressure of 10 kPa Part of the extracted steam is used to heat feedwater in an open feedwa ter heater The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 04 MPa It is subsequently mixed with the feedwater leaving the feedwater heater and the mixture is pumped to the boiler pressure The steam in the condenser is cooled and condensed by the cooling water from a nearby river which enters the adi abatic condenser at a rate of 463 kgs 1 The total power output of the turbine is a 170 MW b 84 MW c 122 MW d 200 MW e 34 MW 2 The temperature rise of the cooling water from the river in the condenser is a 80C b 52C c 96C d 129C e 162C 3 The mass flow rate of steam through the process heater is a 16 kgs b 38 kgs c 52 kgs d 76 kgs e 104 kgs 4 The rate of heat supply from the process heater per unit mass of steam passing through it is a 246 kJkg b 893 kJkg c 1344 kJkg d 1891 kJkg e 2060 kJkg 5 The rate of heat transfer to the steam in the boiler is a 260 MJs b 538 MJs c 395 MJs d 628 MJs e 1254 MJs Design and Essay Problems 10134 Stack gases exhausting from electrical power plants are at approximately 150C Design a basic Rankine cycle that uses water refrigerant134a or ammonia as the working fluid and that produces the maximum amount of work from this energy source while rejecting heat to the ambient air at 40C You are to use a turbine whose efficiency is 92 percent and whose exit quality cannot be less than 85 percent 10135 Design a steam power cycle that can achieve a cycle thermal efficiency of at least 40 percent under the conditions that all turbines have isentropic efficiencies of 85 percent and all pumps have isentropic efficiencies of 60 percent Prepare an engineering report describing your design Your design report must include but is not limited to the following a Discussion of various cycles attempted to meet the goal as well as the positive and negative aspects of your design b System figures and Ts diagrams with labeled states and temperature pressure enthalpy and entropy information for your design c Sample calculations 10136 A natural gasfired furnace in a textile plant is used to provide steam at 130C At times of high demand the fur nace supplies heat to the steam at a rate of 30 MJs The plant also uses up to 6 MW of electrical power purchased from the local power company The plant management is considering converting the existing process plant into a cogeneration plant to meet both their processheat and power requirements Your job is to come up with some designs Designs based on a gas turbine or a steam turbine are to be considered First decide whether a system based on a gas turbine or a steam turbine will best serve the purpose considering the cost and the complex ity Then propose your design for the cogeneration plant com plete with pressures and temperatures and the mass flow rates Show that the proposed design meets the power and process heat requirements of the plant 10137 Design the condenser of a steam power plant that has a thermal efficiency of 40 percent and generates 10 MW of net electric power Steam enters the condenser as saturated vapor at 10 kPa and it is to be condensed outside horizontal tubes through which cooling water from a nearby river flows The temperature rise of the cooling water is limited to 8C and the velocity of the cooling water in the pipes is limited to 6 ms to keep the pressure drop at an acceptable level From prior experience the average heat flux based on the outer surface of the tubes can be taken to be 12000 Wm2 Specify the pipe diameter total pipe length and the arrangement of the pipes to minimize the condenser volume 10138 Several geothermal power plants are in operation in the United States Heat source of a geothermal plant is hot geo thermal water which is free energy An 8MW geothermal power plant is being considered at a location where geothermal water at 160C is available Geothermal water is to serve as the heat source for a closed Rankine power cycle with refrigerant 134a as the working fluid Specify suitable temperatures and FIGURE P10133 4 h3 h4 h9 60466 h2 19220 h5 61073 Pump II 2 FWH 3 9 6 8 h7 h8 h10 26656 kJkg h11 21288 Pump I h1 19181 h6 33029 kJkg Turbine 1 11 7 10 5 Condenser Process heater Boiler ΔT Final PDF to printer cen22672ch10543596indd 595 110917 1148 AM 595 CHAPTER 10 pressures for the cycle and determine the thermal efficiency of the cycle Justify your selections 10139 A 10MW geothermal power plant is being con sidered at a site where geothermal water at 230C is avail able Geothermal water is to be flashed into a chamber to a lower pressure where part of the water evaporates The liquid is returned to the ground while the vapor is used to drive the steam turbine The pressures at the turbine inlet and the turbine exit are to remain above 200 kPa and 8 kPa respectively High pressure flash chambers yield a small amount of steam with high exergy whereas lowerpressure flash chambers yield con siderably more steam but at a lower exergy By trying several pressures determine the optimum pressure of the flash cham ber to maximize the power production per unit mass of geother mal water withdrawn Also determine the thermal efficiency for each case assuming 10 percent of the power produced is used to drive the pumps and other auxiliary equipment 10140 A photographic equipment manufacturer uses a flow of 64500 lbmh of steam in its manufacturing process Currently the spent steam at 38 psia and 224F is exhausted to the atmosphere Do the preliminary design of a system to use the energy in the waste steam economically If electric ity is produced it can be generated about 8000 hyr and its value is 008kWh If the energy is used for space heating the value is also 008kWh but it can only be used about 3000 hyr only during the heating season If the steam is condensed and the liquid H2O is recycled through the process its value is 070100 gal Make all assumptions as realistic as possible Sketch the system you propose Make a separate list of required components and their specifications capacity efficiency etc The final result will be the calculated annual dollar value of the energy use plan actually a saving because it will replace electricity or heat andor water that would other wise have to be purchased 10141 Contact your power company and obtain informa tion on the thermodynamic aspects of their most recently built power plant If it is a conventional power plant find out why it is preferred over a highly efficient combined power plant FIGURE P10139 Flash chamber Turbine 230C Geothermal water Final PDF to printer cen22672ch10543596indd 596 110917 1148 AM Final PDF to printer cen22672ch11597642indd 597 110917 1148 AM 597 CHAPTER 11 R EF R I G E R AT IO N CYCL E S A major application area of thermodynamics is refrigeration which is the transfer of heat from a lowertemperature region to a higher temperature one Devices that produce refrigeration are called refrigerators and the cycles on which they operate are called refrigeration cycles The most frequently used refrigeration cycle is the vaporcompression refrigeration cycle in which the refrigerant is vaporized and condensed alter nately and is compressed in the vapor phase Another wellknown refrigera tion cycle is the gas refrigeration cycle in which the refrigerant remains in the gaseous phase throughout Other refrigeration cycles discussed in this chapter are cascade refrigeration where more than one refrigeration cycle is used and absorption refrigeration where the refrigerant is dissolved in a liquid before it is compressed OBJECTIVES The objectives of Chapter 11 are to Introduce the concepts of refrigerators and heat pumps and the measure of their performance Analyze the ideal vapor compression refrigeration cycle Analyze the actual vapor compression refrigeration cycle Perform secondlaw analysis of vaporcompression refrigeration cycle Review the factors involved in selecting the right refrigerant for an application Discuss the operation of refrigeration and heat pump systems Evaluate the performance of innovative vaporcompression refrigeration systems Analyze gas refrigeration systems Introduce the concepts of absorptionrefrigeration systems Final PDF to printer 598 REFRIGERATION CYCLES cen22672ch11597642indd 598 110917 1148 AM 111 REFRIGERATORS AND HEAT PUMPS We all know from experience that heat flows in the direction of decreasing temperature that is from hightemperature regions to lowtemperature ones This heattransfer process occurs in nature without requiring any devices The reverse process however cannot occur by itself The transfer of heat from a lowtemperature region to a hightemperature one requires special devices called refrigerators Refrigerators are cyclic devices and the working fluids used in the refrig eration cycles are called refrigerants A refrigerator is shown schematically in Fig 111a Here QL is the magnitude of the heat removed from the refrig erated space at temperature TL QH is the magnitude of the heat rejected to the warm space at temperature TH and Wnetin is the net work input to the refrig erator As discussed in Chap 6 QL and QH represent magnitudes and thus are positive quantities Another device that transfers heat from a lowtemperature medium to a hightemperature one is the heat pump Refrigerators and heat pumps are essentially the same devices they differ in their objectives only The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it Discharging this heat to a highertemperature medium is merely a necessary part of the operation not the purpose The objective of a heat pump however is to maintain a heated space at a high temperature This is accomplished by absorbing heat from a lowtemperature source such as well water or cold outside air in winter and supplying this heat to a warmer medium such as a house Fig 111b The performance of refrigerators and heat pumps is expressed in terms of the coefficient of performance COP defined as COP R Desired output Required input Cooling effect Work input Q L W netin 111 COP HP Desired output Required input Heating effect Work input Q H W netin 112 These relations can also be expressed in the rate form by replacing the quanti ties QL QH and Wnetin with Q L Q H and W netin respectively Notice that both COPR and COPHP can be greater than 1 A comparison of Eqs 111 and 112 reveals that COP HP COP R 1 113 for fixed values of QL and QH This relation implies that COPHP 1 since COPR is a positive quantity That is a heat pump functions at worst as a resistance heater supplying as much energy to the house as it consumes In reality however part of QH is lost to the outside air through piping and other devices and COPHP may drop below unity when the outside air temperature is too low When this happens the system normally switches to the fuel natural gas propane oil etc or resistanceheating mode The cooling capacity of a refrigeration systemthat is the rate of heat removal from the refrigerated spaceis often expressed in terms of tons of refrigeration The capacity of a refrigeration system that can freeze FIGURE 111 The objective of a refrigerator is to remove heat QL from the cold medium the objective of a heat pump is to supply heat QH to a warm medium Warm house Cold refrigerated space Cold environment a Refrigerator b Heat pump QH desired output QH QL desired output QL Wnetin required input Wnetin required input Warm environment R HP Final PDF to printer 599 CHAPTER 11 cen22672ch11597642indd 599 110917 1148 AM 1 ton 2000 lbm of liquid water at 0C 32F into ice at 0C in 24 h is said to be 1 ton One ton of refrigeration is equivalent to 211 kJmin or 200 Btumin The cooling load of a typical 200m2 residence is in the 3ton 10kW range 112 THE REVERSED CARNOT CYCLE Recall from Chap 6 that the Carnot cycle is a totally reversible cycle that consists of two reversible isothermal and two isentropic processes It has the maximum thermal efficiency for given temperature limits and it serves as a standard against which actual power cycles can be compared Since it is a reversible cycle all four processes that comprise the Car not cycle can be reversed Reversing the cycle does also reverse the directions of any heat and work interactions The result is a cycle that operates in the counterclockwise direction on a Ts diagram which is called the reversed Carnot cycle A refrigerator or heat pump that oper ates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump Consider a reversed Carnot cycle executed within the saturation dome of a refrigerant as shown in Fig 112 The refrigerant absorbs heat isothermally from a lowtemperature source at TL in the amount of QL process 12 is compressed isentropically to state 3 temperature rises to TH rejects heat iso thermally to a hightemperature sink at TH in the amount of QH process 34 and expands isentropically to state 1 temperature drops to TL The refriger ant changes from a saturated vapor state to a saturated liquid state in the con denser during process 34 The coefficients of performance of Carnot refrigerators and heat pumps are expressed in terms of temperatures as COP RCarnot 1 T H T L 1 114 and COP HPCarnot 1 1 T L T H 115 Notice that both COPs increase as the difference between the two tempera tures decreases that is as TL rises or TH falls The reversed Carnot cycle is the most efficient refrigeration cycle operating between two specified temperature levels Therefore it is natural to look at it first as a prospective ideal cycle for refrigerators and heat pumps If we could we certainly would adapt it as the ideal cycle As explained below however the reversed Carnot cycle is not a suitable model for refrigeration cycles The two isothermal heat transfer processes are not difficult to achieve in practice since maintaining a constant pressure automatically fixes the temperature of a twophase mixture at the saturation value Therefore processes 12 and 34 can be approached closely in actual evaporators and condensers However processes 23 and 41 cannot be approximated closely in practice This is because process 23 involves the compression of a liquidvapor mixture which requires a compressor that will handle two phases and process 41 involves the expansion of highmoisturecontent refrigerant in a turbine FIGURE 112 Schematic of a Carnot refrigerator and Ts diagram of the reversed Carnot cycle QH Warm medium at TH Cold medium at TL QL QH QL 4 3 2 1 T s Evaporator TL Compressor TH Condenser Turbine 2 1 4 3 Final PDF to printer 600 REFRIGERATION CYCLES cen22672ch11597642indd 600 110917 1148 AM It seems as if these problems could be eliminated by executing the reversed Carnot cycle outside the saturation region But in this case we have diffi culty in maintaining isothermal conditions during the heatabsorption and heatrejection processes Therefore we conclude that the reversed Carnot cycle cannot be approximated in actual devices and is not a realistic model for refrigeration cycles However the reversed Carnot cycle can serve as a standard against which actual refrigeration cycles are compared 113 THE IDEAL VAPORCOMPRESSION REFRIGERATION CYCLE Many of the impracticalities associated with the reversed Carnot cycle can be eliminated by vaporizing the refrigerant completely before it is com pressed and by replacing the turbine with a throttling device such as an expansion valve or capillary tube The cycle that results is called the ideal vaporcompression refrigeration cycle and it is shown schematically and on a Ts diagram in Fig 113 The vaporcompression refrigeration cycle is the most widely used cycle for refrigerators airconditioning systems and heat pumps It consists of four processes 12 Isentropic compression in a compressor 23 Constantpressure heat rejection in a condenser 34 Throttling in an expansion device 41 Constantpressure heat absorption in an evaporator In an ideal vaporcompression refrigeration cycle the refrigerant enters the compressor at state 1 as saturated vapor and is compressed isentropically to the condenser pressure The temperature of the refrigerant increases during this isentropic compression process to well above the temperature of the sur rounding medium The refrigerant then enters the condenser as superheated vapor at state 2 and leaves as saturated liquid at state 3 as a result of heat rejection to the surroundings The temperature of the refrigerant at this state is still above the temperature of the surroundings The saturated liquid refrigerant at state 3 is throttled to the evaporator pres sure by passing it through an expansion valve or capillary tube The tem perature of the refrigerant drops below the temperature of the refrigerated space during this process The refrigerant enters the evaporator at state 4 as a lowquality saturated mixture and it completely evaporates by absorbing heat from the refrigerated space The refrigerant leaves the evaporator as saturated vapor and reenters the compressor completing the cycle In a household refrigerator the tubes in the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator The coils behind the refrigerator where heat is dissipated to the kitchen air serve as the con denser Fig 114 Remember that the area under the process curve on a Ts diagram represents the heat transfer for internally reversible processes The area under the process curve 41 represents the heat absorbed by the refrigerant in the evaporator and the area under the process curve 23 represents the heat rejected in the con denser A rule of thumb is that the COP improves by 2 to 4 percent for each C the evaporating temperature is raised or the condensing temperature is lowered FIGURE 113 Schematic and Ts diagram for the ideal vaporcompression refrigeration cycle QH QL 4 3 2 1 T s 4 Saturated vapor Saturated liquid Compressor Condenser Cold refrigerated space Win Expansion valve Win QL Evaporator 4 3 1 2 Warm environment QH Final PDF to printer 601 CHAPTER 11 cen22672ch11597642indd 601 110917 1148 AM Another diagram often used in the analysis of vaporcompression refrigera tion cycles is the Ph diagram as shown in Fig 115 On this diagram three of the four processes appear as straight lines and the heat transfer in the con denser and the evaporator is proportional to the lengths of the corresponding process curves Notice that unlike the ideal cycles discussed before the ideal vapor compression refrigeration cycle is not an internally reversible cycle since it involves an irreversible throttling process This process is maintained in the cycle to make it a more realistic model for the actual vaporcompression refrigeration cycle If the throttling device were replaced by an isentropic tur bine the refrigerant would enter the evaporator at state 4 instead of state 4 As a result the refrigeration capacity would increase by the area under pro cess curve 44 in Fig 113 and the net work input would decrease by the amount of work output of the turbine Replacing the expansion valve with a turbine is not practical however since the added benefits cannot justify the added cost and complexity All four components associated with the vaporcompression refrigeration cycle are steadyflow devices and thus all four processes that make up the cycle can be analyzed as steadyflow processes The kinetic and potential energy changes of the refrigerant are usually small relative to the work and heat transfer terms and therefore they can be neglected Then the steadyflow energy equation on a unitmass basis reduces to q in q out w in w out h e h i 116 The condenser and the evaporator do not involve any work and the compres sor can be approximated as adiabatic Then the COPs of refrigerators and heat pumps operating on the vaporcompression refrigeration cycle can be expressed as COP R q L w netin h 1 h 4 h 2 h 1 117 and COP HP q H w netin h 2 h 3 h 2 h 1 118 where h1 h g P 1 and h3 h f P 3 for the ideal case Vaporcompression refrigeration dates back to 1834 when the Englishman Jacob Perkins received a patent for a closedcycle ice machine using ether or other volatile fluids as refrigerants A working model of this machine was built but it was never produced commercially In 1850 Alexander Twining began to design and build vaporcompression ice machines using ethyl ether which is a commercially used refrigerant in vaporcompression systems Initially vapor compression refrigeration systems were large and were mainly used for ice making brewing and cold storage They lacked automatic controls and were steamengine driven In the 1890s electric motordriven smaller machines equipped with automatic controls started to replace the older units and refrig eration systems began to appear in butcher shops and households By 1930 the continued improvements made it possible to have vaporcompression refrig eration systems that were relatively efficient reliable small and inexpensive FIGURE 114 An ordinary household refrigerator Compressor Condenser coils Kitchen air 25C Capillary tube Evaporator coils Freezer compartment 18C 3C QH QL FIGURE 115 The Ph diagram of an ideal vapor compression refrigeration cycle 1 h 2 3 4 P QH QL Win Final PDF to printer 602 REFRIGERATION CYCLES cen22672ch11597642indd 602 110917 1148 AM EXAMPLE 111 The Ideal VaporCompression Refrigeration Cycle A refrigerator uses refrigerant134a as the working fluid and operates on an ideal vaporcompression refrigeration cycle between 014 and 08 MPa If the mass flow rate of the refrigerant is 005 kgs determine a the rate of heat removal from the refrigerated space and the power input to the compressor b the rate of heat rejection to the environment and c the COP of the refrigerator SOLUTION A refrigerator operates on an ideal vaporcompression refrigeration cycle between two specified pressure limits The rate of refrigeration the power input the rate of heat rejection and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The Ts diagram of the refrigeration cycle is shown in Fig 116 We note that this is an ideal vaporcompression refrigeration cycle and thus the compressor is isentro pic and the refrigerant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor From the refrigerant134a tables the enthalpies of the refrigerant at all four states are determined as follows P 1 014 MPa h 1 h g 014 MPa 23919 kJkg s 1 s g 014 MPa 094467 kJkgK P 2 s 2 08 MPa s 1 h 2 27540 kJkg P 3 08 MPa h 3 h f 08 MPa 9548 kJkg h 4 h 3 throttling h 4 9548 kJkg a The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions Q L m h 1 h 4 005 kgs 23919 9548 kJkg 719 kW and W in m h 2 h 1 005 kgs 27540 23919 kJkg 181 kW b The rate of heat rejection from the refrigerant to the environment is Q H m h 2 h 3 005 kgs 27540 9548 kJkg 900 kW It could also be determined from Q H Q L W in 719 181 900 kW c The coefficient of performance of the refrigerator is COP R Q L W in 719 kW 181 kW 397 That is this refrigerator removes about 4 units of thermal energy from the refrigerated space for each unit of electric energy it consumes Discussion It would be interesting to see what happens if the throttling valve were replaced by an isentropic turbine The enthalpy at state 4s the turbine exit with P4s 014 MPa and s4s s3 035408 kJkgK is 8895 kJkg and the turbine would produce 033 kW of power This would decrease the power input to the refrigerator from 181 to 148 kW and increase the rate of heat removal from the refrigerated space from 719 to 751 kW As a result the COP of the refrigerator would increase from 397 to 507 an increase of 28 percent FIGURE 116 Ts diagram of the ideal vapor compression refrigeration cycle described in Example 111 T s QH 4 1 4s 3 2 014 MPa 08 MPa Win QL Final PDF to printer 603 CHAPTER 11 cen22672ch11597642indd 603 110917 1148 AM 114 ACTUAL VAPORCOMPRESSION REFRIGERATION CYCLE An actual vaporcompression refrigeration cycle differs from the ideal one in several ways owing mostly to the irreversibilities that occur in various com ponents Two common sources of irreversibilities are fluid friction causes pressure drops and heat transfer to or from the surroundings The Ts diagram of an actual vaporcompression refrigeration cycle is shown in Fig 117 In the ideal cycle the refrigerant leaves the evaporator and enters the com pressor as saturated vapor In practice however it may not be possible to control the state of the refrigerant so precisely Instead it is easier to design the system so that the refrigerant is slightly superheated at the compressor inlet This slight overdesign ensures that the refrigerant is completely vapor ized when it enters the compressor Also the line connecting the evaporator to the compressor is usually very long thus the pressure drop caused by fluid friction and heat transfer from the surroundings to the refrigerant can be very significant The result of superheating heat gain in the connecting line and pressure drops in the evaporator and the connecting line is an increase in the specific volume thus an increase in the power input requirements to the com pressor since steadyflow work is proportional to the specific volume The compression process in the ideal cycle is internally reversible and adia batic and thus isentropic The actual compression process however involves fric tional effects which increase the entropy and heat transfer which may increase or decrease the entropy depending on the direction Therefore the entropy of the refrigerant may increase process 12 or decrease process 12 during an actual compression process depending on which effects dominate The compression process 12 may be even more desirable than the isentropic compression process since the specific volume of the refrigerant and thus the work input requirement are smaller in this case Therefore the refrigerant should be cooled during the compression process whenever it is practical and economical to do so In the ideal case the refrigerant is assumed to leave the condenser as saturated liquid at the compressor exit pressure In reality however it is unavoidable to have some pressure drop in the condenser as well as in the lines connecting the condenser to the compressor and to the throttling valve Also it is not easy to execute the condensation process with such precision that the refrigerant is a saturated liquid at the end and it is undesirable to route the refrigerant to the throttling valve before the refrigerant is completely condensed Therefore the refrigerant is subcooled somewhat before it enters the throttling valve We do not mind this at all however since the refrigerant in this case enters the evapo rator with a lower enthalpy and thus can absorb more heat from the refriger ated space The throttling valve and the evaporator are usually located very close to each other so the pressure drop in the connecting line is small FIGURE 117 Schematic and Ts diagram for the actual vaporcompression refrigeration cycle 4 5 2 1 T s 6 7 8 3 2 Compressor Condenser Cold refrigerated space Win Expansion valve QL Evaporator 3 1 2 Warm environment QH 7 8 3 4 EXAMPLE 112 The Actual VaporCompression Refrigeration Cycle Refrigerant134a enters the compressor of a refrigerator as superheated vapor at 014 MPa and 10C at a rate of 005 kgs and leaves at 08 MPa and 50C The refrig erant is cooled in the condenser to 26C and 072 MPa and is throttled to 015 MPa Disregarding any heat transfer and pressure drops in the connecting lines between the Final PDF to printer 604 REFRIGERATION CYCLES cen22672ch11597642indd 604 110917 1148 AM components determine a the rate of heat removal from the refrigerated space and the power input to the compressor b the isentropic efficiency of the compressor and c the coefficient of performance of the refrigerator SOLUTION A refrigerator operating on a vaporcompression cycle is considered The rate of refrigeration the power input the compressor efficiency and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis The Ts diagram of the refrigeration cycle is shown in Fig 118 We note that the refrigerant leaves the condenser as a compressed liquid and enters the compressor as superheated vapor The enthalpies of the refrigerant at various states are determined from the refrigerant tables to be P 1 014 MPa T 1 10C h 1 24637 kJkg P 2 08 MPa T 2 50C h 2 28671 kJkg P 3 072 MPa T 3 26C h 3 h f 26C 8783 kJkg h 4 h 3 throttling h 4 8783 kJkg a The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions Q L m h 1 h 4 005 kgs 24637 8783 kJkg 793 kW and W in m h 2 h 1 005 kgs 28671 24637 kJkg 202 kW b The isentropic efficiency of the compressor is determined from η C h 2s h 1 h 2 h 1 where the enthalpy at state 2s P2s 08 MPa and s2s s1 09724 kJkgK is 28420 kJkg Thus η C 28420 24637 28671 24637 0938 or 938 c The coefficient of performance of the refrigerator is COP R Q L W in 793 kW 202 kW 393 Discussion This problem is identical to the one worked out in Example 111 except that the refrigerant is slightly superheated at the compressor inlet and subcooled at the condenser exit Also the compressor is not isentropic As a result the heat removal rate from the refrigerated space increases by 103 percent but the power input to the com pressor increases even more by 116 percent Consequently the COP of the refrigerator decreases from 397 to 393 FIGURE 118 Ts diagram for Example 112 T s 3 072 MPa 26C 4 QH Win QL 015 MPa 2 08 MPa 50C 2s 1 014 MPa 10C Final PDF to printer 605 CHAPTER 11 cen22672ch11597642indd 605 110917 1148 AM 115 SECONDLAW ANALYSIS OF VAPORCOMPRESSION REFRIGERATION CYCLE Consider the vaporcompression refrigeration cycle operating between a low temperature medium at TL and a hightemperature medium at TH as shown in Fig 119 The maximum COP of a refrigeration cycle operating between temperature limits of TL and TH was given in Eq 114 as COP Rmax COP Rrev COP RCarnot T L T H T L 1 T H T L 1 119 Actual refrigeration cycles are less efficient than the Carnot refrigerator because of the irreversibilities involved But the conclusion we can draw from Eq 119 that the COP is inversely proportional to the temperature ratio THTL is equally valid for actual refrigeration cycles The goal of a secondlaw or exergy analysis of a refrigeration system is to determine the components that can benefit the most by improvements This is done by identifying the locations of greatest exergy destruction and the com ponents with the lowest secondlaw or exergy efficiency Exergy destruction in a component can be determined directly from an exergy balance or indi rectly by first calculating the entropy generation and then using the relation X dest T 0 S gen 1110 where T0 is the environment the deadstate temperature For a refrigera tor T0 is usually the temperature of the hightemperature medium TH for a heat pump it is TL Exergy destructions and the secondlaw efficiencies for major components of a refrigeration system operating on the cycle shown in Fig 119 may be written as follows Compressor X dest12 T 0 S gen12 m T 0 s 2 s 1 1111 η IIComp X recovered X expended W rev W actin m h 2 h 1 T 0 s 2 s 1 m h 2 h 1 ψ 2 ψ 1 h 2 h 1 1 X dest12 W actin 1112 Condenser X dest23 T 0 S gen23 T 0 m s 3 s 2 Q H T H 1113 η IICond X recovered X expended X Q H X 2 X 3 Q H 1 T 0 T H X 2 X 3 Q H 1 T 0 T H m h 2 h 3 T 0 s 2 s 3 1 X dest23 X 2 X 3 1114 Note that when TH T0 which is usually the case for refrigerators ηIICond 0 since there is no recoverable exergy in this case FIGURE 119 The vaporcompression refrigeration cycle considered in the secondlaw analysis 2 3 1 4 Compressor Condenser TH Win Expansion valve QH Evaporator Warm environment Cold environment TL QL Final PDF to printer 606 REFRIGERATION CYCLES cen22672ch11597642indd 606 110917 1148 AM Expansion valve X dest34 T 0 S gen34 m T 0 s 4 s 3 η IIExpValve X recovered X expended 0 X 3 X 4 0 1115 or η IIExpValve 1 X dest34 X expended 1 X 3 X 4 X 3 X 4 0 1116 Evaporator X dest41 T 0 S gen41 T 0 m s 1 s 4 Q L T L η IIEvap X recovered X expended X Q L X 4 X 1 Q L T 0 T L T L X 4 X 1 1117 Q L T 0 T L T L m h 4 h 1 T 0 s 4 s 1 1 X dest41 X 4 X 1 1118 Here X Q L represents the positive of the exergy rate associated with the withdrawal of heat from the lowtemperature medium at TL at a rate Q L Note that the directions of heat and exergy transfer become opposite when TL T0 that is the exergy of the lowtemperature medium increases as it loses heat Also X Q L is equivalent to the power that can be produced by a Carnot heat engine receiving heat from the environment at T0 and rejecting heat to the low temperature medium at TL at a rate of Q L which can be shown to be X Q L Q L T 0 T L T L 1119 From the definition of reversibility this is equivalent to the minimum or reversible power input required to remove heat at a rate of Q L and reject it to the environment at T0 That is W revin W minin X Q L Note that when TL T0 which is often the case for heat pumps ηIIEvap 0 since there is no recoverable exergy in this case The total exergy destruction associated with the cycle is the sum of the exergy destructions X desttotal X dest12 X dest23 X dest34 X dest41 1120 It can be shown that the total exergy destruction associated with a refrigera tion cycle can also be obtained by taking the difference between the exergy supplied power input and the exergy recovered the exergy of the heat with drawn from the lowtemperature medium X desttotal W in X Q L 1121 The secondlaw or exergy efficiency of the cycle can then be expressed as η IIcycle X Q L W in W minin W in 1 X desttotal W in 1122 Final PDF to printer 607 CHAPTER 11 cen22672ch11597642indd 607 110917 1148 AM Substituting W in Q L COP R and X Q L Q L T 0 T L T L into Eq 1122 gives η IIcycle X Q L W in Q L T 0 T L T L Q L COP R COP R T L T H T L COP R COP Rrev 1123 since T0 TH for a refrigeration cycle Thus the secondlaw efficiency is also equal to the ratio of actual and maximum COPs for the cycle This secondlaw efficiency definition accounts for all irreversibilities associated within the refriger ator including the heat transfers with the refrigerated space and the environment EXAMPLE 113 Exergy Analysis of VaporCompression Refrigeration Cycle A vaporcompression refrigeration cycle with refrigerant134a as the working fluid is used to maintain a space at 13C by rejecting heat to ambient air at 27C R134a enters the compressor at 100 kPa superheated by 64C at a rate of 005 kgs The isentropic efficiency of the compressor is 85 percent The refrigerant leaves the con denser at 394C as a saturated liquid Determine a the rate of cooling provided and the COP of the system b the exergy destruction in each basic component c the minimum power input and the secondlaw efficiency of the cycle and d the rate of total exergy destruction SOLUTION A vaporcompression refrigeration cycle is considered The cooling rate the COP the exergy destructions the minimum power input the secondlaw effi ciency and the total exergy destruction are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible Analysis a The Ts diagram of the cycle is given in Fig 1110 The properties of R134a are Tables A11 through A13 P 1 100 kPa T 1 T sat 100 kPa Δ T superheat 264 64 20C h 1 23952 kJkg s 1 09721 kJkgK P 3 P sat 39 4 C 1000 kPa P 2 s 2s P 3 1000 kPa s 1 09721 kJ kgK h 2s 28914 kJ kg P 3 x 3 1000 kPa 0 h 3 10734 kJkg s 3 039196 h 4 h 3 10734 kJkg P 4 h 4 100 kPa 10734 kJkg s 4 04368 kJkgK From the definition of isentropic efficiency η C h 2s h 1 h 2 h 1 085 28914 23952 h 2 23952 h 2 29790 kJkg P 2 1000 kPa h 2 29790 kJkg s 2 09984 kJkgK FIGURE 1110 Temperatureentropy diagram of the vaporcompression refrigeration cycle considered in Example 113 T s 3 394C 4 QH Win QL 100 kPa 2 2s 1 Final PDF to printer 608 REFRIGERATION CYCLES cen22672ch11597642indd 608 110917 1148 AM The refrigeration load the rate of heat rejected and the power input are Q L m h 1 h 4 005 kgs23952 10734 kJkg 6609 kW Q H m h 2 h 3 005 kgs29790 10734 kJkg 9528 kW W in m h 2 h 1 005 kgs29790 23952 kJkg 2919 kW Then the COP of the refrigeration cycle becomes COP R Q L W in 6609 kW 2919 kW 2264 b Noting that the deadstate temperature is T0 TH 27 273 300 K the exergy destruction in each component of the cycle is determined as follows Compressor X dest12 T 0 S gen12 T 0 m s 2 s 1 300 K005 kgs09984 09721 kJkgK 03945 kW Condenser X dest23 T 0 S gen23 T 0 m s 2 s 1 Q H T H 300 K 005 kgs 039196 09984 kJkgK 9528 kW 300 K 04314 kW Expansion valve X dest34 T 0 S gen34 T 0 m s 4 s 3 300 K005 kgs04368 039196 kJkgK 06726 kW Evaporator X dest41 T 0 S gen41 T 0 m s 1 s 4 Q L T L 300 K 005 kgs 09721 04368 kJkgK 6609 kW 260 K 04037 kW c Exergy flow associated with heat transferred from the lowtemperature medium is X Q L Q L T 0 T L T L 6609 kW 300 K 260 K 260 K 1017 kW This is also the minimum or reversible power input for the cycle W minin X Q L 1017 kW The secondlaw efficiency of the cycle is η II X Q L W in 1017 kW 2919 kW 0348 or 348 Final PDF to printer 609 CHAPTER 11 cen22672ch11597642indd 609 110917 1148 AM This efficiency may also be determined from ηII COPRCOPRrev where COP Rrev T L T H T L 13 273 K 27 13 K 6500 Substituting η II COP R COP R rev 2264 6500 0348 or 348 The results are identical as expected d The total exergy destruction is the difference between the exergy expended power input and the exergy recovered the exergy of the heat transferred from the low temperature medium X desttotal W in X Q L 2919 kW 1017 kW 1902 kW The total exergy destruction can also be determined by adding exergy destruction in each component X desttotal X dest12 X dest23 X dest34 X dest41 03945 04314 06726 04037 1902 kW The two results are again identical as expected Discussion The exergy input to the cycle is equal to the actual work input which is 292 kW The same cooling load could have been accomplished by only 348 percent of this power 102 kW if a reversible system were used The difference between the two is the exergy destroyed in the cycle 190 kW The expansion valve appears to be the most irreversible component which accounts for 354 percent of the irreversibilities in the cycle Replacing the expansion valve with a turbine would decrease the irrevers ibilities while decreasing the net power input However this may or may not be practical in an actual system It can be shown that increasing the evaporating temperature and decreasing the condensing temperature would also decrease the exergy destruction in these components 116 SELECTING THE RIGHT REFRIGERANT When designing a refrigeration system there are several refrigerants from which to choose such as chlorofluorocarbons CFCs hydrofluorocarbons HFCs hydrochlorofluorocarbons HCFCs ammonia hydrocarbons pro pane ethane ethylene etc carbon dioxide air in the airconditioning of aircraft and even water in applications above the freezing point The right choice of refrigerant depends on the situation at hand Ethyl ether was the first commercially used refrigerant in vapor compression systems in 1850 followed by ammonia carbon dioxide methyl chloride sulphur dioxide butane ethane propane isobutane gasoline and chlorofluo rocarbons among others The industrial and heavycommercial sectors were very satisfied with ammonia and still are although ammonia is toxic The advantages of ammo nia over other refrigerants are its low cost higher COPs and thus lower energy cost more favorable thermodynamic and transport properties and thus higher heat transfer coefficients requires smaller and lowercost heat exchangers Final PDF to printer 610 REFRIGERATION CYCLES cen22672ch11597642indd 610 110917 1148 AM greater detectability in the event of a leak and no effect on the ozone layer The major drawback of ammonia is its toxicity which makes it unsuitable for domestic use Ammonia is predominantly used in food refrigeration facili ties such as those used for the cooling of fresh fruits vegetables meat and fish refrigeration of beverages and dairy products such as beer wine milk and cheese freezing of ice cream and other foods ice production and low temperature refrigeration in the pharmaceutical and other process industries It is remarkable that the early refrigerants used in the lightcommercial and household sectors such as sulfur dioxide ethyl chloride and methyl chloride were highly toxic The widespread reporting of a few instances of leaks that resulted in serious illnesses and death in the 1920s caused a public outcry to ban or limit the use of these refrigerants creating a need for the development of a safe refrigerant for household use At the request of Frigidaire Corpo ration General Motors research laboratory developed R21 the first mem ber of the CFC family of refrigerants within three days in 1928 Of several CFCs developed the research team settled on R12 as the refrigerant most suitable for commercial use and gave the CFC family the trade name Freon Commercial production of R11 and R12 was started in 1931 by a company jointly formed by General Motors and E I du Pont de Nemours and Co The versatility and low cost of CFCs made them the refrigerants of choice CFCs were also widely used in aerosols foam insulations and the electronics indus try as solvents to clean computer chips R11 is used primarily in largecapacity water chillers serving air conditioning systems in buildings R12 is used in domestic refrigerators and freezers as well as automotive air conditioners R22 is used in window air conditioners heat pumps air conditioners of commercial buildings and large industrial refrigeration systems and it offers strong competition to ammonia The ozone crisis has caused a major stir in the refrigeration and air conditioning industry and has triggered a critical look at the refrigerants in use It was realized in the mid1970s that CFCs allow more ultraviolet radia tion into the earths atmosphere by destroying the protective ozone layer Pro longed exposure to solar ultraviolet radiation can cause serious health effects on the skin eyes and immune system As a result the use of some CFCs is banned by international treaties Fully halogenated CFCs such as R11 R12 and R115 do the most damage to the ozone layer The nonfully haloge nated refrigerants such as R22 an HCFC have about 5 percent of the ozone depleting capability of R12 Note that unlike CFCs and HCFCs HFCs do not contribute to ozone depletion Refrigerants that are friendly to the ozone layer that protects the earth from harmful ultraviolet rays have been developed The oncepopular refrigerant R12 has been replaced by the chlorinefree R134a an HFC R22 is being phased out because it is ozonedepleting R410A and R407C both HFCs are among the common alternatives to R22 in residential and commercial airconditioning and refrigeration applications R410A is the most common refrigerant for new installations due to high efficiency and low global warm ing potential R502 a blend of R115 and R22 was the dominant refrigerant used in commercial refrigeration systems such as those in supermarkets but its use has been discontinued Several replacement options to R502 are available Two important parameters that need to be considered in the selection of a refrigerant are the temperatures of the two media the refrigerated space Final PDF to printer 611 CHAPTER 11 cen22672ch11597642indd 611 110917 1148 AM and the environment with which the refrigerant exchanges heat To have heat transfer at a reasonable rate a temperature difference of 5 to 10C should be maintained between the refrigerant and the medium with which it is exchang ing heat If a refrigerated space is to be maintained at 10C for example the temperature of the refrigerant should remain at about 20C while it absorbs heat in the evaporator The lowest pressure in a refrigeration cycle occurs in the evaporator and this pressure should be above atmospheric pressure to prevent any air leakage into the refrigeration system Therefore a refrigerant should have a saturation pressure of 1 atm or higher at 20C in this case Ammonia and R134a are two such substances The temperature and thus the pressure of the refrigerant on the condenser side depends on the medium to which heat is rejected Lower temperatures in the condenser thus higher COPs can be maintained if the refrigerant is cooled by liquid water instead of air The use of water cooling cannot be justi fied economically however except in large industrial refrigeration systems The temperature of the refrigerant in the condenser cannot fall below the tem perature of the cooling medium about 20C for a household refrigerator and the saturation pressure of the refrigerant at this temperature should be well below its critical pressure if the heat rejection process is to be approximately isothermal If no single refrigerant can meet the temperature requirements then two or more refrigeration cycles with different refrigerants can be used in series Such a refrigeration system is called a cascade system and is dis cussed later in this chapter Other desirable characteristics of a refrigerant include being nontoxic non corrosive nonflammable and chemically stable having a high enthalpy of vaporization minimizes the mass flow rate and of course being available at low cost In the case of heat pumps the minimum temperature and pressure for the refrigerant may be considerably higher since heat is usually extracted from media that are well above the temperatures encountered in refrigeration systems 117 HEAT PUMP SYSTEMS Heat pumps are generally more expensive to purchase and install than other heating systems but they save money in the long run in some areas because they lower the heating bills Despite their relatively higher initial costs the popularity of heat pumps is increasing About onethird of all singlefamily homes built in the United States in recent years are heated by heat pumps The most common energy source for heat pumps is atmospheric air air toair systems although water and soil are also used The major problem with airsource systems is frosting which occurs in humid climates when the temperature falls below 2 to 5C The frost accumulation on the evaporator coils is highly undesirable since it seriously disrupts heat transfer The coils can be defrosted however by reversing the heat pump cycle running it as an air conditioner This results in a reduction in the efficiency of the system Watersource systems usually use well water from depths of up to 80 m in the temperature range of 5 to 18C and they do not have a frosting problem They typically have higher COPs but are more complex and require easy access to a large body of water such as underground water Groundsource systems are Final PDF to printer 612 REFRIGERATION CYCLES cen22672ch11597642indd 612 110917 1148 AM also rather involved since they require long tubing to be placed deep in the ground where the soil temperature is relatively constant The COP of heat pumps usually ranges between 15 and 4 depending on the particular system used and the temperature of the source A new class of recently developed heat pumps that use variablespeed electric motor drives are at least twice as energy efficient as their predecessors Both the capacity and the efficiency of a heat pump fall significantly at low temperatures Therefore most airsource heat pumps require a supple mentary heating system such as electric resistance heaters or an oil or gas furnace Since water and soil temperatures do not fluctuate much supplemen tary heating may not be required for watersource or groundsource systems However the heat pump system must be large enough to meet the maximum heating load Heat pumps and air conditioners have the same mechanical components Therefore it is not economical to have two separate systems to meet the heat ing and cooling requirements of a building One system can be used as a heat pump in winter and an air conditioner in summer This is accomplished by adding a reversing valve to the cycle as shown in Fig 1111 As a result of this modification the condenser of the heat pump located indoors functions as the evaporator of the air conditioner in summer Also the evaporator of the heat pump located outdoors serves as the condenser of the air conditioner This feature increases the competitiveness of the heat pump Such dual purpose units are commonly used in motels Heat pumps are most competitive in areas that have a large cooling load during the cooling season and a relatively small heating load during the heat ing season such as in the southern parts of the United States In these areas the heat pump can meet the entire cooling and heating needs of residential or commercial buildings The heat pump is least competitive in areas where the heating load is very large and the cooling load is small such as in the northern parts of the United States FIGURE 1111 A heat pump can be used to heat a house in winter and to cool it in summer Heat Pump OperationHeating Mode Reversing valve Indoor coil Outdoor coil Expansion valve Highpressure liquid Lowpressure liquidvapor Lowpressure vapor Highpressure vapor Compressor Indoor coil Outdoor coil Heat Pump OperationCooling Mode Reversing valve Compressor Expansion valve Final PDF to printer 613 CHAPTER 11 cen22672ch11597642indd 613 110917 1148 AM 118 INNOVATIVE VAPORCOMPRESSION REFRIGERATION SYSTEMS The simple vaporcompression refrigeration cycle discussed earlier is the most widely used refrigeration cycle and it is adequate for most refrigera tion applications The ordinary vaporcompression refrigeration systems are simple inexpensive reliable and practically maintenancefree when was the last time you serviced your household refrigerator However for large industrial applications efficiency not simplicity is the major concern Also for some applications the simple vaporcompression refrigeration cycle is inadequate and needs to be modified We now discuss a few such modifica tions and refinements Cascade Refrigeration Systems Some industrial applications require moderately low temperatures and the temperature range they involve may be too large for a single vapor compression refrigeration cycle to be practical A large temperature range also means a large pressure range in the cycle and a poor performance for a reciprocating compressor One way of dealing with such situations is to per form the refrigeration process in stages that is to have two or more refrig eration cycles that operate in series Such refrigeration cycles are called cascade refrigeration cycles A twostage cascade refrigeration cycle is shown in Fig 1112 The two cycles are connected through the heat exchanger in the middle which serves as the evaporator for the topping cycle cycle A and the condenser for the bot toming cycle cycle B Assuming the heat exchanger is well insulated and the kinetic and potential energies are negligible the heat transfer from the fluid in the bottoming cycle should be equal to the heat transfer to the fluid in the topping cycle Thus the ratio of mass flow rates through each cycle should be m A h 5 h 8 m B h 2 h 3 m A m B h 2 h 3 h 5 h 8 1124 Also COP Rcascade Q L W netin m B h 1 h 4 m A h 6 h 5 m B h 2 h 1 1125 In the cascade system shown in the figure the refrigerants in both cycles are assumed to be the same This is not necessary however since there is no mixing taking place in the heat exchanger Therefore refrigerants with more desirable characteristics can be used in each cycle In this case there would be a separate saturation dome for each fluid and the Ts diagram for each cycle would be different Also in actual cascade refrigeration systems the two cycles would overlap somewhat since a temperature difference between the two fluids is needed for any heat transfer to take place It is evident from the Ts diagram in Fig 1112 that the compressor work decreases and the amount of heat absorbed from the refrigerated space increases as a result of cascading Therefore cascading improves the COP of a refrigeration system Some refrigeration systems use three or four stages of cascading Final PDF to printer 614 REFRIGERATION CYCLES cen22672ch11597642indd 614 110917 1148 AM FIGURE 1113 Ts diagram of the cascade refrigeration cycle described in Example 114 4 3 2 1 T s 6 7 8 5 h3 5514 h7 9548 h6 27096 kJkg h2 25595 h5 25193 h1 23919 h4 5514 h8 9548 08 MPa 032 MPa 014 MPa A B FIGURE 1112 A twostage cascade refrigeration system with the same refrigerant in both stages Evaporator Condenser Heat 4 5 2 1 T s 6 7 8 3 8 5 QH Condenser Warm environment Decrease in compressor work QH QL Increase in refrigeration capacity Compressor Expansion valve 7 6 Compressor Expansion valve 2 Evaporator A B 4 Heat exchanger A B 1 Cold refrigerated space QL 3 EXAMPLE 114 A TwoStage Cascade Refrigeration Cycle Consider a twostage cascade refrigeration system operating between the pressure limits of 08 and 014 MPa Each stage operates on an ideal vaporcompression refrig eration cycle with refrigerant134a as the working fluid Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 032 MPa In practice the working fluid of the lower cycle is at a higher pressure and temperature in the heat exchanger for effective heat transfer If the mass flow rate of the refrigerant through the upper cycle is 005 kgs determine a the mass flow rate of the refrigerant through the lower cycle b the rate of heat removal from the refrigerated space and the power input to the compres sor and c the coefficient of performance of this cascade refrigerator SOLUTION A cascade refrigeration system operating between the specified pres sure limits is considered The mass flow rate of the refrigerant through the lower cycle the rate of refrigeration the power input and the COP are to be determined Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 The heat exchanger is adiabatic Properties The enthalpies of the refrigerant at all eight states are determined from the refrigerant tables and are indicated on the Ts diagram Final PDF to printer 615 CHAPTER 11 cen22672ch11597642indd 615 110917 1148 AM Analysis The Ts diagram of the refrigeration cycle is shown in Fig 1113 The top ping cycle is labeled cycle A and the bottoming one cycle B For both cycles the refriger ant leaves the condenser as a saturated liquid and enters the compressor as saturated vapor a The mass flow rate of the refrigerant through the lower cycle is determined from the steadyflow energy balance on the adiabatic heat exchanger E out E in m A h 5 m B h 3 m A h 8 m B h 2 m A h 5 h 8 m B h 2 h 3 005 kgs25193 9548 kJkg m B 25595 5514 kJkg m B 00390 kgs b The rate of heat removal by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage The power input to a cascade cycle is the sum of the power inputs to all of the compressors W in W comp Iin W comp IIin m A h 6 h 5 m B h 2 h 1 005 kgs27096 25193 kJkg 0039 kgs25595 23919 kJkg 161 kW c The COP of a refrigeration system is the ratio of the refrigeration rate to the net power input COP R Q L W netin 718 kW 161 kW 446 Discussion This problem was worked out in Example 111 for a singlestage refriger ation system Notice that the COP of the refrigeration system increases from 397 to 446 as a result of cascading The COP of the system can be increased even more by increasing the number of cascade stages Q L m B h 1 h 4 00390 kgs23919 5514 kJkg 718 kW Multistage Compression Refrigeration Systems When the fluid used throughout the cascade refrigeration system is the same the heat exchanger between the stages can be replaced by a mixing chamber called a flash chamber since it has better heattransfer characteristics Such systems are called multistage compression refrigeration systems A two stage compression refrigeration system is shown in Fig 1114 In this system the liquid refrigerant expands in the first expansion valve to the flash chamber pressure which is the same as the compressor inter stage pressure Part of the liquid vaporizes during this process This saturated vapor state 3 is mixed with the superheated vapor from the lowpressure compressor state 2 and the mixture enters the highpressure compressor at state 9 This is in essence a regeneration process The saturated liquid state 7 expands through the second expansion valve into the evaporator where it picks up heat from the refrigerated space The compression process in this system resembles a twostage compres sion with intercooling and the compressor work decreases Care should be exercised in the interpretations of the areas on the Ts diagram in this case because the mass flow rates are different in different parts of the cycle Final PDF to printer 616 REFRIGERATION CYCLES cen22672ch11597642indd 616 110917 1148 AM FIGURE 1114 A twostage compression refrigeration system with a flash chamber 8 5 2 1 T s 7 6 3 4 9 Flash chamber QH Condenser Highpressure compressor Expansion valve Expansion valve Evaporator QL Lowpressure compressor Warm environment Cold refrigerated space 8 7 2 9 4 5 6 3 1 EXAMPLE 115 A TwoStage Refrigeration Cycle with a Flash Chamber Consider a twostage compression refrigeration system operating between the pres sure limits of 08 and 014 MPa The working fluid is refrigerant134a The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 032 MPa Part of the refrigerant evaporates during this flashing process and this vapor is mixed with the refrigerant leaving the lowpressure compressor The mix ture is then compressed to the condenser pressure by the highpressure compressor The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator Assuming the refrigerant leaves the evaporator as a saturated vapor and both compressors are isentropic determine a the fraction of the refrigerant that evaporates as it is throttled to the flash chamber b the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser and c the coefficient of performance SOLUTION A twostage compression refrigeration system operating between specified pressure limits is considered The fraction of the refrigerant that evaporates in the flash chamber the refrigeration and work input per unit mass and the COP are to be determined FIGURE 1115 Ts diagram of the twostage compression refrigeration cycle described in Example 115 8 7 2 1 T s 4 5 6 9 h7 5514 h6 9548 h4 27449 kJkg h2 25595 h9 25513 h1 23919 h8 5514 h3 25193 3 h5 9548 Final PDF to printer 617 CHAPTER 11 cen22672ch11597642indd 617 110917 1148 AM Assumptions 1 Steady operating conditions exist 2 Kinetic and potential energy changes are negligible 3 The flash chamber is adiabatic Properties The enthalpies of the refrigerant at various states are determined from the refrigerant tables and are indicated on the Ts diagram Analysis The Ts diagram of the refrigeration cycle is shown in Fig 1115 We note that the refrigerant leaves the condenser as saturated liquid and enters the lowpressure compressor as saturated vapor a The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6 which is x 6 h 6 h f h fg 9548 5514 19678 02050 b The amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are q L 1 x 6 h 1 h 8 1 02050 23919 5514 kJkg 1463 kJkg and w in w comp Iin w comp IIin 1 x 6 h 2 h 1 1 h 4 h 9 The enthalpy at state 9 is determined from an energy balance on the mixing chamber E out E in 1 h 9 x 6 h 3 1 x 6 h 2 h 9 02050 25193 1 02050 25595 25513 kJkg Also s9 09417 kJkgK Thus the enthalpy at state 4 08 MPa s4 s9 is h4 27449 kJkg Substituting w in 1 02050 25595 23919 kJkg 27449 25513 kJkg 3268 kJkg c The coefficient of performance is COP R q L w in 1463 kJkg 3268 kJkg 448 Discussion This problem was worked out in Example 111 for a singlestage refrig eration system COP 397 and in Example 114 for a twostage cascade refrigeration system COP 446 Notice that the COP of the refrigeration system increased consider ably relative to the singlestage compression but did not change much relative to the two stage cascade compression Multipurpose Refrigeration Systems with a Single Compressor Some applications require refrigeration at more than one temperature This could be accomplished by using a separate throttling valve and a separate compressor for each evaporator operating at different temperatures However Final PDF to printer 618 REFRIGERATION CYCLES cen22672ch11597642indd 618 110917 1148 AM such a system is bulky and probably uneconomical A more practical and eco nomical approach would be to route all the exit streams from the evaporators to a single compressor and let it handle the compression process for the entire system Consider for example an ordinary refrigeratorfreezer unit A simpli fied schematic of the unit and the Ts diagram of the cycle are shown in Fig 1116 Most refrigerated goods have a high water content and the refrig erated space must be maintained above the ice point to prevent freezing The freezer compartment however is maintained at about 18C Therefore the refrigerant should enter the freezer at about 25C to have heat transfer at a reasonable rate in the freezer If a single expansion valve and evaporator were used the refrigerant would have to circulate in both compartments at about 25C which would cause ice formation in the neighborhood of the evapora tor coils and dehydration of the produce This problem can be eliminated by throttling the refrigerant to a higher pressure hence temperature for use in the refrigerated space and then throttling it to the minimum pressure for use in the freezer The entire refrigerant leaving the freezer compartment is subse quently compressed by a single compressor to the condenser pressure Liquefaction of Gases The liquefaction of gases has always been an important area of refrigera tion since many important scientific and engineering processes at cryogenic temperatures temperatures below about 100C depend on liquefied gases Some examples of such processes are the separation of oxygen and nitrogen from air preparation of liquid propellants for rockets the study of material properties at low temperatures and the study of some exciting phenomena such as superconductivity FIGURE 1116 Schematic and Ts diagram for a refrigeratorfreezer unit with one compressor QH QLF 4 3 2 1 T s A 6 QLR 5 Freezer Compressor 2 QLF Expansion valve 4 Expansion valve QLR 1 3 6 Refrigerator Alternative path A 5 QH Kitchen air Condenser Final PDF to printer 619 CHAPTER 11 cen22672ch11597642indd 619 110917 1148 AM At temperatures above the criticalpoint value a substance exists in the gas phase only The critical temperatures of helium hydrogen and nitrogen three commonly used liquefied gases are 268 240 and 147C respectively Therefore none of these substances exist in liquid form at atmospheric condi tions Furthermore low temperatures of this magnitude cannot be obtained by ordinary refrigeration techniques Then the question that needs to be answered in the liquefaction of gases is this How can we lower the temperature of a gas below its criticalpoint value Several cycles some complex and others simple are used successfully for the liquefaction of gases Next we discuss the LindeHampson cycle which is shown schematically and on a Ts diagram in Fig 1117 Makeup gas is mixed with the uncondensed portion of the gas from the previous cycle and the mixture at state 2 is compressed by a multistage compressor to state 3 The compression process approaches an isothermal process due to intercooling The highpressure gas is cooled in an after cooler by a cooling medium or by a separate external refrigeration system to state 4 The gas is further cooled in a regenerative counterflow heat exchanger by the uncondensed portion of gas from the previous cycle to state 5 and it is throttled to state 6 which is a saturated liquidvapor mix ture state The liquid state 7 is collected as the desired product and the vapor state 8 is routed through the regenerator to cool the highpressure gas approaching the throttling valve Finally the gas is mixed with fresh makeup gas and the cycle is repeated This and other refrigeration cycles used for the liquefaction of gases can also be used for the solidification of gases 119 GAS REFRIGERATION CYCLES As explained in Sec 112 the Carnot cycle the standard of comparison for power cycles and the reversed Carnot cycle the standard of comparison for refrigeration cycles are identical except that the reversed Carnot cycle oper ates in the reverse direction This suggests that the power cycles discussed in earlier chapters can be used as refrigeration cycles by simply reversing them In fact the vaporcompression refrigeration cycle is essentially a modi fied Rankine cycle operating in reverse Another example is the reversed Stirling cycle which is the cycle on which Stirling refrigerators operate In this section we discuss the reversed Brayton cycle better known as the gas refrigeration cycle Consider the gas refrigeration cycle shown in Fig 1118 The surroundings are at T0 and the refrigerated space is to be maintained at TL The gas is com pressed during process 12 The highpressure hightemperature gas at state 2 is then cooled at constant pressure to T0 by rejecting heat to the surroundings This is followed by an expansion process in a turbine during which the gas temperature drops to T4 Can we achieve the cooling effect by using a throt tling valve instead of a turbine Finally the cool gas absorbs heat from the refrigerated space until its temperature rises to T1 All the processes described are internally reversible and the cycle executed is the ideal gas refrigeration cycle In actual gas refrigeration cycles the com pression and expansion processes deviate from the isentropic ones and T3 is higher than T0 unless the heat exchanger is infinitely large FIGURE 1117 LindeHampson system for liquefying gases 4 5 2 1 T s 7 8 3 Multistage compressor Q 6 Heat exchanger Liquid removed Vapor recirculated Makeup gas Regenerator 9 4 5 6 7 8 3 2 1 9 Final PDF to printer 620 REFRIGERATION CYCLES cen22672ch11597642indd 620 110917 1148 AM On a Ts diagram the area under process curve 41 represents the heat removed from the refrigerated space and the enclosed area 12341 rep resents the net work input The ratio of these areas is the COP for the cycle which may be expressed as COP R q L w netin q L w compin w turbout 1126 where q L h 1 h 4 w turbout h 3 h 4 w compin h 2 h 1 The gas refrigeration cycle deviates from the reversed Carnot cycle because the heat transfer processes are not isothermal In fact the gas temperature varies considerably during heat transfer processes Consequently the gas refrigeration cycles have lower COPs relative to the vaporcompression refrigeration cycles or the reversed Carnot cycle This is also evident from the Ts diagram in Fig 1119 The reversed Carnot cycle consumes a fraction of the net work rectangular area 1A3B but produces a greater amount of refrig eration triangular area under B1 Despite their relatively low COPs the gas refrigeration cycles have two desir able characteristics They involve simple lighter components which make them suitable for aircraft cooling and they can incorporate regeneration which makes them suitable for liquefaction of gases and cryogenic applications An opencycle aircraft cooling system is shown in Fig 1120 Atmospheric air is compressed by a compressor cooled by the surrounding air and expanded in a turbine The cool air leaving the turbine is then directly routed to the cabin The regenerative gas cycle is shown in Fig 1121 Regenerative cooling is achieved by inserting a counterflow heat exchanger into the cycle With out regeneration the lowest turbine inlet temperature is T0 the temperature of the surroundings or any other cooling medium With regeneration the FIGURE 1118 Simple gas refrigeration cycle 2 3 2 1 T s 4 QH QL Compressor Wnetin Heat exchanger Heat exchanger Cold refrigerated space QL Turbine Warm environment QH 1 3 4 FIGURE 1120 An opencycle aircraft cooling system Compressor Warm air in Cool air out Wnetin Turbine Q Heat exchanger 3 4 2 1 FIGURE 1119 A reversed Carnot cycle produces more refrigeration area under B1 with less work input area 1A3B 3 2 1 T s 4 Gas refrigeration cycle A B Reversed Carnot cycle Final PDF to printer 621 CHAPTER 11 cen22672ch11597642indd 621 110917 1148 AM FIGURE 1122 Ts diagram of the ideal gas refrigera tion cycle described in Example 116 3 2 1 T F s 4 QH QL Tmax Tmin 80 0 FIGURE 1121 Gas refrigeration cycle with regeneration 4 2 1 T s 5 QH QL 3 6 Q Heat exchanger 1 4 Wnetin 2 6 Regenerator Cold refrigerated space Heat exchanger Warm environment Compressor Turbine 3 5 QL QH EXAMPLE 116 The Simple Ideal Gas Refrigeration Cycle An ideal gas refrigeration cycle using air as the working medium is to maintain a refrigerated space at 0F while rejecting heat to the surrounding medium at 80F The pressure ratio of the compressor is 4 Determine a the maximum and minimum temperatures in the cycle b the coefficient of performance and c the rate of refrig eration for a mass flow rate of 01 lbms SOLUTION An ideal gas refrigeration cycle using air as the working fluid is con sidered The maximum and minimum temperatures the COP and the rate of refrigera tion are to be determined Assumptions 1 Steady operating conditions exist 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energy changes are negligible Analysis The Ts diagram of the gas refrigeration cycle is shown in Fig 1122 We note that this is an ideal gascompression refrigeration cycle and thus both the compres sor and the turbine are isentropic and the air is cooled to the environment temperature before it enters the turbine a The maximum and minimum temperatures in the cycle are determined from the isentropic relations of ideal gases for the compression and expansion processes From Table A17E T 1 460 R h 1 10990 Btulbm and P r1 07913 high pressure gas is further cooled to T4 before expanding in the turbine Lowering the turbine inlet temperature automatically lowers the turbine exit temperature which is the minimum temperature in the cycle Extremely low temperatures can be achieved by repeating this process Final PDF to printer 622 REFRIGERATION CYCLES cen22672ch11597642indd 622 110917 1148 AM P r2 P 2 P 1 P r1 4 07913 3165 h 2 1635 Btulbm T 2 683 R or 223F T 3 540 R h 3 12906 Btulbm and P r3 13860 P r4 P 4 P 3 P r3 025 1386 03456 h 4 867 Btulbm T 4 363 R or 97F Therefore the highest and the lowest temperatures in the cycle are 223 and 97F respectively b The COP of this ideal gas refrigeration cycle is COP R q L w netin q L w compin w turbout where q L h 1 h 4 1099 867 232 Btulbm w turbout h 3 h 4 12906 867 4236 Btulbm w compin h 2 h 1 1635 1099 536 Btulbm Thus COP R 232 536 4236 206 c The rate of refrigeration is Q refrig m q L 01 lbms 232 Btulbm 232 Btus Discussion It is worth noting that an ideal vaporcompression cycle working under similar conditions would have a COP greater than 3 1110 ABSORPTION REFRIGERATION SYSTEMS Another form of refrigeration that becomes economically attractive when there is a source of inexpensive thermal energy at a temperature of 100 to 200C is absorption refrigeration Some examples of inexpensive thermal energy sources include geothermal energy solar energy and waste heat from cogeneration or process steam plants and even natural gas when it is available at a relatively low price As the name implies absorption refrigeration systems involve the absorp tion of a refrigerant by a transport medium The most widely used absorption refrigeration system is the ammoniawater system where ammonia NH3 serves as the refrigerant and water H2O as the transport medium Other absorption refrigeration systems include waterlithium bromide and water lithium chloride systems where water serves as the refrigerant The latter two systems are limited to applications such as airconditioning where the mini mum temperature is above the freezing point of water To understand the basic principles involved in absorption refrigeration we examine the NH3H2O system shown in Fig 1123 The ammoniawater refrigeration machine was patented by the Frenchman Ferdinand Carre in Final PDF to printer 623 CHAPTER 11 cen22672ch11597642indd 623 110917 1148 AM 1859 Within a few years the machines based on this principle were being built in the United States primarily to make ice and store food You will immediately notice from the figure that this system looks very much like the vaporcompression system except that the compressor has been replaced by a complex absorption mechanism consisting of an absorber a pump a genera tor a regenerator a valve and a rectifier Once the pressure of NH3 is raised by the components in the box this is the only thing they are set up to do it is cooled and condensed in the condenser by rejecting heat to the surroundings is throttled to the evaporator pressure and absorbs heat from the refrigerated space as it flows through the evaporator So there is nothing new there Here is what happens in the box Ammonia vapor leaves the evaporator and enters the absorber where it dissolves and reacts with water to form NH3 H2O This is an exothermic reaction thus heat is released during this process The amount of NH3 that can be dissolved in H2O is inversely proportional to the tem perature Therefore it is necessary to cool the absorber to keep its temperature as low as possible and thus to maximize the amount of NH3 dissolved in water The liquid NH3 H2O solution which is rich in NH3 is then pumped to the generator Heat is transferred to the solution from a source to vaporize some of the solu tion The vapor which is rich in NH3 passes through a rectifier which separates the water and returns it to the generator The highpressure pure NH3 vapor then continues its journey through the rest of the cycle The hot NH3 H2O solution which is weak in NH3 then passes through a regenerator where it transfers some heat to the rich solution leaving the pump and is throttled to the absorber pressure Compared with vaporcompression systems absorption refrigeration sys tems have one major advantage A liquid is compressed instead of a vapor FIGURE 1123 Ammonia absorption refrigeration cycle Cold refrigerated space Q Wpump Qcool Expansion valve Expansion valve Pump Cooling water Evaporator Pure NH3 Pure NH3 Rectifier H2O Solar energy NH3 H2O Absorber Regenerator Condenser QL Generator NH3 H2O QH Warm environment Qgen Final PDF to printer 624 REFRIGERATION CYCLES cen22672ch11597642indd 624 110917 1148 AM The steadyflow work is proportional to the specific volume and thus the work input for absorption refrigeration systems is very small on the order of 1 percent of the heat supplied to the generator and often neglected in the cycle analysis The operation of these systems is based on heat transfer from an external source Therefore absorption refrigeration systems are often clas sified as heatdriven systems The absorption refrigeration systems are much more expensive than the vaporcompression refrigeration systems They are more complex and occupy more space they are much less efficient thus requiring much larger cooling towers to reject the waste heat and they are more difficult to service since they are less common Therefore absorption refrigeration systems should be considered only when the unit cost of thermal energy is low and is projected to remain low relative to electricity Absorption refrigeration systems are pri marily used in large commercial and industrial installations The COP of absorption refrigeration systems is defined as COP absorption Desired output Required input Q L Q gen W pump Q L Q gen 1127 The maximum COP of an absorption refrigeration system is determined by assuming that the entire cycle is totally reversible ie the cycle involves no irreversibilities and any heat transfer is through a differential temperature difference The refrigeration system would be reversible if the heat from the source Qgen were transferred to a Carnot heat engine and the work output of this heat engine W ηthrevQgen is supplied to a Carnot refrigerator to remove heat from the refrigerated space Note that QL W COPRrev ηthrevQgen COPRrev Then the overall COP of an absorption refrigeration system under reversible conditions becomes Fig 1124 COP revabsorption Q L Q gen η threv COP Rrev 1 T 0 T s T L T 0 T L 1128 where TL T0 and Ts are the thermodynamic temperatures of the refrigerated space the environment and the heat source respectively Any absorption refrigeration system that receives heat from a source at Ts and removes heat from the refrigerated space at TL while operating in an environment at T0 has a lower COP than the one determined from Eq 1128 For example when the source is at 120C the refrigerated space is at 10C and the environment is at 25C the maximum COP that an absorption refrigeration system can have is 18 The COP of actual absorption refrigeration systems is usually less than 1 Airconditioning systems based on absorption refrigeration called absorp tion chillers perform best when the heat source can supply heat at a high temperature with little temperature drop The absorption chillers are typically rated at an input temperature of 116C 240F The chillers perform at lower temperatures but their cooling capacity decreases sharply with decreasing source temperature about 125 percent for each 6C 10F drop in the source temperature For example the capacity goes down to 50 percent when the supply water temperature drops to 93C 200F In that case one needs to double the size and thus the cost of the chiller to achieve the same cooling The COP of the chiller is affected less by the decline of the source tempera ture The COP drops by 25 percent for each 6C 10F drop in the source FIGURE 1124 Determining the maximum COP of an absorption refrigeration system T0 Environment Reversible heat engine Qgen W threv Qgen QL COPRrev W TL Refrigerated space Reversible refrigerator W th rev Qgen 1 Qgen QL COPRrevW W COPrevabsorption 1 T0 Ts TL T0 TL Qgen QL T0 Ts TL T0 TL Source Ts Environment T0 Final PDF to printer 625 CHAPTER 11 cen22672ch11597642indd 625 110917 1148 AM temperature The nominal COP of singlestage absorption chillers at 116C 240F is 065 to 070 Therefore for each ton of refrigeration a heat input of 12000 Btuh065 18460 Btuh is required At 88C 190F the COP drops by 125 percent and thus the heat input increases by 125 percent for the same cooling effect Therefore the economic aspects must be evaluated carefully before any absorption refrigeration system is considered especially when the source temperature is below 93C 200F Another absorption refrigeration system that is quite popular with campers is a propanefired system invented by two Swedish undergraduate students In this system the pump is replaced by a third fluid hydrogen which makes it a truly portable unit EXAMPLE 117 A Reversible Absorption Refrigerator A reversible absorption refrigerator consists of a reversible heat engine and a revers ible refrigerator Fig 1125 The system removes heat from a cooled space at 15C at a rate of 70 kW The refrigerator operates in an environment at 25C If the heat is supplied to the cycle by condensing saturated steam at 150C determine a the rate at which the steam condenses and b the power input to the reversible refrigerator c If the COP of an actual absorption chiller at the same temperature limits has a COP of 08 determine the secondlaw efficiency of this chiller SOLUTION A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator The rate at which the steam condenses the power input to the reversible refrigerator and the secondlaw efficiency of an actual chiller are to be determined Properties The enthalpy of vaporization of water at 150C is h fg 21138 kJ kg Table A4 Analysis a The thermal efficiency of the reversible heat engine is η threv 1 T 0 T s 1 25 27315 K 150 27315 K 02954 The COP of the reversible refrigerator is COP Rrev T L T 0 T L 15 27315 K 25 27315 K 15 27315 K 6454 The COP of the reversible absorption refrigerator is COP absrev η threv COP Rrev 02954 6454 1906 The heat input to the reversible heat engine is Q in Q L COP absrev 70 kW 1906 3672 kW Then the rate at which the steam condenses becomes m s Q in h fg 3672 kJ s 21138 kJ kg 00174 kgs b The power input to the refrigerator is equal to the power output from the heat engine W inR W outHE η threv Q in 02954 3672 kW 109 kW FIGURE 1125 Schematic for Example 117 T0 25C Environment Rev HE TL 15C Refrigerated space Rev Ref 70 kW Source Ts 150C Environment T0 25C Final PDF to printer 626 REFRIGERATION CYCLES cen22672ch11597642indd 626 110917 1148 AM c The secondlaw efficiency of an actual absorption chiller with a COP value of 08 is η II COP actual COP absrev 08 1906 0420 or 420 Discussion The COPs of absorption refrigeration systems are usually lower and their secondlaw efficiencies are usually higher compared with vaporcompression refrigera tion systems See the values in Examples 113 and 117 All the refrigeration systems discussed previously involve many moving parts and bulky complex components Then this question comes to mind Is it really necessary for a refrigeration system to be so complex Can we not achieve the same effect in a more direct way The answer to this question is yes It is pos sible to use electric energy more directly to produce cooling without involving any refrigerants and moving parts Next we discuss one such system called a thermoelectric refrigerator Consider two wires made from different metals joined at both ends junc tions forming a closed circuit Ordinarily nothing will happen However when one of the ends is heated something interesting happens A current flows continuously in the circuit as shown in Fig 1126 This is called the Seebeck effect in honor of Thomas Seebeck who made this discovery in 1821 The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit and a device that operates on this circuit is called a thermoelectric device The Seebeck effect has two major applications temperature measurement and power generation When the thermoelectric circuit is broken as shown in Fig 1127 the current ceases to flow and we can measure the driving force the electromotive force or the voltage generated in the circuit with a voltme ter The voltage generated is a function of the temperature difference and the TOPIC OF SPECIAL INTEREST Thermoelectric Power Generation and Refrigeration Systems This section can be skipped without a loss in continuity FIGURE 1127 When a thermoelectric circuit is broken a potential difference is generated Metal B Metal A I 0 V FIGURE 1126 When one of the junctions of two dissimilar metals is heated a current I flows through the closed circuit Metal B I I Metal A FIGURE 1128 Schematic of a simple thermoelectric power generator Wnet Lowtemperature sink TL QH QL I I Hot junction Cold junction Hightemperature source TH Final PDF to printer 627 CHAPTER 11 cen22672ch11597642indd 627 110917 1148 AM materials of the two wires used Therefore temperature can be measured by simply measuring voltages The two wires used to measure the temperature in this manner form a thermocouple which is the most versatile and most widely used temperature measurement device A common Ttype thermocouple for example consists of copper and constantan wires and it produces about 40 μV per C difference The Seebeck effect also forms the basis for thermoelectric power genera tion The schematic diagram of a thermoelectric generator is shown in Fig 1128 Heat is transferred from a hightemperature source to the hot junc tion in the amount of QH and it is rejected to a lowtemperature sink from the cold junction in the amount of QL The difference between these two quantities is the net electrical work produced that is We QH QL It is evident from Fig 1128 that the thermoelectric power cycle closely resembles an ordinary heat engine cycle with electrons serving as the working fluid Therefore the thermal efficiency of a thermoelectric generator operating between the tem perature limits of TH and TL is limited by the efficiency of a Carnot cycle operating between the same temperature limits Thus in the absence of any irreversibilities such as I2R heating where R is the total electrical resistance of the wires the thermoelectric generator will have the Carnot efficiency The major drawback of thermoelectric generators is their low efficiency The future success of these devices depends on finding materials with more desir able characteristics For example the voltage output of thermoelectric devices has been increased several times by switching from metal pairs to semicon ductors A practical thermoelectric generator using ntype heavily doped to create excess electrons and ptype heavily doped to create a deficiency of electrons materials connected in series is shown in Fig 1129 Despite their low efficiencies thermoelectric generators have definite weight and reliability advantages and are currently used in rural areas and in space applications For example silicongermaniumbased thermoelectric generators have been pow ering Voyager spacecraft since 1980 and are expected to continue generating power for many more years If Seebeck had been fluent in thermodynamics he would probably have tried reversing the direction of flow of electrons in the thermoelectric circuit by externally applying a potential difference in the reverse direction to create a refrigeration effect But this honor belongs to Jean Charles Athanase Peltier who discovered this phenomenon in 1834 He noticed during his experiments that when a small current was passed through the junction of two dissimi lar wires the junction was cooled as shown in Fig 1130 This is called the Peltier effect and it forms the basis for thermoelectric refrigeration A practical thermoelectric refrigeration circuit using semiconductor materials is shown in Fig 1131 Heat is absorbed from the refrigerated space in the amount of QL and rejected to the warmer environment in the amount of QH The difference between these two quantities is the net electrical work that needs to be supplied that is We QH QL At present thermoelectric refrig erators cannot compete with vaporcompression refrigeration systems because of their low coefficient of performance They are available in the market how ever and are preferred in some applications because of their small size sim plicity quietness and reliability FIGURE 1129 A thermoelectric power generator Source Sink Hot plate Cold plate QH QL I p n p n p n Wnet FIGURE 1131 A thermoelectric refrigerator I Warm environment Refrigerated space Hot plate Cold plate QH QL p n p n p n FIGURE 1130 When a current is passed through the junction of two dissimilar materials the junction is cooled Heat rejected Heat absorbed Final PDF to printer cen22672ch11597642indd 628 110917 1148 AM 628 REFRIGERATION CYCLES SUMMARY The transfer of heat from lowertemperature regions to higher temperature ones is called refrigeration Devices that produce refrigeration are called refrigerators and the cycles on which they operate are called refrigeration cycles The working flu ids used in refrigerators are called refrigerants Refrigerators used for the purpose of heating a space by transferring heat from a cooler medium are called heat pumps The performance of refrigerators and heat pumps is expressed in terms of coefficient of performance COP defined as COP R Desired output Required output Cooling effect Work input Q L W netin COP HP Desired output Required output Heating effect Work input Q H W netin The standard of comparison for refrigeration cycles is the reversed Carnot cycle A refrigerator or heat pump that oper ates on the reversed Carnot cycle is called a Carnot refrigera tor or a Carnot heat pump and their COPs are COP RCarnot 1 T H T L 1 COP HPCarnot 1 1 T L T H The most widely used refrigeration cycle is the vapor compression refrigeration cycle In an ideal vapor compression refrigeration cycle the refrigerant enters the compressor as a saturated vapor and is cooled to the saturated liquid state in the condenser It is then throttled to the evaporator pressure and vaporizes as it absorbs heat from the refrigerated space Very low temperatures can be achieved by operating two or more vaporcompression systems in series called cascading The COP of a refrigeration system also increases as a result of cascading Another way of improving the performance of a vaporcompression refrigeration system is by using multistage compression with regenerative cooling A refrigerator with a single compressor can provide refrigeration at several tem peratures by throttling the refrigerant in stages The vapor compression refrigeration cycle can also be used to liquefy gases after some modifications The power cycles can be used as refrigeration cycles by simply reversing them Of these the reversed Brayton cycle which is also known as the gas refrigeration cycle is used to cool aircraft and to obtain very low cryogenic temperatures after it is modified with regeneration The work output of the turbine can be used to reduce the work input requirements to the compressor Thus the COP of a gas refrigeration cycle is COP R q L w netin q L w compin w turbout Another form of refrigeration that becomes economically attractive when there is a source of inexpensive thermal energy at a temperature of 100 to 200C is absorption refrigeration where the refrigerant is absorbed by a transport medium and compressed in liquid form The most widely used absorption refrigeration system is the ammoniawater system where ammonia serves as the refrigerant and water as the transport medium The work input to the pump is usually very small and the COP of absorption refrigeration systems is defined as COP absorption Desired output Required output Q L Q gen W pump Q L Q gen The maximum COP an absorption refrigeration system can have is determined by assuming totally reversible conditions which yields COP revabsorption η threv COP Rrev 1 T 0 T s T L T 0 T L where T0 TL and Ts are the thermodynamic temperatures of the environment the refrigerated space and the heat source respectively REFERENCES AND SUGGESTED READINGS 1 ASHRAE Handbook of Fundamentals Atlanta GA American Society of Heating Refrigerating and Air Conditioning Engineers 1985 2 Heat Pump SystemsA Technology Review OECD Report Paris 1982 3 B Nagengast A Historical Look at CFC Refrigerants ASHRAE Journal Vol 30 No 11 November 1988 pp 3739 4 W F Stoecker Growing Opportunities for Ammonia Refrigeration Proceedings of the Meeting of the International Institute of Ammonia Refrigeration Austin Texas 1989 5 W F Stoecker and J W Jones Refrigeration and Air Conditioning 2nd ed New York McGrawHill 1982 Final PDF to printer cen22672ch11597642indd 629 110917 1148 AM 629 CHAPTER 11 PROBLEMS Problems designated by a C are concept questions and stu dents are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software The Reversed Carnot Cycle 111C Why do we study the reversed Carnot cycle even though it is not a realistic model for refrigeration cycles 112C Why is the reversed Carnot cycle executed within the saturation dome not a realistic model for refrigeration cycles 113 A steadyflow Carnot refrigeration cycle uses refrigerant134a as the working fluid The refrigerant changes from saturated vapor to saturated liquid at 60C in the con denser as it rejects heat The evaporator pressure is 180 kPa Show the cycle on a Ts diagram relative to saturation lines and determine a the coefficient of performance b the amount of heat absorbed from the refrigerated space and c the net work input Answers a 358 b 109 kJkg c 304 kJkg 114E Refrigerant134a enters the condenser of a steady flow Carnot refrigerator as a saturated vapor at 90 psia and it leaves with a quality of 005 The heat absorption from the refrigerated space takes place at a pressure of 30 psia Show the cycle on a Ts diagram relative to saturation lines and determine a the coefficient of performance b the quality at the beginning of the heatabsorption process and c the net work input Ideal and Actual VaporCompression Refrigeration Cycles 115C Does the ideal vaporcompression refrigeration cycle involve any internal irreversibilities 116C Why is the throttling valve not replaced by an isentro pic turbine in the ideal vaporcompression refrigeration cycle 117C In a refrigeration system would you recommend con densing the refrigerant134a at a pressure of 07 or 10 MPa if heat is to be rejected to a cooling medium at 15C Why 118C Does the area enclosed by the cycle on a Ts diagram represent the net work input for the reversed Carnot cycle How about for the ideal vaporcompression refrigeration cycle 119C Consider two vaporcompression refrigeration cycles The refrigerant enters the throttling valve as a saturated liquid at 30C in one cycle and as subcooled liquid at 30C in the other one The evaporator pressure for both cycles is the same Which cycle do you think will have a higher COP 1110C It is proposed to use water instead of refrigerant 134a as the working fluid in airconditioning applications where the minimum temperature never falls below the freezing point Would you support this proposal Explain 1111C The COP of vaporcompression refrigeration cycles improves when the refrigerant is subcooled before it enters the throttling valve Can the refrigerant be subcooled indefinitely to maximize this effect or is there a lower limit Explain 1112 A 10kW cooling load is to be served by operating an ideal vaporcompression refrigeration cycle with its evaporator at 400 kPa and its condenser at 800 kPa Calculate the refrig erant mass flow rate and the compressor power requirement when refrigerant134a is used 1113E An icemaking machine operates on the ideal vapor compression cycle using refrigerant134a The refrigerant enters the compressor as saturated vapor at 20 psia and leaves the condenser as saturated liquid at 80 psia Water enters the ice machine at 55F and leaves as ice at 25F For an ice pro duction rate of 15 lbmh determine the power input to the ice machine 169 Btu of heat needs to be removed from each l bm of water at 55F to turn it into ice at 25F 1114 An air conditioner using refrigerant134a as the work ing fluid and operating on the ideal vaporcompression refrig eration cycle is to maintain a space at 22C while operating its condenser at 1000 kPa Determine the COP of the system when a temperature difference of 2C is allowed for the trans fer of heat in the evaporator 1115 An ideal vaporcompression refrigeration cycle using refrigerant134a as the working fluid is used to cool a brine solution to 5C This solution is pumped to various buildings for the purpose of airconditioning The refrigerant evaporates at 10C with a total mass flow rate of 7 kgs and condenses at 600 kPa Determine the COP of the cycle and the total cool ing load 1116E A refrigerator operates on the ideal vapor compression refrigeration cycle and uses refrigerant134a as the working fluid The condenser operates at 300 psia and the evaporator at 20F If an adiabatic reversible expansion device were available and used to expand the liquid leaving the con denser how much would the COP improve by using this device instead of the throttle device Answer 165 percent 1117 A refrigerator uses refrigerant134a as the working fluid and operates on the ideal vaporcompression refrigera tion cycle except for the compression process The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70C If the compressor con sumes 450 W of power determine a the mass flow rate of the refrigerant b the condenser pressure and c the COP of the refrigerator Answers a 000644 kgs b 800 kPa c 203 Final PDF to printer cen22672ch11597642indd 630 110917 1148 AM 630 REFRIGERATION CYCLES FIGURE P1122 12 MPa 65C 60 kPa 34C Water 18C 26C 42C 2 3 1 4 Compressor Condenser Win Expansion valve QL Evaporator Qin 1118 An ideal vaporcompression refrigeration cycle that uses refrigerant134a as its working fluid maintains a con denser at 800 kPa and the evaporator at 20C Determine this systems COP and the amount of power required to service a 150 kW cooling load Answers 383 392 kW 180 psia This unit serves a 45000 Btuh cooling load Deter mine the mass flow rate of the refrigerant and the power that this unit will require 1120E Repeat Prob 1119E using appropriate software if ammonia is used in place of refrigerant134a 1121 A refrigerator uses refrigerant134a as the working fluid and operates on the vaporcompression refrigeration cycle The evaporator and condenser pressures are 200 kPa and 1400 kPa respectively The isentropic efficiency of the com pressor is 88 percent The refrigerant enters the compressor at a rate of 0025 kgs superheated by 101C and leaves the condenser subcooled by 44C Determine a the rate of cool ing provided by the evaporator the power input and the COP Determine b the same parameters if the cycle operated on the ideal vaporcompression refrigeration cycle between the same pressure limits 1122 A commercial refrigerator with refrigerant134a as the working fluid is used to keep the refrigerated space at 30C by rejecting its waste heat to cooling water that enters the con denser at 18C at a rate of 025 kgs and leaves at 26C The refrigerant enters the condenser at 12 MPa and 65C and leaves at 42C The inlet state of the compressor is 60 kPa and 34C and the compressor is estimated to gain a net heat of 450 W from the surroundings Determine a the quality of the refrig erant at the evaporator inlet b the refrigeration load c the COP of the refrigerator and d the theoretical maximum refrigeration load for the same power input to the compressor FIGURE P1117 120 kPa x 034 QL Cold environment 70C Warm environment 2 3 1 4 Compressor Condenser Win Expansion valve Evaporator QH FIGURE P1118 QH Win QL Compressor Condenser Cold refrigerated space Expansion valve Evaporator 4 3 1 2 Warm environment 1119E A refrigerator uses refrigerant134a as its working fluid and operates on the ideal vaporcompression refrigera tion cycle The refrigerant evaporates at 5F and condenses at 1123 The manufacturer of an air conditioner claims a seasonal energy efficiency ratio SEER of 16 BtuhW for one of its units This unit operates on the normal vapor compression refrigeration cycle and uses refrigerant22 as the working fluid This SEER is for the operating conditions when the evaporator saturation temperature is 5C and the Final PDF to printer cen22672ch11597642indd 631 110917 1148 AM 631 CHAPTER 11 condenser saturation temperature is 45C Selected data for refrigerant22 are provided in the following table T C Psat kPa hf kJkg hg kJkg sg kJkgK 5 4212 3876 2481 09344 45 1728 101 2619 08682 a Sketch the hardware and the Ts diagram for this air conditioner b Determine the heat absorbed by the refrigerant in the evaporator per unit mass of refrigerant22 in kJkg c Determine the work input to the compressor and the heat rejected in the condenser per unit mass of refrigerant22 in kJkg 1124 An actual refrigerator operates on the vapor compression refrigeration cycle with refrigerant22 as the working fluid The refrigerant evaporates at 15C and con denses at 40C The isentropic efficiency of the compres sor is 83 percent The refrigerant is superheated by 5C at the compressor inlet and subcooled by 5C at the exit of the condenser Determine a the heat removed from the cooled space and the work input in kJkg and the COP of the cycle Determine b the same parameters if the cycle operated on the ideal vaporcompression refrigeration cycle between the same evaporating and condensing temperatures The properties of R22 in the case of actual operation are h1 40249 kJkg h2 45400 kJkg h3 24319 kJkg The proper ties of R22 in the case of ideal operation are h1 39904 kJkg h2 44071 kJkg h3 24980 kJkg Note state 1 compres sor inlet state 2 compressor exit state 3 condenser exit state 4 evaporator inlet SecondLaw Analysis of VaporCompression Refrigeration Cycle 1125C How is the secondlaw efficiency of a refrigerator operating on the vaporcompression refrigeration cycle defined Provide two alternative definitions and explain each term 1126C How is the secondlaw efficiency of a heat pump operating on the vaporcompression refrigeration cycle defined Provide two alternative definitions and show that one can be derived from the other 1127C Consider an isentropic compressor in a vapor compression refrigeration cycle What are the isentropic effi ciency and secondlaw efficiency of this compressor Justify your answers Is the secondlaw efficiency of a compressor necessarily equal to its isentropic efficiency Explain 1128 A space is kept at 15C by a vaporcompression refrigeration system in an ambient at 25C The space gains heat steadily at a rate of 3500 kJh and the rate of heat rejec tion in the condenser is 5500 kJh Determine the power input in kW the COP of the cycle and the secondlaw efficiency of the system 1129 Bananas are to be cooled from 28C to 12C at a rate of 1330 kgh by a refrigerator that operates on a vapor compression refrigeration cycle The power input to the refrig erator is 86 kW Determine a the rate of heat absorbed from the bananas in kJh and the COP b the minimum power input to the refrigerator and c the secondlaw efficiency and the exergy destruction for the cycle The specific heat of bananas above freezing is 335 kJkgC Answers a 71300 kJh 230 b 0541 kW c 63 percent 806 kW 1130 A vaporcompression refrigeration system absorbs heat from a space at 0C at a rate of 24000 Btuh and rejects heat to water in the condenser The water experiences a temperature rise of 12C in the condenser The COP of the system is esti mated to be 205 Determine a the power input to the system in kW b the mass flow rate of water through the condenser and c the secondlaw efficiency and the exergy destruction for the refrigerator Take T0 20C and cpwater 418 kJkgC 1131 A room is kept at 5C by a vaporcompression refrig eration cycle with R134a as the refrigerant Heat is rejected to cooling water that enters the condenser at 20C at a rate of 013 kgs and leaves at 28C The refrigerant enters the con denser at 12 MPa and 50C and leaves as a saturated liquid If the compressor consumes 19 kW of power determine a the refrigeration load in Btuh and the COP b the secondlaw efficiency of the refrigerator and the total exergy destruction in the cycle and c the exergy destruction in the condenser Take T0 20C and cpwater 418 kJkgC Answers a 8350 Btuh 129 b 120 percent 167 kW c 0303 kW FIGURE P1131 12 MPa 50C 12 MPa sat liquid Water 20C 013 kgs 28C Compressor Condenser Win Expansion valve QL Evaporator 1132 A refrigerator operates on the ideal vapor compression refrigeration cycle with refrigerant134a as the working fluid The refrigerant evaporates at 10C and condenses at 579C Final PDF to printer cen22672ch11597642indd 632 110917 1148 AM 632 REFRIGERATION CYCLES The refrigerant absorbs heat from a space at 5C and rejects heat to ambient air at 25C Determine a the cooling load in kJkg and the COP b the exergy destruction in each compo nent of the cycle and the total exergy destruction in the cycle and c the secondlaw efficiency of the compressor the evap orator and the cycle 1133E A refrigerator operating on the vaporcompression refrigeration cycle using refrigerant134a as the refrigerant is considered The temperatures of the cooled space and the ambient air are at 10F and 80F respectively R134a enters the compressor at 20 psia as a saturated vapor and leaves at 140 psia and 160F The refrigerant leaves the condenser as a saturated liquid The rate of cooling provided by the system is 45000 Btuh Determine a the mass flow rate of R134a and the COP b the exergy destruction in each component of the cycle and the secondlaw efficiency of the compressor and c the secondlaw efficiency of the cycle and the total exergy destruction in the cycle Selecting the Right Refrigerant 1134C When selecting a refrigerant for a certain applica tion what qualities would you look for in the refrigerant 1135C A refrigerant134a refrigerator is to maintain the refrigerated space at 10C Would you recommend an evapo rator pressure of 012 or 014 MPa for this system Why 1136C Consider a refrigeration system using refrigerant 134a as the working fluid If this refrigerator is to operate in an environment at 30C what is the minimum pressure to which the refrigerant should be compressed Why 1137 A refrigerator that operates on the ideal vapor compression cycle with refrigerant134a is to maintain the refrigerated space at 10C while rejecting heat to the envi ronment at 25C Select reasonable pressures for the evapora tor and the condenser and explain why you chose those values 1138 A heat pump that operates on the ideal vapor compression cycle with refrigerant134a is used to heat a house and maintain it at 26C by using underground water at 14C as the heat source Select reasonable pressures for the evaporator and the condenser and explain why you chose those values Heat Pump Systems 1139C Do you think a heat pump system will be more costeffective in New York or in Miami Why 1140C What is a watersource heat pump How does the COP of a watersource heat pump system compare to that of an airsource system 1141 A heat pump operates on the ideal vaporcompression refrigeration cycle and uses refrigerant134a as the working fluid The condenser operates at 1000 kPa and the evaporator at 200 kPa Determine this systems COP and the rate of heat sup plied to the evaporator when the compressor consumes 6 kW 1142 Refrigerant134a enters the condenser of a residential heat pump at 800 kPa and 50C at a rate of 0022 kgs and leaves at 750 kPa subcooled by 3C The refrigerant enters the com pressor at 200 kPa superheated by 4C Determine a the isen tropic efficiency of the compressor b the rate of heat supplied to the heated room and c the COP of the heat pump Also determine d the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor compression cycle between the pressure limits of 200 and 800 kPa FIGURE P1142 750 kPa 800 kPa 50C 2 3 1 4 Compressor Condenser Win Expansion valve Evaporator QH QL 1143E A heat pump that operates on the ideal vapor compression cycle with refrigerant134a is used to heat a house and maintain it at 75F by using underground water at 50F as the heat source The house is losing heat at a rate of 80000 Btuh The evaporator and condenser pressures are 50 and 120 psia respectively Determine the power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater Answers 328 hp 282 hp 1144E The liquid leaving the condenser of a 100000 Btuh heat pump using refrigerant134a as the working fluid is subcooled by 95F The condenser operates at 160 psia and the evaporator at 50 psia How does this subcooling change the power required to drive the compressor as compared to an ideal vaporcompression refrigeration cycle Answers 411 kW 431 kW 1145E Reconsider Prob 1144E What is the effect on the compressor power requirement when the vapor entering the compressor is superheated by 10F and the condenser operates ideally 1146 A heat pump using refrigerant134a heats a house by using underground water at 8C as the heat source The house is losing heat at a rate of 60000 kJh The refrigerant enters the compressor at 280 kPa and 0C and it leaves at 1 MPa and 60C The refrigerant exits the condenser at 30C Determine a the power input to the heat pump b the rate of Final PDF to printer cen22672ch11597642indd 633 110917 1148 AM 633 CHAPTER 11 heat absorption from the water and c the increase in electric power input if an electric resistance heater is used instead of a heat pump Answers a 355 kW b 1312 kW c 1312 kW 1147 Reconsider Prob 1146 Using appropriate software investigate the effect of varying the compressor isentropic efficiency over the range 60 to 100 per cent Plot the power input to the compressor and the electric power saved by using a heat pump rather than electric resis tance heating as functions of compressor efficiency and dis cuss the results 1148 A heat pump using refrigerant134a as a refrigerant operates its condenser at 800 kPa and its evaporator at 125C It operates on the ideal vaporcompression refrigeration cycle except for the compressor which has an isentropic efficiency of 85 percent How much do the compressor irreversibilities reduce this heat pumps COP as compared to an ideal vapor compression refrigeration cycle Answer 131 percent 1149 Reconsider Prob 1148 What is the effect on the COP when the vapor entering the compressor is superheated by 2C and the compressor has no irreversibilities Innovative Refrigeration Systems 1150C What is cascade refrigeration What are the advan tages and disadvantages of cascade refrigeration 1151C How does the COP of a cascade refrigeration sys tem compare to the COP of a simple vaporcompression cycle operating between the same pressure limits 1152C Consider a twostage cascade refrigeration cycle and a twostage compression refrigeration cycle with a flash cham ber Both cycles operate between the same pressure limits and use the same refrigerant Which system would you favor Why 1153C Can a vaporcompression refrigeration system with a single compressor handle several evaporators operating at different pressures How 1154C In the liquefaction process why are gases com pressed to very high pressures 1155C A certain application requires maintaining the refrigerated space at 32C Would you recommend a simple refrigeration cycle with refrigerant134a or a twostage cas cade refrigeration cycle with a different refrigerant at the bot toming cycle Why 1156 A twostage compression refrigeration system oper ates with refrigerant134a between the pressure limits of 14 and 010 MPa The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 04 MPa The refrigerant leaving the lowpressure compressor at 04 MPa is also routed to the flash chamber The vapor in the flash chamber is then compressed to the condenser pressure by the highpressure compressor and the liquid is throttled to the evaporator pressure Assuming the refrigerant leaves the evap orator as saturated vapor and both compressors are isentropic determine a the fraction of the refrigerant that evaporates as it is throttled to the flash chamber b the rate of heat removed from the refrigerated space for a mass flow rate of 025 kgs through the condenser and c the coefficient of performance 1157 Repeat Prob 1156 for a flash chamber pressure of 06 MPa 1158 Reconsider Prob 1156 Using appropriate software investigate the effect of the various refrigerants for compressor efficiencies of 80 90 and 100 per cent Compare the performance of the refrigeration system with different refrigerants 1159 Consider a twostage cascade refrigeration system operating between the pressure limits of 14 MPa and 160 kPa with refrigerant134a as the working fluid Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where the pressure in the upper and lower cycles are 04 and 05 MPa respectively In both cycles the refrigerant is a saturated liquid at the condenser exit and a saturated vapor at the compressor inlet and the isentropic effi ciency of the compressor is 80 percent If the mass flow rate of the refrigerant through the lower cycle is 025 kgs determine a the mass flow rate of the refrigerant through the upper cycle b the rate of heat removal from the refrigerated space and c the COP of this refrigerator Answers a 0384 kgs b 420 kW c 212 FIGURE P1159 Win Win 5 Condenser Evaporator Compressor Expansion valve 7 Compressor Expansion valve 3 2 6 Condenser Evaporator 4 1 Heat 8 QH QL Final PDF to printer cen22672ch11597642indd 634 110917 1148 AM 634 REFRIGERATION CYCLES 1160 A twostage compression refrigeration system with an adiabatic liquidvapor separation unit as shown in Fig P1160 uses refrigerant134a as the working fluid The system oper ates the evaporator at 40C the condenser at 800 kPa and the separator at 101C This system is to serve a 30kW cooling load Determine the mass flow rate through each of the two compressors the power used by the compressors and the systems COP The refrigerant is saturated liquid at the inlet of each expansion valve and saturated vapor at the inlet of each compressor and the compressors are isentropic Answers 0160 kgs 0230 kgs 109 kW 274 system The refrigerant is saturated liquid at the exit of the condenser and saturated vapor at the exit of each evaporator and the compressor is isentropic Answers 658 kW 451 kW 324 FIGURE P1162 Pressure reducing valve Evaporator 1 Evaporator 2 7 1 m2 m1 m1 m2 5 Condenser 2 3 4 6 FIGURE P1160 Condenser 2 Separator Evaporator 1 4 3 5 7 8 m2 m6 6 Compressor Compressor 1161E A twostage compression refrigeration system with an adiabatic liquidvapor separation unit like that in Fig P1160 uses refrigerant134a as the working fluid The system operates the evaporator at 60 psia the condenser at 300 psia and the separator at 120 psia The compressors use 25 kW of power Determine the rate of cooling provided by the evaporator and the COP of this cycle The refrigerant is satu rated liquid at the inlet of each expansion valve and saturated vapor at the inlet of each compressor and the compressors are isentropic 1162 A twoevaporator compression refrigeration system as shown in Fig P1162 uses refrigerant134a as the work ing fluid The system operates evaporator 1 at 0C evaporator 2 at 264C and the condenser at 800 kPa The refrigerant is circulated through the compressor at a rate of 01 kgs and the lowtemperature evaporator serves a cooling load of 8 kW Determine the cooling rate of the hightemperature evapora tor the power required by the compressor and the COP of the 1163E A twoevaporator compression refrigeration system like that in Fig P1162 uses refrigerant134a as the working fluid The system operates evaporator 1 at 30 psia evapora tor 2 at 10 psia and the condenser at 180 psia The cooling load for evaporator 1 is 9000 Btuh and that for evaporator 2 is 24000 Btuh Determine the power required to operate the compressor and the COP of this system The refrigerant is sat urated liquid at the exit of the condenser and saturated vapor at the exit of each evaporator and the compressor is isentropic 1164E Repeat Prob 1163E if the 30 psia evaporator is to be replaced with a 60 psia evaporator to serve a 15000 Btuh cooling load 1165 Consider a twostage cascade refrigeration cycle with a flash chamber as shown in the figure with refrigerant134a as the working fluid The evaporator temperature is 10C and the condenser pressure is 1600 kPa The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 045 MPa Part of the refrigerant evap orates during this flashing process and this vapor is mixed with the refrigerant leaving the lowpressure compressor The mixture is then compressed to the condenser pressure by the highpressure compressor The liquid in the flash chamber is throttled to the evaporator pressure and cools the refriger ated space as it vaporizes in the evaporator The mass flow Final PDF to printer cen22672ch11597642indd 635 110917 1148 AM 635 CHAPTER 11 rate of the refrigerant through the lowpressure compressor is 011 kgs Assuming the refrigerant leaves the evaporator as a saturated vapor and the isentropic efficiency is 86 percent for both compressors determine a the mass flow rate of the refrigerant through the highpressure compressor b the rate of refrigeration supplied by the system and c the COP of this refrigerator Also determine d the rate of refrigeration and the COP if this refrigerator operated on a singlestage vapor compression cycle between the same evaporating temperature and condenser pressure with the same compressor efficiency and the same flow rate as calculated in part a 1171C How do we achieve very low temperatures with gas refrigeration cycles 1172 An ideal gas refrigeration system operates with air as the working fluid Air is at 100 kPa and 20C before compres sion and 500 kPa and 30C before expansion The system is to provide 15 kW of cooling Calculate the rate at which air is circulated in this system as well as the rates of heat addition and rejection Use constant specific heats at room temperature 1173 Air enters the compressor of an ideal gas refrigeration cycle at 7C and 35 kPa and the turbine at 37C and 160 kPa The mass flow rate of air through the cycle is 02 kgs Assum ing variable specific heats for air determine a the rate of refrigeration b the net power input and c the coefficient of performance Answers a 159 kW b 864 kW c 184 1174 Repeat Prob 1173 for a compressor isentropic effi ciency of 80 percent and a turbine isentropic efficiency of 85 percent 1175 Reconsider Prob 1174 Using appropriate software study the effects of compressor and turbine isentropic efficiencies as they are varied from 70 to 100 percent on the rate of refrigeration the net power input and the COP Plot the Ts diagram of the cycle for the isentro pic case 1176E An ideal gas refrigeration cycle uses air as the work ing fluid The air is at 5 psia and 10F as it enters the com pressor with a compression ratio of 4 The temperature at the turbine entrance is 100F Determine this cycles COP Use constant specific heats at room temperature 1177E Rework Prob 1176E when the compressor isentro pic efficiency is 87 percent the turbine isentropic efficiency is 94 percent and the pressure drop across each heat exchanger is 1 psia Answer 0364 1178 A gas refrigeration cycle with a pressure ratio of 4 uses helium as the working fluid The temperature of the helium is 6C at the compressor inlet and 50C at the turbine inlet Assuming isentropic efficiencies of 88 percent for both the turbine and the compressor determine a the minimum temperature in the cycle b the coefficient of performance and c the mass flow rate of the helium for a refrigeration rate of 25 kW 1179 A gas refrigeration system using air as the working fluid has a pressure ratio of 5 Air enters the compressor at 0C The highpressure air is cooled to 35C by rejecting heat to the surroundings The refrigerant leaves the turbine at 80C and then it absorbs heat from the refrigerated space before entering the regenerator The mass flow rate of air is 04 kgs Assum ing isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature determine a the effectiveness of the regenerator b the rate of heat removal from the refrigerated space and c the COP of the cycle Also determine d the refrigeration load and the COP if this system operated on the FIGURE P1165 QL Condenser Highpressure compressor Expansion valve Expansion valve Evaporator Flash chamber Lowpressure compressor Win Win 2 9 4 3 8 1 7 6 5 QH Gas Refrigeration Cycle 1166C How does the ideal gas refrigeration cycle differ from the Carnot refrigeration cycle 1167C How does the ideal gas refrigeration cycle differ from the Brayton cycle 1168C Devise a refrigeration cycle that works on the reversed Stirling cycle Also determine the COP for this cycle 1169C How is the ideal gas refrigeration cycle modified for aircraft cooling 1170C In gas refrigeration cycles can we replace the tur bine with an expansion valve as we did in vaporcompression refrigeration cycles Why Final PDF to printer cen22672ch11597642indd 636 110917 1148 AM 636 REFRIGERATION CYCLES simple gas refrigeration cycle Use the same compressor inlet temperature as given the same turbine inlet temperature as calculated and the same compressor and turbine efficiencies Answers a 0434 b 214 kW c 0478 d 247 kW 0599 Absorption Refrigeration Systems 1182C What is absorption refrigeration How does an absorption refrigeration system differ from a vapor compression refrigeration system 1183C What are the advantages and disadvantages of absorption refrigeration 1184C Can water be used as a refrigerant in air conditioning applications Explain 1185C What are the functions of the rectifier and the regen erator in an absorption refrigeration system 1186C In absorption refrigeration cycles why is the fluid in the absorber cooled and the fluid in the generator heated 1187C How is the coefficient of performance of an absorp tion refrigeration system defined 1188E Heat is supplied to an absorption refrigeration sys tem from a geothermal well at 250F at a rate of 105 Btuh The environment is at 80F and the refrigerated space is maintained at 0F If the COP of the system is 055 determine the rate at which this system can remove heat from the refrigerated space 1189 An absorption refrigeration system that receives heat from a source at 95C and maintains the refrigerated space at 0C is claimed to have a COP of 31 If the environmental tem perature is 19C can this claim be valid Justify your answer 1190 An absorption refrigeration system receives heat from a source at 120C and maintains the refrigerated space at 4C If the temperature of the environment is 25C what is the maximum COP this absorption refrigeration system can have 1191 Heat is supplied to an absorption refrigeration system from a geothermal well at 110C at a rate of 5 105 kJh The environment is at 25C and the refrigerated space is main tained at 18C Determine the maximum rate at which this system can remove heat from the refrigerated space Answer 658 105 kJh Special Topic Thermoelectric Power Generation and Refrigeration Systems 1192C What is a thermoelectric circuit 1193C Describe the Seebeck and the Peltier effects 1194C Consider a circular copper wire formed by connect ing the two ends of a copper wire The connection point is now heated by a burning candle Do you expect any current to flow through the wire 1195C An iron wire and a constantan wire are formed into a closed circuit by connecting the ends Now both junctions are heated and are maintained at the same temperature Do you expect any electric current to flow through this circuit 1196C A copper wire and a constantan wire are formed into a closed circuit by connecting the ends Now one junction is heated by a burning candle while the other is maintained at FIGURE P1179 Heat exchanger 1 4 5 Compressor QH QL 2 3 6 Regenerator Heat exchanger Turbine 1180 An ideal gas refrigeration system with two stages of compression with intercooling as shown in Fig P1180 oper ates with air entering the first compressor at 90 kPa and 24C Each compression stage has a pressure ratio of 3 and the two intercoolers can cool the air to 5C Calculate the coef ficient of performance of this system and the rate at which air must be circulated through this system to service a 45000 kJh cooling load Use constant specific heats at room temperature Answers 156 0124 kgs FIGURE P1180 5 QH2 6 QL 1 3 QH1 4 2 Compressor Compressor Turbine 1181 Reconsider Prob 1180 How will the answers change when the isentropic efficiency of each compressor is 85 percent and the isentropic efficiency of the turbine is 95 percent Final PDF to printer cen22672ch11597642indd 637 110917 1148 AM 637 CHAPTER 11 room temperature Do you expect any electric current to flow through this circuit 1197C How does a thermocouple work as a temperature measurement device 1198C Why are semiconductor materials preferable to met als in thermoelectric refrigerators 1199C Is the efficiency of a thermoelectric generator lim ited by the Carnot efficiency Why 11100E A thermoelectric generator receives heat from a source at 340F and rejects the waste heat to the environment at 90F What is the maximum thermal efficiency this thermo electric generator can have Answer 313 percent 11101 A thermoelectric refrigerator removes heat from a refrigerated space at 5C at a rate of 130 W and rejects it to an environment at 20C Determine the maximum coefficient of performance this thermoelectric refrigerator can have and the minimum required power input Answers 107 121 W 11102 A thermoelectric cooler has a COP of 015 and removes heat from a refrigerated space at a rate of 180 W Determine the required power input to the thermoelectric cooler in W 11103E A thermoelectric cooler has a COP of 018 and the power input to the cooler is 18 hp Determine the rate of heat removed from the refrigerated space in Btumin 11104 It is proposed to run a thermoelectric generator in conjunction with a solar pond that can supply heat at a rate of 7 106 kJh at 90C The waste heat is to be rejected to the environment at 22C What is the maximum power this thermoelectric generator can produce 11105 A thermoelectric refrigerator is powered by a 12V car battery that draws 3 A of current when running The refrig erator resembles a small ice chest and is claimed to cool nine canned drinks 0350 L each from 25 to 3C in 12 h Deter mine the average COP of this refrigerator in a wellinsulated cup holder Assuming an average COP of 02 in the cooling mode determine a the average rate of heat removal from the drink b the average rate of heat supply to the coffee and c the electric power drawn from the battery of the car all in W Review Problems 11107 Rooms with floor areas of up to 15 m2 are cooled adequately by window air conditioners whose cooling capac ity is 5000 Btuh Assuming the COP of the air conditioner to be 35 determine the rate of heat gain of the room in Btuh when the air conditioner is running continuously to maintain a constant room temperature 11108 Consider a steadyflow Carnot refrigeration cycle that uses refrigerant134a as the working fluid The maximum and minimum temperatures in the cycle are 30 and 20C respectively The quality of the refrigerant is 015 at the begin ning of the heat absorption process and 080 at the end Show the cycle on a Ts diagram relative to saturation lines and determine a the coefficient of performance b the con denser and evaporator pressures and c the net work input 11109 Consider an iceproducing plant that operates on the ideal vaporcompression refrigeration cycle and uses refrigerant 134a as the working fluid The refrigeration cycle operating conditions require an evaporator pressure of 140 kPa and the condenser pressure of 1200 kPa Cooling water flows through the water jacket surrounding the condenser and is sup plied at the rate of 200 kgs The cooling water has a 10C tem perature rise as it flows through the water jacket To produce ice potable water is supplied to the chiller section of the refrig eration cycle For each kg of ice produced 333 kJ of energy must be removed from the potable water supply a Sketch the hardware for all three working fluids of this refrigerantice making system and the Ts diagram for refrigeration cycle b Determine the mass flow rate of the refrigerant in kgs c Determine the mass flow rate of the potable water supply in kgs 11110 A heat pump that operates on the ideal vapor compression cycle with refrigerant134a is used to heat a house The mass flow rate of the refrigerant is 025 kgs The condenser and evaporator pressures are 1400 and 320 kPa respectively Show the cycle on a Ts diagram with respect to saturation lines and determine a the rate of heat supply to the house b the volume flow rate of the refrigerant at the com pressor inlet and c the COP of this heat pump 11111 A heat pump operates on the ideal vapor compression refrigeration cycle and uses refrigerant22 as the working fluid The operating conditions for this heat pump are evapora tor saturation temperature of 5C and the condenser satura tion temperature of 45C Selected data for refrigerant22 are provided in the following table FIGURE P11105 11106E Thermoelectric coolers that plug into the cigarette lighter of a car are commonly available One such cooler is claimed to cool a 12oz 0771lbm drink from 78 to 38F or to heat a cup of coffee from 75 to 130F in about 15 min Final PDF to printer cen22672ch11597642indd 638 110917 1148 AM 638 REFRIGERATION CYCLES T C Psat kPa hf kJkg hg kJkg sg kJkgK 5 4212 3876 2481 09344 45 1728 101 2619 08682 For R22 at P 1728 kPa and s 09344 kJkgK T 6815C and h 2837 kJkg Also take cpair 1005 kJkgK a Sketch the hardware and the Ts diagram for this heat pump application b Determine the COP for this unit c The condenser of this unit is located inside the air handler of an office If the air flowing through the air handler is limited to a 20C temperature rise determine the ratio of volume flow rate of air to mass flow rate of R22 through the air handler in m3 airminkg R22s 11112 A large refrigeration plant is to be maintained at 15C and it requires refrigeration at a rate of 100 kW The condenser of the plant is to be cooled by liquid water which experiences a temperature rise of 8C as it flows over the coils of the condenser Assuming the plant operates on the ideal vaporcompression cycle using refrigerant134a between the pressure limits of 120 and 700 kPa determine a the mass flow rate of the refrigerant b the power input to the compres sor and c the mass flow rate of the cooling water 11113 Reconsider Prob 11112 Using appropriate software investigate the effect of evaporator pressure on the COP and the power input Let the evaporator pressure vary from 120 to 380 kPa Plot the COP and the power input as functions of evaporator pressure and discuss the results 11114 Repeat Prob 11112 assuming the compressor has an isentropic efficiency of 75 percent Also determine the rate of exergy destruction associated with the compression process in this case Take T0 25C 11115 An air conditioner with refrigerant134a as the working fluid is used to keep a room at 26C by rejecting the waste heat to the outside air at 34C The room is gaining heat through the walls and the windows at a rate of 250 kJ min while the heat generated by the computer TV and lights amounts to 900 W An unknown amount of heat is also gener ated by the people in the room The condenser and evaporator pressures are 1200 and 500 kPa respectively The refrigerant is saturated liquid at the condenser exit and saturated vapor at the compressor inlet If the refrigerant enters the compres sor at a rate of 100 Lmin and the isentropic efficiency of the compressor is 75 percent determine a the temperature of the refrigerant at the compressor exit b the rate of heat generation by the people in the room c the COP of the air conditioner and d the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions Answers a 545C b 0665 kW c 587 d 157 Lmin 11116 A refrigerator using refrigerant134a as the working fluid operates the condenser at 700 kPa and the evaporator at 10C This refrigerator freezes water while rejecting heat to the ambient air at 22C The compressor has an isentropic effi ciency of 85 percent Determine the process that causes the greatest amount of exergy destruction 11117 Rework Prob 11116 with a 27C subcooling at the exit of the condenser 11118 An air conditioner operates on the vapor compression refrigeration cycle with refrigerant134a as the refrigerant The air conditioner is used to keep a space at 21C while rejecting the waste heat to the ambient air at 37C The refrigerant enters the compressor at 180 kPa superheated by 27C at a rate of 006 kgs and leaves the compressor at 1200 kPa and 60C R134a is subcooled by 63C at the exit of the condenser Determine a the rate of cooling provided to the space in Btuh and the COP b the isentropic efficiency and the exergy efficiency of the compressor c the exergy destruction in each component of the cycle and the total exergy destruction in the cycle and d the minimum power input and the secondlaw efficiency of the cycle 11119 Consider a twostage compression refrigeration sys tem operating between the pressure limits of 14 and 012 MPa The working fluid is refrigerant134a The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 05 MPa Part of the refrigerant evapo rates during this flashing process and this vapor is mixed with the refrigerant leaving the lowpressure compressor The mixture is then compressed to the condenser pressure by the highpressure compressor The liquid in the flash chamber is throttled to the evaporator pressure and it cools the refrig erated space as it vaporizes in the evaporator Assuming the FIGURE P11115 1200 kPa 500 kPa 2 3 1 4 Compressor Win Expansion valve Evaporator QL 26C Condenser QH 34C Final PDF to printer cen22672ch11597642indd 639 110917 1148 AM 639 CHAPTER 11 refrigerant leaves the evaporator as saturated vapor and both compressors are isentropic determine a the fraction of the refrigerant that evaporates as it is throttled to the flash cham ber b the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser and c the coefficient of per formance Answers a 0290 b 116 kJkg 427 kJkg c 272 11120E A twoevaporator compression refrigeration sys tem as shown in Fig P11120E uses refrigerant134a as the working fluid The system operates evaporator 1 at 30F evap orator 2 at 295F and the condenser at 160 psia The cooling load of evaporator 1 is double that of evaporator 2 Determine the cooling load of both evaporators per unit of flow through the compressor as well as the COP of this system The refrig erant is saturated liquid at the exit of the condenser and satu rated vapor at the exit of each evaporator and the compressor is isentropic 1009C and the condenser at 900 kPa Determine the system COP when the heat exchanger provides 551C of subcooling at the throttle valve entrance Assume the refrigerant leaves the evaporator as a saturated vapor and the compressor is isentro pic Answer 460 FIGURE P11120E Pressure reducing valve Evaporator 1 Evaporator 2 7 1 m2 m1 m1 m2 5 Condenser 2 3 4 6 FIGURE P11122 2 1 6 Condenser Evaporator 4 3 5 Compressor 11121E Reconsider Prob 11120E The refrigeration sys tem of that problem cools one reservoir at 15F and one at 40F while rejecting heat to a reservoir at 80F Which process has the highest exergy destruction 11122 The refrigeration system of Fig P11122 is another variation of the basic vaporcompression refrigeration system which attempts to reduce the compression work In this sys tem a heat exchanger is used to superheat the vapor entering the compressor while subcooling the liquid exiting from the condenser Consider a system of this type that uses refrigerant 134a as its refrigerant and operates the evaporator at 11123 Repeat Prob 11122 if the heat exchanger provides 951C of subcooling 11124 An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle Air enters the compressor at 30C and 100 kPa and is compressed to 250 kPa Air is cooled to 85C before it enters the turbine Assuming both the turbine and the compressor to be isentro pic determine the temperature of the air leaving the turbine and entering the cabin Answer 25C 11125 Consider a regenerative gas refrigeration cycle using helium as the working fluid Helium enters the compressor at 100 kPa and 10C and is compressed to 300 kPa Helium is then cooled to 20C by water It then enters the regenerator where it is cooled further before it enters the turbine Helium leaves the refrigerated space at 25C and enters the regenera tor Assuming both the turbine and the compressor to be isentro pic determine a the temperature of the helium at the turbine inlet b the coefficient of performance of the cycle and c the net power input required for a mass flow rate of 045 kgs 11126 A gas refrigeration system using air as the working fluid has a pressure ratio of 5 Air enters the compressor at 0C The highpressure air is cooled Final PDF to printer cen22672ch11597642indd 640 110917 1148 AM 640 REFRIGERATION CYCLES to 35C by rejecting heat to the surroundings The refrigerant leaves the turbine at 80C and enters the refrigerated space where it absorbs heat before entering the regenerator The mass flow rate of air is 04 kgs Assuming isentropic efficien cies of 80 percent for the compressor and 85 percent for the turbine and using variable specific heats determine a the effectiveness of the regenerator b the rate of heat removal from the refrigerated space and c the COP of the cycle Also determine d the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle Use the same compressor inlet temperature as given the same turbine inlet temperature as calculated and the same compressor and tur bine efficiencies This problem is solved using appropriate software 11128 Using appropriate software investigate the effect of the evaporator pressure on the COP of an ideal vaporcompression refrigeration cycle with R134a as the working fluid Assume the condenser pressure is kept con stant at 14 MPa while the evaporator pressure is varied from 100 kPa to 500 kPa Plot the COP of the refrigeration cycle against the evaporator pressure and discuss the results 11129 Using appropriate software investigate the effect of the condenser pressure on the COP of an ideal vaporcompression refrigeration cycle with R134a as the working fluid Assume the evaporator pressure is kept con stant at 150 kPa while the condenser pressure is varied from 400 to 1400 kPa Plot the COP of the refrigeration cycle against the condenser pressure and discuss the results 11130 An absorption refrigeration system is to remove heat from the refrigerated space at 2C at a rate of 28 kW while operating in an environment at 25C Heat is to be supplied from a solar pond at 95C What is the minimum rate of heat supply required Answer 123 kW 11131 Reconsider Prob 11130 Using appropriate software investigate the effect of the source temperature on the minimum rate of heat supply Let the source temperature vary from 50 to 250C Plot the minimum rate of heat supply as a function of source temperature and discuss the results 11132 Derive a relation for the COP of the twostage refrig eration system with a flash chamber as shown in Fig 1114 in terms of the enthalpies and the quality at state 6 Consider a unit mass in the condenser Fundamentals of Engineering FE Exam Problems 11133 A refrigerator removes heat from a refrigerated space at 0C at a rate of 15 kJs and rejects it to an environ ment at 20C The minimum required power input is a 102 W b 110 W c 140 W d 150 W e 1500 W 11134 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R134a as the working fluid The refrigerant enters the compressor as saturated vapor at 160 kPa exits at 800 kPa and 50C and leaves the con denser as saturated liquid at 800 kPa The coefficient of perfor mance of this refrigerator is a 26 b 10 c 42 d 32 e 44 11135 A refrigerator operates on the ideal vapor compression refrigeration cycle with R134a as the working fluid between the pressure limits of 120 and 800 kPa If the rate of heat removal from the refrigerated space is 32 kJs the mass flow rate of the refrigerant is a 019 kgs b 015 kgs c 023 kgs d 028 kgs e 081 kgs FIGURE P11126 Heat exchanger 1 4 5 Compressor QH QL 2 3 6 Regenerator Heat exchanger Turbine FIGURE P11127 8 1 QL 3 QH1 QH2 QH3 2 5 4 6 7 Compressor Compressor Compressor Turbine 11127 An ideal gas refrigeration system with three stages of compression with intercooling operates with air entering the first compressor at 50 kPa and 30C Each compressor in this system has a pressure ratio of 7 and the air temperature at the outlet of all intercoolers is 15C Calculate the COP of this system Use constant specific heats at room temperature Final PDF to printer cen22672ch11597642indd 641 110917 1148 AM 641 CHAPTER 11 11136 Consider a heat pump that operates on the reversed Carnot cycle with R134a as the working fluid executed under the saturation dome between the pressure limits of 140 and 800 kPa R134a changes from saturated vapor to saturated liquid during the heatrejection process The net work input for this cycle is a 28 kJkg b 34 kJkg c 49 kJkg d 144 kJkg e 275 kJkg 11137 A heat pump operates on the ideal vapor compression refrigeration cycle with R134a as the working fluid between the pressure limits of 032 and 12 MPa If the mass flow rate of the refrigerant is 0193 kgs the rate of heat supply by the heat pump to the heated space is a 33 kW b 23 kW c 26 kW d 31 kW e 45 kW 11138 An ideal vaporcompression refrigeration cycle with R134a as the working fluid operates between the pressure limits of 120 kPa and 700 kPa The mass fraction of the refrig erant that is in the liquid phase at the inlet of the evaporator is a 069 b 063 c 058 d 043 e 035 11139 Consider a heat pump that operates on the ideal vaporcompression refrigeration cycle with R134a as the working fluid between the pressure limits of 024 and 12 MPa The coefficient of performance of this heat pump is a 59 b 53 c 49 d 42 e 38 11140 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 and 280 kPa Air is cooled to 35C before entering the turbine The lowest temperature of this cycle is a 58C b 26C c 5C d 11C e 24C 11141 Consider an ideal gas refrigeration cycle using helium as the working fluid Helium enters the compressor at 100 kPa and 17C and compressed to 400 kPa Helium is then cooled to 20C before it enters the turbine For a mass flow rate of 02 kgs the net power input required is a 283 kW b 405 kW c 647 kW d 937 kW e 113 kW 11142 An absorption airconditioning system is to remove heat from the conditioned space at 20C at a rate of 90 kJs while operating in an environment at 35C Heat is to be sup plied from a geothermal source at 140C The minimum rate of heat supply is a 13 kJs b 18 kJs c 30 kJs d 37 kJs e 90 kJs Design and Essay Problems 11143 Develop and discuss techniques that apply the prin ciple of regeneration to improve the performance of vapor compression refrigeration systems 11144 The heat supplied by a heat pump used to maintain a buildings temperature is often supplemented by another source of direct heat The fraction of the total heat required that is supplied by supplemental heat increases as the temperature of the environmental air which serves as the low temperature sink decreases Develop a supplemental heat schedule as a function of the environmental air temperature that minimizes the total supplemental and heatpump energy required to ser vice the building 11145 Design a thermoelectric refrigerator that is capable of cooling a canned drink in a car The refrigerator is to be powered by the cigarette lighter of the car Draw a sketch of your design Semiconductor components for building thermo electric power generators or refrigerators are available from several manufacturers Using data from one of these manu facturers determine how many of these components you need in your design and estimate the coefficient of performance of your system A critical problem in the design of thermoelectric refrigerators is the effective rejection of waste heat Discuss how you can enhance the rate of heat rejection without using any devices with moving parts such as a fan 11146 Solar or photovoltaic PV cells convert sunlight to electricity and are commonly used to power calculators sat ellites remote communication systems and even pumps The conversion of light to electricity is called the photoelectric effect It was first discovered in 1839 by Frenchman Edmond Becquerel and the first PV module which consisted of several cells connected to each other was built in 1954 by Bell Labo ratories The PV modules today have conversion efficiencies of about 12 to 15 percent Noting that the solar energy incident on a normal surface on earth at noontime is about 1000 Wm2 during a clear day PV modules on a 1m2 surface can provide as much as 150 W of electricity The annual average daily solar energy incident on a horizontal surface in the United States ranges from about 2 to 6 kWhm2 A PVpowered pump is to be used in Arizona to pump water for wildlife from a depth of 180 m at an average rate of 400 Lday Assuming a reasonable efficiency for the pumping system which can be defined as the ratio of the increase in the poten tial energy of the water to the electrical energy consumed by the pump and taking the conversion efficiency of the PV cells to be 013 to be on the conservative side determine the size of the PV module that needs to be installed in m2 11147 The temperature in a car parked in the sun can approach 100C when the outside air temperature is just 25C and it is desirable to ventilate the parked car to avoid such high temperatures However the ventilating fans may run down the battery if they are powered by it To avoid that happening it is proposed to use the PV cells to power the fans It is determined that the air in the car should be replaced once every minute to avoid excessive rise in the interior temperature Determine if this can be accomplished by installing PV cells on part of the roof of the car Also find out if any car is currently ventilated this way Final PDF to printer cen22672ch11597642indd 642 110917 1148 AM 642 REFRIGERATION CYCLES 11148 It is proposed to use a solarpowered thermoelectric system installed on the roof to cool residential buildings The system consists of a thermoelectric refrigerator that is pow ered by a thermoelectric power generator whose top surface is a solar collector Discuss the feasibility and the cost of such a system and determine if the proposed system installed on one side of the roof can meet a significant portion of the cooling requirements of a typical house in your area in the cycle and estimate the required mass flow rate of refrigerant134a for a net power output of 50 kW Also esti mate the surface area of the pond for this level of continuous power production Assume that the solar energy is incident on the pond at a rate of 500 W per m2 of pond area at noontime and that the pond is capable of storing 15 percent of the inci dent solar energy in the storage zone 11150 A company owns a refrigeration system whose refrigeration capacity is 200 tons 1 ton of refrigeration 211 kJmin and you are to design a forcedair cooling system for fruits whose diameters do not exceed 7 cm under the follow ing conditions The fruits are to be cooled from 28C to an average temperature of 8C The air temperature is to remain above 2C and below 10C at all times and the velocity of air approaching the fruits must remain under 2 ms The cool ing section can be as wide as 35 m and as high as 2 m Assuming reasonable values for the average fruit density specific heat and porosity the fraction of air volume in a box recommend reasonable values for a the air velocity approach ing the cooling section b the productcooling capacity of the system in kgfruith and c the volume flow rate of air 11151 In the 1800s before the development of modern air conditioning it was proposed to cool air for buildings with the following procedure using a large pistoncylinder device John Gorrie Pioneer of Cooling and Ice Making ASHRAE Journal 33 no 1 Jan 1991 1 Pull in a charge of outdoor air 2 Compress it to a high pressure 3 Cool the charge of air using outdoor air 4 Expand it back to atmospheric pressure 5 Discharge the charge of air into the space to be cooled Suppose the goal is to cool a room 6 m 10 m 25 m Out door air is at 30C and it has been determined that 10 air changes per hour supplied to the room at 10C could provide adequate cooling Do a preliminary design of the system and do calculations to see if it would be feasible You may make optimistic assumptions for the analysis a Sketch the system showing how you will drive it and how step 3 will be accomplished b Determine what pressure will be required step 2 c Estimate guess how long step 3 will take and what size will be needed for the pistoncylinder to provide the required air changes and temperature d Determine the work required in step 2 for one cycle and per hour e Discuss any problems you see with the concept of your design Include discussion of any changes that may be required to offset optimistic assumptions FIGURE P11148 Thermoelectric generator Solar energy Waste heat Thermoelectric refrigerator Electric current Sun FIGURE P11147 Solar energy Solar panels Solarpowered exhaust fan 11149 Consider a solar pond power plant operating on a closed Rankine cycle Using refrigerant134a as the work ing fluid specify the operating temperatures and pressures Final PDF to printer cen22672ch12643674indd 643 103117 0134 PM 643 CHAPTER 12 T H E R M O DY N AMIC P R O P E RT Y R EL ATIO N S I n the preceding chapters we made extensive use of the property tables We tend to take the property tables for granted but thermodynamic laws and principles are of little use to engineers without them In this chapter we focus our attention on how the property tables are prepared and how some unknown properties can be determined from limited available data It will come as no surprise that some properties such as temperature pressure volume and mass can be measured directly Other properties such as density and specific volume can be determined from these using some simple relations However properties such as internal energy enthalpy and entropy are not so easy to determine because they cannot be measured directly or related to easily measurable properties through some simple relations There fore it is essential that we develop some fundamental relations between com monly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties By the nature of the material this chapter makes extensive use of partial derivatives Therefore we start by reviewing them Then we develop the Maxwell relations which form the basis for many thermodynamic relations Next we discuss the Clapeyron equation which enables us to determine the enthalpy of vaporization from P v and T measurements alone and we develop general relations for cv cp du dh and ds that are valid for all pure substances under all conditions Then we discuss the JouleThomson coef ficient which is a measure of the temperature change with pressure during a throttling process Finally we develop a method of evaluating the Δh Δu and Δs of real gases through the use of generalized enthalpy and entropy depar ture charts OBJECTIVES The objectives of Chapter 12 are to Develop fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties Develop the Maxwell relations which form the basis for many thermodynamic relations Develop the Clapeyron equation and determine the enthalpy of vaporization from P v and T measurements alone Develop general relations for cv cp du dh and ds that are valid for all pure substances Discuss the JouleThomson coefficient Develop a method of evaluating the Δ h Δ u and Δ s of real gases through the use of generalized enthalpy and entropy departure charts Final PDF to printer 644 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 644 103117 0134 PM 121 A LITTLE MATHPARTIAL DERIVATIVES AND ASSOCIATED RELATIONS Many of the expressions developed in this chapter are based on the state pos tulate which expresses that the state of a simple compressible substance is completely specified by any two independent intensive properties All other properties at that state can be expressed in terms of those two properties Mathematically speaking z zx y where x and y are the two independent properties that fix the state and z represents any other property Most basic thermodynamic relations involve differentials Therefore we start by reviewing the derivatives and various relations among derivatives to the extent necessary in this chapter Consider a function f that depends on a single variable x that is f fx Figure 121 shows such a function that starts out flat but gets rather steep as x increases The steepness of the curve is a measure of the degree of depen dence of f on x In our case the function f depends on x more strongly at larger x values The steepness of a curve at a point is measured by the slope of a line tangent to the curve at that point and it is equivalent to the derivative of the function at that point defined as df dx lim Δx0 Δf Δx lim Δx 0 f x Δx f x Δx 121 Therefore the derivative of a function fx with respect to x represents the rate of change of f with x FIGURE 121 The derivative of a function at a specified point represents the slope of the function at that point fx fx fx x x x x f x Slope x EXAMPLE 121 Approximating Differential Quantities by Differences The cp of ideal gases depends on temperature only and it is expressed as cpT dhTdT Determine the cp of air at 300 K using the enthalpy data from Table A17 and compare it to the value listed in Table A2b SOLUTION The cp value of air at a specified temperature is to be determined using enthalpy data Analysis The cp value of air at 300 K is listed in Table A2b to be 1005 kJkgK This value could also be determined by differentiating the function hT with respect to T and evaluating the result at T 300 K However the function hT is not avail able But we can still determine the cp value approximately by replacing the differ entials in the cpT relation by differences in the neighborhood of the specified point Fig 122 c p 300 K dh T dT T 300 K Δh T ΔT T 300 K h 305 K h 295 K 305 295 K 30522 29517 kJkg 305 295 K 1005 kJkgK FIGURE 122 Schematic for Example 121 hT kJkg T K 30522 29517 295 300 305 Slope cpT Final PDF to printer 645 CHAPTER 12 cen22672ch12643674indd 645 103117 0134 PM Partial Differentials Now consider a function that depends on two or more variables such as z zx y This time the value of z depends on both x and y It is sometimes desirable to examine the dependence of z on only one of the variables This is done by allowing one variable to change while holding the others constant and observing the change in the function The variation of zx y with x when y is held constant is called the partial derivative of z with respect to x and it is expressed as z x y lim Δx0 Δz Δx y lim Δx0 zx Δx y zx y Δx 122 This is illustrated in Fig 123 The symbol represents differential changes just like the symbol d They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables whereas represents the partial differential change due to the variation of a single variable Note that the changes indicated by d and are identical for independent variables but not for dependent variables For example xy dx but zy dz In our case dz zx zy Also note that the value of the partial derivative zxy in general is different at different y values To obtain a relation for the total differential change in zx y for simultane ous changes in x and y consider a small portion of the surface zx y shown in Fig 124 When the independent variables x and y change by Δx and Δy respectively the dependent variable z changes by Δz which can be expressed as Δz zx Δx y Δy zx y Adding and subtracting zx y Δy we get Δz zx Δx y Δy zx y Δy zx y Δy zx y or Δz zx Δx y Δy zx y Δy Δx Δx zx y Δy zx y Δy Δy Taking the limits as Δx 0 and Δy 0 and using the definitions of partial derivatives we obtain dz z x y dx z y x dy 123 FIGURE 123 Geometric representation of partial derivative zxy z x y z x y Discussion Note that the calculated cp value is identical to the listed value There fore differential quantities can be viewed as differences They can even be replaced by differences whenever necessary to obtain approximate results The widely used finite difference numerical method is based on this simple principle FIGURE 124 Geometric representation of total derivative dz for a function zx y x y z zx x y y x x y y x y y x x y zx y Final PDF to printer 646 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 646 103117 0134 PM Equation 123 is the fundamental relation for the total differential of a dependent variable in terms of its partial derivatives with respect to the inde pendent variables This relation can easily be extended to include more inde pendent variables EXAMPLE 122 Total Differential Versus Partial Differential Consider air at 300 K and 086 m3kg The state of air changes to 302 K and 087 m3kg as a result of some disturbance Using Eq 123 estimate the change in the pressure of air SOLUTION The temperature and specific volume of air change slightly during a process The resulting change in pressure is to be determined Assumption Air is an ideal gas Analysis Strictly speaking Eq 123 is valid for differential changes in variables However it can also be used with reasonable accuracy if these changes are small The changes in T and v respectively can be expressed as dT ΔT 302 300 K 2 K and dv Δv 087 086 m 3 kg 001 m 3 kg An ideal gas obeys the relation Pv RT Solving for P yields P RT v Note that R is a constant and P PT v Applying Eq 123 and using average values for T and v dP P T v dT P v T dv R dT v RT dv v 2 0287 kPa m 3 kgK 2 K 0865 m 3 kg 301 K001 m 3 kg 0865 m 3 kg 2 0664 kPa 1155 kPa 0491 kPa Therefore the pressure will decrease by 0491 kPa as a result of this disturbance Notice that if the temperature had remained constant dT 0 the pressure would decrease by 1155 kPa as a result of the 001 m3kg increase in specific volume However if the specific volume had remained constant dv 0 the pressure would increase by 0664 kPa as a result of the 2K rise in temperature Fig 125 That is P T v dT P v 0664 kPa P v T dv P T 1155 kPa and dP P v P T 0664 1155 0491 kPa FIGURE 125 Geometric representation of the disturbance discussed in Example 122 P kPa Pv 0664 PT 1155 dP 0491 T K 302 300 086 087 v m3kg Final PDF to printer 647 CHAPTER 12 cen22672ch12643674indd 647 103117 0134 PM Partial Differential Relations Now let us rewrite Eq 123 as dz M dx N dy 124 where M z x y and N z y x Taking the partial derivative of M with respect to y and of N with respect to x yields M y x 2 z x y and N x y 2 z y x The order of differentiation is immaterial for properties since they are contin uous point functions and have exact differentials Therefore the two relations above are identical M y x N x y 125 This is an important relation for partial derivatives and it is used in calculus to test whether a differential dz is exact or inexact In thermodynamics this relation forms the basis for the development of the Maxwell relations dis cussed in the next section Finally we develop two important relations for partial derivativesthe rec iprocity and the cyclic relations The function z zx y can also be expressed as x xy z if y and z are taken to be the independent variables Then the total differential of x becomes from Eq 123 dx x dy z dy x z y dz 126 Eliminating dx by combining Eqs 123 and 126 we have dz z x y x y z z y x dy x z y z x y dz Rearranging z x y x y z z y x dy 1 x z y z x y dz 127 The variables y and z are independent of each other and thus can be varied independently For example y can be held constant dy 0 and z can be Discussion Of course we could have solved this problem easily and exactly by evaluating the pressure from the idealgas relation P RTv at the final state 302 K and 087 m3kg and the initial state 300 K and 086 m3kg and taking their difference This yields 0491 kPa which is exactly the value obtained above Thus the small finite quantities 2 K 001 m3kg can be approximated as differential quantities with reasonable accuracy Final PDF to printer 648 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 648 103117 0134 PM varied over a range of values dz 0 Therefore for this equation to be valid at all times the terms in the brackets must equal zero regardless of the values of y and z Setting the terms in each bracket equal to zero gives x z y z x y 1 x z y 1 z x y 128 z x y x y z z y x x y z y z x z x y 1 129 The first relation is called the reciprocity relation and it shows that the inverse of a partial derivative is equal to its reciprocal Fig 126 The second relation is called the cyclic relation and it is frequently used in thermodynamics FIGURE 126 Demonstration of the reciprocity relation for the function z 2xy 3y2z 0 Function z 2xy 3y2z 0 1 z 1 Thus 2xy 3y2 1 2y 3y2 1 y 2 x 3y2z z 2y 3y2 1 2y y y z x x z x z z x y EXAMPLE 123 Verification of Cyclic and Reciprocity Relations Using the idealgas equation of state verify a the cyclic relation and b the reci procity relation at constant P SOLUTION The cyclic and reciprocity relations are to be verified for an ideal gas Analysis The idealgas equation of state Pv RT involves the three variables P v and T Any two of these can be taken as the independent variables with the remaining one being the dependent variable a Replacing x y and z in Eq 129 with P v and T respectively we can express the cyclic relation for an ideal gas as P v T v T P T P v 1 where P Pv T RT v P v T RT v 2 v vP T RT P v T P R P T TP v Pv R T P v v R Substituting yields RT v 2 R P v R RT Pv 1 which is the desired result b The reciprocity rule for an ideal gas at P constant can be expressed as v T P 1 T v P Performing the differentiations and substituting we have R P 1 P R R P R P Thus the proof is complete Final PDF to printer 649 CHAPTER 12 cen22672ch12643674indd 649 103117 0134 PM 122 THE MAXWELL RELATIONS The equations that relate the partial derivatives of properties P v T and s of a simple compressible system to each other are called the Maxwell relations They are obtained from the four Gibbs equations by exploiting the exactness of the differentials of thermodynamic properties Two of the Gibbs relations were derived in Chap 7 and expressed as du T ds P dv 1210 dh T ds v dP 1211 The other two Gibbs relations are based on two new combination properties the Helmholtz function a and the Gibbs function g defined as a u Ts 1212 g h Ts 1213 Differentiating we get da du T ds s dT dg dh T ds s dT Simplifying the preceding relations by using Eqs 1210 and 1211 we obtain the other two Gibbs relations for simple compressible systems da s dT P dv 1214 dg s dT v dP 1215 A careful examination of the four Gibbs relations reveals that they are of the form dz M dx N dy 124 with M y x N x y 125 since u h a and g are properties and thus have exact differentials Applying Eq 125 to each of them we obtain T v s P s v 1216 T P s v s P 1217 s v T P T v 1218 s P T v T P 1219 These are called the Maxwell relations Fig 127 They are extremely valu able in thermodynamics because they provide a means of determining the FIGURE 127 Maxwell relations are extremely valuable in thermodynamic analysis s s T T P P P s T v T P s P P T s v v s v T v v Final PDF to printer 650 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 650 103117 0134 PM change in entropy which cannot be measured directly by simply measuring the changes in properties P v and T Note that the Maxwell relations given above are limited to simple compressible systems However other similar relations can be written just as easily for nonsimple systems such as those involving electrical magnetic and other effects EXAMPLE 124 Verification of the Maxwell Relations Verify the validity of the last Maxwell relation Eq 1219 for steam at 250C and 300 kPa SOLUTION The validity of the last Maxwell relation is to be verified for steam at a specified state Analysis The last Maxwell relation states that for a simple compressible substance the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure If we had explicit analytical relations for the entropy and specific volume of steam in terms of other properties we could easily verify this by performing the indicated derivations However all we have for steam are tables of properties listed at certain intervals Therefore the only course we can take to solve this problem is to replace the differential quantities in Eq 1219 with corresponding finite quantities using property values from the tables Table A6 in this case at or about the specified state s P T v T P Δs ΔP T 250C Δv ΔT P 300 kPa s 400 kPa s 200 kPa 400 200 kPa T 250C v 300C v 200C 300 200 C P 300 kPa 73804 77100 kJ kgK 400 200 kPa 087535 071643 m 3 kg 300 200 C 000165 m 3 kgK 000159 m 3 kgK since kJ kPam3 and K C for temperature differences The two values are within 4 percent of each other This difference is due to replacing the differential quantities with relatively large finite quantities Based on the close agreement between the two values the steam seems to satisfy Eq 1219 at the specified state Discussion This example shows that the entropy change of a simple compressible system during an isothermal process can be determined from a knowledge of the easily measurable properties P v and T alone 123 THE CLAPEYRON EQUATION The Maxwell relations have farreaching implications in thermodynamics and are often used to derive useful thermodynamic relations The Clapeyron equation is one such relation and it enables us to determine the enthalpy change Final PDF to printer 651 CHAPTER 12 cen22672ch12643674indd 651 103117 0134 PM associated with a phase change such as the enthalpy of vaporization hfg from a knowledge of P v and T data alone Consider the third Maxwell relation Eq 1218 P T v s v T During a phasechange process the pressure is the saturation pressure which depends on the temperature only and is independent of the specific volume That is Psat f Tsat Therefore the partial derivative PTv can be expressed as a total derivative dPdTsat which is the slope of the satura tion curve on a PT diagram at a specified saturation state Fig 128 This slope is independent of the specific volume and thus it can be treated as a constant during the integration of Eq 1218 between two saturation states at the same temperature For an isothermal liquidvapor phasechange process for example the integration yields s g s f dP dT sat v g v f 1220 or dP dT sat s fg v fg 1221 During this process the pressure also remains constant Therefore from Eq 1211 dh T ds v dP 0 f g dh f g T ds h fg T s fg Substituting this result into Eq 1221 we obtain dP dT sat h fg T v fg 1222 which is called the Clapeyron equation after the French engineer and physicist E Clapeyron 17991864 This is an important thermodynamic relation since it enables us to determine the enthalpy of vaporization hfg at a given temperature by simply measuring the slope of the saturation curve on a PT diagram and the specific volume of saturated liquid and saturated vapor at the given temperature The Clapeyron equation is applicable to any phasechange process that occurs at constant temperature and pressure It can be expressed in a general form as dP dT sat h 12 T v 12 1223 where the subscripts 1 and 2 indicate the two phases EXAMPLE 125 Estimating Boiling Temperature with the Clapeyron Equation Two grams of a saturated liquid are converted to a saturated vapor by being heated in a weighted pistoncylinder device arranged to maintain the pressure at 200 kPa FIGURE 128 The slope of the saturation curve on a PT diagram is constant at a constant T or P P T T const Liquid Solid Vapor P sat T Final PDF to printer 652 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 652 103117 0134 PM The Clapeyron equation can be simplified for liquidvapor and solidvapor phase changes by utilizing some approximations At low pressures vg vf and thus vfg vg By treating the vapor as an ideal gas we have vg RTP Substituting these approximations into Eq 1222 we find dP dT sat P h fg R T 2 or dP P sat h fg R dT T 2 sat For small temperature intervals hfg can be treated as a constant at some average value Then integrating this equation between two saturation states yields ln P 2 P 1 sat h fg R 1 T 1 1 T 2 sat 1224 This equation is called the ClapeyronClausius equation and it can be used to determine the variation of saturation pressure with temperature It can also be used in the solidvapor region by replacing hfg with hig the enthalpy of sublimation of the substance Fig 129 During the phase conversion the volume of the system increases by 1000 cm3 5 kJ of heat are required and the temperature of the substance stays constant at 80C Estimate the boiling temperature of this substance when its pressure is 180 kPa SOLUTION A substance is heated in a pistoncylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature The boiling temperature of this substance at a different pressure is to be estimated Analysis From the Clapeyron equation dP dT sat h fg T v fg 5 kJ 1 kPa m 3 1 kJ 0002 kg 80 27315 K 1 10 3 m 3 0002 kg 1416 kPa K Using the finite difference approximation dP dT sat P 2 P 1 T 2 T 1 sat Solving for T2 T 2 T 1 P 2 P 1 dP dT 80 27315 K 180 200 kPa 1416 kPa K 3517 K Discussion The Clapeyron equation can also be used to determine the enthalpy of vaporization of a substance at a given temperature when saturation temperature and pres sure values as well as specific volume are available FIGURE 129 Schematic for Example 125 Weight 200 kPa 80C 2 grams sat liquid Q Final PDF to printer 653 CHAPTER 12 cen22672ch12643674indd 653 103117 0134 PM 124 GENERAL RELATIONS FOR du dh ds cv AND cp The state postulate established that the state of a simple compressible system is completely specified by two independent intensive properties Therefore at least theoretically we should be able to calculate all the properties of a system at any state once two independent intensive properties are available This is certainly good news for properties that cannot be measured directly such as internal energy enthalpy and entropy However the calculation of these properties from measurable ones depends on the availability of simple and accurate relations between the two groups In this section we develop general relations for changes in internal energy enthalpy and entropy in terms of pressure specific volume temperature and specific heats alone We also develop some general relations involving specific heats The relations developed will enable us to determine the changes in these properties The property values at specified states can be determined only after the selection of a reference state the choice of which is quite arbitrary EXAMPLE 126 Extrapolating Tabular Data with the Clapeyron Equation Estimate the saturation pressure of refrigerant134a at 50F using the data available in the refrigerant tables SOLUTION The saturation pressure of refrigerant134a is to be determined using other tabulated data Analysis Table A11E lists saturation data at temperatures 40F and above There fore we should either resort to other sources or use extrapolation to obtain saturation data at lower temperatures Equation 1224 provides an intelligent way to extrapolate ln P 2 P 1 sat h fg R 1 T 1 1 T 2 sat In our case T1 40F and T2 50F For refrigerant134a R 001946 BtulbmR Also from Table A11E at 40F we read hfg 97104 Btulbm and P1 Psat 40F 7432 psia Substituting these values into Eq 1224 gives ln P 2 7432 psia 97104 Btulbm 001946 BtulbmR 1 420 R 1 410 R P 2 556 psia Therefore according to Eq 1224 the saturation pressure of refrigerant134a at 50F is 556 psia The actual value obtained from another source is 5506 psia Thus the value predicted by Eq 1224 is in error by about 1 percent which is quite acceptable for most purposes If we had used linear extrapolation instead we would have obtained 5134 psia which is in error by 7 percent Final PDF to printer 654 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 654 103117 0134 PM Internal Energy Changes We choose the internal energy to be a function of T and v that is u uT v and take its total differential Eq 123 du u T v dT u v T dv Using the definition of cv we have du c v dT u v T dv 1225 Now we choose the entropy to be a function of T and v that is s sT v and take its total differential ds s T v dT s v T dv 1226 Substituting this into the T ds relation du T ds P dv yields du T s T v dT T s v T P dv 1227 Equating the coefficients of dT and dv in Eqs 1225 and 1227 gives s T v c v T u v T T s v T P 1228 Using the third Maxwell relation Eq 1218 we get u v T T P T v P Substituting this into Eq 1225 we obtain the desired relation for du du c v dT T P T v P dv 1229 The change in internal energy of a simple compressible system associated with a change of state from T1 v1 to T2 v2 is determined by integration u 2 u 1 T 1 T 2 c v dt v 1 v 2 T P T v P dv 1230 Enthalpy Changes The general relation for dh is determined in exactly the same manner This time we choose the enthalpy to be a function of T and P that is h hT P and take its total differential dh h T p dT h P T dP Using the definition of cp we have dh c p dT h P T dP 1231 Final PDF to printer 655 CHAPTER 12 cen22672ch12643674indd 655 103117 0134 PM Now we choose the entropy to be a function of T and P that is we take s sT P and take its total differential ds s T p dT s P T dP 1232 Substituting this into the T ds relation dh T ds v dP gives dh T s T p dT v T s P T dP 1233 Equating the coefficients of dT and dP in Eqs 1231 and 1233 we obtain s T P c p T h P T v T s P T 1234 Using the fourth Maxwell relation Eq 1219 we have h P T v T v T P Substituting this into Eq 1231 we obtain the desired relation for dh dh c p dT v T v T P dP 1235 The change in enthalpy of a simple compressible system associated with a change of state from T1 P1 to T2 P2 is determined by integration h 2 h 1 T 1 T 2 c p dT P 1 P 2 v T v T P dP 1236 In reality one need only determine either u2 u1 from Eq 1230 or h2 h1 from Eq 1236 depending on which is more suitable to the data at hand The other can easily be determined by using the definition of enthalpy h u Pv h 2 h 1 u 2 u 1 P 2 v 2 P 1 v 1 1237 Entropy Changes Next we develop two general relations for the entropy change of a simple compressible system The first relation is obtained by replacing the first partial derivative in the total differential ds Eq 1226 with Eq 1228 and the second partial deriva tive with the third Maxwell relation Eq 1218 yielding ds c v T dT P T v dv 1238 and s 2 s 1 T 1 T 2 c v T dT v 1 v 2 P T v dv 1239 Final PDF to printer 656 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 656 103117 0134 PM The second relation is obtained by replacing the first partial derivative in the total differential of ds Eq 1232 with Eq 1234 and the second partial derivative with the fourth Maxwell relation Eq 1219 yielding ds c p T dT v T P dP 1240 and s 2 s 1 T 1 T 2 c p T dT P 1 P 2 v T P dP 1241 Either relation can be used to determine the entropy change The proper choice depends on the available data Specific Heats cv and cp Recall that the specific heats of an ideal gas depend on temperature only For a general pure substance however the specific heats depend on specific volume or pressure as well as the temperature Next we develop some general relations to relate the specific heats of a substance to pressure specific vol ume and temperature At low pressures gases behave as ideal gases and their specific heats essentially depend on temperature only These specific heats are called zero pressure or idealgas specific heats denoted cv0 and cp0 and they are rela tively easier to determine Thus it is desirable to have some general relations that enable us to calculate the specific heats at higher pressures or lower specific volumes from a knowledge of cv0 or cp0 and the PvT behavior of the substance Such relations are obtained by applying the test of exactness Eq 125 on Eqs 1238 and 1240 which yields c v v T T 2 P T 2 v 1242 and c p P T T 2 v T 2 P 1243 The deviation of cp from cp0 with increasing pressure for example is deter mined by integrating Eq 1243 from zero pressure to any pressure P along an isothermal path c p c p0 T T 0 P 2 v T 2 P dP 1244 The integration on the righthand side requires a knowledge of the PvT behavior of the substance alone The notation indicates that v should be differentiated twice with respect to T while P is held constant The resulting expression should be integrated with respect to P while T is held constant Another desirable general relation involving specific heats is one that relates the two specific heats cp and cv The advantage of such a relation is obvious We will need to determine only one specific heat usually cp and calculate the other one using that relation and the PvT data of the substance We start the Final PDF to printer 657 CHAPTER 12 cen22672ch12643674indd 657 103117 0134 PM development of such a relation by equating the two ds relations Eqs 1238 and 1240 and solving for dT dT T P T v c p c v dv T v T P c p c v dP Choosing T Tv P and differentiating we get dT T v P dv T dP v dP Equating the coefficient of either dv or dP of the preceding two equations gives the desired result c p c v T v T P P T v 1245 An alternative form of this relation is obtained by using the cyclic relation P T v T v P v P T 1 P T v v T P P v T Substituting the result into Eq 1245 gives c p c v T v T P 2 P v T 1246 This relation can be expressed in terms of two other thermodynamic proper ties called the volume expansivity β and the isothermal compressibility α which are defined as Fig 1210 β 1 v v T P 1247 and α 1 v v P T 1248 Substituting these two relations into Eq 1246 we obtain a third general relation for cp cv c p c v vT β 2 α 1249 It is called the Mayer relation in honor of the German physician and physi cist J R Mayer 18141878 We can draw several conclusions from this equation 1 The isothermal compressibility α is a positive quantity for all substances in all phases The volume expansivity could be negative for some substances such as liquid water below 4C but its square is always positive or zero The temperature T in this relation is thermodynamic temperature which is also positive Therefore we conclude that the constantpressure specific heat is always greater than or equal to the constantvolume specific heat c p c v 1250 FIGURE 1210 The coefficient of volume expansion is a measure of the change in volume of a substance with temperature at constant pressure 20C 100 kPa 1 kg 21C 100 kPa 1 kg 20C 100 kPa 1 kg 21C 100 kPa 1 kg P P a A substance with a large b A substance with a small v v Final PDF to printer 658 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 658 103117 0134 PM 2 The difference between cp and cv approaches zero as the absolute temperature approaches zero 3 The two specific heats are identical for truly incompressible substances since v constant The difference between the two specific heats is very small and is usually disregarded for substances that are nearly incompressible such as liquids and solids EXAMPLE 127 Internal Energy Change of a van der Waals Gas Derive a relation for the internal energy change as a gas that obeys the van der Waals equation of state Assume that in the range of interest cv varies according to the rela tion cv c1 c2T where c1 and c2 are constants SOLUTION A relation is to be obtained for the internal energy change of a van der Waals gas Analysis The change in internal energy of any simple compressible system in any phase during any process can be determined from Eq 1230 u 2 u 1 T 1 T 2 c v dT v 1 v 2 T P T v P dv The van der Waals equation of state is P RT v b a v 2 Then P T v R v b Thus T P T v P RT v b RT v b a v 2 a v 2 Substituting gives u 2 u 1 T 1 T 2 c 1 c 2 T dT v 1 v 2 a v 2 dv Integrating yields u 2 u 1 c 1 T 2 T 1 c 2 2 T 2 2 T 1 2 a 1 v 1 1 v 2 which is the desired relation EXAMPLE 128 Internal Energy as a Function of Temperature Alone Show that the internal energy of a an ideal gas and b an incompressible substance is a function of temperature only u uT SOLUTION It is to be shown that u uT for ideal gases and incompressible substances Final PDF to printer 659 CHAPTER 12 cen22672ch12643674indd 659 103117 0134 PM Analysis The differential change in the internal energy of a general simple compress ible system is given by Eq 1229 as du c v dT T P T v P dv a For an ideal gas Pv RT Then T P T v P T R v P P P 0 Thus du c v dT To complete the proof we need to show that cv is not a function of v either This is done with the help of Eq 1242 c v v T T 2 P T 2 v For an ideal gas P RTv Then P T v R v and 2 P T 2 v R v T v 0 Thus c v v T 0 which states that cv does not change with specific volume That is cv is not a function of specific volume either Therefore we conclude that the internal energy of an ideal gas is a function of temperature only Fig 1211 b For an incompressible substance v constant and thus dv 0 Also from Eq 1249 cp cv c since α β 0 for incompressible substances Then Eq 1229 reduces to du c dT Again we need to show that the specific heat c depends on temperature only and not on pressure or specific volume This is done with the help of Eq 1243 c p P T T 2 v T 2 P 0 since v constant Therefore we conclude that the internal energy of a truly incom pressible substance depends on temperature only EXAMPLE 129 The Specific Heat Difference of an Ideal Gas Show that cp cv R for an ideal gas SOLUTION It is to be shown that the specific heat difference for an ideal gas is equal to its gas constant FIGURE 1211 The internal energies and specific heats of ideal gases and incompressible substances depend on temperature only u uT cv cvT cp cpT Air u uT c cT Lake Final PDF to printer 660 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 660 103117 0134 PM Analysis This relation is easily proved by showing that the righthand side of Eq 1246 is equivalent to the gas constant R of the ideal gas c p c v T v T P 2 P v T P RT v P v T RT v 2 P v v RT P v T P 2 R P 2 Substituting T v T P 2 P v T T R P 2 P v R Therefore c p c v R 125 THE JOULETHOMSON COEFFICIENT When a fluid passes through a restriction such as a porous plug a capillary tube or an ordinary valve its pressure decreases As we have shown in Chap 5 the enthalpy of the fluid remains approximately constant during such a throt tling process You will remember that a fluid may experience a large drop in its temperature as a result of throttling which forms the basis of operation for refrigerators and air conditioners This is not always the case however The temperature of the fluid may remain unchanged or it may even increase during a throttling process Fig 1212 The temperature behavior of a fluid during a throttling h constant pro cess is described by the JouleThomson coefficient defined as μ T P h 1251 Thus the JouleThomson coefficient is a measure of the change in temperature with pressure during a constantenthalpy process Notice that if μ JT 0 temperature increases 0 temperature remains constant 0 temperature decreases during a throttling process A careful look at its defining equation reveals that the JouleThomson coeffi cient represents the slope of h constant lines on a TP diagram Such diagrams can be easily constructed from temperature and pressure measurements alone during throttling processes A fluid at a fixed temperature and pressure T and P thus fixed enthalpy is forced to flow through a porous plug and its tem perature and pressure downstream T and P are measured The experiment is repeated for different sizes of porous plugs each giving a different set of T and P Plotting the temperatures against the pressures gives us an h constant line on a FIGURE 1212 The temperature of a fluid may increase decrease or remain constant during a throttling process T1 20C T2 20C P1 800 kPa P2 200 kPa Final PDF to printer 661 CHAPTER 12 cen22672ch12643674indd 661 103117 0134 PM TP diagram as shown in Fig 1213 Repeating the experiment for different sets of inlet pressure and temperature and plotting the results we can construct a TP diagram for a substance with several h constant lines as shown in Fig 1214 Some constantenthalpy lines on the TP diagram pass through a point of zero slope or zero JouleThomson coefficient The line that passes through these points is called the inversion line and the temperature at a point where a constantenthalpy line intersects the inversion line is called the inversion temperature The temperature at the intersection of the P 0 line ordinate and the upper part of the inversion line is called the maximum inversion temperature Notice that the slopes of the h constant lines are negative μJT 0 at states to the right of the inversion line and posi tive μJT 0 to the left of the inversion line A throttling process proceeds along a constantenthalpy line in the direction of decreasing pressure that is from right to left Therefore the temperature of a fluid increases during a throttling process that takes place on the right hand side of the inversion line However the fluid temperature decreases dur ing a throttling process that takes place on the lefthand side of the inversion line It is clear from this diagram that a cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion temperature This presents a problem for substances whose maximum inversion temperature is well below room temperature For hydrogen for example the maximum inversion temperature is 68C Thus hydrogen must be cooled below this temperature if any further cooling is to be achieved by throttling Next we would like to develop a general relation for the JouleThomson coefficient in terms of the specific heats pressure specific volume and tem perature This is easily accomplished by modifying the generalized relation for enthalpy change Eq 1235 dh c p dT v T v T P dP For an h constant process we have dh 0 Then this equation can be rear ranged to give 1 c p v T v T P T P h μ JT 1252 which is the desired relation Thus the JouleThomson coefficient can be determined from a knowledge of the constantpressure specific heat and the PvT behavior of the substance Of course it is also possible to predict the constant pressure specific heat of a substance by using the JouleThomson coefficient which is relatively easy to determine together with the PvT data for the substance EXAMPLE 1210 JouleThomson Coefficient of an Ideal Gas Show that the JouleThomson coefficient of an ideal gas is zero SOLUTION It is to be shown that μJT 0 for an ideal gas Analysis For an ideal gas v RTP and thus v T P R P FIGURE 1213 The development of an h constant line on a TP diagram T P Exit states Inlet state h constant line 2 2 2 2 2 1 P1 P2 T2 varied P1 T1 fixed FIGURE 1214 Constantenthalpy lines of a substance on a TP diagram Maximum inversion temperature Inversion line h const T P JT 0 JT 0 Final PDF to printer 662 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 662 103117 0134 PM Substituting this into Eq 1252 yields μ JT 1 c p v T v T P 1 c p v T R P 1 c p v v 0 Discussion This result is not surprising since the enthalpy of an ideal gas is a function of temperature only h hT which requires that the temperature remain constant when the enthalpy remains constant Therefore a throttling process cannot be used to lower the temperature of an ideal gas Fig 1215 126 THE Δh Δu AND Δs OF REAL GASES We have mentioned many times that gases at low pressures behave as ideal gases and obey the relation Pv RT The properties of ideal gases are relatively easy to evaluate since the properties u h cv and cp depend on temperature only At high pressures however gases deviate considerably from idealgas behavior and it becomes necessary to account for this deviation In Chap 3 we accounted for the deviation in properties P v and T by either using more complex equations of state or evaluating the compressibility factor Z from the compressibility charts Now we extend the analysis to evaluate the changes in the enthalpy internal energy and entropy of nonideal real gases using the general relations for du dh and ds developed earlier Enthalpy Changes of Real Gases The enthalpy of a real gas in general depends on the pressure as well as on the temperature Thus the enthalpy change of a real gas during a process can be evaluated from the general relation for dh Eq 1236 h 2 h 1 T 1 T 2 c p dT P 1 P 2 v T v T P dP where P1 T1 and P2 T2 are the pressures and temperatures of the gas at the initial and the final states respectively For an isothermal process dT 0 and the first term vanishes For a constantpressure process dP 0 and the second term vanishes Properties are point functions and thus the change in a property between two specified states is the same no matter which process path is followed This fact can be exploited to greatly simplify the integration of Eq 1236 Consider for example the process shown on a Ts diagram in Fig 1216 The enthalpy change during this process h2 h1 can be determined by performing the integrations in Eq 1236 along a path that consists of two isothermal T1 constant and T2 constant lines and one isobaric P0 constant line instead of the actual process path as shown in Fig 1216 Although this approach increases the number of integrations it also simpli fies them since one property remains constant now during each part of the process The pressure P0 can be chosen to be very low or zero so that the gas can be treated as an ideal gas during the P0 constant process Using FIGURE 1215 The temperature of an ideal gas remains constant during a throttling process since h constant and T constant lines on a TP diagram coincide T P1 P2 P h constant line FIGURE 1216 An alternative process path to evaluate the enthalpy changes of real gases T s Actual process path Alternative process path T2 T1 1 1 2 2 P0 0 P2 P1 Final PDF to printer 663 CHAPTER 12 cen22672ch12643674indd 663 103117 0134 PM a superscript asterisk to denote an idealgas state we can express the enthalpy change of a real gas during process 12 as h 2 h 1 h 2 h 2 h 2 h 1 h 1 h 1 1253 where from Eq 1236 h 2 h 2 0 P 2 P 2 v T v T P T T 2 dP P 0 P 2 v T v T P T T 2 dP 1254 h 2 h 1 T 1 T 2 c p dT 0 T 1 T 2 c p0 T dT 1255 h 1 h 1 0 P 1 P 1 v T v T P T T 1 dP P 0 P 1 v T v T P T T 1 dP 1256 The difference between h and h is called the enthalpy departure and it represents the variation of the enthalpy of a gas with pressure at a fixed temperature The calculation of enthalpy departure requires a knowledge of the PvT behavior of the gas In the absence of such data we can use the relation Pv ZRT where Z is the compressibility factor Substituting v ZRTP and simplifying Eq 1256 we can write the enthalpy departure at any temperature T and pressure P as h h T R T 2 0 P Z T P dP P The preceding equation can be generalized by expressing it in terms of the reduced coordinates using T TcrTR and P PcrPR After some manipulations the enthalpy departure can be expressed in a nondimensionalized form as Z h h h T R u T cr T R 2 0 P R Z T R P R dln P R 1257 where Zh is called the enthalpy departure factor The integral in the preced ing equation can be performed graphically or numerically by employing data from the compressibility charts for various values of PR and TR The values of zh are presented in graphical form as a function of PR and TR in Fig A29 This graph is called the generalized enthalpy departure chart and it is used to determine the deviation of the enthalpy of a gas at a given P and T from the enthalpy of an ideal gas at the same T By replacing h with hideal for clar ity Eq 1253 for the enthalpy change of a gas during a process 12 can be rewritten as h 2 h 1 h 2 h 1 ideal R u T cr Z h 2 Z h 1 1258 or h 2 h 1 h 2 h 1 ideal R T cr Z h 2 Z h 1 1259 where the values of zh are determined from the generalized enthalpy departure chart and h 2 h 1 ideal is determined from the idealgas tables Notice that the last terms on the righthand side are zero for an ideal gas Final PDF to printer 664 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 664 103117 0134 PM Internal Energy Changes of Real Gases The internal energy change of a real gas is determined by relating it to the enthalpy change through the definition h u P v u Z R u T u 2 u 1 h 2 h 1 R u Z 2 T 2 Z 1 T 1 1260 Entropy Changes of Real Gases The entropy change of a real gas is determined by following an approach similar to that used above for the enthalpy change There is some difference in derivation however owing to the dependence of the idealgas entropy on pressure as well as the temperature The general relation for ds was expressed as Eq 1241 s 2 s 1 T 1 T 2 c p T dT P 1 P 2 v T P dP where P1 T1 and P2 T2 are the pressures and temperatures of the gas at the initial and the final states respectively The thought that comes to mind at this point is to perform the integrations in the previous equation first along a T1 constant line to zero pressure then along the P 0 line to T2 and finally along the T2 constant line to P2 as we did for the enthalpy This approach is not suitable for entropychange calculations however since it involves the value of entropy at zero pressure which is infinity We can avoid this diffi culty by choosing a different but more complex path between the two states as shown in Fig 1217 Then the entropy change can be expressed as s 2 s 1 s 2 s b s b s 2 s 2 s 1 s 1 s a s a s 1 1261 States 1 and 1 are identical T1 T 1 and P1 P 1 and so are states 2 and 2 The gas is assumed to behave as an ideal gas at the imaginary states 1 and 2 as well as at the states between the two Therefore the entropy change during process 12 can be determined from the entropychange relations for ideal gases The calculation of entropy change between an actual state and the cor responding imaginary idealgas state is more involved however and requires the use of generalized entropy departure charts as explained below Consider a gas at a pressure P and temperature T To determine how much different the entropy of this gas would be if it were an ideal gas at the same temperature and pressure we consider an isothermal process from the actual state P T to zero or close to zero pressure and back to the imaginary idealgas state P T denoted by superscript as shown in Fig 1217 The entropy change during this isothermal process can be expressed as s P s P T s P s 0 T s 0 s P T 0 P v T P dP P 0 v T P dP where v ZRTP and v videal RTP Performing the differentiations and rearranging we obtain FIGURE 1217 An alternative process path to evaluate the entropy changes of real gases during process 12 s T 2 b a 1 1 2 Alternative process path Actual process path P2 P1 P0 T2 T1 Final PDF to printer 665 CHAPTER 12 cen22672ch12643674indd 665 103117 0134 PM s P s P T 0 P 1 Z R P RT P Z T P dP By substituting T TcrTR and P PcrPR and rearranging the entropy depar ture can be expressed in a nondimensionalized form as Z s s s TP R u 0 P R Z 1 T R Z T R P R dln P R 1262 The difference s s TP is called the entropy departure and Zs is called the entropy departure factor The integral in the preceding equation can be performed by using data from the compressibility charts The values of Zs are presented in graphical form as a function of PR and TR in Fig A30 This graph is called the generalized entropy departure chart and it is used to determine the deviation of the entropy of a gas at a given P and T from the entropy of an ideal gas at the same P and T Replacing s with sideal for clarity we can rewrite Eq 1261 for the entropy change of a gas during a process 12 as s 2 s 1 s 2 s 1 ideal R u Z s 2 Z s 1 1263 or s 2 s 1 s 2 s 1 ideal R Z s 2 Z s 1 1264 where the values of Zs are determined from the generalized entropy departure chart and the entropy change s2 s1ideal is determined from the idealgas relations for entropy change Notice that the last terms on the righthand side are zero for an ideal gas EXAMPLE 1211 Thermodynamic Analysis with Nonideal Gas Properties Propane is compressed isothermally by a pistoncylinder device from 200F and 200 psia to 800 psia Fig 1218 Using the generalized charts determine the work done and the heat transfer per unit mass of propane SOLUTION Propane is compressed isothermally by a pistoncylinder device The work done and the heat transfer are to be determined using the generalized charts Assumptions 1 The compression process is quasiequilibrium 2 Kinetic and potential energy changes are negligible Analysis The critical temperature and pressure of propane are Tcr 6659 R and Pcr 617 psia Table A1E respectively Propane remains close to its critical tem perature and is compressed to a pressure above its critical value Therefore propane is expected to deviate from the idealgas behavior and thus it should be treated as a real gas The enthalpy departure and the compressibility factors of propane at the initial and the final states are determined from the generalized charts to be Figs A29 and A15 T R 1 T 1 T cr 660 R 6659 R 0991 P R 1 P 1 P cr 200 psia 617 psia 0324 Z h 1 037 and Z 1 088 FIGURE 1218 Schematic for Example 1211 Propane 200 psia 200F Q Final PDF to printer 666 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 666 103117 0134 PM and T R 2 T 2 T cr 660 R 6659 R 0991 P R 2 P 2 P cr 800 psia 617 psia 1297 Z h 2 42 and Z 2 022 As an approximation treating propane as a real gas with Zavg Z1 Z22 088 0222 055 gives Pv ZRT Z avg RT C constant Then the boundary work becomes w bin 1 2 P dv 1 2 C v dv C ln v 2 v 1 Z avg RT ln Z 2 RT P 2 Z 1 RT P 1 Z avg RT ln Z 2 P 1 Z 1 P 2 055004504 BtulbmR660 R ln 022200 psia 088800 psia 453 Btulbm Also h 2 h 1 R T cr Z h 1 Z h 2 h 2 h 1 ideal 004504 BtulbmR6659 R037 42 0 1149 Btulbm u 2 u 1 h 2 h 1 R Z 2 T 2 Z 1 T 1 1149 Btulbm 004504 BtulbmR 022660 R 088660 R 953 Btulbm Then the heat transfer during this process is determined from the closedsystem energy balance equation for the pistoncylinder device to be E in E out Δ E system q in w bin Δu u 2 u 1 q in u 2 u 1 w bin 953 453 1406 Btulbm The negative sign indicates heat rejection Therefore heat transfer out of the system during this process is q out 1406 Btulbm Discussion Note that if the idealgas assumption were used for propane the mag nitudes of boundary work and heat transfer would have been the same 412 Btulbm Therefore the idealgas approximation would underestimate boundary work by 9 percent and the heat transfer by 71 percent 0 Final PDF to printer 667 CHAPTER 12 cen22672ch12643674indd 667 103117 0134 PM SUMMARY Some thermodynamic properties can be measured directly but many others cannot Therefore it is necessary to develop some relations between these two groups so that the properties that cannot be measured directly can be evaluated The derivations are based on the fact that properties are point functions and the state of a simple compressible system is completely specified by any two independent intensive properties The equations that relate the partial derivatives of proper ties P v T and s of a simple compressible substance to each other are called the Maxwell relations They are obtained from the four Gibbs equations expressed as du T ds P dv dh T ds v dP da s dT P dv dg s dT v dP The Maxwell relations are T v s P s v T P s v s P s v T P T v s P T v T P The Clapeyron equation enables us to determine the enthalpy change associated with a phase change from a knowledge of P v and T data alone It is expressed as dP dT sat h fg T v fg For liquidvapor and solidvapor phasechange processes at low pressures it can be approximated as ln P 2 P 1 sat h fg R 1 T 1 1 T 2 sat The changes in internal energy enthalpy and entropy of a sim ple compressible substance can be expressed in terms of pres sure specific volume temperature and specific heats alone as du c v dT T P T v P dv dh c p dT v T v T P dP ds c v T dT P T v dv or ds c p T dT v T P dP For specific heats we have the following general relations c v v T T 2 P T 2 v c p P T T 2 v T 2 P c pT c p0T T 0 P 2 v T 2 P dP c p c v T v T P 2 P v T c p c v vT β 2 α where β is the volume expansivity and α is the isothermal compressibility defined as β 1 v v T P and α 1 v v P T The difference cp cv is equal to R for ideal gases and to zero for incompressible substances The temperature behavior of a fluid during a throttling h constant process is described by the JouleThomson coefficient defined as μ JT T P h The JouleThomson coefficient is a measure of the change in temperature of a substance with pressure during a constant enthalpy process and it can also be expressed as μ JT 1 c p v T v T P The enthalpy internal energy and entropy changes of real gases can be determined accurately by utilizing generalized enthalpy or entropy departure charts to account for the deviation from the idealgas behavior by using the following relations h 2 h 1 h 2 h 1 ideal R u T cr Z h 2 Z h 1 u 2 u 1 h 2 h 1 R u Z 2 T 2 Z 1 T 1 s 2 s 1 s 2 s 1 ideal R u Z s 2 Z s 1 where the values of Zh and Zs are determined from the generalized charts Final PDF to printer 668 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 668 103117 0134 PM REFERENCES AND SUGGESTED READINGS 1 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley 2006 2 K Wark Jr Advanced Thermodynamics for Engineers New York McGrawHill 1995 PROBLEMS Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Partial Derivatives and Associated Relations 121C What is the difference between partial differentials and ordinary differentials 122C Consider the function zx y Plot a differential sur face on xyz coordinates and indicate x dx y dy zx zy and dz 123C Consider a function zx y and its partial derivative zyx Under what conditions is this partial derivative equal to the total derivative dzdy 124C Consider a function zx y and its partial derivative zyx If this partial derivative is equal to zero for all values of x what does it indicate 125C Consider a function zx y and its partial derivative zyx Can this partial derivative still be a function of x 126C Consider a function f x and its derivative dfdx Can this derivative be determined by evaluating dxdf and taking its inverse 127C Consider the function zx y its partial derivatives zxy and zyx and the total derivative dzdx a How do the magnitudes xy and dx compare b How do the magnitudes zy and dz compare c Is there any relation among dz zx and zy 128 Consider air at 350 K and 075 m3kg Using Eq 123 determine the change in pressure corresponding to an increase of a 1 percent in temperature at constant specific volume b 1 percent in specific volume at constant temperature and c 1 percent in both the temperature and specific volume 129 Repeat Prob 128 for helium 1210E Nitrogen gas at 800 R and 50 psia behaves as an ideal gas Estimate the cp and cv of the nitrogen at this state using enthalpy and internal energy data from Table A18E and compare them to the values listed in Table A2Eb Answers 0250 BtulbmR 0179 BtulbmR 1211 Consider an ideal gas at 400 K and 100 kPa As a result of some disturbance the conditions of the gas change to 404 K and 96 kPa Estimate the change in the specific volume of the gas using a Eq 123 and b the idealgas relation at each state 1212 Using the equation of state Pv a RT verify a the cyclic relation and b the reciprocity relation at con stant v 1213 Prove for an ideal gas that a the P constant lines on a Tv diagram are straight lines and b the highpressure lines are steeper than the lowpressure lines The Maxwell Relations 1214E Verify the validity of the last Maxwell relation Eq 1219 for steam at 800F and 400 psia 1215 Verify the validity of the last Maxwell relation Eq 1219 for refrigerant134a at 50C and 07 MPa 1216 Reconsider Prob 1215 Using appropriate software verify the validity of the last Maxwell relation for refrigerant134a at the specified state 1217 Show how you would evaluate T v u a and g from the thermodynamic function h hs P 1218 Using the Maxwell relations determine a relation for sPT for a gas whose equation of state is Pv b RT Answer RP 1219 Using the Maxwell relations determine a rela tion for svT for a gas whose equation of state is P av2 v b RT 1220 Using the Maxwell relations and the idealgas equa tion of state determine a relation for svT for an ideal gas Answer Rv 1221 Prove that P T s k k 1 P T v The Clapeyron Equation 1222C What is the value of the Clapeyron equation in thermodynamics 1223C What approximations are involved in the Clapeyron Clausius equation Final PDF to printer 669 CHAPTER 12 cen22672ch12643674indd 669 103117 0134 PM 1224 Using the Clapeyron equation estimate the enthalpy of vaporization of refrigerant134a at 40C and compare it to the tabulated value 1225 Reconsider Prob 1224 Using appropriate software plot the enthalpy of vaporization of refrigerant134a as a function of temperature over the tempera ture range 20 to 80C by using the Clapeyron equation and the refrigerant134a data in the software Discuss your results 1226 Using the Clapeyron equation estimate the enthalpy of vaporization of water at 300 kPa and compare it to the tabu lated value 1227E Determine the hfg of refrigerant134a at 10F on the basis of a the Clapeyron equation and b the Clapeyron Clausius equation Compare your results to the tabulated hfg value 1228 Using the ClapeyronClausius equation and the triplepoint data of water estimate the sublimation pressure of water at 30C and compare to the value in Table A8 1229 Two grams of a saturated liquid are converted to a saturated vapor by being heated in a weighted pistoncylinder device arranged to maintain the pressure at 200 kPa During the phase conversion the volume of the system increases by 1000 cm3 5 kJ of heat are required and the temperature of the substance stays constant at 80C Estimate the saturation pressure Psat of this substance when its temperature is 100C 1230 Estimate sfg of the substance in Prob 1229 at 80C Answer 708 kJkgK 1231 Show that c pg c pf T h fg T T P v fg P T sat 1232 The saturation table for refrigerant134a lists the following at 40C P 5125 kPa hfg 22586 kJkg and vfg 035993 m3kg Estimate the saturation pressure of refrigerant134a at 50C and 30C General Relations for du dh ds cv and cp 1233C Can the variation of specific heat cp with pressure at a given temperature be determined from a knowledge of PvT data alone 1234 Determine the change in the internal energy of air in kJkg as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state 1235 Determine the change in the enthalpy of air in kJkg as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state Answers 335 kJkg 285 kJkg 1236 Determine the change in the entropy of air in kJkgK as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state 1237 Determine the change in the internal energy of helium in kJkg as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state Answers 872 kJkg 872 kJkg 1238 Determine the change in the enthalpy of helium in kJ kg as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state 1239 Determine the change in the entropy of helium in kJ kgK as it undergoes a change of state from 100 kPa and 20C to 600 kPa and 300C using the equation of state Pv a RT where a 010 m3kg and compare the result to the value obtained by using the ideal gas equation of state Answers 0239 kJkgK 0239 kJkgK 1240 Estimate the volume expansivity β and the isothermal compressibility α of refrigerant134a at 200 kPa and 30C 1241E Estimate the specific heat difference cp cv for liquid water at 1000 psia and 300F Answer 0183 BtulbmR 1242 Derive expressions for a Δu b Δh and c Δs for a gas whose equation of state is Pv a RT for an isothermal process Answers a 0 b aP2 P1 c R ln P2P1 1243 Derive an expression for the specific heat difference cp cv for a an ideal gas b a van der Waals gas and c an incompressible substance 1244 Derive an expression for the specific heat difference of a substance whose equation of state is P RT v b a v v b T 12 where a and b are empirical constants 1245 Derive an expression for the isothermal compressibil ity of a substance whose equation of state is P RT v b a v 2 T where a and b are empirical constants 1246 Derive an expression for the specific heat difference of a substance whose equation of state is P RT v b a v 2 T where a and b are empirical constants Final PDF to printer 670 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 670 103117 0134 PM 1247 Show that c p c v T P T v v T P 1248 Show that the enthalpy of an ideal gas is a function of temperature only and that for an incompressible substance it also depends on pressure 1249 Temperature may alternatively be defined as T u s v Prove that this definition reduces the net entropy change of two constantvolume systems filled with simple compress ible substances to zero as the two systems approach thermal equilibrium 1250 Show that β α P T v 1251 Derive expressions for uPT and hvT in terms of P v and T only 1252 Demonstrate that k c p c v vα v P s 1253 The Helmholtz function of a substance has the form a RT ln v v 0 c T 0 1 T T 0 T T 0 ln T T 0 where T0 and v0 are the temperature and specific volume at a reference state Show how to obtain P h s cv and cp from this expression The JouleThomson Coefficient 1254C What does the JouleThomson coefficient represent 1255C Describe the inversion line and the maximum inver sion temperature 1256C Does the JouleThomson coefficient of a substance change with temperature at a fixed pressure 1257C The pressure of a fluid always decreases during an adiabatic throttling process Is this also the case for the temperature 1258C Will the temperature of helium change if it is throt tled adiabatically from 300 K and 600 kPa to 150 kPa 1259 Estimate the JouleThomson coefficient of refrigerant 134a at 200 kPa and 20C Answer 00235 KkPa 1260 Estimate the JouleThomson coefficient of refrigerant 134a at 07 MPa and 50C 1261E Estimate the JouleThomson coefficient of refrigerant 134a at 30 psia and 20F 1262 Steam is throttled slightly from 1 MPa and 300C Will the temperature of the steam increase decrease or remain the same during this process 1263 What is the most general equation of state for which the JouleThomson coefficient is always zero 1264 Demonstrate that the JouleThomson coefficient is given by μ T 2 c p v T T P 1265 Consider a gas whose equation of state is Pv a RT where a is a positive constant Is it possible to cool this gas by throttling 1266 The equation of state of a gas is given by v RT P a T b where a and b are constants Use this equation of state to derive an equation for the JouleThomson coefficient inver sion line The dh du and ds of Real Gases 1267C What is the enthalpy departure 1268C On the generalized enthalpy departure chart the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero How do you explain this behavior 1269C Why is the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T 1270 What is the error involved in the a enthalpy and b internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas Answers a 50 percent b 49 percent 1271E Determine the enthalpy of nitrogen in Btulbm at 400 R and 2000 psia using a data from the idealgas nitro gen table and b the generalized enthalpy chart Compare your results to the actual value of 1778 Btulbm 1272 Saturated water vapor at 300C is expanded while its pressure is kept constant until its temperature is 700C Calculate the change in the specific enthalpy and entropy using a the departure charts and b the property tables Answers a 973 kJkg 1295 kJkgK b 1129 kJkg 1541 kJkgK 1273 Determine the enthalpy change and the entropy change of oxygen per unit mole as it undergoes a change of state from 220 K and 5 MPa to 300 K and 10 MPa a by assuming idealgas behavior and b by accounting for the deviation from idealgas behavior 1274 Methane is compressed adiabatically by a steadyflow compressor from 08 MPa and 10C to 6 MPa and 175C at Final PDF to printer 671 CHAPTER 12 cen22672ch12643674indd 671 103117 0134 PM a rate of 02 kgs Using the generalized charts determine the required power input to the compressor Answer 799 kW FIGURE P1274 CH4 m 02 kgs 6 MPa 175C 08 MPa 10C W 1275 Carbon dioxide enters an adiabatic nozzle at 8 MPa and 450 K with a low velocity and leaves at 2 MPa and 350 K Using the generalized enthalpy departure chart determine the exit velocity of the carbon dioxide Answer 384 ms 1276 Reconsider Prob 1275 Using appropriate software compare the exit velocity to the nozzle assuming idealgas behavior the generalized chart data and the software data for carbon dioxide 1277E Oxygen is adiabatically and reversibly expanded in a nozzle from 200 psia and 600F to 70 psia Determine the velocity at which the oxygen leaves the nozzle assuming that it enters with negligible velocity treating the oxygen as an ideal gas with temperaturevariable specific heats and using the departure charts Answers 1738 fts 1740 fts FIGURE P1277E O2 200 psia 600F 0 fts 70 psia 1278 A 005m3 wellinsulated rigid tank contains oxygen at 175 K and 6 MPa A paddle wheel placed in the tank is turned on and the temperature of the oxygen rises to 225 K Using the generalized charts determine a the final pressure in the tank and b the paddlewheel work done during this process Answers a 9652 kPa b 423 kJ 1279 Propane is compressed isothermally by a piston cylinder device from 100C and 1 MPa to 4 MPa Using the generalized charts determine the work done and the heat transfer per unit mass of propane 1280 Reconsider Prob 1279 Using appropriate software extend the problem to compare the solutions based on the idealgas assumption generalized chart data and real fluid data Also extend the solution to methane 1281 Reconsider Prob 1279 Determine the exergy destruction associated with the process Assume T0 30C Review Problems 1282 Develop expressions for h u s Pr and vr for an ideal gas whose c p o is given by c p o ai T in a 0 e βT β T e βT 1 where ai a0 n and β are empirical constants 1283 Starting with the relation dh T ds v dP show that the slope of a constantpressure line on an hs diagram a is constant in the saturation region and b increases with tem perature in the superheated region 1284 Using the cyclic relation and the first Maxwell relation derive the other three Maxwell relations 1285 For ideal gases the development of the constant volume specific heat yields u v T 0 Prove this by using the definitions of pressure and temperature T usv and P uvs 1286 Show that c v T v T s P T v and c p T P T s v T P 1287 Temperature and pressure may be defined as T u s v and P u v v Using these definitions prove that for a simple compressible substance s v u P T 1288 For a homogeneous singlephase simple pure sub stance the pressure and temperature are independent proper ties and any property can be expressed as a function of these two properties Taking v vP T show that the change in Final PDF to printer 672 THERMODYNAMIC PROPERTY RELATIONS cen22672ch12643674indd 672 103117 0134 PM specific volume can be expressed in terms of the volume expansivity β and isothermal compressibility α as dv v β dT α dP Also assuming constant average values for β and α obtain a relation for the ratio of the specific volumes v2v1 as a homoge neous system undergoes a process from state 1 to state 2 1289 Repeat Prob 1288 for an isobaric process 1290 Starting with μJT 1cpTvTp v and noting that Pv ZRT where Z ZP T is the compressibility factor show that the position of the JouleThomson coefficient inversion curve on the TP plane is given by the equation ZTP 0 1291 Consider an infinitesimal reversible adiabatic com pression or expansion process By taking s sP v and using the Maxwell relations show that for this process Pvk con stant where k is the isentropic expansion exponent defined as k v P P v s Also show that the isentropic expansion exponent k reduces to the specific heat ratio cpcv for an ideal gas 1292 Estimate the cp of nitrogen at 300 kPa and 400 K using a the relation in Prob 1291 and b its definition Compare your results to the value listed in Table A2b 1293 The volume expansivity of water at 20C is β 0207 106 K1 Treating this value as a constant determine the change in volume of 05 m3 of water as it is heated from 10C to 30C at constant pressure 1294 Steam is throttled from 45 MPa and 300C to 25 MPa Estimate the temperature change of the steam dur ing this process and the average JouleThomson coefficient Answers 263C 131CMPa 1295E Argon gas enters a turbine at 1000 psia and 1000 R with a velocity of 300 fts and leaves at 150 psia and 500 R with a velocity of 450 fts at a rate of 12 lbms Heat is being lost to the surroundings at 75F at a rate of 80 Btus Using the generalized charts determine a the power output of the turbine and b the exergy destruction associated with the process Answers a 922 hp b 122 Btus FIGURE P1295E Ar m 12 lbms W 1000 psia 1000 R 300 fts 80 Btus 150 psia 500 R 450 fts T0 75F 1296E Methane is to be adiabatically and reversibly com pressed from 50 psia and 100F to 500 psia Calculate the specific work required for this compression treating the meth ane as an ideal gas with variable specific heats and using the departure charts FIGURE P1296E Methane 500 psia 50 psia 100F w 1297 Propane at 500 kPa and 100C is compressed in a steadyflow device to 4000 kPa and 500C Calculate the change in the specific entropy of the propane and the specific work required for this compression a treating the propane as an ideal gas with temperaturevariable specific heats and b using the departure charts Answers a 1121 kJkg 1587 kJkgK b 1113 kJkg 1583 kJkgK 1298 Reconsider Prob 1297 Determine the secondlaw efficiency of the compression process Take T0 25C 1299 A rigid tank contains 12 m3 of argon at 100C and 1 MPa Heat is now transferred to argon until the temperature in the tank rises to 0C Using the generalized charts determine a the mass of the argon in the tank b the final pressure and c the heat transfer Answers a 351 kg b 1531 kPa c 1251 kJ 12100 Methane is contained in a pistoncylinder device and is heated at constant pressure of 5 MPa from 100 to 250C Determine the heat transfer work and entropy change per unit mass of the methane using a the idealgas assumption b the generalized charts and c real fluid data This problem is solved using appropriate software 12101 An adiabatic 02m3 storage tank that is initially evacuated is connected to a supply line that carries nitrogen at 225 K and 10 MPa A valve is opened and nitrogen flows into the tank from the supply line The valve is closed when the pressure in the tank reaches 10 MPa Determine the final temperature in the tank a treating nitrogen as an ideal gas Final PDF to printer 673 CHAPTER 12 cen22672ch12643674indd 673 103117 0134 PM and b using generalized charts Compare your results to the actual value of 293 K FIGURE P12101 02 m3 Initially evacuated 225 K 10 MPa N2 Fundamentals of Engineering FE Exam Problems 12102 A substance whose JouleThomson coefficient is negative is throttled to a lower pressure During this process select the correct statement a the temperature of the substance will increase b the temperature of the substance will decrease c the entropy of the substance will remain constant d the entropy of the substance will decrease e the enthalpy of the substance will decrease 12103 Consider the liquidvapor saturation curve of a pure substance on the PT diagram The magnitude of the slope of the tangent line to this curve at a temperature T in Kelvin is a proportional to the enthalpy of vaporization hfg at that temperature b proportional to the temperature T c proportional to the square of the temperature T d proportional to the volume change vfg at that temperature e inversely proportional to the entropy change sfg at that temperature 12104 For a gas whose equation of state is Pv b RT the specified heat difference cp cv is equal to a R b R b c R b d 0 e R1 vb 12105 Based on the generalized charts the error involved in the enthalpy of CO2 at 300 K and 5 MPa if it is assumed to be an ideal gas is a 0 b 9 c 16 d 22 e 27 12106 Based on data from the refrigerant134a tables the JouleThompson coefficient of refrigerant134a at 08 MPa and 60C is approximately a 0 b 5CMPa c 11CMPa d 16CMPa e 25CMPa Design and Essay Problems 12107 There have been several attempts to represent the thermodynamic relations geometrically the best known of these being Koenigs thermodynamic square shown in the fig ure There is a systematic way of obtaining the four Maxwell relations as well as the four relations for du dh dg and da from this figure By comparing these relations to Koenigs dia gram come up with the rules to obtain these eight thermody namic relations from this diagram FIGURE P12107 g P h s u a T v 12108 Several attempts have been made to express the par tial derivatives of the most common thermodynamic properties in a compact and systematic manner in terms of measurable properties The work of P W Bridgman is perhaps the most fruitful of all and it resulted in the wellknown Bridgmans table The 28 entries in that table are sufficient to express the partial derivatives of the eight common properties P T v s u h f and g in terms of the six properties P v T cp β and α which can be measured directly or indirectly with relative ease Obtain a copy of Bridgmans table and explain with examples how it is used 12109 Consider the function z zx y Write an essay on the physical interpretation of the ordinary derivative dzdx and the partial derivative zxy Explain how these two derivatives are related to each other and when they become equivalent Final PDF to printer No text present cen22672ch13675710indd 675 103117 0112 PM 675 CHAPTER 13 GAS M I X T U R ES U p to this point we have limited our consideration to thermodynamic systems that involve a single pure substance such as water Many important thermodynamic applications however involve mixtures of several pure substances rather than a single pure substance Therefore it is important to develop an understanding of mixtures and learn how to handle them In this chapter we deal with nonreacting gas mixtures A nonreacting gas mixture can be treated as a pure substance since it is usually a homogeneous mixture of different gases The properties of a gas mixture obviously depend on the properties of the individual gases called components or constituents as well as on the amount of each gas in the mixture Therefore it is possible to prepare tables of properties for mixtures This has been done for common mixtures such as air It is not practical to prepare property tables for every conceivable mixture composition however since the number of possible compositions is endless Therefore we need to develop rules for determining mixture properties from a knowledge of mixture composition and the proper ties of the individual components We do this first for idealgas mixtures and then for realgas mixtures The basic principles involved are also applicable to liquid or solid mixtures called solutions OBJECTIVES The objectives of Chapter 13 are to Develop rules for determining nonreacting gas mixture properties from knowledge of mixture composition and the properties of the individual components Define the quantities used to describe the composition of a mixture such as mass fraction mole fraction and volume fraction Apply the rules for determining mixture properties of idealgas mixtures and realgas mixtures Predict the PvT behavior of gas mixtures based on Daltons law of additive pressures and Amagats law of additive volumes Final PDF to printer 676 GAS MIXTURES cen22672ch13675710indd 676 103117 0112 PM 131 COMPOSITION OF A GAS MIXTURE MASS AND MOLE FRACTIONS To determine the properties of a mixture we need to know the composition of the mixture as well as the properties of the individual components There are two ways to describe the composition of a mixture either by specifying the number of moles of each component called molar analysis or by specifying the mass of each component called gravimetric analysis Consider a gas mixture composed of k components The mass of the mix ture mm is the sum of the masses of the individual components and the mole number of the mixture Nm is the sum of the mole numbers of the individual components Figs 131 and 132 That is m m i 1 k m i and N m i 1 k N i 131a b The ratio of the mass of a component to the mass of the mixture is called the mass fraction mf and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction y mf i m i m m and y i N i N m 132a b Dividing Eq 131a by mm or Eq 131b by Nm we can easily show that the sum of the mass fractions or mole fractions for a mixture is equal to 1 Fig 133 i 1 k mf i 1 and i 1 k y i 1 The mass of a substance can be expressed in terms of the mole number N and molar mass M of the substance as m NM Then the apparent or average molar mass and the gas constant of a mixture can be expressed as M m m m N m m i N m N i M i N m i 1 k y i M i and R m R u M m 133a b The molar mass of a mixture can also be expressed as M m m m N m m m m i M i 1 m i m m M i 1 i 1 k mf i M i 134 Mass and mole fractions of a mixture are related by mf i m i m m N i M i N m M m y i M i M m 135 Throughout this chapter the subscript m denotes the gas mixture and the subscript i denotes any single component of the mixture FIGURE 131 The mass of a mixture is equal to the sum of the masses of its components H2 6 kg O2 32 kg H2 O2 38 kg FIGURE 132 The number of moles of a nonreacting mixture is equal to the sum of the number of moles of its components H2 3 kmol O2 1 kmol H2 O2 4 kmol FIGURE 133 The sum of the mole fractions of a mixture is equal to 1 H2 O2 yH2 075 yO2 025 100 Final PDF to printer 677 CHAPTER 13 cen22672ch13675710indd 677 103117 0112 PM EXAMPLE 131 The Gas Constant of a Gas Mixture A gas mixture consists of 5 lbmol of H2 and 4 lbmol of N2 as shown in Fig 134 Determine the mass of each gas and the apparent gas constant of the mixture SOLUTION The mole numbers of the constituents of a gas mixture are given The mass of each gas and the apparent gas constant are to be determined Properties The molar masses of H2 and N2 are 20 and 280 lbmlbmol respectively Table A1E Analysis The mass of each component is determined from N H 2 5 lbmol m H 2 N H 2 M H 2 5 lbmol 20 lbm lbmol 10 lbm N N 2 5 lbmol m H 2 N H 2 M H 2 4 lbmol 28 lbm lbmol 112 lbm The total mass and the total number of moles are m m m H 2 m N 2 10 lbm 112 lbm 122 lbm N m N H 2 N N 2 5 lbmol 4 lbmol 9 lbmol The molar mass and the gas constant of the mixture are determined from their definitions M m m m N m 122 lbm 9 lbmol 1356 lbm lbmol and R m R u M m 1986 Btu lbmolR 1356 lbm lbmol 01465 Btu lbmR Discussion The mole fractions of H2 and N2 can be calculated as 0556 and 0444 and the corresponding mass fractions are 0082 and 0918 respectively FIGURE 134 Schematic for Example 131 5 lbmol H2 4 lbmol N2 132 PvT BEHAVIOR OF GAS MIXTURES IDEAL AND REAL GASES An ideal gas is defined as a gas whose molecules are spaced far apart so that the behavior of a molecule is not influenced by the presence of other moleculesa situation encountered at low densities We also mentioned that real gases approximate this behavior closely when they are at a low pressure or high temperature relative to their criticalpoint values The PvT behavior of an ideal gas is expressed by the simple relation Pv RT which is called the idealgas equation of state The PvT behavior of real gases is expressed by more complex equations of state or by Pv ZRT where Z is the compress ibility factor When two or more ideal gases are mixed the behavior of a molecule normally is not influenced by the presence of other similar or dissimilar molecules and therefore a nonreacting mixture of ideal gases also behaves as an ideal gas Air for example is conveniently treated as an ideal gas in the range where nitrogen and oxygen behave as ideal gases When a gas mixture consists of real nonideal gases however the prediction of the PvT behavior of the mixture becomes rather involved Final PDF to printer 678 GAS MIXTURES cen22672ch13675710indd 678 103117 0112 PM The prediction of the PvT behavior of gas mixtures is usually based on two models Daltons law of additive pressures and Amagats law of additive volumes Both models are described and discussed below Daltons law of additive pressures The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume Fig 135 Amagats law of additive volumes The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mix ture temperature and pressure Fig 136 Daltons and Amagats laws hold exactly for idealgas mixtures but only approximately for realgas mixtures This is due to intermolecular forces that may be significant for real gases at high densities For ideal gases these two laws are identical and give identical results Daltons and Amagats laws can be expressed as follows Daltons law P m i 1 k P i T m V m Amagats law V m i 1 k V i T m P m exact for ideal gases approximate for real gases In these relations Pi is called the component pressure and Vi is called the component volume Fig 137 Note that Vi is the volume a component would occupy if it existed alone at Tm and Pm not the actual volume occupied by the component in the mixture In a vessel that holds a gas mixture each component fills the entire volume of the vessel Therefore the volume of each component is equal to the volume of the vessel Also the ratio Pi Pm is called the pressure fraction and the ratio Vi Vm is called the volume fraction of component i IdealGas Mixtures For ideal gases Pi and Vi can be related to yi by using the idealgas relation for both the components and the gas mixture P i T m V m P m N i R u T m V m N m R u T m V m N i N m y i V i T m P m V m N i R u T m P m N m R u T m P m N i N m y i Therefore P i P m V i V m N i N m y i 138 Equation 138 is strictly valid for idealgas mixtures since it is derived by assum ing idealgas behavior for the gas mixture and each of its components The quantity yiPm is called the partial pressure identical to the component pressure for ideal gases and the quantity yiVm is called the partial volume identical to the compo nent volume for ideal gases Note that for an idealgas mixture the mole fraction the pressure fraction and the volume fraction of a component are identical 136 137 FIGURE 136 Amagats law of additive volumes for a mixture of two ideal gases Gas A P T VA Gas B P T VB Gas mixture A B P T VA VB FIGURE 137 The volume a component would occupy if it existed alone at the mixture T and P is called the component volume for ideal gases it is equal to the partial volume yiVm O2 N2 100 kPa 400 K 1 m3 O2 100 kPa 400 K 03 m3 N2 100 kPa 400 K 07 m3 FIGURE 135 Daltons law of additive pressures for a mixture of two ideal gases Gas mixture A B V T PA PB Gas B V T PB Gas A V T PA Final PDF to printer 679 CHAPTER 13 cen22672ch13675710indd 679 103117 0112 PM The composition of an idealgas mixture such as the exhaust gases leaving a combustion chamber is often determined by a volumetric analysis called the Orsat analysis and Eq 138 A sample gas at a known volume pressure and temperature is passed into a vessel containing reagents that absorb one of the gases The volume of the remaining gas is then measured at the original pressure and temperature The ratio of the reduction in volume to the original volume volume fraction represents the mole fraction of that particular gas RealGas Mixtures Daltons law of additive pressures and Amagats law of additive volumes can also be used for real gases often with reasonable accuracy This time how ever the component pressures or component volumes should be evaluated from relations that take into account the deviation of each component from idealgas behavior One way of doing that is to use more exact equations of state van der Waals BeattieBridgeman BenedictWebbRubin etc instead of the idealgas equation of state Another way is to use the compress ibility factor Fig 138 as PV ZN R u T 139 The compressibility factor of the mixture Zm can be expressed in terms of the compressibility factors of the individual gases Zi by applying Eq 139 to both sides of Daltons law or Amagats law expression and simplifying We obtain Z m i 1 k y i Z i 1310 where Zi is determined either at Tm and Vm Daltons law or at Tm and Pm Amagats law for each individual gas It may seem that using either law gives the same result but it does not The compressibilityfactor approach in general gives more accurate results when the Zis in Eq 1310 are evaluated by using Amagats law instead of Daltons law This is because Amagats law involves the use of mixture pres sure Pm which accounts for the influence of intermolecular forces between the molecules of different gases Daltons law disregards the influence of dis similar molecules in a mixture on each other As a result it tends to underpre dict the pressure of a gas mixture for a given Vm and Tm Therefore Daltons law is more appropriate for gas mixtures at low pressures Amagats law is more appropriate at high pressures Note that there is a significant difference between using the compressibility factor for a single gas and for a mixture of gases The compressibility factor predicts the PvT behavior of single gases rather accurately as discussed in Chap 3 but not for mixtures of gases When we use compressibility factors for the components of a gas mixture we account for the influence of like mol ecules on each other the influence of dissimilar molecules remains largely unaccounted for Consequently a property value predicted by this approach may be considerably different from the experimentally determined value Another approach for predicting the PvT behavior of a gas mixture is to treat the gas mixture as a pseudopure substance Fig 139 One such method proposed by W B Kay in 1936 and called Kays rule involves the use of a pseudocritical pressure P crm ʹ and pseudocritical temperature T crm ʹ for FIGURE 138 One way of predicting the PvT behavior of a realgas mixture is to use the compressibility factor Pm Vm Zm Nm Ru Tm Zm yi Zi k i 1 FIGURE 139 Another way of predicting the PvT behavior of a realgas mixture is to treat it as a pseudopure substance with critical properties P cr ʹ and T cr ʹ Pseudopure substance Pcrm yi Pcri k i1 Tcrm yi Tcri k i1 Final PDF to printer 680 GAS MIXTURES cen22672ch13675710indd 680 103117 0112 PM the mixture defined in terms of the critical pressures and temperatures of the mixture components as P crm ʹ i 1 k y i P cri and T crm ʹ i 1 k y i T cri 1311a b The compressibility factor of the mixture Zm is then easily determined by using these pseudocritical properties The result obtained by using Kays rule is accurate to within about 10 percent over a wide range of temperatures and pressures which is acceptable for most engineering purposes Another way of treating a gas mixture as a pseudopure substance is to use a more accurate equation of state such as the van der Waals Beattie Bridgeman or BenedictWebbRubin equation for the mixture and to determine the constant coefficients in terms of the coefficients of the components In the van der Waals equation for example the two constants for the mixture are determined from a m i 1 k y i a i 12 2 and b m i 1 k y i b i 1312a b where expressions for ai and bi are given in Chap 3 EXAMPLE 132 PvT Behavior of Nonideal Gas Mixtures A rigid tank contains 2 kmol of N2 and 6 kmol of CO2 gases at 300 K and 15 MPa Fig 1310 Estimate the volume of the tank on the basis of a the idealgas equation of state b Kays rule c compressibility factors and Amagats law and d compressibility factors and Daltons law SOLUTION The composition of a mixture in a rigid tank is given The volume of the tank is to be determined using four different approaches Assumptions Stated in each section Analysis a When the mixture is assumed to behave as an ideal gas the volume of the mixture is easily determined from the idealgas relation for the mixture V m N m R u T m P m 8 kmol8314 kPa m 3 kmolK300 K 15000 kPa 1330 m 3 since N m N N 2 N CO 2 2 6 8 kmol b To use Kays rule we need to determine the pseudocritical temperature and pseudocritical pressure of the mixture by using the criticalpoint properties of N2 and CO2 from Table A1 However first we need to determine the mole fraction of each component y N 2 N N 2 N m 2 kmol 8 kmol 025 and y CO 2 N CO 2 N m 6 kmol 8 kmol 075 T crm ʹ y i T cri y N 2 T crN 2 y CO 2 T crCO 2 0251262 K 0753042 K 2597 K P crm ʹ y i P cri y N 2 P crN 2 y CO 2 P crCO 2 025339 MPa 075739 MPa 639 MPa FIGURE 1310 Schematic for Example 132 2 kmol N2 6 kmol CO2 300 K 15 MPa Vm Final PDF to printer 681 CHAPTER 13 cen22672ch13675710indd 681 103117 0112 PM Then T R T m T crm 300 K 2597 K 116 P R P m P crm 15 MPa 639 MPa 235 Z m 049 Fig A15b Thus V m Z m N m R u T m P m Z m V ideal 0491330 m 3 0652 m 3 c When Amagats law is used in conjunction with compressibility factors Zm is determined from Eq 1310 But first we need to determine the Z of each component on the basis of Amagats law N 2 T R N 2 T m T crN 2 300 K 1262 K 238 P R N 2 P m P crN 2 15 MPa 339 MPa 442 Z N 2 102 Fig A15b CO 2 T R CO 2 T m T crCO 2 300 K 3042 K 099 P R CO 2 P m P crCO 2 15 MPa 739 MPa 203 Z CO 2 030 Fig A15b Mixture Z m y i Z i y N 2 Z N 2 y CO 2 Z CO 2 025102 075030 048 Thus V m Z m N m R u T m P m Z m V ideal 0481330 m 3 0638 m 3 The compressibility factor in this case turned out to be almost the same as the one determined by using Kays rule d When Daltons law is used in conjunction with compressibility factors Zm is again determined from Eq 1310 However this time the Z of each component is to be determined at the mixture temperature and volume which is not known There fore an iterative solution is required We start the calculations by assuming that the volume of the gas mixture is 1330 m3 the value determined by assuming idealgas behavior The TR values in this case are identical to those obtained in part c and remain constant The pseudoreduced volume is determined from its definition in Chap 3 v R N 2 v N 2 R u T crN 2 P crN 2 V m N N 2 R u T crN 2 P crN 2 133 m 3 2 kmol 8314 kPa m 3 kmolK 1262 K 3390 kPa 215 Final PDF to printer 682 GAS MIXTURES cen22672ch13675710indd 682 103117 0112 PM 133 PROPERTIES OF GAS MIXTURES IDEAL AND REAL GASES Consider a gas mixture that consists of 2 kg of N2 and 3 kg of CO2 The total mass an extensive property of this mixture is 5 kg How did we do it Well we simply added the mass of each component This example suggests a simple way of evaluating the extensive properties of a nonreacting ideal or realgas mixture Just add the contributions of each component of the mixture Fig 1311 Then the total internal energy enthalpy and entropy of a gas mixture can be expressed respectively as U m i 1 k U i i 1 k m i u i i 1 k N i u i kJ 1313 H m i 1 k H i i 1 k m i h i i 1 k N i h i kJ 1314 S m i 1 k S i i 1 k m i s i i 1 k N i s i kJK 1315 By following a similar logic the changes in internal energy enthalpy and entropy of a gas mixture during a process can be expressed respectively as Δ U m i 1 k Δ U i i 1 k m i Δ u i i 1 k N i Δ u i kJ 1316 Similarly v R CO 2 133 m 3 6 kmol 8314 kPa m 3 kmolK3042 K7390 kPa 0648 From Fig A15 we read Z N 2 099 and Z CO 2 056 Thus Z m y N 2 Z N 2 y CO 2 Z CO 2 025 099 075 056 067 and V m Z m N m R T m P m Z m V ideal 0671330 m 3 0891 m 3 This is 33 percent lower than the assumed value Therefore we should repeat the calculations using the new value of Vm When the calculations are repeated we obtain 0738 m3 after the second iteration 0678 m3 after the third iteration and 0648 m3 after the fourth iteration This value does not change with more iterations Therefore V m 0648 m 3 Discussion Notice that the results obtained in parts b c and d are very close But they are very different from the idealgas values Therefore treating a mixture of gases as an ideal gas may yield unacceptable errors at high pressures FIGURE 1311 The extensive properties of a mixture are determined by simply adding the properties of the components 2 kmol A 6 kmol B UA 1000 kJ UB 1800 kJ Um 2800 kJ Final PDF to printer 683 CHAPTER 13 cen22672ch13675710indd 683 103117 0112 PM Δ H m i 1 k Δ H i i 1 k m i Δ h i i 1 k N i Δ h i kJ 1317 Δ S m i 1 k Δ S i i 1 k m i Δ s i i 1 k N i Δ s i kJK 1318 Now reconsider the same mixture and assume that both N2 and CO2 are at 25C The temperature an intensive property of the mixture is as you would expect also 25C Notice that we did not add the component temperatures to determine the mixture temperature Instead we used some kind of averaging scheme a characteristic approach for determining the intensive properties of a mixture The internal energy enthalpy and entropy of a mixture per unit mass or per unit mole of the mixture can be determined by dividing the pre ceding equations by the mass or the mole number of the mixture mm or Nm We obtain Fig 1312 u m i 1 k mf i u i kJkg and u m i 1 k y i u i kJkmol 1319 h m i 1 k mf i h i kJkg and h m i 1 k y i h i kJkmol 1320 s m i 1 k mf i s i kJkgK and s m i 1 k y i s i kJkmolK 1321 Similarly the specific heats of a gas mixture can be expressed as c vm i 1 k mf i c vi kJkgK and c vm i 1 k y i c vi kJkmolK 1322 c pm i 1 k mf i c pi kJkgK and c pm i 1 k y i c pi kJkmolK 1323 Notice that properties per unit mass involve mass fractions mfi and proper ties per unit mole involve mole fractions yi The relations given above are exact for idealgas mixtures and approximate for realgas mixtures In fact they are also applicable to nonreacting liquid and solid solutions especially when they form an ideal solution The only major difficulty associated with these relations is the determination of properties for each individual gas in the mixture The analysis can be simplified greatly however by treating the individual gases as ideal gases if doing so does not introduce a significant error IdealGas Mixtures The gases that comprise a mixture are often at a high temperature and low pressure relative to the criticalpoint values of individual gases In such cases the gas mixture and its components can be treated as ideal gases with neg ligible error Under the idealgas approximation the properties of a gas are not influenced by the presence of other gases and each gas component in the mixture behaves as if it exists alone at the mixture temperature Tm and mixture FIGURE 1312 The intensive properties of a mixture are determined by weighted averaging 2 kmol A 3 kmol B uA 500 kJkmol uB 600 kJkmol um 560 kJkmol Final PDF to printer 684 GAS MIXTURES cen22672ch13675710indd 684 103117 0112 PM volume Vm This principle is known as the GibbsDalton law which is an extension of Daltons law of additive pressures Also the h u cv and cp of an ideal gas depend on temperature only and are independent of the pressure or the volume of the idealgas mixture The partial pressure of a component in an idealgas mixture is simply Pi yiPm where Pm is the mixture pressure Evaluation of Δu or Δh of the components of an idealgas mixture during a process is relatively easy since it requires only a knowledge of the initial and final temperatures Care should be exercised however in evaluating the Δs of the components since the entropy of an ideal gas depends on the pressure or volume of the component as well as on its temperature The entropy change of individual gases in an idealgas mixture during a process can be determined from Δ s i s i2 s i1 R i ln P i2 P i1 c pi ln T i2 T i1 R i ln P i2 P i1 1324 or Δ s i s i2 s i1 R u ln P i2 P i1 c pi ln T i2 T i1 R u ln P i2 P i1 1325 where Pi2 yi2Pm2 and Pi1 yi1Pm1 Notice that the partial pressure Pi of each component is used in the evaluation of the entropy change not the mixture pressure Pm Fig 1313 FIGURE 1313 Partial pressures not the mixture pressure are used in the evaluation of entropy changes of idealgas mixtures Partial pressure of component i at state 2 Partial pressure of component i at state 1 si si2 si1 Ri ln Pi2 Pi1 EXAMPLE 133 Expansion of an Ideal Gas Mixture in a Turbine A mixture of oxygen O2 carbon dioxide CO2 and helium He gases with mass frac tions of 00625 0625 and 03125 respectively enter an adiabatic turbine at 1000 kPa and 600 K steadily and expand to 100 kPa pressure Fig 1314 The isentropic efficiency of the turbine is 90 percent For gas components assuming constant specific heats at room temperature determine a the work output per unit mass of mixture and b the exergy destruction and the secondlaw efficiency of the turbine Take the environment temperature to be T0 25C SOLUTION The mass fractions of the components of a gas mixture that expands in an adiabatic turbine are given The work output the exergy destruction and the secondlaw efficiency are to be determined Assumptions All gases will be modeled as ideal gases with constant specific heats Analysis a The mass fractions of mixture components are given to be mf CO 2 00625 mf CO 2 0625 and mfHe 03125 The specific heats of these gases at room temperature are Table A2a cv kJkgK cp kJkgK O2 0658 0918 CO2 0657 0846 He 31156 51926 Then the constantpressure and constantvolume specific heats of the mixture become FIGURE 1314 Schematic for Example 133 1000 kPa 600 K 100 kPa O2 CO2 He mixture w Final PDF to printer 685 CHAPTER 13 cen22672ch13675710indd 685 103117 0112 PM c p mf O 2 c p O 2 mf CO 2 c p CO 2 mf He c pHe 00625 0918 0625 0846 03125 51926 2209 kJkgK c v mf O 2 c v O 2 mf CO 2 c v CO 2 mf He c vHe 00625 0658 0625 0657 03125 31156 1425 kJkgK The apparent gas constant of the mixture and the specific heat ratio are R c p c v 2209 1425 07836 kJkgK k c p c v 2209 kJkgK 1425 kJkgK 1550 The temperature at the end of the expansion for the isentropic process is T 2s T 1 P 2 P 1 k1k 600 K 100 kPa 1000 kPa 055155 2650 K Using the definition of turbine isentropic efficiency the actual outlet tempera ture is T 2 T 1 η T T 1 T 2s 600 K 090 600 265 K 2985 K Noting that the turbine is adiabatic and thus there is no heat transfer the actual work output is determined to be w out h 1 h 2 c p T 1 T 2 2209 kJkgK 600 2985 K 6660 kJkg b The entropy change of the gas mixture and the exergy destruction in the turbine are s 2 s 1 c p ln T 2 T 1 R ln P 2 P 1 2209 kJkgK ln 2985 K 600 K 07836 kJkgK ln 100 kPa 1000 kPa 02658 kJkgK The expended exergy is the sum of the work output of the turbine exergy recovered and the exergy destruction exergy wasted x expended x recovered x dest w out x dest 6660 792 7452 kJkg The secondlaw efficiency is the ratio of the recovered to expended exergy η II x recovered x expended w out x expended 6600 kJkg 7452 kJkg 0894 or 894 Discussion The secondlaw efficiency is a measure of thermodynamic perfection A process that generates no entropy and thus destroys no exergy always has a secondlaw efficiency of 100 percent x dest T 0 s gen T 0 s 2 s 1 298 K02658 kJkgK 792 kJkg Final PDF to printer 686 GAS MIXTURES cen22672ch13675710indd 686 103117 0112 PM EXAMPLE 134 Exergy Destruction During Mixing of Ideal Gases An insulated rigid tank is divided into two compartments by a partition as shown in Fig 1315 One compartment contains 3 kmol of O2 and the other compartment contains 5 kmol of CO2 Both gases are initially at 25C and 200 kPa Now the partition is removed and the two gases are allowed to mix Assuming the surroundings are at 25C and both gases behave as ideal gases determine the entropy change and exergy destruction associated with this process SOLUTION A rigid tank contains two gases separated by a partition The entropy change and exergy destroyed after the partition is removed are to be determined Assumptions Both gases and their mixture are ideal gases Analysis We take the entire contents of the tank both compartments as the system This is a closed system since no mass crosses the boundary during the process We note that the volume of a rigid tank is constant and there is no energy transfer as heat or work Also both gases are initially at the same temperature and pressure When two ideal gases initially at the same temperature and pressure are mixed by removing a partition between them the mixture will also be at the same temperature and pressure Can you prove it Will this be true for nonideal gases Therefore the temperature and pressure in the tank will still be 25C and 200 kPa respectively after the mixing The entropy change of each component gas can be determined from Eqs 1318 and 1325 Δ S m Δ S i N i Δ s i N i c pi ln T i2 T i1 R u ln P i2 P i1 R u N i ln y i2 P m2 P i1 R u N i ln y i2 since Pm2 Pi1 200 kPa It is obvious that the entropy change is independent of the composition of the mixture in this case and that it depends on only the mole fraction of the gases in the mixture What is not so obvious is that if the same gas in two different chambers is mixed at constant temperature and pressure the entropy change is zero Substituting the known values the entropy change becomes N m N O 2 N CO 2 3 5 kmol 8 kmol y O 2 N O 2 N m 3 kmol 8 kmol 0375 y CO 2 N CO 2 N m 5 kmol 8 kmol 0625 Δ S m R u N O 2 ln y O 2 N CO 2 ln y CO 2 8314 kJkmolK3 kmolln 0375 5 kmolln 0625 440 kJK The exergy destruction associated with this mixing process is determined from X destroyed T 0 S gen T 0 Δ S sys 298 K440 kJK 131 MJ Discussion This large value of exergy destruction shows that mixing processes are highly irreversible 0 FIGURE 1315 Schematic for Example 134 O2 25C 200 kPa CO2 25C 200 kPa Final PDF to printer 687 CHAPTER 13 cen22672ch13675710indd 687 103117 0112 PM RealGas Mixtures When the components of a gas mixture do not behave as ideal gases the anal ysis becomes more complex because the properties of real nonideal gases such as u h cv and cp depend on the pressure or specific volume as well as on the temperature In such cases the effects of deviation from idealgas behavior on the mixture properties should be accounted for Consider two nonideal gases contained in two separate compartments of an adiabatic rigid tank at 100 kPa and 25C The partition separating the two gases is removed and the two gases are allowed to mix What do you think the final pressure in the tank will be You are probably tempted to say 100 kPa which would be true for ideal gases However this is not true for nonideal gases because of the influence of the molecules of different gases on each other deviation from Daltons law Fig 1316 When realgas mixtures are involved it may be necessary to account for the effect of nonideal behavior on the mixture properties such as enthalpy and entropy One way of doing that is to use compressibility factors in conjunction with generalized equations and charts developed in Chap 12 for real gases Consider the following T ds relation for a gas mixture d h m T m d s m v m d P m It can also be expressed as d mf i h i T m d mf i s i mf i v i d P m or mf i d h i T m d s i v i d P m 0 which yields d h i T m d s i v i d P m 1326 This is an important result because Eq 1326 is the starting equation in the development of the generalized relations and charts for enthalpy and entropy It suggests that the generalized property relations and charts for real gases devel oped in Chap 12 can also be used for the components of realgas mixtures But the reduced temperature TR and reduced pressure PR for each component should be evaluated by using the mixture temperature Tm and mixture pressure Pm This is because Eq 1326 involves the mixture pressure Pm not the component pressure Pi The approach just described is somewhat analogous to Amagats law of additive volumes evaluating mixture properties at the mixture pressure and temperature which holds exactly for idealgas mixtures and approximately for realgas mixtures Therefore the mixture properties determined with this approach are not exact but they are sufficiently accurate What if the mixture volume and temperature are specified instead of the mixture pressure and temperature Well there is no need to panic Just evalu ate the mixture pressure using Daltons law of additive pressures and then use this value which is only approximate as the mixture pressure Another way of evaluating the properties of a realgas mixture is to treat the mixture as a pseudopure substance having pseudocritical properties FIGURE 1316 It is difficult to predict the behavior of nonidealgas mixtures because of the influence of dissimilar molecules on each other Real gas mixture A B 25C 1 m3 102 kPa Real gas A 25C 04 m3 100 kPa Real gas B 25C 06 m3 100 kPa Final PDF to printer 688 GAS MIXTURES cen22672ch13675710indd 688 103117 0112 PM determined in terms of the critical properties of the component gases by using Kays rule The approach is quite simple and the accuracy is usually acceptable EXAMPLE 135 Cooling of a Nonideal Gas Mixture Air is a mixture of N2 O2 and small amounts of other gases and it can be approximated as 79 percent N2 and 21 percent O2 on a mole basis During a steadyflow process air is cooled from 220 to 160 K at a constant pressure of 10 MPa Fig 1317 Determine the heat transfer during this process per kmol of air using a the idealgas approximation b Kays rule and c Amagats law SOLUTION Air at a low temperature and high pressure is cooled at constant pressure The heat transfer is to be determined using three different approaches Assumptions 1 This is a steadyflow process since there is no change with time at any point and thus ΔmCV 0 and ΔECV 0 2 The kinetic and potential energy changes are negligible Analysis We take the cooling section as the system This is a control volume since mass crosses the system boundary during the process We note that heat is transferred out of the system The critical properties are Tcr 1262 K and Pcr 339 MPa for N2 and Tcr 1548 K and Pcr 508 MPa for O2 Both gases remain above their critical temperatures but they are also above their critical pressures Therefore air will probably deviate from idealgas behavior and thus it should be treated as a realgas mixture The energy balance for this steadyflow system can be expressed on a unitmole basis as e in e out Δ e system 0 e in e out h 1 h 2 q out q out h 1 h 2 y N 2 h 1 h 2 N 2 y O 2 h 1 h 2 O 2 where the enthalpy change for either component can be determined from the general ized enthalpy departure chart Fig A29 and Eq 1258 h 1 h 2 h 1ideal h 2ideal R u T cr Z h 1 Z h 2 The first two terms on the righthand side of this equation represent the idealgas enthalpy change of the component The terms in parentheses represent the deviation from the idealgas behavior and their evaluation requires a knowledge of reduced pressure PR and reduced temperature TR which are calculated at the mixture tempera ture Tm and mixture pressure Pm a If the N2 and O2 mixture is assumed to behave as an ideal gas the enthalpy of the mixture will depend on temperature only and the enthalpy values at the initial and the final temperatures can be determined from the idealgas tables of N2 and O2 Tables A18 and A19 T 1 200 K h 1 idealN 2 6391 kJkmol h 1 idealO 2 6404 kJkmol T 2 160 K h 2 idealN 2 4648 kJkmol h 2 idealO 2 4657 kJkmol 0 FIGURE 1317 Schematic for Example 135 T1 220 K T2 160 K P1 10 MPa P2 10 MPa Heat Air 79 N2 21 O2 Final PDF to printer 689 CHAPTER 13 cen22672ch13675710indd 689 103117 0112 PM q out y N 2 h 1 h 2 N 2 y O 2 h 1 h 2 O 2 079 6391 4648 kJkmol 021 6404 4657 kJkmol 1744 kJkmol b Kays rule is based on treating a gas mixture as a pseudopure substance whose critical temperature and pressure are T crm y i T cri y N 2 T crN 2 y O 2 T crO 2 0791262 K 0211548 K 1322 K and P crm y i P cri y N 2 P crN 2 y O 2 P crO 2 079339 MPa 021508 MPa 374 MPa Then T R1 T m1 T crm 220 K 1322 K 166 P R P m P crm 10 MPa 374 MPa 267 Z h 1 m 10 T R2 T m2 T crm 160 K 1322 K 121 Z h 2 m 26 Also h m 1 ideal y N 2 h 1 idealN 2 y O 2 h 1 idealO 2 079 6391 kJkmol 021 6404 kJkmol 6394 kJkmol h m 2 ideal y N 2 h 2 idealN 2 y O 2 h 2 idealO 2 079 4648 kJkmol 021 4657 kJkmol 4650 kJkmol Therefore q out h m 1 ideal h m 2 ideal R u T cr Z h 1 Z h 2 m 6394 4650 kJkmol 8314 kJkmolK 1322 K 10 26 3503 kJkmol c The reduced temperatures and pressures for both N2 and O2 at the initial and final states and the corresponding enthalpy departure factors are from Fig A29 N 2 T R 1 N 2 T m1 T crN 2 220 K 1262 K 174 P R N 2 P m P crN 2 10 MPa 339 MPa 295 T R 2 N 2 T m2 T crN 2 160 K 1262 K 127 Z h 1 N 2 09 Z h 1 N 2 24 Final PDF to printer 690 GAS MIXTURES cen22672ch13675710indd 690 103117 0112 PM O 2 T R 1 O 2 T m1 T crO 2 220 K 1548 K 142 P R O 2 P m P crO 2 10 MPa 508 MPa 197 T R 1 O 2 T m2 T crO 2 160 K 1548 K 103 From Eq 1258 h 1 h 2 N 2 h 1ideal h 2ideal N 2 R u T cr Z h 1 Z h 2 N 2 6391 4648 kJkmol 8314 kJkmolK 1262 K 09 24 3317 kJkmol h 1 h 2 O 2 h 1ideal h 2ideal O 2 R u T cr Z h 1 Z h 2 O 2 6404 4657 kJkmol 8314 kJkmolK 1548 K 13 40 5222 kJkmol Therefore q out y N 2 h 1 h 2 N 2 y O 2 h 1 h 2 O 2 0793317 kJkmol 021 5222 kJkmol 3717 kJkmol Discussion This result is about 6 percent greater than the result obtained in part b by using Kays rule But it is more than twice the result obtained by assuming the mixture to be an ideal gas Z h 1 O 2 13 Z h 2 O 2 40 When two gases or two miscible liquids are brought into contact they mix and form a homogeneous mixture or solution without requiring any work input That is the natural tendency of miscible substances brought into contact is to mix with each other These are irreversible processes and thus it is impossi ble for the reverse process of separation to occur spontaneously For example pure nitrogen and oxygen gases readily mix when brought into contact but a mixture of nitrogen and oxygen such as air never separates into pure nitrogen and oxygen when left unattended Mixing and separation processes are commonly used in practice Separation processes require a work or more generally exergy input and minimizing this required work input is an important part of the design process of separa tion plants Dissimilar molecules in a mixture affect each other and therefore the influence of composition on the properties must be taken into consider ation in any thermodynamic analysis In this section we analyze the general TOPIC OF SPECIAL INTEREST Chemical Potential and the Separation Work of Mixtures This section can be skipped without a loss in continuity Final PDF to printer 691 CHAPTER 13 cen22672ch13675710indd 691 103117 0112 PM mixing processes with particular emphasis on ideal solutions and determine the entropy generation and exergy destruction We then consider the reverse process of separation and determine the minimum or reversible work input needed The specific Gibbs function or Gibbs free energy g is defined as the combi nation property g h Ts Using the relation dh v dP T ds the differential change of the Gibbs function of a pure substance is obtained by differentiation as dg v dP s dT or dG V dP S dT pure substance 1327 For a mixture the total Gibbs function is a function of two independent inten sive properties as well as the composition and thus it can be expressed as G GP T N1 N2 Ni Its differential is dG G P TN dP G T PN dT i G N i PT N j d N i mixture 1328 where the subscript Nj indicates that the mole numbers of all components in the mixture other than component i are to be held constant during differen tiation For a pure substance the last term drops out since the composition is fixed and the preceding equation must reduce to the one for a pure substance Comparing Eqs 1327 and 1328 gives dG V dP S dT i μ i d N i or d g v dP s dT i m i d y i 1329 where yi NiNm is the mole fraction of component i Nm is the total number of moles of the mixture and μ i G N i PT N j g i h i T s i for component i of a mixture 1330 is the chemical potential of component i which is the differential change in the Gibbs function of the mixture in a specified phase per differential change of component i in the same phase as pressure temperature and the amounts of all other components are held constant The symbol tilde as in v h and s is used to denote the partial molar properties of the components Note that the summation term in Eq 1329 is zero for a singlecomponent system and thus the chemical potential of a pure system in a given phase is equivalent to the molar Gibbs function Fig 1318 since G Ng Nμ where μ G N PT g h T s pure substance 1331 Therefore the difference between the chemical potential and the Gibbs function is due to the effect of dissimilar molecules in a mixture on each other It is because of this molecular effect that the volume of the mixture of two miscible liquids may be more or less than the sum of the initial volumes of the individual liquids Likewise the total enthalpy of the mixture of two com ponents at the same pressure and temperature in general is not equal to the sum of the total enthalpies of the individual components before mixing the difference being the enthalpy or heat of mixing which is the heat released or absorbed as two or more components are mixed isothermally For example the FIGURE 1318 For a pure substance the chemical potential is equivalent to the Gibbs function Mixture Pure substance i himixture Tsimixture gimixture h Ts g Final PDF to printer 692 GAS MIXTURES cen22672ch13675710indd 692 103117 0112 PM volume of an ethyl alcoholwater mixture is a few percent less than the sum of the volumes of the individual liquids before mixing Also when water and flour are mixed to make dough the temperature of the dough rises noticeably due to the enthalpy of mixing released For reasons explained above the partial molar properties of the components denoted by a tilde should be used in the evaluation of the extensive proper ties of a mixture instead of the specific properties of the pure components For example the total volume enthalpy and entropy of a mixture should be determined from respectively V i N i v i H i N i h i and S i N i s i mixture 1332 instead of V i N i v i H i N i h i and S i N i s i 1333 Then the changes in these extensive properties during mixing become Δ V mixing i N i v i v i Δ H mixing i N i h i h i Δ S mixing i N i s i s i 1334 where ΔHmixing is the enthalpy of mixing and ΔSmixing is the entropy of mixing Fig 1319 The enthalpy of mixing is negative for exothermic mixing processes positive for endothermic mixing processes and zero for isothermal mixing processes during which no heat is absorbed or released Note that mixing is an irreversible process and thus the entropy of mixing must be a positive quantity during an adiabatic process The specific volume enthalpy and entropy of a mixture are determined from v i y i v i h i y i h i and s i y i s i 1335 where yi is the mole fraction of component i in the mixture Reconsider Eq 1329 for dG Recall that properties are point functions and they have exact differentials Therefore the test of exactness can be applied to the righthand side of Eq 1329 to obtain some important relations For the differential dz M dx N dy of a function zx y the test of exactness is expressed as Myx Nxy When the amount of component i in a mixture is varied at constant pressure or temperature while other components indicated by j are held constant Eq 1329 simplifies to dG S dT μ i d N i for P constant and N j constant 1336 dG V dP μ i d N i for T constant and N j constant 1337 Applying the test of exactness to both of these relations gives μ i T PN S N i TP N j s i and μ i P TN V N i TP N j v i 1338 FIGURE 1319 The amount of heat released or absorbed during a mixing process is called the enthalpy or heat of mixing which is zero for ideal solutions A B yB yA A B mixture Mixing chamber Δhmixing yAhAmixture hA yBhBmixture hB Final PDF to printer 693 CHAPTER 13 cen22672ch13675710indd 693 103117 0112 PM where the subscript N indicates that the mole numbers of all components and thus the composition of the mixture are to remain constant Taking the chemi cal potential of a component to be a function of temperature pressure and composition and thus μi μi P T y1 y2 yj its total differential can be expressed as d μ i d g i μ i P Ty dP μ i T Py dT i μ i y i PT y i d y i 1339 where the subscript y indicates that the mole fractions of all components and thus the composition of the mixture is to remain constant Substituting Eqs 1338 into the preceding relation gives d μ i v i dP s i dT i μ i y i PT y i d y i 1340 For a mixture of fixed composition undergoing an isothermal process it simplifies to d μ i v i dP T constant y i constant 1341 IdealGas Mixtures and Ideal Solutions When the effect of dissimilar molecules in a mixture on each other is negligible the mixture is said to be an ideal mixture or ideal solution and the chemical potential of a component in such a mixture equals the Gibbs function of the pure component Many liquid solutions encountered in practice especially dilute ones satisfy this condition very closely and can be considered to be ideal solutions with negligible error As expected the ideal solution approxi mation greatly simplifies the thermodynamic analysis of mixtures In an ideal solution a molecule treats the molecules of all components in the mixture the same wayno extra attraction or repulsion for the molecules of other compo nents This is usually the case for mixtures of similar substances such as those of petroleum products Very dissimilar substances such as water and oil wont even mix at all to form a solution For an idealgas mixture at temperature T and total pressure P the partial molar volume of a component i is v i vi RuTP Substituting this relation into Eq 1341 gives d μ i R u T P dP R u T d ln P R u T d ln P i T constant y i constant ideal gas 1342 since from Daltons law of additive pressures Pi yiP for an idealgas mixture and d ln P i d ln y i P dln y i ln P d ln P y i constant 1343 Integrating Eq 1342 at constant temperature from the total mixture pressure P to the component pressure Pi of component i gives μ i T P i μ i T P R u T ln P i P μ i T P R u T ln y i ideal gas 1344 Final PDF to printer 694 GAS MIXTURES cen22672ch13675710indd 694 103117 0112 PM For yi 1 ie a pure substance of component i alone the last term in the preceding equation drops out and we end up with μiT Pi μiT P which is the value for the pure substance i Therefore the term μiT P is simply the chemical potential of the pure substance i when it exists alone at total mix ture pressure and temperature which is equivalent to the Gibbs function since the chemical potential and the Gibbs function are identical for pure substances The term μiT P is independent of mixture composition and mole fractions and its value can be determined from the property tables of pure substances Then Eq 1344 can be rewritten more explicitly as μ imixtureideal T P i μ ipure T P R u T ln y i 1345 Note that the chemical potential of a component of an idealgas mixture depends on the mole fraction of the components as well as the mixture tem perature and pressure and is independent of the identity of the other constitu ent gases This is not surprising since the molecules of an ideal gas behave like they exist alone and are not influenced by the presence of other molecules Eq 1345 is developed for an idealgas mixture but it is also applicable to mix tures or solutions that behave the same waythat is mixtures or solutions in which the effects of molecules of different components on each other are negligible The class of such mixtures is called ideal solutions or ideal mixtures as discussed before The idealgas mixture described is just one category of ideal solutions Another major category of ideal solutions is the dilute liquid solutions such as the saline water It can be shown that the enthalpy of mixing and the volume change due to mixing are zero for ideal solutions see Wark 1995 That is Δ V mixtureideal i N i v i v i 0 and Δ H mixtureideal i N i h i h i 0 1346 Then it follows that v i v i and h i h i That is the partial molar volume and the partial molar enthalpy of a component in a solution equal the specific vol ume and enthalpy of that component when it existed alone as a pure substance at the mixture temperature and pressure Therefore the specific volume and enthalpy of individual components do not change during mixing if they form an ideal solution Then the specific volume and enthalpy of an ideal solution can be expressed as Fig 1320 v mixtureideal i y i v i i y i v ipure and h mixtureideal i y i h i i y i h ipure 1347 Note that this is not the case for entropy and the properties that involve entropy such as the Gibbs function even for ideal solutions To obtain a relation for the entropy of a mixture we differentiate Eq 1345 with respect to temperature at constant pressure and mole fraction μ imixture T P i T Py μ ipure T P T Py R u ln y i 1348 We note from Eq 1338 that the two partial derivatives above are simply the negative of the partial molar entropies Substituting s imixtureideal T P i s ipure T P R u ln y i ideal solution 1349 FIGURE 1320 The specific volume and enthalpy of individual components do not change during mixing if they form an ideal solution this is not the case for entropy Vmixingideal 0 Hmixingideal 0 vimixture vipure vmixture yivipure hmixture yihipure himixture hipure i i Final PDF to printer 695 CHAPTER 13 cen22672ch13675710indd 695 103117 0112 PM Note that ln yi is a negative quantity since yi 1 and thus Ru ln yi is always positive Therefore the entropy of a component in a mixture is always greater than the entropy of that component when it exists alone at the mixture tem perature and pressure Then the entropy of mixing of an ideal solution is deter mined by substituting Eq 1349 into Eq 1334 to be Δ S mixingideal i N i s i s i R u i N i ln y i ideal solution 1350a or dividing by the total number of moles of the mixture Nm Δ s mixingideal i y i s i s i R u i y i ln y i per unit mole of mixture 1350b Minimum Work of Separation of Mixtures The entropy balance for a steadyflow system simplifies to Sin Sout Sgen 0 Noting that entropy can be transferred by heat and mass only the entropy generation during an adiabatic mixing process that forms an ideal solution becomes S gen S out S in Δ S mixing R u i N i ln y i ideal solution 1351a or s gen s out s in Δ s mixing R u i y i ln y i per unit mole of mixture 1351b Also noting that Xdestroyed T0 Sgen the exergy destroyed during this and any other process is obtained by multiplying the entropy generation by the temperature of the environment T0 It gives X destroyed T 0 S gen R u T 0 i N i ln y i ideal solution 1352a or x destroyed T 0 s gen R u T 0 i y i ln y i per unit mole of mixture 1352b Exergy destroyed represents the wasted work potentialthe work that would be produced if the mixing process occurred reversibly For a reversible or thermodynamically perfect process the entropy generation and thus the exergy destroyed is zero Also for reversible processes the work output is a maximum or the work input is a minimum if the process does not occur natu rally and requires input The difference between the reversible work and the actual useful work is due to irreversibilities and is equal to the exergy destruc tion Therefore Xdestroyed Wrev Wactual Then it follows that for a naturally occurring process during which no work is produced the reversible work is equal to the exergy destruction Fig 1321 Therefore for the adiabatic mix ing process that forms an ideal solution the reversible work total and per unit mole of mixture is from Eq 1352 W rev R u T 0 i N i ln y i and w rev R u T 0 i y i ln y i 1353 FIGURE 1321 For a naturally occurring process during which no work is produced or consumed the reversible work is equal to the exergy destruction Wrev Xdestruction T0Sgen A B A B mixture Mixing chamber T0 Final PDF to printer 696 GAS MIXTURES cen22672ch13675710indd 696 103117 0112 PM A reversible process by definition is a process that can be reversed without leaving a net effect on the surroundings This requires that the direction of all interactions be reversed while their magnitudes remain the same when the process is reversed Therefore the work input during a reversible separation process must be equal to the work output during the reverse process of mix ing A violation of this requirement will be a violation of the second law of thermodynamics The required work input for a reversible separation process is the minimum work input required to accomplish that separation since the work input for reversible processes is always less than the work input of cor responding irreversible processes Then the minimum work input required for the separation process can be expressed as W minin R u T 0 i N i ln y i and w minin R u T 0 i y i ln y i 1354 It can also be expressed in the rate form as W minin R u T 0 i N i ln y i N m R u T 0 i y i ln y i kW 1355 where W minin is the minimum power input required to separate a solution that approaches at a rate of N m kmols or m m N mMm kgs into its components The work of separation per unit mass of mixture can be determined from w minin w minin M m where Mm is the apparent molar mass of the mixture The minimum work relations above are for complete separation of the com ponents in the mixture The required work input will be less if the exiting streams are not pure The reversible work for incomplete separation can be determined by calculating the minimum separation work for the incoming mixture and the minimum separation works for the outgoing mixtures and then taking their difference Reversible Mixing Processes The mixing processes that occur naturally are irreversible and all the work potential is wasted during such processes For example when the fresh water from a river mixes with the saline water in an ocean an opportunity to produce work is lost If this mixing is done reversibly through the use of semiperme able membranes for example some work can be produced The maximum amount of work that can be produced during a mixing process is equal to the minimum amount of work input needed for the corresponding separation process Fig 1322 That is W maxoutmixing W mininseparation 1356 Therefore the minimum work input relations given above for separation can also be used to determine the maximum work output for mixing The minimum work input relations are independent of any hardware or pro cess Therefore the relations developed above are applicable to any separation process regardless of actual hardware system or process and they can be used for a wide range of separation processes including the desalination of sea or brackish water FIGURE 1322 Under reversible conditions the work consumed during separation is equal to the work produced during the reverse process of mixing a Mixing A B yB yA A B mixture Mixing chamber Wmaxout 5 kJkg mixture b Separation A B yB yA A B mixture Separation unit Wminin 5 kJkg mixture Final PDF to printer 697 CHAPTER 13 cen22672ch13675710indd 697 103117 0112 PM SecondLaw Efficiency The secondlaw efficiency is a measure of how closely a process approximates a corresponding reversible process and it indicates the range available for poten tial improvements Noting that the secondlaw efficiency ranges from 0 for a totally irreversible process to 100 percent for a totally reversible process the secondlaw efficiency for separation and mixing processes can be defined as η IIseparation W minin W actin w minin w actin and η IImixing W actout W maxout w actout w maxout 1357 where W actin is the actual power input or exergy consumption of the separa tion plant and W actout is the actual power produced during mixing Note that the secondlaw efficiency is always less than 1 since the actual separation process requires a greater amount of work input because of irreversibilities There fore the minimum work input and the secondlaw efficiency provide a basis for comparison of actual separation processes to the idealized ones and for assessing the thermodynamic performance of separation plants A secondlaw efficiency for mixing processes can also be defined as the actual work produced during mixing divided by the maximum work potential available This definition does not have much practical value however since no effort is used to produce work during most mixing processes and thus the secondlaw efficiency is zero Special Case Separation of a TwoComponent Mixture Consider a mixture of two components A and B whose mole fractions are yA and yB Noting that yB 1 yA the minimum work input required to separate 1 kmol of this mixture at temperature T0 completely into pure A and pure B is from Eq 1354 w minin R u T 0 y A ln y A y B ln y B kJkmol mixture 1358a or W minin R u T 0 N A ln y A N B ln y B kJ 1358b or from Eq 1355 W minin N m R u T 0 y A ln y A y B ln y B m m R m T 0 y A ln y A y B ln y B kW 1358c Some separation processes involve the extraction of just one of the compo nents from a large amount of mixture so that the composition of the remaining mixture remains practically the same Consider a mixture of two components A and B whose mole fractions are yA and yB respectively The minimum work required to separate 1 kmol of pure component A from the mixture of Nm NA NB kmol with NA 1 is determined by subtracting the mini mum work required to separate the remaining mixture RuT0NA 1 ln yA NB ln yB from the minimum work required to separate the initial mixture Wminin RuT0NA ln yA NB ln yB It gives Fig 1323 w minin R u T 0 ln y A R u T 0 ln 1 y A kJkmol A 1359 FIGURE 1323 The minimum work required to separate a twocomponent mixture for the two limiting cases A B yA yB 1 kmol A B yA yB Separation unit Separation unit pure A 1 kmol A B pure A pure B a Separating 1 kmol of A from a large body of mixture b Complete separation of 1 kmol mixture into its components A and B wminin RuT0 ln yA kJkmol A wminin RuT0 yA ln yA yB ln yB kJkmol mixture Final PDF to printer 698 GAS MIXTURES cen22672ch13675710indd 698 103117 0112 PM The minimum work needed to separate a unit mass 1 kg of component A is determined from Eq 1359 by replacing Ru with RA or by dividing the relation by the molar mass of component A since RA RuMA Eq 1359 also gives the maximum amount of work that can be done as one unit of pure component A mixes with a large amount of A B mixture An Application Desalination Processes The potable water needs of the world are increasing steadily due to population growth rising living standards industrialization and irrigation in agriculture There are over 10000 desalination plants in the world with a total desalted water capacity of over 5 billion gallons a day Saudi Arabia is the largest user of desalination with about 25 percent of the world capacity and the United States is the secondlargest user with 10 percent The major desalination meth ods are distillation and reverse osmosis The relations can be used directly for desalination processes by taking the water the solvent to be component A and the dissolved salts the solute to be component B Then the minimum work needed to produce 1 kg of pure water from a large reservoir of brackish or sea water at temperature T0 in an environment at T0 is from Eq 1359 Desalination w minin R w T 0 ln1 y w kJkg pure water 1360 where Rw 04615 kJkgK is the gas constant of water and yw is the mole fraction of water in brackish or seawater This relation also gives the maximum amount of work that can be produced as 1 kg of fresh water from a river for example mixes with seawater whose water mole fraction is yw The reversible work associated with liquid flow can also be expressed in terms of pressure difference ΔP and elevation difference Δz potential energy as wminin ΔPρ g Δz where ρ is the density of the liquid Combining these relations with Eq 1360 gives Δ P min ρ w minin ρ R w T 0 ln1 y w kPa 1361 and Δ z min w minin g R w T 0 ln1 y w g m 1362 where ΔPmin is the osmotic pressure which represents the pressure differ ence across a semipermeable membrane that separates fresh water from the saline water under equilibrium conditions ρ is the density of saline water and Δzmin is the osmotic rise which represents the vertical distance the saline water would rise when separated from the fresh water by a membrane that is permeable to water molecules alone again at equilibrium For desalination processes ΔPmin represents the minimum pressure that the saline water must be compressed in order to force the water molecules in saline water through the membrane to the freshwater side during a reverse osmosis desalination pro cess Alternately Δzmin represents the minimum height above the freshwater level that the saline water must be raised to produce the required osmotic pres sure difference across the membrane to produce fresh water The Δzmin also represents the height that the water with dissolved organic matter inside the Final PDF to printer 699 CHAPTER 13 cen22672ch13675710indd 699 103117 0112 PM roots will rise through a tree when the roots are surrounded by fresh water with the roots acting as semipermeable membranes The reverse osmosis process with semipermeable membranes is also used in dialysis machines to purify the blood of patients with failed kidneys EXAMPLE 136 Obtaining Fresh Water from Seawater Fresh water is to be obtained from seawater at 15C with a salinity of 348 percent on mass basis or TDS 34800 ppm Determine a the mole fractions of the water and the salts in the seawater b the minimum work input required to separate 1 kg of seawater completely into pure water and pure salts c the minimum work input required to obtain 1 kg of fresh water from the sea and d the minimum gauge pressure that the seawater must be raised if fresh water is to be obtained by reverse osmosis using semipermeable membranes SOLUTION Fresh water is to be obtained from seawater The mole fractions of seawater the minimum works of separation needed for two limiting cases and the required pressurization of seawater for reverse osmosis are to be determined Assumptions 1 The seawater is an ideal solution since it is dilute 2 The total dissolved solids in water can be treated as table salt NaCl 3 The environment temperature is also 15C Properties The molar masses of water and salt are Mw 180 kgkmol and Ms 5844 kgkmol The gas constant of pure water is Rw 04615 kJkgK Table A1 The density of seawater is 1028 kgm3 Analysis a Noting that the mass fractions of salts and water in seawater are mfs 00348 and mfw 1 mfs 09652 the mole fractions are determined from Eqs 134 and 135 to be M m 1 mf i M i 1 mf s M s mf w M w 1 00348 5844 09652 180 1844 kgkmol y w mf w M m M w 09652 1844 kgkmol 180 kgkmol 09888 y s 1 y w 1 09888 00112 112 b The minimum work input required to separate 1 kg of seawater completely into pure water and pure salts is w minin R u T 0 y A ln y A y B ln y B R u T 0 y w ln y w y s ln y s 8314 kJkmolK 28815 K 09888 ln 09888 00112 ln 00112 1472 kJkmol Therefore it takes a minimum of 798 kJ of work input to separate 1 kg of seawater into 00348 kg of salt and 09652 kg nearly 1 kg of fresh water c The minimum work input required to produce 1 kg of fresh water from seawater is w minin R w T 0 ln 1 y w 04615 kJkgK28815 K ln1 09888 150 kJkg fresh water w minin w minin M m 1472 kJkmol 1844 kgkmol 798 kJkg seawater FIGURE 1324 The osmotic pressure and the osmotic rise of saline water Saline water Δz Membrane P2 P1 Pure water ΔP P2 P1 Final PDF to printer 700 GAS MIXTURES cen22672ch13675710indd 700 103117 0112 PM Note that it takes about five times more work to separate 1 kg of seawater com pletely into fresh water and salt than it does to produce 1 kg of fresh water from a large amount of seawater d The osmotic pressure in this case is Δ P min ρ m R w T 0 ln1 y w 1028 kgm 3 04615 kPa m 3 kgK28815 K ln1 09888 1540 kPa which is equal to the minimum gauge pressure to which seawater must be compressed if the fresh water is to be discharged at the local atmospheric pressure As an alterna tive to pressurizing the minimum height above the freshwater level that the seawater must be raised to produce fresh water is Fig 1324 Δ z min w minin g 150 kJkg 981 ms 2 1 kg ms 2 1 N 1000 Nm 1 kJ 153 m Discussion The minimum separation works determined above also represent the max imum works that can be produced during the reverse process of mixing Therefore 798 kJ of work can be produced when 00348 kg of salt is mixed with 09652 kg of water revers ibly to produce 1 kg of saline water and 150 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly Therefore the power that can be generated as a river with a flow rate of 105 m3s mixes reversibly with seawater through semipermeable membranes is Fig 1325 W maxout ρV w maxout 1000 kgm 3 10 5 m 3 s150 kJkg 1 MW 10 3 kJs 15 10 5 MW which shows the tremendous amount of power potential wasted as the rivers discharge into the seas SUMMARY A mixture of two or more gases of fixed chemical composi tion is called a nonreacting gas mixture The composition of a gas mixture is described by specifying either the mole fraction or the mass fraction of each component defined as mf i m i m m and y i N i N m where m m i 1 k m i and N m i 1 k N i The apparent or average molar mass and gas constant of a mixture are expressed as M m m m N m i 1 k y i M i and R m R u M m Also mf i y i M i M m and M m 1 i 1 k mf i M i Daltons law of additive pressures states that the pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume Amagats law of additive volumes states that the volume of a gas mixture is equal to the sum of the vol umes each gas would occupy if it existed alone at the mixture FIGURE 1325 Power can be produced by mixing solutions of different concentrations reversibly Fresh and saline water mixing irreversibly Fresh river water Seawater salinity 348 Fresh and saline water mixing reversibly through semipermeable membranes and producing power z 153 m Final PDF to printer 701 CHAPTER 13 cen22672ch13675710indd 701 103117 0112 PM REFERENCES AND SUGGESTED READINGS 1 A Bejan Advanced Engineering Thermodynamics 3rd ed New York Wiley Interscience 2006 2 Y A Çengel Y Cerci and B Wood Second Law Analysis of Separation Processes of Mixtures ASME International Mechanical Engineering Congress and Exposition Nashville Tennessee 1999 3 Y Cerci Y A Çengel and B Wood The Minimum Separation Work for Desalination Processes ASME International Mechanical Engineering Congress and Exposition Nashville Tennessee 1999 4 K Wark Jr Advanced Thermodynamics for Engineers New York McGrawHill 1995 temperature and pressure Daltons and Amagats laws hold exactly for idealgas mixtures but only approximately for realgas mixtures They can be expressed as Daltons law P m i 1 k P i T m V m Amagats law V m i 1 k V i T m P m Here Pi is called the component pressure and Vi is called the component volume Also the ratio PiPm is called the pressure fraction and the ratio ViVm is called the volume and fraction of component i For ideal gases Pi and Vi can be related to yi by P i P m V i V m N i N m y i The quantity yiPm is called the partial pressure and the quantity yiVm is called the partial volume The PvT behavior of real gas mixtures can be predicted by using generalized compress ibility charts The compressibility factor of the mixture can be expressed in terms of the compressibility factors of the individual gases as Z m i 1 k y i Z i where Zi is determined either at Tm and Vm Daltons law or at Tm and Pm Amagats law for each individual gas The PvT behavior of a gas mixture can also be predicted approximately by Kays rule which involves treating a gas mixture as a pure substance with pseudocritical properties determined from P crm i 1 k y i P cri and T crm i 1 k y i T cri The extensive properties of a gas mixture in general can be determined by summing the contributions of each compo nent of the mixture The evaluation of intensive properties of a gas mixture however involves averaging in terms of mass or mole fractions U m i 1 k U i i 1 k m i u i i 1 k N i u i H m i 1 k H i i 1 k m i h i i 1 k N i h i S m i 1 k S i i 1 k m i s i i 1 k N i s i and u m i 1 k mf i u i and u m i 1 k y i u i h m i 1 k mf i h i and h m i 1 k y i h i s m i 1 k mf i s i and s m i 1 k y i s i c vm i 1 k mf i c vi and c vm i 1 k y i c vi c pm i 1 k mf i c pi and c pm i 1 k y i c pi These relations are exact for idealgas mixtures and approxi mate for realgas mixtures The properties or property changes of individual components can be determined by using ideal gas or realgas relations developed in earlier chapters Final PDF to printer 702 GAS MIXTURES cen22672ch13675710indd 702 103117 0112 PM PROBLEMS Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Composition of Gas Mixtures 131C What are mass and mole fractions 132C Consider a mixture of several gases of identical masses Will all the mass fractions be identical How about the mole fractions 133C The sum of the mole fractions for an idealgas mix ture is equal to 1 Is this also true for a realgas mixture 134C Somebody claims that the mass and mole fractions for a mixture of CO2 and N2O gases are identical Is this true Why 135C Consider a mixture of two gases Can the apparent molar mass of this mixture be determined by simply taking the arithmetic average of the molar masses of the individual gases When will this be the case 136C What is the apparent molar mass for a gas mixture Does the mass of every molecule in the mixture equal the apparent molar mass 137C What is the apparent gas constant for a gas mixture Can it be larger than the largest gas constant in the mixture 138 The composition of moist air is given on a molar basis to be 78 percent N2 20 percent O2 and 2 percent water vapor Determine the mass fractions of the constituents of air 139 A gas mixture has the following composition on a mole basis 60 percent N2 and 40 percent CO2 Determine the gravimet ric analysis of the mixture its molar mass and the gas constant 1310 Repeat Prob 139 by replacing N2 with O2 1311 A gas mixture consists of 20 percent O2 30 percent N2 and 50 percent CO2 on mass basis Determine the volumet ric analysis of the mixture and the apparent gas constant 1312 A gas mixture consists of 4 kg of O2 5 kg of N2 and 7 kg of CO2 Determine a the mass fraction of each compo nent b the mole fraction of each component and c the aver age molar mass and gas constant of the mixture 1313 Using the definitions of mass and mole fractions derive a relation between them 1314 Consider a mixture of two gases A and B Show that when the mass fractions mfA and mfB are known the mole fractions can be determined from y A M B M A 1 mf A 1 M B and y B 1 y A where MA and MB are the molar masses of A and B PvT Behavior of Gas Mixtures 1315C Is a mixture of ideal gases also an ideal gas Give an example 1316C Express Daltons law of additive pressures Does this law hold exactly for idealgas mixtures How about nonidealgas mixtures 1317C Express Amagats law of additive volumes Does this law hold exactly for idealgas mixtures How about nonidealgas mixtures 1318C Explain how a realgas mixture can be treated as a pseudopure substance using Kays rule 1319C How is the PvT behavior of a component in an idealgas mixture expressed How is the PvT behavior of a component in a realgas mixture expressed 1320C What is the difference between the component pres sure and the partial pressure When are these two equivalent 1321C What is the difference between the component vol ume and the partial volume When are these two equivalent 1322C In a gas mixture which component will have the higher partial pressurethe one with the higher mole number or the one with the larger molar mass 1323C Consider a rigid tank that contains a mixture of two ideal gases A valve is opened and some gas escapes As a result the pressure in the tank drops Will the partial pressure of each component change How about the pressure fraction of each component 1324C Consider a rigid tank that contains a mixture of two ideal gases The gas mixture is heated and the pressure and temperature in the tank rise Will the partial pressure of each component change How about the pressure fraction of each component 1325C Is this statement correct The temperature of an idealgas mixture is equal to the sum of the temperatures of each individual gas in the mixture If not how would you correct it 1326C Is this statement correct The volume of an idealgas mixture is equal to the sum of the volumes of each individual gas in the mixture If not how would you correct it 1327C Is this statement correct The pressure of an ideal gas mixture is equal to the sum of the partial pressures of each individual gas in the mixture If not how would you correct it 1328 A gas mixture at 300 K and 200 kPa consists of 1 kg of CO2 and 3 kg of CH4 Determine the partial pressure of each gas and the apparent molar mass of the gas mixture 1329 A 03m3 rigid tank contains 06 kg of N2 and 04 kg of O2 at 300 K Determine the partial pressure of each gas and the total pressure of the mixture Answers 178 kPa 104 kPa 282 kPa Final PDF to printer 703 CHAPTER 13 cen22672ch13675710indd 703 103117 0112 PM 1330 Separation units often use membranes absorbers and other devices to reduce the mole fraction of selected con stituents in gaseous mixtures Consider a mixture of hydro carbons that consists of 60 percent by volume methane 30 percent ethane and 10 percent propane After passing through a separator the mole fraction of the propane is reduced to 1 percent The mixture pressure before and after the separa tion is 100 kPa Determine the change in the partial pressures of all the constituents in the mixture 1331 A mixture of gases consists of 30 percent hydrogen 40 percent helium and 30 percent nitrogen by volume Calculate the mass fractions and apparent molecular weight of this mixture 1332 The mass fractions of a mixture of gases are 15 per cent nitrogen 5 percent helium 60 percent methane and 20 percent ethane Determine the mole fractions of each con stituent the mixtures apparent molecular weight the partial pressure of each constituent when the mixture pressure is 1200 kPa and the apparent specific heats of the mixture when the mixture is at the room temperature 1333 The volumetric analysis of a mixture of gases is 30 percent oxygen 40 percent nitrogen 10 percent carbon diox ide and 20 percent methane Calculate the apparent specific heats and molecular weight of this mixture of gases Answers 1105 kJkgK 0812 kJkgK 2840 kgkmol FIGURE P1333 30 O2 40 N2 10 CO2 20 CH4 by volume 1334 An engineer has proposed mixing extra oxygen with normal air in internal combustion engines to control some of the exhaust products If an additional 5 percent by volume of oxygen is mixed with standard atmospheric air how will this change the mixtures molecular weight 1335 A rigid tank contains 05 kmol of Ar and 2 kmol of N2 at 250 kPa and 280 K The mixture is now heated to 400 K Determine the volume of the tank and the final pressure of the mixture 1336 A mixture of gases consists of 09 kg of oxygen 07 kg of carbon dioxide and 02 kg of helium This mixture is maintained at 100 kPa and 27C Determine the apparent molecular weight of this mixture the volume it occupies the partial volume of the oxygen and the partial pressure of the helium Answers 191 kgkmol 235 m3 0702 m3 532 kPa 1337 One liter of a liquid whose specific volume is 00003 m3kg is mixed with 2 liters of a liquid whose specific volume is 000023 m3kg in a container whose total volume is 3 liters What is the density of the resulting mixture in kgm3 1338E One poundmass of a gas whose density is 0001 lbmft3 is mixed with 2 lbm of a gas whose density is 0002 lbmft3 such that the pressure and temperature of the gases do not change Determine the resulting mixtures vol ume in ft3 and specific volume in ft3lbm 1339 A 30 percent by mass ethane and 70 percent meth ane mixture is to be blended in a 100m3 tank at 130 kPa and 25C If the tank is initially evacuated to what pressure should ethane be added before methane is added FIGURE P1339 70 CH4 30 C2H6 by mass 100 m3 130 kPa 25C 1340E The dry stack gas of an electricalgeneration station boiler has the following Orsat analysis 15 percent CO2 15 percent O2 and 1 percent CO This gas passes through a 10ft2 crosssection metering duct at a velocity of 20 fts at standard atmospheric pressure and 200F Determine the gas mixtures mass flow rate Answer 129 lbms 1341 A mixture of air and methane is formed in the inlet manifold of a natural gasfueled internal combustion engine The mole fraction of the methane is 15 percent This engine is operated at 3000 rpm and has a 5L displacement Determine the mass flow rate of this mixture in the manifold where the pressure and temperature are 80 kPa and 20C Answer 665 kgmin 1342 A rigid tank that contains 2 kg of N2 at 25C and 550 kPa is connected to another rigid tank that contains 4 kg of O2 at 25C and 150 kPa The valve connecting the two tanks is opened and the two gases are allowed to mix If the final mixture temperature is 25C determine the volume of each tank and the final mixture pressure Answers 0322 m3 207 m3 204 kPa FIGURE P1342 O2 4 kg 25C 150 kPa N2 2 kg 25C 550 kPa 1343E A rigid tank contains 1 lbmol of argon gas at 400 R and 750 psia A valve is now opened and 3 lbmol of N2 gas is allowed to enter the tank at 340 R and 1200 psia The final mixture tem perature is 360 R Determine the pressure of the mixture using a the idealgas equation of state and b the compressibility chart and Daltons law Answers a 2700 psia b 2472 psia Final PDF to printer 704 GAS MIXTURES cen22672ch13675710indd 704 103117 0112 PM 1344 A volume of 03 m3 of O2 at 200 K and 8 MPa is mixed with 05 m3 of N2 at the same temperature and pressure forming a mixture at 200 K and 8 MPa Determine the vol ume of the mixture using a the idealgas equation of state b Kays rule and c the compressibility chart and Amagats law 1345E The volumetric analysis of a mixture of gases is 30 percent oxygen 40 percent nitrogen 10 percent carbon dioxide and 20 percent methane This mixture flows through a 1indiameter pipe at 1500 psia and 70F with a velocity of 10 fts Determine the volumetric and mass flow rates of this mixture a treating it as an idealgas mixture b using a com pressibility factor based on Amagats law of additive volumes and c using Kays pseudocritical pressure and temperature Properties of Gas Mixtures 1346C Is the total internal energy of an idealgas mixture equal to the sum of the internal energies of each individual gas in the mixture Answer the same question for a realgas mixture 1347C Is the specific internal energy of a gas mixture equal to the sum of the specific internal energies of each individual gas in the mixture 1348C Answer Prob 1346C and 1347C for entropy 1349C When evaluating the entropy change of the com ponents of an idealgas mixture do we have to use the partial pressure of each component or the total pressure of the mixture 1350C Suppose we want to determine the enthalpy change of a realgas mixture undergoing a process The enthalpy change of each individual gas is determined by using the generalized enthalpy chart and the enthalpy change of the mixture is deter mined by summing them Is this an exact approach Explain 1351 A mixture that is 15 percent carbon dioxide 5 percent carbon monoxide 10 percent oxygen and 70 percent nitrogen by volume undergoes an adiabatic compression process having a compression ratio of 8 If the initial state of the mixture is 300 K and 100 kPa determine the makeup of the mixture on a mass basis and the internal energy change per unit mass of mixture 1352 The volumetric analysis of a mixture of gases is 30 percent oxygen 40 percent nitrogen 10 percent carbon dioxide and 20 percent methane This mixture is heated from 20C to 200C while flowing through a tube in which the pres sure is maintained at 150 kPa Determine the heat transfer to the mixture per unit mass of the mixture FIGURE P1352 150 kPa 200C 150 kPa 20C 30 O2 40 N2 10 CO2 20 CH4 by volume q 1353 A mixture of nitrogen and carbon dioxide has a car bon dioxide mass fraction of 50 percent This mixture is heated at constant pressure in a closed system from 120 kPa and 30C to 200C Calculate the work produced during this heating in kJkg Answer 413 kJkg 1354E The mass fractions of a mixture of gases are 15 percent nitrogen 5 percent helium 60 percent methane and 20 percent ethane This mixture is compressed from 20 psia and 100F in an isentropic process to 200 psia Determine the final mixture tem perature and the work required per unit mass of the mixture 1355 A mixture of gases consists of 01 kg of oxygen 1 kg of carbon dioxide and 05 kg of helium This mixture is heated from 10C to 260C while its pressure is maintained constant at 350 kPa Determine the change in the volume of the mixture and the total heat transferred to the mixture Answers 0896 m3 552 kJkg 1356 An insulated tank that contains 1 kg of O2 at 15C and 300 kPa is connected to a 2m3 uninsulated tank that contains N2 at 50C and 500 kPa The valve connecting the two tanks is opened and the two gases form a homogeneous mixture at 25C Determine a the final pressure in the tank b the heat trans fer and c the entropy generated during this process Assume T0 25C Answers a 445 kPa b 187 kJ c 0962 kJK FIGURE P1356 O2 1 kg 15C 300 kPa N2 2 m3 50C 500 kPa 1357 Reconsider Prob 1356 Using appropriate soft ware compare the results obtained assuming idealgas behavior with constant specific heats at the average temperature and using realgas data obtained from the software by assuming variable specific heats over the temperature range 1358 An insulated rigid tank is divided into two compart ments by a partition One compartment contains 7 kg of oxygen gas at 40C and 100 kPa and the other compartment contains 4 kg of nitrogen gas at 20C and 150 kPa Now the partition is removed and the two gases are allowed to mix Determine a the mixture temperature and b the mixture pressure after equilibrium has been established O2 7 kg 40C 100 kPa N2 4 kg 20C 150 kPa Partition FIGURE P1358 Final PDF to printer 705 CHAPTER 13 cen22672ch13675710indd 705 103117 0112 PM 1359 A mixture of hydrocarbon gases is composed of 60 per cent methane 25 percent propane and 15 percent butane by weight This mixture is compressed from 100 kPa and 20C to 800 kPa in a reversible isothermal steadyflow compressor Calculate the work and heat transfer for this compression per unit mass of the mixture FIGURE P1359 800 kPa 100 kPa 20C q 60 CH4 25 C3H8 15 C4H10 by mass w 1360E A mixture of 65 percent N2 and 35 percent CO2 gases on a mass basis enters the nozzle of a turbojet engine at 60 psia and 1400 R with a low velocity and it expands to a pressure of 12 psia If the isentropic efficiency of the nozzle is 88 percent determine a the exit temperature and b the exit velocity of the mixture Assume constant specific heats at room temperature 1361E Reconsider Prob 1360E Using appropriate software first solve the stated problem and then for all other conditions being the same resolve the problem to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 2200 fts at the nozzle exit 1362 An equimolar mixture of helium and argon gases is to be used as the working fluid in a closedloop gasturbine cycle The mixture enters the turbine at 25 MPa and 1300 K and expands isentropically to a pressure of 200 kPa Determine the work output of the turbine per unit mass of the mixture FIGURE P1362 25 MPa 1300 K w 200 kPa He Ar turbine 1363 The combustion of a hydrocarbon fuel with air results in a mixture of products of combustion having the composi tion on a volume basis as follows 489 percent carbon dioxide 650 percent water vapor 1220 percent oxygen and 7641 percent nitrogen Determine the average molar mass of the mixture the average specific heat at constant pressure of the mix ture at 600 K in kJkmolK and the partial pressure of the water vapor in the mixture for a mixture pressure of 200 kPa 1364 In a liquidoxygen plant it is proposed that the pressure and temperature of air that is initially at 9000 kPa and 10C be adiabatically reduced to 50 kPa and 73C Using Kays rule and the departure charts determine whether this is possible If so then how much work per unit mass will this process produce FIGURE P1364 21 O2 79 N2 by mole 9000 kPa 10C 1365E A gaseous mixture consists of 75 percent methane and 25 percent ethane by mass 2 million cubic feet of this mixture is trapped in a geological formation as natural gas at 300F and 1300 psia This natural gas is pumped 6000 ft to the surface At the surface the gas pressure is 20 psia and its temperature is 200F Using Kays rule and the enthalpydeparture charts cal culate the work required to pump this gas Answer 186 108 Btu 1366 A mixture of hydrogen and oxygen has a hydrogen mass fraction of 033 Determine the difference in the entropy of the mixture between a state of 750 kPa 150C and another state of 150 kPa 150C in kJkgK 1367 A pistoncylinder device contains a mixture of 05 kg of H2 and 12 kg of N2 at 100 kPa and 300 K Heat is now transferred to the mixture at constant pressure until the volume is doubled Assuming constant specific heats at the average temperature deter mine a the heat transfer and b the entropy change of the mixture 1368E During the expansion process of the ideal Otto cycle the gas is a mixture whose volumetric composition is 25 percent nitrogen 7 percent oxygen 28 percent water and 40 percent carbon dioxide Calculate the thermal efficiency of this cycle when the air at the beginning of the compression is at 12 psia and 55F the compression ratio is 7 and the maxi mum cycle temperature is 1600F Model the heataddition and heatrejection processes using constant gas properties that are the average of the air and expansion gas properties 1369E Reconsider Prob 1368E How does the ther mal efficiency of the cycle compare to that predicted by air standard analysis 1370 The gas passing through the turbine of a simple ideal Brayton cycle has the volumetric composition 30 percent Final PDF to printer 706 GAS MIXTURES cen22672ch13675710indd 706 103117 0112 PM nitrogen 10 percent oxygen 40 percent carbon dioxide and 20 percent water Calculate the thermal efficiency of this cycle when the air enters the compressor at 100 kPa and 20C the pressure ratio is 8 and the temperature at the turbine inlet is 1000C Model the heataddition and heatrejection processes using constant gas properties that are the average of the air and expansion gas properties Answer 373 percent 1371 Reconsider Prob 1370 How does the thermal effi ciency of the cycle compare to that predicted by airstandard analysis 1372 A pistoncylinder device contains 6 kg of H2 and 21 kg of N2 at 160 K and 5 MPa Heat is now transferred to the device and the mixture expands at constant pressure until the temperature rises to 200 K Determine the heat transfer during this process by treating the mixture a as an ideal gas and b as a nonideal gas and using Amagats law Answers a 4273 kJ b 4745 kJ FIGURE P1372 Heat 6 kg H2 21 kg N2 160 K 5 MPa 1373 Reconsider Prob 1372 Determine the total entropy change and exergy destruction associated with the process by treating the mixture a as an ideal gas and b as a nonideal gas and using Amagats law Assume constant specific heats at room temperature and take T0 20C Special Topic Chemical Potential and the Separation Work of Mixtures 1374C What is an ideal solution Comment on the vol ume change enthalpy change entropy change and chemical potential change during the formation of ideal and nonideal solutions 1375C It is common experience that two gases brought into contact mix by themselves In the future could it be possible to invent a process that will enable a mixture to separate into its components by itself without any work or exergy input 1376C A 2L liquid is mixed with 3 L of another liquid forming a homogeneous liquid solution at the same tempera ture and pressure Can the volume of the solution be more or less than the 5 L Explain 1377C A 2L liquid at 20C is mixed with 3 L of another liquid at the same temperature and pressure in an adiabatic container forming a homogeneous liquid solution Someone claims that the temperature of the mixture rose to 22C after mixing Another person refutes the claim saying that this would be a violation of the first law of thermodynamics Who do you think is right 1378 Brackish water at 18C with total dissolved solid con tent of TDS 780 ppm a salinity of 0078 percent on mass basis is to be used to produce fresh water with negligible salt content at a rate of 175 Ls Determine the minimum power input required Also determine the minimum height to which the brackish water must be pumped if fresh water is to be obtained by reverse osmosis using semipermeable membranes 1379 A river is discharging into the ocean at a rate of 150000 m3s Determine the amount of power that can be gen erated if the river water mixes with the ocean water reversibly Take the salinity of the ocean to be 25 percent on mass basis and assume both the river and the ocean are at 15C 1380 Reconsider Prob 1379 Using appropriate software investigate the effect of the salinity of the ocean on the maximum power generated Let the salinity vary from 0 to 5 percent Plot the power produced versus the salinity of the ocean and discuss the results 1381E Fresh water is to be obtained from brackish water at 65F with a salinity of 012 percent on mass basis or TDS 1200 ppm Determine a the mole fractions of the water and the salts in the brackish water b the minimum work input required to separate 1 lbm of brackish water completely into pure water and pure salts and c the minimum work input required to obtain 1 lbm of fresh water 1382 Fresh water is obtained from seawater at a rate of 15 m3s by a desalination plant that consumes 115 MW of power and has a secondlaw efficiency of 20 percent Determine the power that can be produced if the fresh water produced is mixed with the seawater reversibly 1383E Is it possible for an adiabatic liquidvapor separator to separate wet steam at 100 psia and 90 percent quality so that the pressure of the outlet streams is greater than 100 psia 1384 A desalination plant produces fresh water from sea water at 10C with a salinity of 32 percent on mass basis at a rate of 12 m3s while consuming 85 MW of power The salt content of the fresh water is negligible and the amount of fresh water produced is a small fraction of the seawater used Determine the secondlaw efficiency of this plant Review Problems 1385 Air has the following composition on a mole basis 21 percent O2 78 percent N2 and 1 percent Ar Determine the gravimetric analysis of air and its molar mass Answers 232 percent O2 754 percent N2 14 percent Ar 2896 kgkmol 1386 The products of combustion of a hydrocarbon fuel and air are composed of 8 kmol CO2 9 kmol H2O 4 kmol O2 and 94 kmol N2 If the mixture pressure is 101 kPa determine the partial pressure of the water vapor in the product gas mix ture and the temperature at which the water vapor would begin to condense when the products are cooled at constant pressure Final PDF to printer 707 CHAPTER 13 cen22672ch13675710indd 707 103117 0112 PM 1387 A mixture of gases is assembled by first filling an evacu ated 015m3 tank with neon until the pressure is 35 kPa Oxygen is added next until the pressure increases to 105 kPa Finally nitrogen is added until the pressure increases to 140 kPa During each step of the tanks filling the contents are maintained at 60C Determine the mass of each constituent in the resulting mixture the apparent molecular weight of the mixture and the fraction of the tank volume occupied by nitrogen 1388 A mixture of carbon dioxide and nitrogen flows through a converging nozzle The mixture leaves the nozzle at a temperature of 500 K with a velocity of 360 ms If the velocity is equal to the speed of sound at the exit temperature determine the required makeup of the mixture on a mass basis 1389 A pistoncylinder device contains products of com bustion from the combustion of a hydrocarbon fuel with air The combustion process results in a mixture that has the composition on a volume basis as follows 489 percent car bon dioxide 650 percent water vapor 1220 percent oxygen and 7641 percent nitrogen This mixture is initially at 1800 K and 1 MPa and expands in an adiabatic reversible process to 200 kPa Determine the work done on the piston by the gas in kJkg of mixture Treat the water vapor as an ideal gas 1390 A mixture of gases consists of 1 kmol of carbon diox ide 1 kmol of nitrogen and 03 kmol of oxygen Determine the total amount of work required to compress this mixture isothermally from 10 kPa and 27C to 100 kPa 1391 A rigid tank contains 2 kmol of N2 and 6 kmol of CH4 gases at 200 K and 12 MPa Estimate the volume of the tank using a the idealgas equation of state b Kays rule and c the compressibility chart and Amagats law 1392 A mixture of ideal gases has a specific heat ratio of k 135 and an apparent molecular weight of M 32 kgkmol Determine the work in kJkg required to compress this mix ture isentropically in a closed system from 100 kPa and 35C to 700 kPa Answer 150 kJkg FIGURE P1392 Gas mixture k 135 M 32 kgkmol 100 kPa 35C 1393 A rigid tank contains a mixture of 4 kg of He and 8 kg of O2 at 170 K and 7 MPa Heat is now transferred to the tank and the mixture temperature rises to 220 K Treating the He as an ideal gas and the O2 as a nonideal gas determine a the final pressure of the mixture and b the heat transfer 1394 A springloaded pistoncylinder device contains a mixture of gases whose pressure fractions are 25 percent Ne 50 percent O2 and 25 percent N2 The piston diameter and spring are selected for this device such that the volume is 01 m3 when the pressure is 200 kPa and 10 m3 when the pres sure is 1000 kPa Initially the gas is added to this device until the pressure is 200 kPa and the temperature is 10C The device is now heated until the pressure is 500 kPa Calculate the total work and heat transfer for this process Answers 118 kJ 569 kJ FIGURE P1394 Q 25 Ne 50 O2 25 N2 by pressure 01 m3 10C 200 kPa 1395 Reconsider Prob 1394 The pistoncylinder device is filled with a mixture whose mass is 55 percent nitro gen and 45 percent carbon dioxide Initially this mixture is at 200 kPa and 45C The gas is heated until the volume has dou bled Calculate the total work and heat transfer for this process 1396 Reconsider Prob 1395 Calculate the total work and heat transfer required to triple the initial pressure of the mix ture as it is heated in the springloaded pistoncylinder device 1397 A mixture of gases consists of 01 kg of oxygen 1 kg of carbon dioxide and 05 kg of helium This mixture is expanded from 1000 kPa and 327C to 100 kPa in an adia batic steadyflow turbine of 90 percent isentropic efficiency Calculate the secondlaw efficiency and the exergy destruction during this expansion process Take T0 25C Answers 894 percent 79 kJkg FIGURE P1397 O2 CO2 He mixture 100 kPa 1000 kPa 327C Final PDF to printer 708 GAS MIXTURES cen22672ch13675710indd 708 103117 0112 PM 1398 Using appropriate software write a program to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given and to determine the mass fractions of the components when the mole fractions are given Run the program for a sample case and give the results 1399 Using appropriate software write a program to determine the apparent gas constant constant volume specific heat and internal energy of a mixture of three ideal gases when the mass fractions and other properties of the constituent gases are given Run the program for a sample case and give the results 13100 Using Daltons law show that Z m i 1 k y i Z i for a realgas mixture of k gases where Z is the compressibility factor 13101 Two mass streams of two different ideal gases are mixed in a steadyflow chamber while receiving energy by heat transfer from the surroundings The mixing process takes place at constant pressure with no work and negligible changes in kinetic and potential energies Assume the gases have con stant specific heats a Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the mass flow rates specific heats and temperatures of the three mass streams b Obtain an expression for the exit volume flow rate in terms of the rate of heat transfer to the mixing chamber mixture pressure universal gas constant and the specific heats and molar masses of the inlet gases and exit mixture c For the special case of adiabatic mixing show that the exit volume flow rate is a function of the two inlet volume flow rates and the specific heats and molar masses of the inlets and exit d For the special case of adiabatic mixing of the same ideal gases show that the exit volume flow rate is a function of the two inlet volume flow rates FIGURE P13101 Qin 1 2 3 Steadyflow chamber Surroundings Fundamentals of Engineering FE Exam Problems 13102 An idealgas mixture consists of 3 kmol of N2 and 6 kmol of CO2 The mass fraction of CO2 in the mixture is a 0241 b 0333 c 0500 d 0667 e 0759 13103 An idealgas mixture whose apparent molar mass is 20 kgkmol consists of N2 and three other gases If the mole fraction of nitrogen is 055 its mass fraction is a 015 b 023 c 039 d 055 e 077 13104 An idealgas mixture consists of 2 kmol of N2 and 4 kmol of CO2 The apparent gas constant of the mixture is a 0215 kJkgK b 0225 kJkgK c 0243 kJkgK d 0875 kJkgK e 124 kJkgK 13105 A rigid tank is divided into two compartments by a partition One compartment contains 3 kmol of N2 at 400 kPa and the other compartment contains 7 kmol of CO2 at 200 kPa Now the partition is removed and the two gases form a homo geneous mixture at 250 kPa The partial pressure of N2 in the mixture is a 75 kPa b 90 kPa c 125 kPa d 175 kPa e 250 kPa 13106 A 60L rigid tank contains an idealgas mixture of 5 g of N2 and 5 g of CO2 at a specified pressure and tem perature If N2 were separated from the mixture and stored at mixture temperature and pressure its volume would be a 30 L b 37 L c 42 L d 49 L e 60 L 13107 An idealgas mixture consists of 3 kg of Ar and 6 kg of CO2 gases The mixture is now heated at constant volume from 250 K to 350 K The amount of heat transfer is a 374 kJ b 436 kJ c 488 kJ d 525 kJ e 664 kJ 13108 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20C and 150 kPa while the other compart ment contains 5 kmol of H2 gas at 35C and 300 kPa Now the partition between the two gases is removed and the two gases form a homogeneous idealgas mixture The temperature of the mixture is a 25C b 30C c 22C d 32C e 34C 13109 A pistoncylinder device contains an idealgas mix ture of 3 kmol of He gas and 7 kmol of Ar gas at 70C and 400 kPa Now the gas expands at constant pressure until its volume doubles The amount of heat transfer to the gas mixture is a 286 MJ b 71 MJ c 30 MJ d 15 MJ e 66 MJ 13110 An idealgas mixture of helium and argon gases with identical mass fractions enters a turbine at 1500 K and 1 MPa at a rate of 012 kgs and expands isentropically to 100 kPa The power output of the turbine is a 253 kW b 310 kW c 341 kW d 463 kW e 550 kW 13111 An idealgas mixture consists of 60 percent helium and 40 percent argon gases by mass The mixture is now expanded isentropically in a turbine from 400C and 12 MPa Final PDF to printer 709 CHAPTER 13 cen22672ch13675710indd 709 103117 0112 PM to a pressure of 200 kPa The mixture temperature at the tur bine exit is a 56C b 195C c 130C d 112C e 400C Design and Essay Problems 13112 The simple additive rule may not be appropriate for the volume of binary mixtures of gases Prove this for a pair of gases of your choice at several different temperatures and pressures using Kays rule and the principle of corresponding states 13113 You have a rigid tank equipped with a pressure gauge Describe a procedure by which you could use this tank to blend ideal gases in prescribed molefraction portions 13114 Prolonged exposure to mercury even at relatively low but toxic concentrations in the air is known to cause per manent mental disorders insomnia and pain and numbness in the hands and the feet among other things Therefore the max imum allowable concentration of mercury vapor in the air at workplaces is regulated by federal agencies These regulations require that the average level of mercury concentration in the air does not exceed 01 mgm3 Consider a mercury spill that occurs in an airtight storage room at 20C in San Francisco during an earthquake Calculate the highest level of mercury concentration in the air that can occur in the storage room in mgm3 and determine if it is within the safe level The vapor pressure of mercury at 20C is 0173 Pa Propose some guidelines to safeguard against the formation of toxic concentrations of mercury vapor in air in storage rooms and laboratories 13115 A pressurized mixture of nitrogen and argon is sup plied to a directional control nozzle on a space satellite Plot the gas velocity at the nozzle exit as a function of the argon mass fraction with fixed pressure and temperature at the entrance and pressure at the exit The force produced by this nozzle is proportional to the product of the mass flow rate and velocity at the exit Is there an optimal argon mass fraction that produces the greatest force Final PDF to printer No text present cen22672ch14711746indd 711 103117 0113 PM 711 OBJECTIVES The objectives of Chapter 14 are to Differentiate between dry air and atmospheric air Define and calculate the specific and relative humidity of atmospheric air Calculate the dewpoint temperature of atmospheric air Relate the adiabatic saturation temperature and wetbulb temperatures of atmospheric air Use the psychrometric chart as a tool to determine the properties of atmospheric air Apply the principles of the conservation of mass and energy to various air conditioning processes GASVAP O R M I X T UR E S AN D A I R C O N D ITIO N IN G A t temperatures below the critical temperature the gas phase of a sub stance is frequently referred to as a vapor The term vapor implies a gaseous state that is close to the saturation region of the substance raising the possibility of condensation during a process In Chap 13 we discussed mixtures of gases that are usually above their critical temperatures Therefore we were not concerned about any of the gases condensing during a process Not having to deal with two phases greatly simplified the analysis When we are dealing with a gasvapor mixture how ever the vapor may condense out of the mixture during a process forming a twophase mixture This may complicate the analysis considerably There fore a gasvapor mixture needs to be treated differently from an ordinary gas mixture Several gasvapor mixtures are encountered in engineering In this chap ter we consider the airwater vapor mixture which is the most commonly encountered gasvapor mixture in practice We also discuss airconditioning which is the primary application area of airwater vapor mixtures 14 CHAPTER Final PDF to printer 712 GASVAPOR MIXTURES cen22672ch14711746indd 712 103117 0113 PM 141 DRY AND ATMOSPHERIC AIR Air is a mixture of nitrogen oxygen and small amounts of some other gases Air in the atmosphere normally contains some water vapor or moisture and is referred to as atmospheric air By contrast air that contains no water vapor is called dry air It is often convenient to treat air as a mixture of water vapor and dry air since the composition of dry air remains relatively constant but the amount of water vapor changes as a result of condensation and evaporation from oceans lakes rivers showers and even the human body Although the amount of water vapor in the air is small it plays a major role in human com fort Therefore it is an important consideration in airconditioning applications The temperature of air in airconditioning applications ranges from about 10 to about 50C In this range dry air can be treated as an ideal gas with a con stant cp value of 1005 kJkgK 0240 BtulbmR with negligible error under 02 percent as illustrated in Fig 141 Taking 0C as the reference tempera ture the enthalpy and enthalpy change of dry air can be determined from h dry air c p T 1005 kJkgCT kJkg 141a and Δ h dry air c p ΔT 1005 kJkgC ΔT kJkg 141b where T is the air temperature in C and ΔT is the change in temperature In airconditioning processes we are concerned with the changes in enthalpy Δh which is independent of the reference point selected It would also be very convenient to treat the water vapor in the air as an ideal gas and you would probably be willing to sacrifice some accuracy for such convenience Well it turns out that we can have the convenience without much sacrifice At 50C the saturation pressure of water is 123 kPa At pressures below this value water vapor can be treated as an ideal gas with negligible error under 02 percent even when it is a saturated vapor Therefore water vapor in air behaves as if it existed alone and obeys the idealgas relation Pv RT Then the atmospheric air can be treated as an idealgas mixture whose pressure is the sum of the partial pressure of dry air Pa and that of water vapor Pv P P a P v kPa 142 The partial pressure of water vapor is usually referred to as the vapor pressure It is the pressure water vapor would exert if it existed alone at the temperature and volume of atmospheric air Since water vapor is an ideal gas the enthalpy of water vapor is a function of temperature only that is h hT This can also be observed from the Ts diagram of water given in Fig A9 and Fig 142 where the constant enthalpy lines coincide with constanttemperature lines at temperatures below 50C Therefore the enthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at the same temperature That is h v T low P h g T 143 Throughout this chapter the subscript a denotes dry air and the subscript v denotes water vapor FIGURE 141 The cp of air can be assumed to be constant at 1005 kJkgC in the temperature range 10 to 50C with an error under 02 percent Dry air T C cp kJkgC 10 0 10 20 30 40 50 10038 10041 10045 10049 10054 10059 10065 FIGURE 142 At temperatures below 50C the h constant lines coincide with the T constant lines in the superheated vapor region of water T C s h const 50 Final PDF to printer 713 CHAPTER 14 cen22672ch14711746indd 713 103117 0113 PM The enthalpy of water vapor at 0C is 25009 kJkg The average cp value of water vapor in the temperature range 10 to 50C can be taken to be 182 kJkgC Then the enthalpy of water vapor can be determined approxi mately from h g T 25009 182T kJkg T in C 144 or h g T 10609 0435T Btulbm T in F 145 in the temperature range 10 to 50C or 15 to 120F with negligible error as shown in Fig 143 142 SPECIFIC AND RELATIVE HUMIDITY OF AIR The amount of water vapor in the air can be specified in various ways Probably the most logical way is to specify directly the mass of water vapor present in a unit mass of dry air This is called absolute or specific humidity also called humidity ratio and is denoted by ω ω m v m a kg water vaporkg dry air 146 The specific humidity can also be expressed as ω m v m a P v V R v T P a V R a T P v R v P a R a 0622 P v P a 147 or ω 0622 P v P P v kg water vaporkg dry air 148 where P is the total pressure Consider 1 kg of dry air By definition dry air contains no water vapor and thus its specific humidity is zero Now let us add some water vapor to this dry air The specific humidity will increase As more vapor or moisture is added the specific humidity will keep increasing until the air can hold no more moisture At this point the air is said to be saturated with moisture and it is called saturated air Any moisture introduced into saturated air will con dense The amount of water vapor in saturated air at a specified temperature and pressure can be determined from Eq 148 by replacing Pv with Pg the saturation pressure of water at that temperature Fig 144 The amount of moisture in the air has a definite effect on how comfortable we feel in an environment However the comfort level depends more on the amount of moisture the air holds mv relative to the maximum amount of moisture the air can hold at the same temperature mg The ratio of these two quantities is called the relative humidity ϕ Fig 145 ϕ m v m g P v V R v T P g V R v T P v P g 149 FIGURE 143 In the temperature range 10 to 50C the hg of water can be determined from Eq 144 with negligible error Water vapor 10 0 10 20 30 40 50 24821 25009 25192 25374 25556 25735 25913 24827 25009 25191 25373 25555 25737 25919 06 00 01 01 01 02 06 hg kJkg T C Table A4 Eq 144 Difference kJkg FIGURE 144 For saturated air the vapor pressure is equal to the saturation pressure of water Air 25C 100 kPa PsatH2O 25C 31698 kPa Pv 0 Pv 31698 kPa Pv 31698 kPa dry air unsaturated air saturated air Final PDF to printer 714 GASVAPOR MIXTURES cen22672ch14711746indd 714 103117 0113 PM where P g P sat T 1410 Combining Eqs 148 and 149 we can also express the relative humidity as ϕ ωP 0622 ω P g and ω 0622ϕ P g P ϕ P g 1411a b The relative humidity ranges from 0 for dry air to 1 for saturated air Note that the amount of moisture air can hold depends on its temperature There fore the relative humidity of air changes with temperature even when its spe cific humidity remains constant Atmospheric air is a mixture of dry air and water vapor and thus the enthalpy of air is expressed in terms of the enthalpies of the dry air and the water vapor In most practical applications the amount of dry air in the air water vapor mixture remains constant but the amount of water vapor changes Therefore the enthalpy of atmospheric air is expressed per unit mass of dry air instead of per unit mass of the airwater vapor mixture The total enthalpy an extensive property of atmospheric air is the sum of the enthalpies of dry air and the water vapor H H a H v m a h a m v h v Dividing by ma gives h H m a h a m v m a h v h a ω h v or h h a ω h g kJkg dry air 1412 since hv hg Fig 146 Also note that the ordinary temperature of atmospheric air is frequently referred to as the drybulb temperature to differentiate it from other forms of temperatures that shall be discussed FIGURE 145 Specific humidity is the actual amount of water vapor in 1 kg of dry air whereas relative humidity is the ratio of the actual amount of moisture in the air at a given temperature to the maximum amount of moisture air can hold at the same temperature Air 25C 1 atm ma 1 kg mv 001 kg mv max 002 kg Specific humidity ω 001 Relative humidity ϕ 50 kg H2O kg dry air FIGURE 146 The enthalpy of moist atmospheric air is expressed per unit mass of dry air not per unit mass of moist air Dry air 1 kg ha moisture ω kg hg 1 ω kg of moist air h ha g kJkg dry air ωh FIGURE 147 Schematic for Example 141 Room 5 m 5 m 3 m T 25C P ϕ 100 kPa 75 EXAMPLE 14 1 The Amount of Water Vapor in Room Air A 5m 5m 3m room shown in Fig 147 contains air at 25C and 100 kPa at a relative humidity of 75 percent Determine a the partial pressure of dry air b the specific humidity c the enthalpy per unit mass of the dry air and d the masses of the dry air and water vapor in the room SOLUTION The relative humidity of air in a room is given The dry air pressure specific humidity enthalpy and the masses of dry air and water vapor in the room are to be determined Assumptions The dry air and the water vapor in the room are ideal gases Properties The constantpressure specific heat of air at room temperature is cp 1005 kJkgK Table A2a For water at 25C we have Psat 31698 kPa and hg 25465 kJkg Table A4 Final PDF to printer 715 CHAPTER 14 cen22672ch14711746indd 715 103117 0113 PM 143 DEWPOINT TEMPERATURE If you live in a humid area you are probably used to waking up most summer mornings and finding the grass wet You know it did not rain the night before So what happened Well the excess moisture in the air simply condensed on the cool surfaces forming what we call dew In summer a considerable amount of water vaporizes during the day As the temperature falls during the Analysis a The partial pressure of dry air can be determined from Eq 142 P a P P v where P v ϕ P g ϕ P sat 25C 07531698 kPa 238 kPa Thus P a 100 238 kPa 9762 kPa b The specific humidity of air is determined from Eq 148 ω 0622 P v P P v 0622238 kPa 100 238 kPa 00152 kg H 2 Okg dry air c The enthalpy of air per unit mass of dry air is determined from Eq 1412 h h a ω h v c p T ω h g 1005 kJkgC25C 0015225465 kJkg 638 kJkg dry air The enthalpy of water vapor 25465 kJkg could also be determined from the approximation given by Eq 144 h g 25C 25009 18225 25464 kJkg which is almost identical to the value obtained from Table A4 d Both the dry air and the water vapor fill the entire room completely Therefore the volume of each gas is equal to the volume of the room V a V v V room 5 m5 m3 m 75 m3 The masses of the dry air and the water vapor are determined from the idealgas rela tion applied to each gas separately m a P a V a R a T 9762 kPa75 m 3 0287 kPa m 3 kgK298 K 8561 kg m v P v V v R v T 238 kPa75 m 3 04615 kPa m 3 kgK298 K 130 kg The mass of the water vapor in the air could also be determined from Eq 146 m v ω m a 001528561 kg 130 kg Final PDF to printer 716 GASVAPOR MIXTURES cen22672ch14711746indd 716 103117 0113 PM night so does the moisture capacity of air which is the maximum amount of moisture air can hold What happens to the relative humidity during this process After a while the moisture capacity of air equals its moisture con tent At this point air is saturated and its relative humidity is 100 percent Any further drop in temperature results in the condensation of some of the moisture and this is the beginning of dew formation The dewpoint temperature Tdp is defined as the temperature at which con densation begins when the air is cooled at constant pressure In other words Tdp is the saturation temperature of water corresponding to the vapor pressure T dp T sat P v 1413 This is also illustrated in Fig 148 As the air cools at constant pressure the vapor pressure Pv remains constant Therefore the vapor in the air state 1 undergoes a constantpressure cooling process until it strikes the saturated vapor line state 2 The temperature at this point is Tdp and if the temperature drops any further some vapor condenses out As a result the amount of vapor in the air decreases which results in a decrease in Pv The air remains satu rated during the condensation process and thus follows a path of 100 percent relative humidity the saturated vapor line The ordinary temperature and the dewpoint temperature of saturated air are identical You have probably noticed that when you buy a cold canned drink from a vending machine on a hot and humid day dew forms on the can The forma tion of dew on the can indicates that the temperature of the drink is below the dewpoint temperature of the surrounding air Fig 149 The dewpoint temperature of room air can be determined easily by cooling some water in a metal cup by adding small amounts of ice and stirring The temperature of the outer surface of the cup when dew starts to form on the surface is the dewpoint temperature of the air FIGURE 149 When the temperature of a cold drink is below the dewpoint temperature of the surrounding air it sweats Moist air Liquid water droplets dew T Tdp FIGURE 1410 Schematic for Example 142 Cold outdoors 10C Air 20C 75 Typical temperature distribution 16C 18C 20C 20C 20C18C 16C EXAMPLE 142 Fogging of the Windows in a House In cold weather condensation often occurs on the inner surfaces of the windows due to the lower air temperatures near the window surface Consider a house shown in Fig 1410 that contains air at 20C and 75 percent relative humidity At what win dow temperature will the moisture in the air start condensing on the inner surfaces of the windows SOLUTION The interior of a house is maintained at a specified temperature and humidity The window temperature at which fogging starts is to be determined Properties The saturation pressure of water at 20C is Psat 23392 kPa Table A4 Analysis The temperature distribution in a house in general is not uniform When the outdoor temperature drops in winter so does the indoor temperature near the walls and the windows Therefore the air near the walls and the windows remains at a lower temperature than at the inner parts of a house even though the total pressure and the vapor pressure remain constant throughout the house As a result the air near the walls and the windows undergoes a Pv constant cooling process until the moisture in the air starts condensing This happens when the air reaches its dewpoint tempera ture Tdp which is determined from Eq 1413 to be T dp T sat P v FIGURE 148 Constantpressure cooling of moist air and the dewpoint temperature on the Ts diagram of water T s T1 Tdp 2 1 Pv const Final PDF to printer 717 CHAPTER 14 cen22672ch14711746indd 717 103117 0113 PM 144 ADIABATIC SATURATION AND WETBULB TEMPERATURES Relative humidity and specific humidity are frequently used in engineering and atmospheric sciences and it is desirable to relate them to easily mea surable quantities such as temperature and pressure One way of determin ing the relative humidity is to determine the dewpoint temperature of air as discussed in the last section Knowing the dewpoint temperature we can determine the vapor pressure Pv and thus the relative humidity This approach is simple but not quite practical Another way of determining the absolute or relative humidity is related to an adiabatic saturation process shown schematically and on a Ts diagram in Fig 1411 The system consists of a long insulated channel that contains a pool of water A steady stream of unsaturated air that has a specific humid ity of ω1 unknown and a temperature of T1 is passed through this chan nel As the air flows over the water some water evaporates and mixes with the airstream The moisture content of air increases during this process and its temperature decreases since part of the latent heat of vaporization of the water that evaporates comes from the air If the channel is long enough the airstream exits as saturated air ϕ 100 percent at temperature T2 which is called the adiabatic saturation temperature If makeup water is supplied to the channel at the rate of evaporation at tem perature T2 the adiabatic saturation process described above can be analyzed as a steadyflow process The process involves no heat or work interactions and the kinetic and potential energy changes can be neglected Then the con servation of mass and conservation of energy relations for this twoinlet one exit steadyflow system reduce to the following Mass balance m a 1 m a 2 m a The mass flow rate of dry air remains constant m w 1 m f m w 2 The mass flow rate of vapor in the air increases by an amount equal to the rate of evaporation m f or m a ω 1 m f m a ω 2 Thus m f m a ω 2 ω 1 where P v ϕ P g 20C 075 23392 kPa 1754 kPa Thus T dp T sat 1754 kPa 154C Discussion Note that the inner surface of the window should be maintained above 154C if condensation on the window surfaces is to be avoided FIGURE 1411 The adiabatic saturation process and its representation on a Ts diagram of water T s 2 1 Adiabatic saturation temperature Dewpoint temperature Pv1 Unsaturated air T1 1 1 Saturated air T2 2 2 100 1 2 Liquid water at T2 Liquid water ω ω ϕ ϕ Final PDF to printer 718 GASVAPOR MIXTURES cen22672ch14711746indd 718 103117 0113 PM Energy balance E in E out since Q 0 and W 0 m a h 1 m f h f 2 m a h 2 or m a h 1 m a ω 2 ω 1 h f 2 m a h 2 Dividing by m a gives h 1 ω 2 ω 1 h f 2 h 2 or c p T 1 ω 1 h g 1 ω 2 ω 1 h f 2 c p T 2 ω 2 h g 2 which yields ω 1 c p T 2 T 1 ω 2 h f g 2 h g 1 h f 2 1414 where from Eq 1411b ω 2 0622 P g 2 P 2 P g 2 1415 since ϕ2 100 percent Thus we conclude that the specific humidity and relative humidity of air can be determined from Eqs 1414 and 1415 by measuring the pressure and temperature of air at the inlet and the exit of an adiabatic saturator If the air entering the channel is already saturated then the adiabatic satura tion temperature T2 will be identical to the inlet temperature T1 in which case Eq 1414 yields ω1 ω2 In general the adiabatic saturation temperature is between the inlet and dewpoint temperatures The adiabatic saturation process discussed above provides a means of determining the absolute or relative humidity of air but it requires a long channel or a spray mechanism to achieve saturation conditions at the exit A more practical approach is to use a thermometer whose bulb is covered with a cotton wick saturated with water and to blow air over the wick as shown in Fig 1412 The temperature measured in this manner is called the wetbulb temperature Twb and it is commonly used in airconditioning applications The basic principle involved is similar to that in adiabatic saturation When unsaturated air passes over the wet wick some of the water in the wick evapo rates As a result the temperature of the water drops creating a temperature difference which is the driving force for heat transfer between the air and the water After a while the heat loss from the water by evaporation equals the heat gain from the air and the water temperature stabilizes The thermom eter reading at this point is the wetbulb temperature The wetbulb tempera ture can also be measured by placing the wetwicked thermometer in a holder attached to a handle and rotating the holder rapidly that is by moving the thermometer instead of the air A device that works on this principle is called FIGURE 1412 A simple arrangement to measure the wetbulb temperature Ordinary thermometer Wetbulb thermometer Air flow Wick Liquid water Final PDF to printer 719 CHAPTER 14 cen22672ch14711746indd 719 103117 0113 PM a sling psychrometer and is shown in Fig 1413 Usually a drybulb ther mometer is also mounted on the frame of this device so that both the wet and drybulb temperatures can be read simultaneously Advances in electronics made it possible to measure humidity directly in a fast and reliable way It appears that sling psychrometers and wetwicked thermometers are about to become things of the past Today handheld elec tronic humidity measurement devices based on the capacitance change in a thin polymer film as it absorbs water vapor are capable of sensing and digitally displaying the relative humidity within 1 percent accuracy in a matter of seconds In general the adiabatic saturation temperature and the wetbulb tempera ture are not the same However for airwater vapor mixtures at atmospheric pressure the wetbulb temperature happens to be approximately equal to the adiabatic saturation temperature Therefore the wetbulb temperature Twb can be used in Eq 1414 in place of T2 to determine the specific humidity of air FIGURE 1413 Sling psychrometer Drybulb thermometer Wetbulb thermometer wick Wetbulb thermometer EXAMPLE 143 The Specific and Relative Humidity of Air The dry and the wetbulb temperatures of atmospheric air at 1 atm 101325 kPa pressure are measured with a sling psychrometer and determined to be 25 and 15C respectively Determine a the specific humidity b the relative humidity and c the enthalpy of the air SOLUTION Dry and wetbulb temperatures are given The specific humidity relative humidity and enthalpy are to be determined Properties The saturation pressure of water is 17057 kPa at 15C and 31698 kPa at 25C Table A4 The constantpressure specific heat of air at room temperature is cp 1005 kJkgK Table A2a Analysis a The specific humidity ω1 is determined from Eq 1414 ω 1 c p T 2 T 1 ω 2 h f g 2 h g 1 h f 2 where T2 is the wetbulb temperature and ω2 is ω 2 0622 P g 2 P 2 P g 2 062217057 kPa 101325 17057 kPa 001065 kg H 2 Okg dry air Thus ω 1 1005 kJkgC 15 25C 00106524654 kJkg 25465 62982 kJkg 000653 kg H 2 Okg dry air b The relative humidity ϕ1 is determined from Eq 1411a to be ϕ 1 ω 1 P 2 0622 ω 1 P g 1 000653101325 kPa 0622 00065331698 kPa 0332 or 332 Final PDF to printer 720 GASVAPOR MIXTURES cen22672ch14711746indd 720 103117 0113 PM 145 THE PSYCHROMETRIC CHART The state of the atmospheric air at a specified pressure is completely speci fied by two independent intensive properties The rest of the properties can be calculated easily from the previous relations The sizing of a typical air conditioning system involves numerous such calculations which may eventu ally get on the nerves of even the most patient engineers Therefore there is clear motivation to computerize calculations or to do these calculations once and to present the data in the form of easily readable charts Such charts are called psychrometric charts and they are used extensively in air conditioning applications A psychrometric chart for a pressure of 1 atm 101325 kPa or 14696 psia is given in Fig A31 in SI units and in Fig A31E in English units Psychrometric charts at other pressures for use at considerably higher elevations than sea level are also available The basic features of the psychrometric chart are illustrated in Fig 1414 The drybulb temperatures are shown on the horizontal axis and the specific humidity is shown on the vertical axis Some charts also show the vapor pressure on the vertical axis since at a fixed total pressure P there is a one toone correspondence between the specific humidity ω and the vapor pres sure Pv as can be seen from Eq 148 On the left end of the chart there is a curve called the saturation line instead of a straight line All the saturated air states are located on this curve Therefore it is also the curve of 100 per cent relative humidity Other constant relativehumidity curves have the same general shape Lines of constant wetbulb temperature have a downhill appearance to the right Lines of constant specific volume in m3kg dry air look similar except they are steeper Lines of constant enthalpy in kJkg dry air lie very nearly parallel to the lines of constant wetbulb temperature Therefore the constant wetbulbtemperature lines are used as constantenthalpy lines in some charts For saturated air the drybulb wetbulb and dewpoint temperatures are identical Fig 1415 Therefore the dewpoint temperature of atmospheric air at any point on the chart can be determined by drawing a horizontal line a line of ω constant or Pv constant from the point to the saturated curve The temperature value at the intersection point is the dewpoint temperature The psychrometric chart also serves as a valuable aid in visualizing the airconditioning processes An ordinary heating or cooling process for example appears as a horizontal line on this chart if no humidification or dehumidification is involved that is ω constant Any deviation from a horizontal line indicates that moisture is added or removed from the air dur ing the process c The enthalpy of air per unit mass of dry air is determined from Eq 1412 h 1 h a 1 ω 1 h v 1 c p T 1 ω 1 h g 1 1005 kJkgC25C 00065325465 kJkg 418 kJkg dry air Discussion The previous property calculations can be performed easily using pro grams with builtin psychrometric functions FIGURE 1414 Schematic for a psychrometric chart Drybulb temperature Specific humidity ω Saturation line ϕ 100 ϕ const Twb const h const v const FIGURE 1415 For saturated air the drybulb wet bulb and dewpoint temperatures are identical Saturation line Tdp 15C Tdb 15C Twb 15C 15C 15C Final PDF to printer 721 CHAPTER 14 cen22672ch14711746indd 721 103117 0113 PM 146 HUMAN COMFORT AND AIRCONDITIONING Human beings have an inherent weaknessthey want to feel comfortable They want to live in an environment that is neither hot nor cold neither humid nor dry However comfort does not come easily since the desires of the human body and the weather usually are not quite compatible Achieving comfort requires a constant struggle against the factors that cause discomfort such as high or low temperatures and high or low humidity As engineers it is our duty to help people feel comfortable Besides it keeps us employed FIGURE 1416 Schematic for Example 144 T 35C Tdp Twb h ϕ 40 v ω EXAMPLE 144 The Use of the Psychrometric Chart Consider a room that contains air at 1 atm 35C and 40 percent relative humidity Using the psychrometric chart determine a the specific humidity b the enthalpy c the wetbulb temperature d the dewpoint temperature and e the specific vol ume of the air SOLUTION The relative humidity of air in a room is given The specific humid ity enthalpy wetbulb temperature dewpoint temperature and specific volume of the air are to be determined using the psychrometric chart Analysis At a given total pressure the state of atmospheric air is completely specified by two independent properties such as the drybulb temperature and the relative humid ity Other properties are determined by directly reading their values at the specified state a The specific humidity is determined by drawing a horizontal line from the speci fied state to the right until it intersects with the ω axis as shown in Fig 1416 At the intersection point we read ω 00142 kg H 2 Okg dry air b The enthalpy of air per unit mass of dry air is determined by drawing a line par allel to the h constant lines from the specific state until it intersects the enthalpy scale giving h 715 kJkg dry air c The wetbulb temperature is determined by drawing a line parallel to the Twb constant lines from the specified state until it intersects the saturation line giving T wb 24C d The dewpoint temperature is determined by drawing a horizontal line from the specified state to the left until it intersects the saturation line giving T dp 194C e The specific volume per unit mass of dry air is determined by noting the distances between the specified state and the v constant lines on both sides of the point The specific volume is determined by visual interpolation to be v 0893 m 3 kg dry air Discussion Values read from the psychrometric chart inevitably involve reading errors and thus are of limited accuracy Final PDF to printer 722 GASVAPOR MIXTURES cen22672ch14711746indd 722 103117 0113 PM It did not take long for people to realize that they could not change the weather in an area All they can do is change it in a confined space such as a house or a workplace Fig 1417 In the past this was partially accom plished by fire and simple indoor heating systems Today modern air conditioning systems can heat cool humidify dehumidify clean and even deodorize the airin other words condition the air to peoples desires Air conditioning systems are designed to satisfy the needs of the human body therefore it is essential that we understand the thermodynamic aspects of the body The human body can be viewed as a heat engine whose energy input is food As with any other heat engine the human body generates waste heat that must be rejected to the environment if the body is to continue operating The rate of heat generation depends on the level of the activity For an aver age adult male it is about 87 W when sleeping 115 W when resting or doing office work 230 W when bowling and 440 W when doing heavy physical work The corresponding numbers for an adult female are about 15 percent less This difference is due to the body size not the body temperature The deepbody temperature of a healthy person is maintained constant at about 37C A body will feel comfortable in environments in which it can dissipate this waste heat comfortably Fig 1418 Heat transfer is proportional to the temperature difference Therefore in cold environments a body loses more heat than it normally generates which results in a feeling of discomfort The body tries to minimize the energy defi cit by cutting down the blood circulation near the skin causing a pale look This lowers the skin temperature which is about 34C for an average person and thus the heat transfer rate A low skin temperature causes discomfort The hands for example feel painfully cold when the skin temperature reaches 10C 50F We can also reduce the heat loss from the body either by putting barriers additional clothes blankets etc in the path of heat or by increasing the rate of heat generation within the body by exercising For example the comfort level of a resting person dressed in warm winter clothing in a room at 10C 50F is roughly equal to the comfort level of an identical person doing moderate work in a room at about 23C 10F Or we can just cuddle up and put our hands between our legs to reduce the surface area through which heat flows In hot environments we have the opposite problemwe do not seem to be dissipating enough heat from our bodies and we feel as if we are going to burst We dress lightly to make it easier for heat to get away from our bodies and we reduce the level of activity to minimize the rate of waste heat genera tion in the body We also turn on the fan to continuously replace the warmer air layer that forms around our bodies as a result of body heat with the cooler air in other parts of the room When doing light work or walking slowly about half of the rejected body heat is dissipated through perspiration as latent heat while the other half is dissipated through convection and radiation as sensible heat When resting or doing office work most of the heat about 70 percent is dissipated in the form of sensible heat whereas when doing heavy physi cal work most of the heat about 60 percent is dissipated in the form of latent heat The body helps out by perspiring or sweating more As this sweat evaporates it absorbs latent heat from the body and cools it Perspiration is FIGURE 1417 We cannot change the weather but we can change the climate in a confined space by airconditioning Ryan McVayGetty Images RF FIGURE 1418 A body feels comfortable when it can freely dissipate its waste heat and no more 23C Waste heat 37C Final PDF to printer 723 CHAPTER 14 cen22672ch14711746indd 723 103117 0113 PM not much help however if the relative humidity of the environment is close to 100 percent Prolonged sweating without any fluid intake causes dehydra tion and reduced sweating which may lead to a rise in body temperature and a heat stroke Another important factor that affects human comfort is heat transfer by radiation between the body and the surrounding surfaces such as walls and windows The suns rays travel through space by radiation You warm up in front of a fire even if the air between you and the fire is quite cold Likewise in a warm room you feel chilly if the ceiling or the wall surfaces are at a con siderably lower temperature This is due to direct heat transfer between your body and the surrounding surfaces by radiation Radiant heaters are com monly used for heating hardtoheat places such as car repair shops The comfort of the human body depends primarily on three factors the drybulb temperature relative humidity and air motion The temperature of the environment is the single most important index of comfort Most people feel comfortable when the environment temperature is between 22 and 27C 72 and 80F The relative humidity also has a considerable effect on com fort since it affects the amount of heat a body can dissipate through evapora tion Relative humidity is a measure of airs ability to absorb more moisture High relative humidity slows down heat rejection by evaporation and low relative humidity speeds it up Most people prefer a relative humidity of 40 to 60 percent Air motion also plays an important role in human comfort It removes the warm moist air that builds up around the body and replaces it with fresh air Therefore air motion improves heat rejection by both convection and evapo ration Air motion should be strong enough to remove heat and moisture from the vicinity of the body but gentle enough to be unnoticed Most people feel comfortable at an airspeed of about 15 mmin Veryhighspeed air motion causes discomfort instead of comfort For example an environment at 10C 50F with 48 kmh winds feels as cold as an environment at 7C 20F with 3 kmh winds as a result of the bodychilling effect of the air motion the windchill factor Other factors that affect comfort are air cleanliness odor noise and radiation effect 147 AIRCONDITIONING PROCESSES Maintaining a living space or an industrial facility at the desired tem perature and humidity requires some processes called airconditioning processes These processes include simple heating raising the tempera ture simple cooling lowering the temperature humidifying adding moisture and dehumidifying removing moisture Sometimes two or more of these processes are needed to bring the air to a desired temperature and humidity level Various airconditioning processes are illustrated on the psychrometric chart in Fig 1419 Notice that simple heating and cooling processes appear as horizontal lines on this chart since the moisture content of the air remains constant ω constant during these processes Air is commonly heated and humidified in winter and cooled and dehumidified in summer Notice how these processes appear on the psychrometric chart FIGURE 1419 Various airconditioning processes Cooling Heating Humidifying Dehumidifying Cooling and dehumidifying Heating and humidifying Final PDF to printer 724 GASVAPOR MIXTURES cen22672ch14711746indd 724 103117 0113 PM Most airconditioning processes can be modeled as steadyflow processes and thus the mass balance relation m in m out can be expressed for dry air and water as Mass balance for dry air in m a out m a kgs 1416 Mass balance for water in m w out m w or in m a ω out m a ω 1417 Disregarding the kinetic and potential energy changes the steadyflow energy balance relation Ėin Ėout can be expressed in this case as Q in W in in m h Q out W out out m h 1418 The work term usually consists of the fan work input which is small relative to the other terms in the energy balance relation Next we examine some commonly encountered processes in airconditioning Simple Heating and Cooling ω constant Many residential heating systems consist of a stove a heat pump or an electric resistance heater The air in these systems is heated by circulating it through a duct that contains the tubing for the hot gases or the electric resistance wires as shown in Fig 1420 The amount of moisture in the air remains constant during this process since no moisture is added to or removed from the air That is the specific humidity of the air remains constant ω constant dur ing a heating or cooling process with no humidification or dehumidifica tion Such a heating process proceeds in the direction of increasing drybulb temperature following a line of constant specific humidity on the psychromet ric chart which appears as a horizontal line Notice that the relative humidity of air decreases during a heating process even if the specific humidity ω remains constant This is because the rela tive humidity is the ratio of the moisture content to the moisture capacity of air at the same temperature and moisture capacity increases with tempera ture Therefore the relative humidity of heated air may be well below com fortable levels causing dry skin respiratory difficulties and an increase in static electricity A cooling process at constant specific humidity is similar to the heating process discussed above except the drybulb temperature decreases and the relative humidity increases during such a process as shown in Fig 1421 Cooling can be accomplished by passing the air over some coils through which a refrigerant or chilled water flows The conservation of mass equations for a heating or cooling process that involves no humidification or dehumidification reduce to m a1 m a2 m a for dry air and ω1 ω2 for water Neglecting any fan work that may be present the conservation of energy equation in this case reduces to Q m a h 2 h 1 or q h 2 h 1 where h1 and h2 are enthalpies per unit mass of dry air at the inlet and the exit of the heating or cooling section respectively FIGURE 1420 During simple heating specific humidity remains constant but relative humidity decreases Heating coils Heat Air T2 2 1 T1 1 1 2 1 ω ϕ ω ω ϕ ϕ FIGURE 1421 During simple cooling specific humidity remains constant but relative humidity increases 1 2 12C 30C ω constant cooling 2 80 1 30 ϕ ϕ Final PDF to printer 725 CHAPTER 14 cen22672ch14711746indd 725 103117 0113 PM FIGURE 1422 Schematic for Example 145 Cooling coils 1 atm 100F Air 070 1 1 2 ϕ ϕ EXAMPLE 145 Cooling of Air Humid air at 1 atm 100F and 70 percent relative humidity is cooled at constant pressure to the dewpoint temperature Fig 1422 Determine the cooling in Btu lbm dry air required for this process SOLUTION Humid air at a specified state is cooled at constant pressure to the dewpoint temperature The cooling required for this process is to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process m a1 m a2 m a 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible Analysis The amount of moisture in the air remains constant ω1 ω2 as it flows through the cooling section since the process involves no humidification or dehu midification The inlet and exit states of the air are completely specified and the total pressure is 1 atm The properties of the air at the inlet state are determined from the psychrometric chart Fig A31E to be h 1 567 Btulbm dry air ω 1 00296 lbm H 2 Olbm dry air ω 2 T dp1 884F The exit state enthalpy is P 1 atm T 2 T dp1 884 F ϕ 2 1 h 2 538 Btulbm dry air From the energy balance on air in the cooling section q out h 1 h 2 567 538 29 Btulbm dry air Discussion Air is cooled by 116C during this process The specific humidity remains constant during a simple cooling process and is represented by a horizontal line in the psychrometric chart Heating with Humidification Problems associated with the low relative humidity resulting from simple heating can be eliminated by humidifying the heated air This is accomplished by passing the air first through a heating section process 12 and then through a humidifying section process 23 as shown in Fig 1423 The location of state 3 depends on how the humidification is accom plished If steam is introduced in the humidification section this will result in humidification with additional heating T3 T2 If humidification is accomplished by spraying water into the airstream instead part of the latent heat of vaporization comes from the air which results in the cooling of the heated airstream T3 T2 Air should be heated to a higher temperature in the heating section in this case to make up for the cooling effect during the humidification process FIGURE 1423 Heating with humidification Air Heating coils 2 1 3 2 1 2 3 Heating section Humidifying section Humidifier ω ω ω ω Final PDF to printer 726 GASVAPOR MIXTURES cen22672ch14711746indd 726 103117 0113 PM FIGURE 1424 Schematic and psychrometric chart for Example 146 Air 1 2 3 V1 45 m3min 10C 22C 25C 1 2 3 T1 10C T2 22C T3 25C Humidifier Heating coils 1 30 3 60 1 30 ϕ ϕ3 60 ϕ ϕ EXAMPLE 146 Heating and Humidification of Air An airconditioning system is to take in outdoor air at 10C and 30 percent relative humidity at a steady rate of 45 m3min and to condition it to 25C and 60 percent relative humidity The outdoor air is first heated to 22C in the heating section and then humidified by the injection of hot steam in the humidifying section Assuming the entire process takes place at a pressure of 100 kPa determine a the rate of heat supply in the heating section and b the mass flow rate of the steam required in the humidifying section SOLUTION Outdoor air is first heated and then humidified by steam injection The rate of heat transfer and the mass flow rate of steam are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible Properties The constantpressure specific heat of air at room temperature is cp 1005 kJkgK and its gas constant is Ra 0287 kJkgK Table A2a The saturation pressure of water is 12281 kPa at 10C and 31698 kPa at 25C The enthalpy of saturated water vapor is 25192 kJkg at 10C and 25410 kJkg at 22C Table A4 Analysis We take the system to be the heating or the humidifying section as appro priate The schematic of the system and the psychrometric chart of the process are shown in Fig 1424 We note that the amount of water vapor in the air remains constant in the heating section ω1 ω2 but increases in the humidifying section ω3 ω2 a Applying the mass and energy balances on the heating section gives Dry air mass balance m a 1 m a 2 m a Water mass balance m a 1 ω 1 m a 2 ω 2 ω 1 ω 2 Energy balance Q in m a h 1 m a h 2 Q in m a h 2 h 1 The psychrometric chart offers great convenience in determining the properties of moist air However its use is limited to a specified pressure only which is 1 atm 101325 kPa for the one given in the appendix At pressures other than 1 atm either other charts for that pressure or the relations developed earlier should be used In our case the choice is clear P v 1 ϕ 1 P g 1 ϕ P sat 10C 0312281 kPa 0368 kPa P a 1 P 1 P v 1 100 0368 kPa 99632 kPa v 1 R a T 1 P a 0287 kPa m 3 kgK283 K 99632 kPa 0815 m 3 kg dry air m a V 1 v 1 45 m 3 min 0815 m 3 kg 552 kgmin h 1 c p T 1 ω 1 h g 1 1005 kJkgC10C 0002325192 kJkg 158 kJkg dry air h 2 c p T 2 ω 2 h g 2 1005 kJkgC22C 0002325410 kJkg 280 kJkg dry air ω 1 0622 P v 1 P 1 P v 1 06220368 kPa 100 0368 kPa 00023 kg H 2 Okg dry air Final PDF to printer 727 CHAPTER 14 cen22672ch14711746indd 727 103117 0113 PM since ω2 ω1 Then the rate of heat transfer to air in the heating section becomes Q in m a h 2 h 1 552 kgmin 280 158 kJkg 673 kJmin b The mass balance for water in the humidifying section can be expressed as m a 2 ω 2 m w m a 3 ω 3 or m w m a ω 3 ω 2 where ω 3 0622 ϕ 3 P g 3 P 3 ϕ 3 P g 3 062206031698 kPa 100 06031698 kPa 001206 kg H 2 Okg dry air Thus m w 552 kgmin001206 00023 0539 kgmin Discussion The result 0539 kgmin corresponds to a water requirement of close to one ton a day which is significant Cooling with Dehumidification The specific humidity of air remains constant during a simple cooling pro cess but its relative humidity increases If the relative humidity reaches unde sirably high levels it may be necessary to remove some moisture from the air that is to dehumidify it This requires cooling the air below its dewpoint temperature The cooling process with dehumidifying is illustrated schematically and on the psychrometric chart in Fig 1425 in conjunction with Example 147 Hot moist air enters the cooling section at state 1 As it passes through the cooling coils its temperature decreases and its relative humidity increases at constant specific humidity If the cooling section is sufficiently long air reaches its dew point state x saturated air Further cooling of air results in the condensation of part of the moisture in the air Air remains saturated during the entire condensation process which follows a line of 100 percent relative humidity until the final state state 2 is reached The water vapor that condenses out of the air during this process is removed from the cooling sec tion through a separate channel The condensate is usually assumed to leave the cooling section at T2 The cool saturated air at state 2 is usually routed directly to the room where it mixes with the room air In some cases however the air at state 2 may be at the right specific humidity but at a very low temperature In such cases air is passed through a heating section where its temperature is raised to a more comfortable level before it is routed to the room FIGURE 1425 Schematic and psychrometric chart for Example 147 Air Cooling coils 2 1 14C 30C 1 2 1 80 2 100 T2 14C 2 100 T1 30C 1 80 V1 10 m3min Condensate removal 14C Condensate x ϕ ϕ ϕ ϕ Final PDF to printer 728 GASVAPOR MIXTURES cen22672ch14711746indd 728 103117 0113 PM EXAMPLE 147 Cooling and Dehumidification of Air Air enters a window air conditioner at 1 atm 30C and 80 percent relative humidity at a rate of 10 m3min and it leaves as saturated air at 14C Part of the moisture in the air that condenses during the process is also removed at 14C Determine the rates of heat and moisture removal from the air SOLUTION Air is cooled and dehumidified by a window air conditioner The rates of heat and moisture removal are to be determined Assumptions 1 This is a steadyflow process and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and the water vapor are ideal gases 3 The kinetic and potential energy changes are negligible Properties The enthalpy of saturated liquid water at 14C is 588 kJkg Table A4 Also the inlet and the exit states of the air are completely specified and the total pres sure is 1 atm Therefore we can determine the properties of the air at both states from the psychrometric chart to be h 1 854 kJkg dry air h 2 393 kJkg dry air ω 1 00216 kg H 2 Okg dry air ω 2 00100 kg H 2 Okg dry air v 1 0889 m 3 kg dry air Analysis We take the cooling section to be the system The schematic of the system and the psychrometric chart of the process are shown in Fig 1425 We note that the amount of water vapor in the air decreases during the process ω2 ω1 due to dehumidification Applying the mass and energy balances on the cooling and dehu midification section gives Dry air mass balance m a 1 m a 2 m a Water mass balance m a 1 ω 1 m a 2 ω 2 m w m w m a ω 1 ω 2 Energy balance in m h Q out out m h Q out m h 1 h 2 m w h w Then m a V 1 v 1 10 m 3 min 0889 m 3 kg dry air 1125 kgmin Q out 1125 kgmin 854 393 kJkg 0131 kgmin588 kJkg 511 kJmin Therefore this airconditioning unit removes moisture and heat from the air at rates of 0131 kgmin and 511 kJmin respectively m w 1125 kgmin00216 00100 0131 kgmin Evaporative Cooling Conventional cooling systems operate on a refrigeration cycle and they can be used in any part of the world But they have a high initial and operating cost In desert hot and dry climates we can avoid the high cost of cooling by using evaporative coolers also known as swamp coolers Evaporative cooling is based on a simple principle As water evaporates the latent heat of vaporization is absorbed from the water body and the sur rounding air As a result both the water and the air are cooled during the Final PDF to printer 729 CHAPTER 14 cen22672ch14711746indd 729 103117 0113 PM process This approach has been used for thousands of years to cool water A porous jug or pitcher filled with water is left in an open shaded area A small amount of water leaks out through the porous holes and the pitcher sweats In a dry environment this water evaporates and cools the remaining water in the pitcher Fig 1426 You have probably noticed that on a hot dry day the air feels a lot cooler when the yard is watered This is because water absorbs heat from the air as it evaporates An evaporative cooler works on the same principle The evapo rative cooling process is shown schematically and on a psychrometric chart in Fig 1427 Hot dry air at state 1 enters the evaporative cooler where it is sprayed with liquid water Part of the water evaporates during this process by absorbing heat from the airstream As a result the temperature of the air stream decreases and its humidity increases state 2 In the limiting case the air leaves the evaporative cooler saturated at state 2 This is the lowest tem perature that can be achieved by this process The evaporative cooling process is essentially identical to the adiabatic satu ration process since the heat transfer between the airstream and the surround ings is usually negligible Therefore the evaporative cooling process follows a line of constant wetbulb temperature on the psychrometric chart Note that this will not exactly be the case if the liquid water is supplied at a temperature different from the exit temperature of the airstream Since the constantwet bulbtemperature lines almost coincide with the constantenthalpy lines the enthalpy of the airstream can also be assumed to remain constant That is T wb constant 1419 and h constant 1420 during an evaporative cooling process This is a reasonably accurate approxi mation and it is commonly used in airconditioning calculations FIGURE 1426 Water in a porous jug left in an open breezy area cools as a result of evaporative cooling Water that leaks out Hot dry air FIGURE 1427 Evaporative cooling Hot dry air 2 1 1 2 2 Cool moist air Liquid water Twb const h const EXAMPLE 148 Evaporative Cooling with Soaked Head Cover Desert dwellers often wrap their heads with a watersoaked porous cloth Fig 1428 On a desert where the pressure is 1 atm temperature is 120F and relative humidity is 10 percent what is the temperature of this cloth SOLUTION Desert dwellers often wrap their heads with a watersoaked porous cloth The temperature of this cloth on a desert with a specified temperature and rela tive humidity is to be determined Assumptions Air leaves the head covering as saturated Analysis Since the cloth behaves like the wick on a wetbulb thermometer the temperature of the cloth will become the wetbulb temperature If we assume the liquid water is supplied at a temperature not much different from the exit temperature of the airstream the evaporative cooling process follows a line of constant wetbulb temperature on the psychrometric chart That is T wb constant Final PDF to printer 730 GASVAPOR MIXTURES cen22672ch14711746indd 730 103117 0113 PM FIGURE 1428 Head wrap discussed in Example 148 GlowimagesGetty Images RF The wetbulb temperature at 1 atm 120F and 10 percent relative humidity is deter mined from the psychrometric chart to be T 2 T wb 737F Discussion Note that for saturated air the dry and the wetbulb temperatures are identical Therefore the lowest temperature to which air can be cooled is the wetbulb temperature Also note that the temperature of air drops by as much as 46F in this case by evaporative cooling Adiabatic Mixing of Airstreams Many airconditioning applications require the mixing of two airstreams This is particularly true for large buildings most production and process plants and hospitals which require that the conditioned air be mixed with a cer tain fraction of fresh outside air before it is routed into the living space The mixing is accomplished by simply merging the two airstreams as shown in Fig 1429 The heat transfer with the surroundings is usually small and thus the mixing processes can be assumed to be adiabatic Mixing processes normally involve no work interactions and the changes in kinetic and potential energies if any are negligible Then the mass and energy balances for the adiabatic mixing of two airstreams reduce to Mass of dry air m a 1 m a 2 m a 3 1421 Mass of water vapor ω 1 m a 1 ω 3 m a 3 ω 3 m a 3 1422 Energy m a 1 h 1 m a 2 h 2 m a 2 h 3 1423 Eliminating m a 3 from the preceding relations we obtain m a 1 m a 2 ω 2 ω 3 ω 3 ω 1 h 2 h 3 h 3 h 1 1424 This equation has an instructive geometric interpretation on the psychrometric chart It shows that the ratio of ω2 ω3 to ω3 ω1 is equal to the ratio of m a 1 to m a 2 The states that satisfy this condition are indicated by the dashed line AB The ratio of h2 h3 to h3 h1 is also equal to the ratio of m a 1 to m a 2 and the states that satisfy this condition are indicated by the dashed line CD The only state that satisfies both conditions is the intersection point of these two dashed lines which is located on the straight line connecting states 1 and 2 Thus we conclude that when two airstreams at two different states states 1 and 2 are mixed adiabatically the state of the mixture state 3 lies on the straight line connecting states 1 and 2 on the psychrometric chart and the ratio of the dis tances 23 and 31 is equal to the ratio of mass flow rates m a 1 and m a 2 The concave nature of the saturation curve and the conclusion above lead to an interesting possibility When states 1 and 2 are located close to the satura tion curve the straight line connecting the two states will cross the saturation curve and state 3 may lie to the left of the saturation curve In this case some water will inevitably condense during the mixing process FIGURE 1429 When two airstreams at states 1 and 2 are mixed adiabatically the state of the mixture lies on the straight line connecting the two states 1 2 3 A C h2 h3 h1 D B 2 1 3 Mixing section h1 h2 h3 h2 h3 h3 h1 2 3 2 3 3 1 1 2 3 1 ω ω ω ω ω ω ω ω ω ω Final PDF to printer 731 CHAPTER 14 cen22672ch14711746indd 731 103117 0113 PM EXAMPLE 149 Mixing of Conditioned Air with Outdoor Air Saturated air leaving the cooling section of an airconditioning system at 14C at a rate of 50 m3min is mixed adiabatically with the outside air at 32C and 60 percent relative humidity at a rate of 20 m3min Assuming that the mixing process occurs at a pressure of 1 atm determine the specific humidity the relative humidity the drybulb temperature and the volume flow rate of the mixture SOLUTION Conditioned air is mixed with outside air at specified rates The spe cific and relative humidities the drybulb temperature and the flow rate of the mix ture are to be determined Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The mixing section is adiabatic Properties The properties of each inlet stream are determined from the psychro metric chart to be h 1 394 kJkg dry air ω 1 0010 kg H 2 Okg dry air v 1 0826 m 3 kg dry air and h 2 790 kJkg dry air ω 2 00182 kg H 2 Okg dry air v 2 0889 m 3 kg dry air Analysis We take the mixing section of the streams as the system The schematic of the system and the psychrometric chart of the process are shown in Fig 1430 We note that this is a steadyflow mixing process The mass flow rates of dry air in each stream are m a 1 V 1 v 1 50 m 3 min 0826 m 3 kg dry air 605 kgmin m a 2 V 2 v 2 20 m 3 min 0889 m 3 kg dry air 225 kgmin From the mass balance of dry air m a 3 m a 1 m a 2 605 225 kgmin 83 kgmin The specific humidity and the enthalpy of the mixture can be determined from Eq 1424 m a 1 m a 2 ω 2 ω 3 ω 3 ω 1 h 2 h 3 h 3 h 1 605 225 00182 ω 3 ω 3 0010 790 h 3 h 3 394 which yield ω 3 00122 kg H 2 Okg dry air h 3 501 kJkg dry air FIGURE 1430 Schematic and psychrometric chart for Example 149 V2 20 m3min Saturated air T1 14C V1 50 m3min 2 1 3 Mixing section P 1 atm T2 32C 2 60 V3 3 3 T3 2 3 1 14C 32C 1 100 ϕ 2 60 ϕ ϕ ω ϕ Final PDF to printer 732 GASVAPOR MIXTURES cen22672ch14711746indd 732 103117 0113 PM Wet Cooling Towers Power plants large airconditioning systems and some industries generate large quantities of waste heat that is often rejected to cooling water from nearby lakes or rivers In some cases however the cooling water supply is limited or thermal pollution is a serious concern In such cases the waste heat must be rejected to the atmosphere with cooling water recirculating and serving as a transport medium for heat transfer between the source and the sink the atmosphere One way of achieving this is through the use of wet cooling towers A wet cooling tower is essentially a semienclosed evaporative cooler An induceddraft counterflow wet cooling tower is shown schematically in Fig 1431 Air is drawn into the tower from the bottom and leaves through the top Warm water from the condenser is pumped to the top of the tower and is sprayed into this airstream The purpose of spraying is to expose a large surface area of water to the air As the water droplets fall under the influence of gravity a small fraction of water usually a few percent evaporates and cools the remaining water The temperature and the moisture content of the air increase during this process The cooled water collects at the bottom of the tower and is pumped back to the condenser to absorb additional waste heat Makeup water must be added to the cycle to replace the water lost by evaporation and air draft To minimize water carried away by the air drift eliminators are installed in the wet cooling towers above the spray section The air circulation in the cooling tower described is provided by a fan and therefore it is classified as a forceddraft cooling tower Another popular type of cooling tower is the naturaldraft cooling tower which looks like a large chimney and works like an ordinary chimney The air in the tower has a high watervapor content and thus it is lighter than the outside air Consequently the light air in the tower rises and the heavier outside air fills the vacant space creating an airflow from the bottom of the tower to the top The flow rate of air is controlled by the conditions of the atmospheric air Naturaldraft These two properties fix the state of the mixture Other properties of the mixture are determined from the psychrometric chart T 3 190C ϕ 3 89 v 3 0844 m 3 kg dry air Finally the volume flow rate of the mixture is determined from V 3 m a 3 v 3 83 kgmin0844 m 3 kg 701 m 3 min Discussion Notice that the volume flow rate of the mixture is approximately equal to the sum of the volume flow rates of the two incoming streams This is typical in airconditioning applications FIGURE 1431 An induceddraft counterflow cooling tower Cool water Air exit Warm water Air inlet Fan Final PDF to printer 733 CHAPTER 14 cen22672ch14711746indd 733 103117 0113 PM cooling towers do not require any external power to induce the air but they cost a lot more to build than forceddraft cooling towers The naturaldraft cooling towers are hyperbolic in profile as shown in Fig 1432 and some are over 100 m high The hyperbolic profile is for greater structural strength not for any thermodynamic reason The idea of a cooling tower started with the spray pond where the warm water is sprayed into the air and is cooled by the air as it falls into the pond as shown in Fig 1433 Some spray ponds are still in use today However they require 25 to 50 times the area of a cooling tower water loss due to air drift is high and they are unprotected against dust and dirt We could also dump the waste heat into a still cooling pond which is basically a large artificial lake open to the atmosphere Fig 1434 Heat transfer from the pond surface to the atmosphere is very slow however and we would need about 20 times the area of a spray pond in this case to achieve the same cooling EXAMPLE 1410 Cooling of a Power Plant by a Cooling Tower Cooling water leaves the condenser of a power plant and enters a wet cooling tower at 35C at a rate of 100 kgs Water is cooled to 22C in the cooling tower by air that enters the tower at 1 atm 20C and 60 percent relative humidity and leaves saturated at 30C Neglecting the power input to the fan determine a the volume flow rate of air into the cooling tower and b the mass flow rate of the required makeup water SOLUTION Warm cooling water from a power plant is cooled in a wet cooling tower The flow rates of makeup water and air are to be determined Assumptions 1 Steady operating conditions exist and thus the mass flow rate of dry air remains constant during the entire process 2 Dry air and the water vapor are ideal gases 3 The kinetic and potential energy changes are negligible 4 The cooling tower is adiabatic Properties The enthalpy of saturated liquid water is 9228 kJkg at 22C and 14664 kJkg at 35C Table A4 From the psychrometric chart h 1 422 kJkg dry air h 2 1000 kJkg dry air ω 1 00087 kg H 2 Okg dry air ω 2 00273 kg H 2 Okg dry air v 1 0842 m 3 kg dry air Analysis We take the entire cooling tower to be the system which is shown sche matically in Fig 1435 We note that the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process The water lost through evaporation must be made up later in the cycle to maintain steady operation a Applying the mass and energy balances on the cooling tower gives Dry air mass balance m a 1 m a 2 m a Water mass balance m 3 m a 1 ω 1 m 4 m a 2 ω 2 FIGURE 1432 Two natural draft cooling towers on a roadside Yunus Çengel FIGURE 1433 A spray pond Yunus Çengel FIGURE 1434 A cooling pond Yunus Çengel Final PDF to printer 734 GASVAPOR MIXTURES cen22672ch14711746indd 734 103117 0113 PM or m 3 m 4 m a ω 2 ω 1 m makeup Energy balance in m h out m h m a 1 h 1 m 3 h 3 m a 2 h 2 m 4 h 4 or m 3 h 3 m a h 2 h 1 m 3 m makeup h 4 Solving for m a gives m a m 3 h 3 h 4 h 2 h 1 ω 2 ω 1 h 4 Substituting m a 100 kgs 14664 9228 kJkg 1000 422 kJkg 00273 000879228 kJkg 969 kgs Then the volume flow rate of air into the cooling tower becomes V 1 m a v 1 969 kgs0842 m 3 kg 816 m 3 s b The mass flow rate of the required makeup water is determined from m makeup m a ω 2 ω 1 969 kgs00273 00087 180 kgs Discussion Note that over 98 percent of the cooling water is saved and recirculated in this case SUMMARY In this chapter we discussed the airwater vapor mixture which is the most commonly encountered gasvapor mixture in prac tice The air in the atmosphere normally contains some water vapor and it is referred to as atmospheric air By contrast air that contains no water vapor is called dry air In the temperature range encountered in airconditioning applications both the dry air and the water vapor can be treated as ideal gases The enthalpy change of dry air during a process can be determined from Δ h dry air c p ΔT 1005 kJkgC ΔT The atmospheric air can be treated as an idealgas mixture whose pressure is the sum of the partial pressure of dry air Pa and that of the water vapor Pv P P a P v The enthalpy of water vapor in the air can be taken to be equal to the enthalpy of the saturated vapor at the same temperature h v T low P h g T 25009 182T kJkg T in C 10609 0435T Btulbm T in F in the temperature range 10 to 50C 15 to 120F The mass of water vapor present per unit mass of dry air is called the specific or absolute humidity ω ω m v m a 0622 P v P P v kg H 2 Okg dry air where P is the total pressure of air and Pv is the vapor pres sure There is a limit on the amount of vapor the air can hold FIGURE 1435 Schematic for Example 1410 Air 35C 100 kgs System boundary 4 1 3 2 1 atm 20C 1 60 30C 2 100 V1 Makeup water 22C Cool water Warm water ϕ ϕ Final PDF to printer cen22672ch14711746indd 735 103117 0113 PM 735 CHAPTER 14 at a given temperature Air that is holding as much moisture as it can at a given temperature is called saturated air The ratio of the amount of moisture air holds mv to the maximum amount of moisture air can hold at the same temperature mg is called the relative humidity ϕ ϕ m v m g P v P g where Pg Psat T The relative and specific humidities can also be expressed as ϕ ωP 0622 ω P g and ω 0622ϕ P g P ϕ P g Relative humidity ranges from 0 for dry air to 1 for saturated air The enthalpy of atmospheric air is expressed per unit mass of dry air instead of per unit mass of the airwater vapor mix ture as h h a ω h g kJkg dry air The ordinary temperature of atmospheric air is referred to as the drybulb temperature to differentiate it from other forms of temperatures The temperature at which condensa tion begins if the air is cooled at constant pressure is called the dewpoint temperature Tdp T dp T sat P v Relative humidity and specific humidity of air can be deter mined by measuring the adiabatic saturation temperature of air which is the temperature air attains after flowing over water in a long adiabatic channel until it is saturated ω 1 c p T 2 T 1 ω 2 h f g 2 h g 1 h f 2 where ω 2 0622 P g 2 P 2 P g 2 and T2 is the adiabatic saturation temperature A more prac tical approach in airconditioning applications is to use a thermometer whose bulb is covered with a cotton wick sat urated with water and to blow air over the wick The tem perature measured in this manner is called the wetbulb temperature Twb and it is used in place of the adiabatic satura tion temperature The properties of atmospheric air at a speci fied total pressure are presented in the form of easily readable charts called psychrometric charts The lines of constant enthalpy and the lines of constant wetbulb temperature are very nearly parallel on these charts The needs of the human body and the conditions of the environment are not quite compatible Therefore it often becomes necessary to change the conditions of a living space to make it more comfortable Maintaining a liv ing space or an industrial facility at the desired tempera ture and humidity may require simple heating raising the temperature simple cooling lowering the temperature humidifying adding moisture or dehumidifying remov ing moisture Sometimes two or more of these processes are needed to bring the air to the desired temperature and humidity level Most airconditioning processes can be modeled as steady flow processes and therefore they can be analyzed by apply ing the steadyflow mass for both dry air and water and energy balances Dry air mass in m a out m a Water mass in m w out m w or in m a ω out m a ω Energy Q in W in in m h Q out W out out m h The changes in kinetic and potential energies are assumed to be negligible During a simple heating or cooling process the specific humidity remains constant but the temperature and the rela tive humidity change Sometimes air is humidified after it is heated and some cooling processes include dehumidification In dry climates air can be cooled via evaporative cooling by passing it through a section where it is sprayed with water In locations with limited cooling water supply large amounts of waste heat can be rejected to the atmosphere with minimum water loss through the use of cooling towers Final PDF to printer cen22672ch14711746indd 736 103117 0113 PM 736 GASVAPOR MIXTURES REFERENCES AND SUGGESTED READINGS 1 ASHRAE 1981 Handbook of Fundamentals Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers 1981 2 S M Elonka Cooling Towers Power March 1963 3 W F Stoecker and J W Jones Refrigeration and Air Conditioning 2nd ed New York McGrawHill 1982 4 L D Winiarski and B A Tichenor Model of Natural Draft Cooling Tower Performance Journal of the Sanitary Engineering Division Proceedings of the American Society of Civil Engineers August 1970 FIGURE P1419 800 kPa 100 kPa 20C 90 Humid air PROBLEMS Dry and Atmospheric Air Specific and Relative Humidity 141C What is the difference between dry air and atmo spheric air 142C What is vapor pressure 143C What is the difference between the specific humidity and the relative humidity 144C Can the water vapor in air be treated as an ideal gas Explain 145C Explain how vapor pressure of the ambient air is determined when the temperature total pressure and the rela tive humidity of air are given 146C Is the relative humidity of saturated air necessarily 100 percent 147C Moist air is passed through a cooling section where it is cooled and dehumidified How do a the specific humidity and b the relative humidity of air change during this process 148C How will a the specific humidity and b the rela tive humidity of the air contained in a wellsealed room change as it is heated 149C How will a the specific humidity and b the rela tive humidity of the air contained in a wellsealed room change as it is cooled 1410C Consider a tank that contains moist air at 3 atm and whose walls are permeable to water vapor The surrounding air at 1 atm pressure also contains some moisture Is it possible for the water vapor to flow into the tank from surroundings Explain 1411C Is it possible to obtain saturated air from unsatu rated air without adding any moisture Explain 1412C Why are the chilled water lines always wrapped with vapor barrier jackets 1413C How would you compare the enthalpy of water vapor at 20C and 2 kPa with the enthalpy of water vapor at 20C and 05 kPa 1414 A tank contains 15 kg of dry air and 017 kg of water vapor at 30C and 100 kPa total pressure Determine a the specific humidity b the relative humidity and c the volume of the tank 1415 Repeat Prob 1414 for a temperature of 40C 1416 An 8m3 tank contains saturated air at 30C 105 kPa Determine a the mass of dry air b the specific humidity and c the enthalpy of the air per unit mass of the dry air 1417 Determine the masses of dry air and the water vapor contained in a 90m3 room at 93 kPa 15C and 50 percent relative humidity Answers 100 kg 0578 kg 1418E A room contains air at 85F and 135 psia at a rela tive humidity of 60 percent Determine a the partial pressure of dry air b the specific humidity and c the enthalpy per unit mass of dry air Answers a 131 psia b 00169 lbm H2Olbm dry air c 390 Btulbm dry air 1419 Humid air at 100 kPa 20C and 90 percent relative humidity is compressed in a steadyflow isentropic compres sor to 800 kPa What is the relative humidity of the air at the compressor outlet Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch14711746indd 737 103117 0113 PM 737 CHAPTER 14 1420E Humid air at 100 psia and 400F and a humidity ratio of 0025 lbm H2Olbm dry air is expanded to 15 psia in an isentropic nozzle How much of the initial water vapor has been converted to liquid water at the nozzle outlet DewPoint Adiabatic Saturation and WetBulb Temperatures 1421C What is the dewpoint temperature 1422C In summer the outer surface of a glass filled with iced water often sweats How can you explain this sweating 1423C In some climates cleaning the ice off the windshield of a car is a common chore on winter mornings Explain how ice forms on the windshield during some nights even when there is no rain or snow 1424C Andy and Wendy both wear glasses On a cold win ter day Andy comes from the cold outside and enters the warm house while Wendy leaves the house and goes outside Whose glasses are more likely to be fogged Explain 1425C When are the drybulb and dewpoint temperatures identical 1426C When are the adiabatic saturation and wetbulb tem peratures equivalent for atmospheric air 1427 A house contains air at 25C and 65 percent relative humidity Will any moisture condense on the inner surfaces of the windows when the temperature of the window drops to 10C 1428E A thirsty woman opens the refrigerator and picks up a cool canned drink at 40F Do you think the can will sweat as she enjoys the drink in a room at 70F and 38 percent rela tive humidity 1429 The air in a room has a drybulb temperature of 26C and a wetbulb temperature of 21C Assuming a pres sure of 100 kPa determine a the specific humidity b the relative humidity and c the dewpoint temperature Answers a 00138 kg H2Okg dry air b 644 percent c 188C 1430 Reconsider Prob 1429 Determine the required properties using appropriate software What would the property values be at a pressure of 300 kPa 1431E The air in a room has a drybulb temperature of 75F and a wetbulb temperature of 60F Assuming a pres sure of 143 psia determine a the specific humidity b the relative humidity and c the dewpoint temperature Answers a 00079 lbm H2Olbm dry air b 417 percent c 502F 1432 The dry and wetbulb temperatures of atmospheric air at 95 kPa are 25 and 17C respectively Determine a the specific humidity b the relative humidity and c the enthalpy of the air in kJkg dry air 1433 Atmospheric air at 35C flows steadily into an adi abatic saturation device and leaves as a saturated mixture at 25C Makeup water is supplied to the device at 25C Atmospheric pressure is 98 kPa Determine the relative humid ity and specific humidity of the air Psychrometric Chart 1434C How do constantenthalpy and constantwetbulb temperature lines compare on the psychrometric chart 1435C At what states on the psychrometric chart are the drybulb wetbulb and dewpoint temperatures identical 1436C How is the dewpoint temperature at a specified state determined on the psychrometric chart 1437C Can the enthalpy values determined from a psychro metric chart at sea level be used at higher elevations 1438 Atmospheric air at a pressure of 1 atm and drybulb temperature of 30C has a relative humidity of 80 percent Using the psychrometric chart determine a the wetbulb temperature b the humidity ratio c the enthalpy d the dewpoint temperature and e the water vapor pressure 1439E Atmospheric air at a pressure of 1 atm and drybulb temperature of 90F has a wetbulb temperature of 85F Using the psychrometric chart determine a the relative humidity b the humidity ratio c the enthalpy d the dewpoint tem perature and e the water vapor pressure FIGURE P1439E Air 1 atm 90F Twb 85F 1440E Reconsider Prob 1439E Determine the adiabatic saturation temperature of the humid air Answer 85F 1441 The air in a room has a pressure of 1 atm a dry bulb temperature of 24C and a wetbulb temperature of 17C Using the psychrometric chart determine a the spe cific humidity b the enthalpy in kJkg dry air c the rela tive humidity d the dewpoint temperature and e the specific volume of the air in m3kg dry air 1442 Reconsider Prob 1441 Determine the required properties using appropriate software instead of the psychrometric chart What would the property values be at a location at 3000 m altitude 1443 Atmospheric air at a pressure of 1 atm and drybulb temperature of 28C has a dewpoint temperature of 20C Using the psychrometric chart determine a the relative humidity b the humidity ratio c the enthalpy d the wet bulb temperature and e the water vapor pressure 1444 Reconsider Prob 1443 Determine the adiabatic sat uration temperature of the humid air Final PDF to printer cen22672ch14711746indd 738 103117 0113 PM 738 GASVAPOR MIXTURES FIGURE P1444 Air Humidifier 1 atm 28C Tdp 20C Water 100 FIGURE P1468 Cooling coils 1 atm Air 35C 45 18 ms Human Comfort and AirConditioning 1445C What does a modern airconditioning system do besides heating or cooling the air 1446C How does the human body respond to a hot weather b cold weather and c hot and humid weather 1447C How does the air motion in the vicinity of the human body affect human comfort 1448C Consider a tennis match in cold weather where both players and spectators wear the same clothes Which group of people will feel colder Why 1449C Why do you think little babies are more susceptible to cold 1450C What is the radiation effect How does it affect human comfort 1451C How does humidity affect human comfort 1452C What are humidification and dehumidification 1453C What is metabolism What is the range of metabolic rate for an average man Why are we interested in the meta bolic rate of the occupants of a building when we deal with heating and airconditioning 1454C Why is the metabolic rate of women in general lower than that of men What is the effect of clothing on the environmental temperature that feels comfortable 1455C What is sensible heat How is the sensible heat loss from a human body affected by the a skin temperature b environment temperature and c air motion 1456C What is latent heat How is the latent heat loss from the human body affected by the a skin wettedness and b relative humidity of the environment How is the rate of evaporation from the body related to the rate of latent heat loss 1457 A department store expects to have 225 customers and 20 employees at peak times in summer Determine the contribution of people to the total cooling load of the store 1458E In a movie theater in winter 500 people each gen erating sensible heat at a rate of 80 W are watching a movie The heat losses through the walls windows and the roof are estimated to be 130000 Btuh Determine if the theater needs to be heated or cooled 1459 For an infiltration rate of 12 air changes per hour ACH determine sensible latent and total infiltration heat load of a building at sea level in kW that is 20 m long 13 m wide and 3 m high when the outdoor air is at 32C and 35 per cent relative humidity The building is maintained at 24C and 55 percent relative humidity at all times 1460 Repeat Prob 1459 for an infiltration rate of 18 ACH 1461 An average 182 kg or 40 lbm chicken has a basal metabolic rate of 547 W and an average metabolic rate of 102 W 378 W sensible and 642 W latent during normal activity If there are 100 chickens in a breeding room deter mine the rate of total heat generation and the rate of moisture production in the room Take the heat of vaporization of water to be 2430 kJkg 1462 An average person produces 025 kg of moisture while taking a shower and 005 kg while bathing in a tub Consider a family of four who each shower once a day in a bathroom that is not ventilated Taking the heat of vaporization of water to be 2450 kJkg determine the contribution of showers to the latent heat load of the air conditioner per day in summer Simple Heating and Cooling 1463C How do relative and specific humidities change dur ing a simple heating process Answer the same question for a simple cooling process 1464C Why does a simple heating or cooling process appear as a horizontal line on the psychrometric chart 1465 Humid air at 150 kPa 40C and 70 percent relative humidity is cooled at constant pressure in a pipe to its dew point temperature Calculate the heat transfer in kJkg dry air required for this process Answer 68 kJkg dry air 1466E Humid air at 40 psia 50F and 90 percent relative humidity is heated in a pipe at constant pressure to 120F Calculate the relative humidity at the pipe outlet and the amount of heat in Btulbm dry air required 1467 Air enters a heating section at 95 kPa 10C and 30 percent relative humidity at a rate of 6 m3min and it leaves at 25C Determine a the rate of heat transfer in the heat ing section and b the relative humidity of the air at the exit Answers a 106 kJmin b 116 percent 1468 Air enters a 30cmdiameter cooling section at 1 atm 35C and 45 percent relative humidity at 18 ms Heat is removed from the air at a rate of 750 kJmin Determine a the exit temperature b the exit relative humidity of the air and c the exit velocity Answers a 265C b 731 percent c 175 ms Final PDF to printer cen22672ch14711746indd 739 103117 0113 PM 739 CHAPTER 14 1469 Repeat Prob 1468 for a heat removal rate of 950 kJmin 1470E A heating section consists of a 15indiameter duct that houses a 4kW electric resistance heater Air enters the heating section at 147 psia 50F and 40 percent relative humidity at a velocity of 25 fts Determine a the exit tem perature b the exit relative humidity of the air and c the exit velocity Answers a 566F b 314 percent c 254 fts Heating with Humidification 1471C Why is heated air sometimes humidified 1472 Air at 1 atm 15C and 60 percent relative humidity is first heated to 20C in a heating section and then humidi fied by introducing water vapor The air leaves the humidifying section at 25C and 65 percent relative humidity Determine a the amount of steam added to the air and b the amount of heat transfer to the air in the heating section Answers a 00065 kg H2Okg dry air b 51 kJkg dry air 1473E Air at 147 psia 35F and 50 percent relative humid ity is first heated to 65F in a heating section and then humidi fied by introducing water vapor The air leaves the humidifying section at 85F and 55 percent relative humidity Determine a the amount of steam added to the air in lbm H2Olbm dry air and b the amount of heat transfer to the air in the heating section in Btulbm dry air 1474 An airconditioning system operates at a total pres sure of 1 atm and consists of a heating section and a humidi fier that supplies wet steam saturated water vapor at 100C Air enters the heating section at 10C and 70 percent relative humidity at a rate of 35 m3min and it leaves the humidifying section at 20C and 60 percent relative humidity Determine a the temperature and relative humidity of air when it leaves the heating section b the rate of heat transfer in the heating section and c the rate at which water is added to the air in the humidifying section 1477 Atmospheric air at 1 atm 30C and 80 percent rela tive humidity is cooled to 20C while the mixture pressure remains constant Calculate the amount of water in kgkg dry air removed from the air and the cooling requirement in kJkg dry air when the liquid water leaves the system at 22C Answers 00069 kg H2Okg dry air 273 kJkg dry air 1478E Ten thousand cubic feet per hour of atmospheric air at 1 atm and 85F with a dewpoint temperature of 70F are to be cooled to 60F Determine the rate at which conden sate leaves this system and the cooling rate when the conden sate leaves the system at 65F 1479 Air enters a 40cmdiameter cooling section at 1 atm 32C and 70 percent relative humidity at 120 mmin The air is cooled by passing it over a cooling coil through which cold water flows The water experiences a temperature rise of 6C The air leaves the cooling section saturated at 20C Determine a the rate of heat transfer b the mass flow rate of the water and c the exit velocity of the airstream FIGURE P1474 Air Heating coils Humidifier 10C 70 35 m3min P 1 atm 20C 60 Sat vapor 100C 1475 Repeat Prob 1474 for a total pressure of 95 kPa for the airstream Answers a 195C 377 percent b 391 kJmin c 0147 kgmin Cooling with Dehumidification 1476C Why is cooled air sometimes reheated in summer before it is discharged to a room FIGURE P1479 Cooling coils Air T 6C T 20C Saturated 32C 70 120 mmin Water 1480 Reconsider Prob 1479 Using appropriate soft ware develop a general solution to the problem in which the input variables may be supplied and parametric studies performed For each set of input variables for which the pressure is atmospheric show the process on the psychrometric chart 1481 Repeat Prob 1479 for a total pressure of 88 kPa for air Answers a 452 kJmin b 180 kgmin c 114 mmin 1482 On a summer day in New Orleans Louisiana the pressure is 1 atm the temperature is 32C and the relative humidity is 95 percent This air is to be conditioned to 24C and 60 percent relative humidity Determine the amount of cooling in kJ required and water removed in kg per 1000 m3 of dry air processed at the entrance to the system FIGURE P1482 Cooling coils Condensate 1 atm T1 32C 1 95 T1 24C ϕ2 60 ϕ Final PDF to printer cen22672ch14711746indd 740 103117 0113 PM 740 GASVAPOR MIXTURES 1483 Reconsider Prob 1482 How far will the tempera ture of the humid air have to be reduced to produce the desired dehumidification Answer 158C 1484 Atmospheric air from the inside of an automobile enters the evaporator section of the air conditioner at 1 atm 27C and 50 percent relative humidity The air returns to the automobile at 10C and 90 percent relative humidity The passenger compartment has a volume of 2 m3 and five air changes per minute are required to maintain the inside of the automobile at the desired comfort level Sketch the psychro metric diagram for the atmospheric air flowing through the airconditioning process Determine the dewpoint and wet bulb temperatures at the inlet to the evaporator section in C Determine the required heat transfer rate from the atmospheric air to the evaporator fluid in kW Determine the rate of con densation of water vapor in the evaporator section in kgmin 1488 Atmospheric air at 1 atm 32C and 95 percent rela tive humidity is cooled to 24C and 60 percent relative humid ity A simple ideal vaporcompression refrigeration system using refrigerant134a as the working fluid is used to provide the cooling required It operates its evaporator at 4C and its condenser at a saturation temperature of 394C The con denser rejects its heat to the atmospheric air Calculate the exergy destruction in kJ in the total system per 1000 m3 of dry air processed FIGURE P1487 Cooling coils Heating coils Condensate removal T3 17C T1 39C 1 50 1 atm Twb3 108C 1 2 3 ϕ FIGURE P1488 Condensate 24C 60 32C 95 1 atm Evaporator Expansion valve Compressor 1 2 3 4 Condenser FIGURE P1484 Condensate Air Cooling coils 1485 Humid air at 1013 kPa 39C dry bulb and 50 percent relative humidity is cooled at constant pressure to a tempera ture 10C below its dewpoint temperature a Sketch the system hardware and the psychrometric diagram for the process b If it has been determined that the rate of heat transfer from the atmospheric air is 1340 kW what is the inlet volume flow rate of atmospheric air in m3s for this process 1486E Saturated humid air at 70 psia and 200F is cooled to 100F as it flows through a 3indiameter pipe with a veloc ity of 50 fts and at constant pressure Calculate the rate at which liquid water is formed inside this pipe and the rate at which the air is cooled Answers 00670 lbms 832 Btus 1487 Humid air is to be conditioned in a constantpressure process at 1 atm from 39C dry bulb and 50 percent relative humidity to 17C dry bulb and 108C wet bulb The air is first passed over cooling coils to remove all of the moisture neces sary to achieve the final moisture content and then is passed over heating coils to achieve the final state a Sketch the psychometric diagram for the process b Determine the dewpoint temperature of the mixture at the inlet of the cooling coils and at the inlet of the heating coils c Determine the heat removal by the cooling coils the heat addition by the heating coils and the net heat transfer for the entire process all in kJkg dry air Evaporative Cooling 1489C What is evaporative cooling Will it work in humid climates 1490C During evaporation from a water body to air under what conditions will the latent heat of vaporization be equal to the heat transfer from the air 1491C Does an evaporation process have to involve heat transfer Describe a process that involves both heat and mass transfer 1492 Desert dwellers often wrap their heads with a water soaked porous cloth On a desert where the pressure is 1 atm Final PDF to printer cen22672ch14711746indd 741 103117 0113 PM 741 CHAPTER 14 temperature is 45C and relative humidity is 15 percent what is the temperature of this cloth 1493E Air enters an evaporative cooler at 145 psia 93F and 30 percent relative humidity and exits saturated Deter mine the exit temperature of air 1494 Air enters an evaporative or swamp cooler at 147 psi 95F and 20 percent relative humidity and it exits at 80 percent relative humidity Determine a the exit tempera ture of the air and b the lowest temperature to which the air can be cooled by this evaporative cooler 1495 Air enters an evaporative cooler at 1 atm 40C and 20 percent relative humidity at a rate of 7 m3min and it leaves with a relative humidity of 90 percent Determine a the exit temperature of the air and b the required rate of water supply to the evaporative cooler temperature and the volume flow rate of the mixture Answers 00088 kg H2Okg dry air 597 percent 202C 400 m3min FIGURE P14100 2 3 P 1 atm Air T3 1 35C 30 12C 90 3 ω3 ϕ FIGURE P14104E 3 1 2 P 1 atm Air T3 100F 90 3 ft3s 1 ft3s 50F 30 3 ϕ FIGURE P1495 Air Humidifier 1 atm 40C 20 Water mw 90 1496 Air at 1 atm 20C and 70 percent relative humidity is first heated to 35C in a heating section and then passed through an evaporative cooler where its temperature drops to 25C Determine a the exit relative humidity and b the amount of water added to air in kg H2Okg dry air Adiabatic Mixing of Airstreams 1497C Two unsaturated airstreams are mixed adiabatically It is observed that some moisture condenses during the mixing process Under what conditions will this be the case 1498C Consider the adiabatic mixing of two airstreams Does the state of the mixture on the psychrometric chart have to be on the straight line connecting the two states 1499 Saturated humid air at 1 atm and 10C is to be mixed with atmospheric air at 1 atm 32C and 80 percent relative humidity to form air of 70 percent relative humidity Deter mine the proportions at which these two streams are to be mixed and the temperature of the resulting air 14100 Two airstreams are mixed steadily and adiabatically The first stream enters at 35C and 30 percent relative humidity at a rate of 15 m3min while the second stream enters at 12C and 90 percent relative humidity at a rate of 25 m3min Assum ing that the mixing process occurs at a pressure of 1 atm deter mine the specific humidity the relative humidity the drybulb 14101 Repeat Prob 14100 for a total mixingchamber pressure of 90 kPa 14102 A stream of warm air with a drybulb temperature of 36C and a wetbulb temperature of 30C is mixed adia batically with a stream of saturated cool air at 12C The dry air mass flow rates of the warm and cool airstreams are 8 and 10 kgs respectively Assuming a total pressure of 1 atm determine a the temperature b the specific humidity and c the relative humidity of the mixture 14103 Reconsider Prob 14102 Using appropriate software determine the effect of the mass flow rate of a saturated cool airstream on the mixture temperature specific humidity and relative humidity Vary the mass flow rate of saturated cool air from 0 to 16 kgs while keeping the mass flow rate of warm air constant at 8 kgs Plot the mixture temperature specific humidity and relative humidity as func tions of the mass flow rate of cool air and discuss the results 14104E Two humid airstreams are adiabatically mixed at 1 atm pressure to form a third stream The first stream has a temperature of 100F a relative humidity of 90 percent and a volume flow rate of 3 ft3s while the second stream has a temperature of 50F a relative humidity of 30 percent and a volume flow rate of 1 ft3s Calculate the third streams tem perature and relative humidity Final PDF to printer cen22672ch14711746indd 742 103117 0113 PM 742 GASVAPOR MIXTURES 14105E Reconsider Prob 14104E Calculate the rate of entropy generation for this process Answer 340 104 BtusR Wet Cooling Towers 14106C How does a naturaldraft wet cooling tower work 14107C What is a spray pond How does its performance compare to the performance of a wet cooling tower 14108 The cooling water from the condenser of a power plant enters a wet cooling tower at 40C at a rate of 90 kgs The water is cooled to 25C in the cooling tower by air that enters the tower at 1 atm 23C and 60 percent relative humid ity and leaves saturated at 32C Neglecting the power input to the fan determine a the volume flow rate of air into the cool ing tower and b the mass flow rate of the required makeup water 14109 A wet cooling tower is to cool 60 kgs of water from 40 to 33C Atmospheric air enters the tower at 1 atm with dry and wetbulb temperatures of 22 and 16C respectively and leaves at 30C with a relative humidity of 95 percent Using the psychrometric chart determine a the volume flow rate of air into the cooling tower and b the mass flow rate of the required makeup water Answers a 303 m3s b 0605 kgs 14111 A wet cooling tower is to cool 17 kgs of cooling water from 40 to 30C at a location where the atmospheric pressure is 96 kPa Atmospheric air enters the tower at 20C and 70 percent relative humidity and leaves saturated at 35C Neglecting the power input to the fan determine a the vol ume flow rate of air into the cooling tower and b the mass flow rate of the required makeup water Answers a 758 m3s b 0238 kgs 14112 Water at 30C is to be cooled to 22C in a cooling tower which it enters at a rate of 5 kgs Humid air enters this tower at 1 atm and 15C with a relative humidity of 25 percent and leaves at 18C with a relative humidity of 95 percent Determine the mass flow rate of dry air through this tower Answer 629 kgs 14113 Reconsider Prob 14112 How much work poten tial in kJkg dry air is lost in the cooling tower Take T0 15C Review Problems 14114 The air in a room is at 1 atm 32C and 60 percent relative humidity Determine a the specific humidity b the enthalpy in kJkg dry air c the wetbulb temperature d the dewpoint temperature and e the specific volume of the air in m3kg dry air Use the psychrometric chart 14115 Determine the mole fraction of dry air at the surface of a lake whose temperature is 18C The air at the lake surface is saturated and the atmospheric pressure at lake level can be taken to be 100 kPa 14116 Dry air whose molar analysis is 781 percent N2 209 percent O2 and 1 percent Ar flows over a water body until it is saturated If the pressure and temperature of air remain constant at 1 atm and 25C during the process determine a the molar analysis of the saturated air and b the density of air before and after the process What do you conclude from your results 14117 The condensation of the water vapor in compressed air lines is a major concern in industrial facilities and the compressed air is often dehumidified to avoid the problems associated with condensation Consider a compressor that compresses ambient air from the local atmospheric pressure of 92 kPa to a pressure of 800 kPa absolute The compressed air is then cooled to the ambient temperature as it flows through the compressedair lines Disregarding any pressure losses determine if there will be any condensation in the compressed air lines on a day when the ambient air is at 20C and 50 percent relative humidity 14118E Consider a room that is cooled adequately by an air conditioner whose cooling capacity is 7500 Btuh If the room is to be cooled by an evaporative cooler that removes heat at the same rate by evaporation determine how much water needs to be supplied to the cooler per hour at design conditions FIGURE P14109 Air inlet 1 atm Tdb 22C Twb 16C 30C 95 Air exit 60 kgs 40C Makeup water 33C Cool water Warm water 14110E Water at 100F is to be cooled in a cooling tower which it enters at a rate of 10000 lbmh Humid air enters this tower at 1 atm 60F and 20 percent relative humidity with a dry airflow rate of 7000 lbmh and leaves at 75F and 0018 lbm H2Olbm dry air Determine the relative humid ity at which the air leaves the tower and the waters exit temperature Final PDF to printer cen22672ch14711746indd 743 103117 0113 PM 743 CHAPTER 14 14119 The airconditioning costs of a house can be reduced by up to 10 percent by installing the outdoor unit the con denser of the air conditioner at a location shaded by trees and shrubs If the airconditioning costs of a house are 500 a year determine how much the trees will save the homeowner in the 20year life of the system 14120E The US Department of Energy estimates that 190000 barrels of oil would be saved per day if every house hold in the United States raised the thermostat setting in sum mer by 6F 33C Assuming the average cooling season to be 120 days and the cost of oil to be 70barrel determine how much money would be saved per year 14121 A laboratory has a volume of 700 m3 and must have one complete air change per minute when being used Outside atmospheric air at 100 kPa 30C dry bulb and 60 percent rela tive humidity is ducted into the laboratory airconditioning equipment and conditioned to 20C dry bulb and 12C wet bulb the required state for air supplied to the laboratory a Sketch the system hardware and the psychrometric diagram for the process b What outside atmospheric air mass flow rate is required for the air change in kgh c Determine the mass flow rate of water condensed from the atmospheric air in kgmin d The cooling fluid in the airconditioning system is chilled water which has a 15C temperature rise during the heat exchange process Determine the chilled water mass flow rate in kgmin 14122 A 18m3 tank contains saturated air at 20C and 90 kPa Determine a the mass of the dry air b the spe cific humidity and c the enthalpy of the air per unit mass of the dry air Answers a 188 kg b 00166 kg H2Okg dry air c 622 kJkg dry air 14123 Reconsider Prob 14122 Using appropriate software determine the properties of the air at the initial state Study the effect of heating the air at constant volume until the pressure is 110 kPa Plot the required heat transfer in kJ as a function of pressure 14124E Air at 15 psia 60F and 70 percent relative humid ity flows in a 6indiameter duct at a velocity of 27 fts Deter mine a the dewpoint temperature b the volume flow rate of air and c the mass flow rate of dry air 14125 Air flows steadily through an isentropic nozzle The air enters the nozzle at 35C 200 kPa and 50 percent relative humidity If no condensation is to occur during the expansion process determine the pressure temperature and velocity of the air at the nozzle exit 14126 During a summer day in El Paso Texas outdoor air is at 1 atm 40C and 20 percent relative humidity Water at 20C is evaporated into this air to produce air at 25C and 80 percent rela tive humidity How much water in kg H2Okg dry air is required and how much cooling in kJkg dry air has been produced 14127 Reconsider Prob 14126 If the system is operated as an adiabatic system and the air produced by this system has a relative humidity of 80 percent what is the temperature of the air produced Answer 246C 14128E A typical winter day in Fairbanks Alaska has a pressure of 1 atm a temperature of 32F and a relative humid ity of 60 percent What is the relative humidity inside a home where this air has been heated to 70F FIGURE P14126 Air Water Humidifier 1 atm 40C 20 25C 20C 80 FIGURE P14128E Heating coils 1 atm Air 32F 60 70F 14129E Reconsider Prob 14128E The relative humidity of the air in a home is to be restored to 60 percent by evaporat ing 60F water into the air How much heat in Btu is required to do this in a home of 16000 ft3 volume 14130 Air enters a cooling section at 97 kPa 35C and 30 percent relative humidity at a rate of 6 m3min where it is cooled until the moisture in the air starts condensing Deter mine a the temperature of the air at the exit and b the rate of heat transfer in the cooling section 14131 Outdoor air enters an airconditioning system at 10C and 70 percent relative humidity at a steady rate of 26 m3min and it leaves at 25C and 45 percent relative humidity The outdoor air is first heated to 18C in the heat ing section and then humidified by the injection of hot steam in the humidifying section Assuming the entire process takes place at a pressure of 1 atm determine a the rate of heat sup ply in the heating section and b the mass flow rate of steam required in the humidifying section 14132 Atmospheric air enters an airconditioning system at 30C and 70 percent relative humidity with a volume flow rate of 4 m3min and is cooled to 20C and 20 percent relative humidity at a pressure of 1 atm The system uses refrigerant 134a as the cooling fluid that enters the cooling section at 350 kPa with a quality of 20 percent and leaves as a saturated Final PDF to printer cen22672ch14711746indd 744 103117 0113 PM 744 GASVAPOR MIXTURES vapor Show the process on the psychrometric chart What is the heat transfer from the air to the cooling coils in kW If any water is condensed from the air how much water will be con densed from the atmospheric air per min Determine the mass flow rate of the refrigerant in kgmin of the air when it leaves the heating section b the rate of heat transfer in the heating section and c the rate of water added to air in the evaporative cooler Answers a 325C 192 percent b 655 kJmin c 0112 kgmin 14136 Reconsider Prob 14135 Using appropriate software study the effect of total pressure in the range 94 to 100 kPa on the results required in the problem Plot the results as functions of total pressure 14137 Repeat Prob 14135 for a total pressure of 96 kPa 14138 Conditioned air at 13C and 90 percent relative humidity is to be mixed with outside air at 34C and 40 percent relative humidity at 1 atm If it is desired that the mixture have a relative humidity of 60 percent determine a the ratio of the dry air mass flow rates of the conditioned air to the outside air and b the temperature of the mixture 14139 Reconsider Prob 14138 Determine the desired quantities using appropriate software instead of the psychrometric chart What would the answers be at a location at an atmospheric pressure of 80 kPa 14140 A naturaldraft cooling tower is to remove waste heat from the cooling water flowing through the condenser of a steam power plant The turbine in the steam power plant receives 42 kgs of steam from the steam generator and 18 percent of the steam entering the turbine is extracted for various feedwater heaters The condensate of the higher pressure feedwater heaters is trapped to the next lowestpressure feedwa ter heater The last feedwater heater operates at 02 MPa and all of the steam extracted for the feedwater heaters is throttled from the last feedwater heater exit to the condenser operating at a pressure of 10 kPa The remainder of the steam produces work in the turbine and leaves the lowestpressure stage of the turbine at 10 kPa with an entropy of 7962 kJkgK The cooling tower supplies the cooling water at 26C to the condenser and cooling water returns from the condenser to the cooling tower at 40C Atmospheric air enters the tower at 1 atm with dry and wet bulb temperatures of 23 and 18C respectively and leaves satu rated at 37C Determine a the mass flow rate of the cooling water b the volume flow rate of air into the cooling tower and c the mass flow rate of the required makeup water This prob lem is solved using appropriate software Fundamentals of Engineering FE Exam Problems 14141 A room contains 65 kg of dry air and 043 kg of water vapor at 25C and 90 kPa total pressure The relative humidity of air in the room is a 299 b 352 c 415 d 600 e 662 14142 A 40m3 room contains air at 30C and a total pres sure of 90 kPa with a relative humidity of 75 percent The mass of dry air in the room is a 247 kg b 299 kg c 399 kg d 414 kg e 523 kg FIGURE P14132 Condensate Air 1 atm 30C 70 4 m3min 20C 20 R134a 350 kPa x 020 Saturated vapor Cooling coils 14133 Humid air at 1013 kPa 36C dry bulb and 65 percent relative humidity is cooled at constant pressure to a temperature 10C below its dewpoint temperature Sketch the psychrometric diagram for the process and determine the heat transfer from the air in kJkg dry air FIGURE P14133 Cooling coils Condensate removal Condensate T2 1013 kPa T1 36C 1 65 2 1 2 ϕ ϕ 14134 An automobile air conditioner uses refrigerant134a as the cooling fluid The evaporator operates at 100 kPa gage and the condenser operates at 15 MPa gage The compres sor requires a power input of 6 kW and has an isentropic effi ciency of 85 percent Atmospheric air at 25C and 60 percent relative humidity enters the evaporator and leaves at 8C and 90 percent relative humidity Determine the volume flow rate of the atmospheric air entering the evaporator of the air condi tioner in m3min 14135 An airconditioning system operates at a total pres sure of 1 atm and consists of a heating section and an evap orative cooler Air enters the heating section at 15C and 55 percent relative humidity at a rate of 30 m3min and it leaves the evaporative cooler at 25C and 45 percent relatively humidity Determine a the temperature and relative humidity Final PDF to printer cen22672ch14711746indd 745 103117 0113 PM 745 CHAPTER 14 14143 A room is filled with saturated moist air at 25C and a total pressure of 100 kPa If the mass of dry air in the room is 100 kg the mass of water vapor is a 052 kg b 197 kg c 296 kg d 204 kg e 317 kg 14144 A room contains air at 30C and a total pressure of 960 kPa with a relative humidity of 75 percent The partial pressure of dry air is a 820 kPa b 858 kPa c 928 kPa d 906 kPa e 720 kPa 14145 The air in a house is at 25C and 65 percent relative humidity Now the air is cooled at constant pressure The tem perature at which the moisture in the air will start condensing is a 74C b 163C c 180C d 113C e 202C 14146 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 100 kPa from 30C and a humidity ratio of 0023 kgkg dry air to 15C and a humidity ratio of 0015 kgkg dry air If the mass flow rate of dry air is 04 kgs the rate of heat removal from the air is a 6 kJs b 8 kJs c 11 kJs d 14 kJs e 16 kJs 14147 Air at a total pressure of 90 kPa 15C and 75 percent relative humidity is heated and humidified to 25C and 75 percent relative humidity by introducing water vapor If the mass flow rate of dry air is 4 kgs the rate at which steam is added to the air is a 0032 kgs b 0013 kgs c 0019 kgs d 00079 kgs e 0 kgs 14148 On the psychrometric chart a cooling and dehumidi fication process appears as a line that is a horizontal to the left b vertical downward c diagonal upwards to the right NE direction d diagonal upwards to the left NW direction e diagonal downwards to the left SW direction 14149 On the psychrometric chart a heating and humidifi cation process appears as a line that is a horizontal to the right b vertical upward c diagonal upwards to the right NE direction d diagonal upwards to the left NW direction e diagonal downwards to the right SE direction 14150 An airstream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature The lowest temperature the airstream can be cooled to is a the drybulb temperature at the given state b the wetbulb temperature at the given state c the dewpoint temperature at the given state d the saturation temperature corresponding to the humidity ratio at the given state e the triplepoint temperature of water Design and Essay Problems 14151 Identify the major sources of heat gain in your house in summer and propose ways of minimizing them and thus reducing the cooling load 14152 The airconditioning needs of a large building can be met by a single central system or by several individual window units Considering that both approaches are commonly used in practice the right choice depends on the situation at hand Identify the important factors that need to be considered in decision making and discuss the conditions under which an airconditioning system that consists of several window units is preferable over a large single central system and vice versa 14153 Design an inexpensive evaporative cooling system suitable for use in your house Show how you would obtain a water spray how you would provide airflow and how you would prevent water droplets from drifting into the living space 14154 The daily change in the temperature of the atmosphere tends to be smaller in locations where the relative humidity is high Demonstrate why this occurs by calculating the change in the temperature of a fixed quantity of air when a fixed quantity of heat is removed from the air Plot this temperature change as a function of the initial relative humidity and be sure that the air temperature reaches or exceeds the dewpoint temperature Do the same when a fixed amount of heat is added to the air 14155 The condensation and even freezing of moisture in building walls without effective vapor retarders are of real con cern in cold climates as they undermine the effectiveness of the insulation Investigate how the builders in your area are coping with this problem whether they are using vapor retard ers or vapor barriers in the walls and where they are located in the walls Prepare a report on your findings and explain the reasoning for the current practice 14156 The operation of a cooling tower is governed by the principles of fluid mechanics heat transfer and mass transfer as well as thermodynamics The laws of thermodynamics do place bounds on the conditions under which satisfactory operation may be expected while the other sciences determine equipment sizes and other factors Use the second law as expressed by the increase in entropy or other appropriate principle and the first law to place bounds on the humid air at its inlet in comparison to the condi tions at the liquidwater inlet Do the same for the humidair outlet conditions as compared to the liquidwater outlet conditions 14157 A hurricane is a large heat engine driven by the exchange of water with humid air Evaporation of ocean water occurs as the air approaches the eye of the storm and con densation occurs as rain near the eye of the storm Develop a plot of the wind speed near the eye of the storm as a function of the amount of water released from the air as rain On this plot indicate the minimum air temperature and relative humid ity necessary to sustain each wind speed Hint As an upper bound all of the energy released by the condensing water would be converted into kinetic energy Final PDF to printer cen22672ch14711746indd 746 103117 0113 PM Final PDF to printer cen22672ch15747790indd 747 110917 1154 AM 747 CHAPTER 15 C H E M I CA L R E ACT I O N S I n the preceding chapters we limited our consideration to nonreacting systemssystems whose chemical composition remains unchanged during a process This was the case even with mixing processes during which a homogeneous mixture is formed from two or more fluids without the occur rence of any chemical reactions In this chapter we specifically deal with sys tems whose chemical composition changes during a process that is systems that involve chemical reactions When dealing with nonreacting systems we need to consider only the sen sible internal energy associated with temperature and pressure changes and the latent internal energy associated with phase changes When deal ing with reacting systems however we also need to consider the chemical internal energy which is the energy associated with the destruction and for mation of chemical bonds between the atoms The energy balance relations developed for nonreacting systems are equally applicable to reacting systems but the energy terms in the latter case should include the chemical energy of the system In this chapter we focus on a particular type of chemical reaction known as combustion because of its importance in engineering Readers should keep in mind however that the principles developed are equally applicable to other chemical reactions We start this chapter with a general discussion of fuels and combustion Then we apply the mass and energy balances to reacting systems In this regard we discuss the adiabatic flame temperature which is the highest tem perature a reacting mixture can attain Finally we examine the secondlaw aspects of chemical reactions OBJECTIVES The objectives of Chapter 15 are to Give an overview of fuels and combustion Apply the conservation of mass to reacting systems to determine balanced reaction equations Define the parameters used in combustion analysis such as airfuel ratio percent theoretical air and dewpoint temperature Calculate the enthalpy of reaction the enthalpy of combustion and the heating values of fuels Apply energy balances to reacting systems for both steadyflow control volumes and fixedmass systems Determine the adiabatic flame temperature for reacting mixtures Evaluate the entropy change of reacting systems Analyze reacting systems from the secondlaw perspective Final PDF to printer 748 CHEMICAL REACTIONS cen22672ch15747790indd 748 110917 1154 AM 151 FUELS AND COMBUSTION Any material that can be burned to release thermal energy is called a fuel Most familiar fuels consist primarily of hydrogen and carbon They are called hydrocarbon fuels and are denoted by the general formula CnHm Hydrocarbon fuels exist in all phases some examples being coal gasoline and natural gas The main constituent of coal is carbon Coal also contains varying amounts of oxygen hydrogen nitrogen sulfur moisture and ash It is difficult to give an exact mass analysis for coal since its composition varies considerably from one geographical area to the next and even within the same geographical loca tion Most liquid hydrocarbon fuels are a mixture of numerous hydrocarbons and are obtained from crude oil by distillation Fig 151 The most volatile hydrocarbons vaporize first forming what we know as gasoline The less vol atile fuels obtained during distillation are kerosene diesel fuel and fuel oil The composition of a particular fuel depends on the source of the crude oil as well as on the refinery Although liquid hydrocarbon fuels are mixtures of many different hydrocar bons they are usually considered to be a single hydrocarbon for convenience For example gasoline is treated as octane C8H18 and diesel fuel as dodecane C12H26 Another common liquid hydrocarbon fuel is methyl alcohol CH3OH which is also called methanol and is used in some gasoline blends The gas eous hydrocarbon fuel natural gas which is a mixture of methane and smaller amounts of other gases is often treated as methane CH4 for simplicity Natural gas is produced from gas wells or oil wells rich in natural gas It is com posed mainly of methane but it also contains small amounts of ethane propane hydrogen helium carbon dioxide nitrogen hydrogen sulfate and water vapor On vehicles it is stored either in the gas phase at pressures of 150 to 250 atm as CNG compressed natural gas or in the liquid phase at 162C as LNG liquefied natural gas Over a million vehicles in the world mostly buses run on natural gas Liquefied petroleum gas LPG is a byproduct of natural gas processing or crude oil refining It consists mainly of propane and thus LPG is usually referred to as propane However it also contains varying amounts of butane propylene and butylenes Propane is commonly used in fleet vehicles taxis school buses and private cars Ethanol is obtained from corn grains and organic waste Methanol is produced mostly from natural gas but it can also be obtained from coal and biomass Both alcohols are commonly used as additives in oxygenated gasoline and reformulated fuels to reduce air pollution Vehicles are a major source of air pollutants such as nitric oxides carbon monoxide and hydrocarbons as well as the greenhouse gas carbon diox ide and thus there is a growing shift in the transportation industry from the traditional petroleumbased fuels such as gasoline and diesel fuel to the cleanerburning alternative fuels friendlier to the environment such as natu ral gas alcohols ethanol and methanol liquefied petroleum gas LPG and hydrogen The use of electric and hybrid cars is also on the rise A comparison of some alternative fuels for transportation to gasoline is given in Table 151 Note that the energy content of alternative fuels per unit volume are lower than that of gasoline or diesel fuel and thus the driving range of a vehicle on a full tank is lower when it is running on an alternative fuel Also when comparing cost a realistic measure is the cost per unit energy rather than cost per unit volume For example methanol at a unit cost of 120L may appear FIGURE 151 Most liquid hydrocarbon fuels are obtained from crude oil by distillation Gasoline Kerosene Diesel fuel Fuel oil Crude oil Final PDF to printer 749 CHAPTER 15 cen22672ch15747790indd 749 110917 1154 AM cheaper than gasoline at 180L but this is not the case since the cost of 10000 kJ of energy is 057 for gasoline and 066 for methanol A chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion The oxidizer most often used in combustion processes is air for obvious reasonsit is free and readily avail able Pure oxygen O2 is used as an oxidizer only in some specialized applica tions such as cutting and welding where air cannot be used Therefore a few words about the composition of air are in order On a mole or a volume basis dry air is composed of 209 percent oxygen 781 percent nitrogen 09 percent argon and small amounts of carbon diox ide helium neon and hydrogen In the analysis of combustion processes the argon in the air is treated as nitrogen and the gases that exist in trace amounts are disregarded Then dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers Therefore each mole of oxygen enter ing a combustion chamber is accompanied by 079021 376 mol of nitro gen Fig 152 That is 1 kmol O 2 376 kmol N 2 476 kmol air 151 During combustion nitrogen behaves as an inert gas and does not react with other elements other than forming a very small amount of nitric oxides However even then the presence of nitrogen greatly affects the outcome of a combustion process since nitrogen usually enters a combustion chamber in large quantities at low temperatures and exits at considerably higher tempera tures absorbing a large proportion of the chemical energy released during combustion Throughout this chapter nitrogen is assumed to remain perfectly inert Keep in mind however that at very high temperatures such as those encountered in internal combustion engines a small fraction of nitrogen reacts with oxygen forming hazardous gases such as nitric oxide Air that enters a combustion chamber normally contains some water vapor or moisture which also deserves consideration For most combustion FIGURE 152 Each kmol of O2 in air is accompanied by 376 kmol of N2 Air 21 O2 79 N2 1 kmol O2 376 kmol N2 TABLE 151 A comparison of some alternative fuels to the traditional petroleumbased fuels used in transportation Fuel Energy content kJL Gasoline equivalence LLgasoline Gasoline Light diesel Heavy diesel LPG Liquefied petroleum gas primarily propane Ethanol or ethyl alcohol Methanol or methyl alcohol CNG Compressed natural gas primarily methane at 200 atm LNG Liquefied natural gas primarily methane 31850 33170 35800 23410 29420 18210 8080 20490 1 096 089 136 108 175 394 155 Amount of fuel whose energy content is equal to the energy content of 1L gasoline Final PDF to printer 750 CHEMICAL REACTIONS cen22672ch15747790indd 750 110917 1154 AM processes the moisture in the air and the H2O that forms during combustion can also be treated as an inert gas like nitrogen At very high temperatures however some water vapor dissociates into H2 and O2 as well as into H O and OH When the combustion gases are cooled below the dewpoint temper ature of the water vapor some moisture condenses It is important to be able to predict the dewpoint temperature since the water droplets often combine with the sulfur dioxide that may be present in the combustion gases forming sulfuric acid which is highly corrosive During a combustion process the components that exist before the reac tion are called reactants and the components that exist after the reaction are called products Fig 153 Consider for example the combustion of 1 kmol of carbon with 1 kmol of pure oxygen forming carbon dioxide C O 2 CO 2 152 Here C and O2 are the reactants since they exist before combustion and CO2 is the product since it exists after combustion Note that a reactant does not have to react chemically in the combustion chamber For example if carbon is burned with air instead of pure oxygen both sides of the combustion equation will include N2 That is the N2 will appear both as a reactant and as a product We should also mention that bringing a fuel into intimate contact with oxy gen is not sufficient to start a combustion process Thank goodness it is not Otherwise the whole world would be on fire now The fuel must be brought above its ignition temperature to start the combustion The minimum igni tion temperatures of various substances in atmospheric air are approximately 260C for gasoline 400C for carbon 580C for hydrogen 610C for carbon monoxide and 630C for methane Moreover the proportions of the fuel and air must be in the proper range for combustion to begin For example natural gas does not burn in air in concentrations less than 5 percent or greater than about 15 percent As you may recall from your chemistry courses chemical equations are balanced on the basis of the conservation of mass principle or the mass balance which can be stated as follows The total mass of each ele ment is conserved during a chemical reaction Fig 154 That is the total mass of each element on the righthand side of the reaction equation the products must be equal to the total mass of that element on the lefthand side the reactants even though the elements exist in different chemical com pounds in the reactants and products Also the total number of atoms of each element is conserved during a chemical reaction since the total number of atoms is equal to the total mass of the element divided by its atomic mass For example both sides of Eq 152 contain 12 kg of carbon and 32 kg of oxygen even though the carbon and the oxygen exist as elements in the reactants and as a compound in the product Also the total mass of reac tants is equal to the total mass of products each being 44 kg It is common practice to round the molar masses to the nearest integer if great accuracy is not required However notice that the total mole number of the reactants 2 kmol is not equal to the total mole number of the products 1 kmol That is the total number of moles is not conserved during a chemical reaction A frequently used quantity in the analysis of combustion processes to quan tify the amounts of fuel and air is the airfuel ratio AF It is usually expressed FIGURE 153 In a steadyflow combustion process the components that enter the reaction chamber are called reactants and the components that exit are called products Reactants Products Reaction chamber FIGURE 154 The mass and number of atoms of each element is conserved during a chemical reaction H2 2 2 kg hydrogen 16 kg oxygen 2 kg hydrogen 16 kg oxygen 1 O2 H2O Final PDF to printer 751 CHAPTER 15 cen22672ch15747790indd 751 110917 1154 AM FIGURE 155 The airfuel ratio AF represents the amount of air used per unit mass of fuel during a combustion process Products 18 kg Fuel 1 kg Air 17 kg Combustion chamber AF 17 EXAMPLE 151 Balancing the Combustion Equation One kmol of octane C8H18 is burned with air that contains 20 kmol of O2 as shown in Fig 156 Assuming the products contain only CO2 H2O O2 and N2 determine the mole number of each gas in the products and the airfuel ratio for this combustion process SOLUTION The amount of fuel and the amount of oxygen in the air are given The amount of the products and the AF are to be determined Assumptions The combustion products contain CO2 H2O O2 and N2 only Properties The molar mass of air is Mair 2897 kgkmol 290 kgkmol Table A1 Analysis The chemical equation for this combustion process can be written as C 8 H 18 20 O 2 376 N 2 x CO 2 y H 2 O z O 2 w N 2 where the terms in the parentheses represent the composition of dry air that contains 1 kmol of O2 and x y z and w represent the unknown mole numbers of the gases in the products These unknowns are determined by applying the mass balance to each of the elementsthat is by requiring that the total mass or mole number of each ele ment in the reactants be equal to that in the products C 8 x x 8 H 18 2y y 9 O 20 2 2x y 2z z 75 N 2 20 376 w w 752 Substituting yields C 8 H 18 20 O 2 376 N 2 8CO 2 9H 2 O 75O 2 752N 2 Note that the coefficient 20 in the balanced equation above represents the number of moles of oxygen not the number of moles of air The latter is obtained by adding 20 376 752 moles of nitrogen to the 20 moles of oxygen giving a total of 952 moles of air The airfuel ratio AF is determined from Eq 153 by taking the ratio of the mass of the air and the mass of the fuel AF m air m fuel NM air NM C NM H 2 20 476 kmol29 kgkmol 8 kmol12 kgkmol 9 kmol2 kgkmol 242 kg airkg fuel That is 242 kg of air is used to burn each kilogram of fuel during this combustion process FIGURE 156 Schematic for Example 151 x CO2 y H2O z O2 w N2 C8H18 1 kmol Air Combustion chamber on a mass basis and is defined as the ratio of the mass of air to the mass of fuel for a combustion process Fig 155 That is AF m air m fuel 153 The mass m of a substance is related to the number of moles N through the relation m NM where M is the molar mass The airfuel ratio can also be expressed on a mole basis as the ratio of the mole numbers of air to the mole numbers of fuel But we will use the former definition The reciprocal of the airfuel ratio is called the fuelair ratio Final PDF to printer 752 CHEMICAL REACTIONS cen22672ch15747790indd 752 110917 1154 AM 152 THEORETICAL AND ACTUAL COMBUSTION PROCESSES It is often instructive to study the combustion of a fuel by assuming that the combustion is complete A combustion process is complete if all the carbon in the fuel burns to CO2 all the hydrogen burns to H2O and all the sulfur if any burns to SO2 That is all the combustible components of a fuel are burned to completion during a complete combustion process Fig 157 Conversely the combustion process is incomplete if the combustion products contain any unburned fuel or components such as C H2 CO or OH Insufficient oxygen is an obvious reason for incomplete combustion but it is not the only one Incomplete combustion occurs even when more oxygen is present in the combustion chamber than is needed for complete combustion This may be attributed to insufficient mixing in the combustion chamber dur ing the limited time that the fuel and the oxygen are in contact Another cause of incomplete combustion is dissociation which becomes important at high temperatures Oxygen has a much greater tendency to combine with hydrogen than it does with carbon Therefore the hydrogen in the fuel normally burns to comple tion forming H2O even when there is less oxygen than needed for complete combustion Some of the carbon however ends up as CO or just as plain C particles soot in the products The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air Thus when a fuel is com pletely burned with theoretical air no uncombined oxygen is present in the product gases The theoretical air is also referred to as the chemically cor rect amount of air or 100 percent theoretical air A combustion process with less than the theoretical air is bound to be incomplete The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel Fig 158 For example the theoretical combustion of methane is CH 4 2 O 2 376 N 2 CO 2 2 H 2 O 752N 2 Notice that the products of the theoretical combustion contain no unburned methane and no C H2 CO OH or free O2 In actual combustion processes it is common practice to use more air than the stoichiometric amount to increase the chances of complete combustion or to control the temperature of the combustion chamber The amount of air in excess of the stoichiometric amount is called excess air The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air For example 50 percent excess air is equivalent to 150 percent theoretical air and 200 percent excess air is equivalent to 300 percent theoretical air Of course the stoichiometric air can be expressed as 0 percent excess air or 100 percent theoretical air Amounts of air less than the stoichiometric amount are called deficiency of air and are often expressed as percent deficiency of air For example 90 percent theoretical air is equiva lent to 10 percent deficiency of air The amount of air used in combustion processes is also expressed in terms of the equivalence ratio which is the ratio of the actual fuelair ratio to the stoichiometric fuelair ratio FIGURE 158 The complete combustion process with no free oxygen in the products is called theoretical combustion CH4 2O2 376N2 CO2 2H2O 752N2 no unburned fuel no free oxygen in products FIGURE 157 A combustion process is complete if all the combustible components of the fuel are burned to completion Fuel CnHm Air Combustion chamber n CO2 Excess O2 N2 m H2O 2 Final PDF to printer 753 CHAPTER 15 cen22672ch15747790indd 753 110917 1154 AM Predicting the composition of the products is relatively easy when the com bustion process is assumed to be complete and the exact amounts of the fuel and air used are known All one needs to do in this case is simply apply the mass balance to each element that appears in the combustion equation with out needing to take any measurements Things are not so simple however when one is dealing with actual combustion processes For one thing actual combustion processes are hardly ever complete even in the presence of excess air Therefore it is impossible to predict the composition of the products on the basis of the mass balance alone Then the only alternative we have is to measure the amount of each component in the products directly A commonly used device to analyze the composition of combustion gases is the Orsat gas analyzer In this device a sample of the combustion gases is collected and cooled to room temperature and pressure at which point its volume is measured The sample is then brought into contact with a chemi cal that absorbs the CO2 The remaining gases are returned to the room tem perature and pressure and the new volume they occupy is measured The ratio of the reduction in volume to the original volume is the volume fraction of the CO2 which is equivalent to the mole fraction if idealgas behavior is assumed Fig 159 The volume fractions of the other gases are determined by repeating this procedure In Orsat analysis the gas sample is collected over water and is kept saturated at all times Therefore the vapor pressure of water remains constant during the entire test For this reason the presence of water vapor in the test chamber is ignored and data are reported on a dry basis However the amount of H2O formed during combustion is easily deter mined by balancing the combustion equation FIGURE 159 Determining the mole fraction of the CO2 in combustion gases by using the Orsat gas analyzer yCO2 VCO2 V 01 1 01 Before 100 kPa 25C Gas sample without CO2 09 liter After 100 kPa 25C Gas sample including CO2 1 liter EXAMPLE 152 Combustion of Coal with Theoretical Air Coal from Pennsylvania which has an ultimate analysis by mass of 8436 percent C 189 percent H2 440 percent O2 063 percent N2 089 percent S and 783 percent ash noncombustibles is burned with a theoretical amount of air Fig 1510 Disre garding the ash content determine the mole fractions of the products and the apparent molar mass of the product gases Also determine the airfuel ratio required for this combustion process SOLUTION Coal with known mass analysis is burned with theoretical amount of air The mole fractions of the product gases their apparent molar mass and the air fuel ratio are to be determined Assumptions 1 Combustion is stoichiometric and thus complete 2 Combustion products contain CO2 H2O SO2 and N2 only ash disregarded 3 Combustion gases are ideal gases Analysis The molar masses of C H2 O2 S and air are 12 2 32 32 and 29 kgkmol respectively Table A1 We now consider 100 kg of coal for simplicity Noting that the mass percentages in this case correspond to the masses of the constituents the mole num bers of the constituent of the coal are determined to be N C m C M C 8436 kg 12 kgkmol 7030 kmol N H 2 m H 2 M H 2 189 kg 2 kgkmol 09450 kmol FIGURE 1510 Schematic for Example 152 Coal Product gases Ash Theoretical air 8436 C 189 H2 440 O2 063 N2 089 S 783 ash Combustion chamber Final PDF to printer 754 CHEMICAL REACTIONS cen22672ch15747790indd 754 110917 1154 AM N O 2 m O 2 M O 2 440 kg 32 kgkmol 01375 kmol N N 2 m N 2 M N 2 063 kg 28 kgkmol 00225 kmol N S m S M S 089 kg 32 kgkmol 00278 kmol Ash consists of the noncombustible matter in coal Therefore the mass of ash content that enters the combustion chamber is equal to the mass content that leaves Disregarding this nonreacting component for simplicity the combustion equation may be written as 703C 0 945H 2 0 1375O 2 0 0225N 2 00278S a th O 2 376 N 2 x CO 2 y H 2 O z SO 2 w N 2 Performing mass balances for the constituents gives C balance x 703 H 2 balance y 0945 S balance z 00278 O 2 balance 01375 a th x 05y z a th 7393 N 2 balance w 00225 376 a th 00225 376 7393 2782 Substituting the balanced combustion equation without the ash becomes 703C 0 945H 2 0 1375O 2 0 0225N 2 00278S 7393 O 2 376 N 2 7 03CO 2 0 945H 2 O 0 0278SO 2 27 82N 2 The mole fractions of the product gases are determined as follows N prod 703 0945 00278 2782 3582 kmol y CO 2 N CO 2 N prod 703 kmol 3582 kmol 01963 y H 2 O N H 2 O N prod 0945 kmol 3582 kmol 002638 y SO 2 N SO 2 N prod 00278 kmol 3582 kmol 0000776 y N 2 N N 2 N prod 2782 kmol 3582 kmol 07767 Then the apparent molar mass of product gases becomes M prod m prod N prod 703 44 0945 18 00278 64 2782 28 kg 3582 kmol 309 kgkmol Finally the airfuel mass ratio is determined from its definition to be AF m air m fuel 7393 476 kmol 29 kgkmol 100 kg 102 kg airkg fuel That is 102 kg of air is supplied for each kg of coal in the furnace Discussion We could also solve this problem by considering just 1 kg of coal and still obtain the same results But we would have to deal with very small fractions in calcula tions in this case Final PDF to printer 755 CHAPTER 15 cen22672ch15747790indd 755 110917 1154 AM FIGURE 1511 Schematic for Example 153 Fuel Air 20C 80 CH4 O2 H2 N2 CO2 Combustion chamber 1 atm CO2 H2O N2 EXAMPLE 153 Combustion of a Gaseous Fuel with Moist Air A certain natural gas has the following volumetric analysis 72 percent CH4 9 per cent H2 14 percent N2 2 percent O2 and 3 percent CO2 This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20C 1 atm and 80 percent relative humidity as shown in Fig 1511 Assuming complete combustion and a total pressure of 1 atm determine the dewpoint temperature of the products SOLUTION A gaseous fuel is burned with the stoichiometric amount of moist air The dew point temperature of the products is to be determined Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO2 and all the hydrogen to H2O 2 The fuel is burned with the stoichiomet ric amount of air and thus there is no free O2 in the product gases 3 Combustion gases are ideal gases Properties The saturation pressure of water at 20C is 23392 kPa Table A4 Analysis We note that the moisture in the air does not react with anything it simply shows up as additional H2O in the products Therefore for simplicity we balance the combustion equation by using dry air and then add the moisture later to both sides of the equation Considering 1 kmol of fuel 0 72CH 4 0 09H 2 0 14N 2 0 02O 2 0 03CO 2 fuel a th O 2 376 N 2 dry air The unknown coefficients in the preceding equation are determined from mass balances on various elements C 072 003 x x 075 H 072 4 009 2 2y y 153 O 2 002 003 a th x y 2 a th 1465 N 2 014 376 a th z z 5648 Next we determine the amount of moisture that accompanies 476ath 4761465 697 kmol of dry air The partial pressure of the moisture in the air is P vair ϕ air P sat 20 C 08023392 kPa 1871 kPa Assuming idealgas behavior the number of moles of the moisture in the air is N vair P vair P total N total 1871 kPa 101325 kPa 697 N vair which yields N vair 0131 kmol The balanced combustion equation is obtained by substituting the coefficients deter mined earlier and adding 0131 kmol of H2O to both sides of the equation 0 72CH 4 0 09H 2 0 14N 2 0 02O 2 0 03CO 2 fuel 1465 O 2 376 N 2 dry air 0131H2O moisture 075 CO 2 1661 H 2 O includes moisture 5648 N 2 x CO 2 y H 2 O z N 2 Final PDF to printer 756 CHEMICAL REACTIONS cen22672ch15747790indd 756 110917 1154 AM EXAMPLE 154 Reverse Combustion Analysis Octane C8H18 is burned with dry air The volumetric analysis of the products on a dry basis is Fig 1512 CO 2 1002 percent O 2 562 percent CO 088 percent N 2 8348 percent Determine a the airfuel ratio b the percentage of theoretical air used and c the amount of H2O that condenses as the products are cooled to 25C at 100 kPa SOLUTION Combustion products whose composition is given are cooled to 25C The AF the percent theoretical air used and the fraction of water vapor that condenses are to be determined Assumptions Combustion gases are ideal gases Properties The saturation pressure of water at 25C is 31698 kPa Table A4 Analysis Note that we know the relative composition of the products but we do not know how much fuel or air is used during the combustion process However they can be determined from mass balances The H2O in the combustion gases will start condensing when the temperature drops to the dewpoint temperature For ideal gases the volume fractions are equivalent to the mole fractions Con sidering 100 kmol of dry products for convenience the combustion equation can be written as x C 8 H 18 a O 2 376 N 2 1002 CO 2 088CO 562O 2 8348 N 2 b H 2 O The unknown coefficients x a and b are determined from mass balances N 2 376a 8348 a 2220 C 8x 1002 088 x 136 H 18x 2b b 1224 O 2 a 1002 044 562 b 2 2220 2220 FIGURE 1512 Schematic for Example 154 C8H18 Air Combustion chamber 1002 CO2 562 O2 088 CO 8348 N2 The dewpoint temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled Again assuming idealgas behavior the partial pressure of the water vapor in the combustion gases is P vprod N vprod N prod P prod 1661 kmol 8059 kmol 101325 kPa 2088 kPa Thus T dp T sat 2088 kPa 609C Discussion If the combustion process were achieved with dry air instead of moist air the products would contain less moisture and the dewpoint temperature in this case would be 595C Final PDF to printer 757 CHAPTER 15 cen22672ch15747790indd 757 110917 1154 AM The O2 balance is not necessary but it can be used to check the values obtained from the other mass balances as we did previously Substituting we get 136 C 8 H 18 222 O 2 376 N 2 10 02CO 2 088CO 562O2 83 48N 2 12 24H 2 O The combustion equation for 1 kmol of fuel is obtained by dividing the preceding equation by 136 C 8 H 18 1632 O 2 376 N 2 737CO2 065CO 413O2 61 38N 2 9H 2 O a The airfuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel Eq 153 AF m air m fuel 1632 476 kmol 29 kgkmol 8 kmol 12 kgkmol 9 kmol 2 kgkmol 1976 kg airkg fuel b To find the percentage of theoretical air used we need to know the theoretical amount of air which is determined from the theoretical combustion equation of the fuel C 8 H 18 a th O 2 376 N 2 8 CO 2 9 H 2 O 376 a th N 2 O 2 balance a th 8 45 a th 125 Then Percentage of theoretical air m airact m airth N airact N airth 1632476 kmol 1250476 kmol 131 That is 31 percent excess air was used during this combustion process Notice that some carbon formed carbon monoxide even though there was considerably more oxy gen than needed for complete combustion c For each kmol of fuel burned 737 065 413 6138 9 8253 kmol of products are formed including 9 kmol of H2O Assuming that the dewpoint tempera ture of the products is above 25C some of the water vapor will condense as the prod ucts are cooled to 25C If Nw kmol of H2O condenses there will be 9 Nw kmol of water vapor left in the products The mole number of the products in the gas phase will also decrease to 8253 Nw as a result By treating the product gases including the remaining water vapor as ideal gases Nw is determined by equating the mole fraction of the water vapor to its pressure fraction N v N prodgas P v P prod 9 N w 8253 N w 31698 kPa 100 kPa N w 659 kmol Therefore the majority of the water vapor in the products 73 percent of it condenses as the product gases are cooled to 25C Final PDF to printer 758 CHEMICAL REACTIONS cen22672ch15747790indd 758 110917 1154 AM 153 ENTHALPY OF FORMATION AND ENTHALPY OF COMBUSTION We mentioned in Chap 2 that the molecules of a system possess energy in various forms such as sensible and latent energy associated with a change of state chemical energy associated with the molecular structure and nuclear energy associated with the atomic structure as illustrated in Fig 1513 In this text we do not intend to deal with nuclear energy We also ignored chemical energy until now since the systems considered in previous chap ters involved no changes in their chemical structure and thus no changes in chemical energy Consequently all we needed to deal with were the sensible and latent energies During a chemical reaction some chemical bonds that bind the atoms into molecules are broken and new ones are formed The chemical energy associ ated with these bonds in general is different for the reactants and the prod ucts Therefore a process that involves chemical reactions involves changes in chemical energies which must be accounted for in an energy balance Fig 1514 Assuming the atoms of each reactant remain intact no nuclear reactions and disregarding any changes in kinetic and potential energies the energy change of a system during a chemical reaction is due to a change in state and a change in chemical composition That is Δ E sys Δ E state Δ E chem 154 Therefore when the products formed during a chemical reaction exit the reac tion chamber at the inlet state of the reactants we have ΔEstate 0 and the energy change of the system in this case is due to the changes in its chemical composition only In thermodynamics we are concerned with the changes in the energy of a system during a process and not the energy values at the particular states Therefore we can choose any state as the reference state and assign a value of zero to the internal energy or enthalpy of a substance at that state When a process involves no changes in chemical composition the reference state chosen has no effect on the results When the process involves chemical reactions however the composition of the system at the end of a process is no longer the same as that at the beginning of the process In this case it becomes necessary to have a common reference state for all substances The chosen reference state is 25C 77F and 1 atm which is known as the standard reference state Property values at the standard reference state are indicated by a superscript such as h and u When analyzing reacting systems we must use property values relative to the standard reference state However it is not necessary to prepare a new set of property tables for this purpose We can use the existing tables by subtract ing the property values at the standard reference state from the values at the specified state The idealgas enthalpy of N2 at 500 K relative to the standard reference state for example is h 500 K h o 14581 8669 5912 kJkmol Consider the formation of CO2 from its elements carbon and oxygen dur ing a steadyflow combustion process Fig 1515 Both the carbon and the oxygen enter the combustion chamber at 25C and 1 atm The CO2 formed during this process also leaves the combustion chamber at 25C and 1 atm FIGURE 1513 The microscopic form of energy of a substance consists of sensible latent chemical and nuclear energies Nuclear energy Chemical energy Latent energy Sensible energy Molecule Molecule Atom Atom FIGURE 1514 When the existing chemical bonds are destroyed and new ones are formed during a combustion process usually a large amount of sensible energy is absorbed or released Broken chemical bond Sensible energy Atom Atom Atom FIGURE 1515 The formation of CO2 during a steady flow combustion process at 25C and 1 atm 1 kmol C 25C 1 atm CO2 393520 kJ 25C 1 atm 1 kmol O2 25C 1 atm Combustion chamber Final PDF to printer 759 CHAPTER 15 cen22672ch15747790indd 759 110917 1154 AM The combustion of carbon is an exothermic reaction a reaction during which chemical energy is released in the form of heat Therefore some heat is trans ferred from the combustion chamber to the surroundings during this process which is 393520 kJkmol CO2 formed When one is dealing with chemical reactions it is more convenient to work with quantities per unit mole than per unit time even for steadyflow processes The process described above involves no work interactions Therefore from the steadyflow energy balance relation the heat transfer during this process must be equal to the difference between the enthalpy of the products and the enthalpy of the reactants That is Q H prod H react 393520 kJkmol 155 Since both the reactants and the products are at the same state the enthalpy change during this process is solely due to the changes in the chemical com position of the system This enthalpy change is different for different reac tions and it is very desirable to have a property to represent the changes in chemical energy during a reaction This property is the enthalpy of reaction hR which is defined as the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction For combustion processes the enthalpy of reaction is usually referred to as the enthalpy of combustion hC which represents the amount of heat released during a steadyflow combustion process when 1 kmol or 1 kg of fuel is burned completely at a specified temperature and pressure Fig 1516 It is expressed as h R h C H prod H react 156 which is 393520 kJkmol for carbon at the standard reference state The enthalpy of combustion of a particular fuel is different at different tempera tures and pressures The enthalpy of combustion is obviously a very useful property for analyz ing the combustion processes of fuels However there are so many different fuels and fuel mixtures that it is not practical to list hC values for all pos sible cases Besides the enthalpy of combustion is not of much use when the combustion is incomplete Therefore a more practical approach would be to have a more fundamental property to represent the chemical energy of an element or a compound at some reference state This property is the enthalpy of formation h f which can be viewed as the enthalpy of a substance at a specified state due to its chemical composition To establish a starting point we assign the enthalpy of formation of all stable elements such as O2 N2 H2 and C a value of zero at the standard reference state of 25C and 1 atm That is h f 0 for all stable elements This is no different from assigning the internal energy of saturated liquid water a value of zero at 001C Perhaps we should clarify what we mean by stable The stable form of an element is simply the chemically stable form of that element at 25C and 1 atm Nitrogen for example exists in diatomic form N2 at 25C and 1 atm Therefore the stable form of nitrogen at the standard reference state is diatomic nitrogen N2 not monatomic nitrogen N If an ele ment exists in more than one stable form at 25C and 1 atm one of the forms FIGURE 1516 The enthalpy of combustion represents the amount of energy released as a fuel is burned during a steadyflow process at a specified state 1 kmol C 25C 1 atm 1 kmol CO2 25C 1 atm 1 kmol O2 25C 1 atm Combustion process hC Q 393520 kJkmol C Final PDF to printer 760 CHEMICAL REACTIONS cen22672ch15747790indd 760 110917 1154 AM should be specified as the stable form For carbon for example the stable form is assumed to be graphite not diamond Now reconsider the formation of CO2 a compound from its elements C and O2 at 25C and 1 atm during a steadyflow process The enthalpy change during this process was determined to be 393520 kJkmol However Hreact 0 since both reactants are elements at the standard reference state and the products consist of 1 kmol of CO2 at the same state Therefore the enthalpy of formation of CO2 at the standard reference state is 393520 kJ kmol Fig 1517 That is h f CO 2 393520 kJkmol The negative sign is due to the fact that the enthalpy of 1 kmol of CO2 at 25C and 1 atm is 393520 kJ less than the enthalpy of 1 kmol of C and 1 kmol of O2 at the same state In other words 393520 kJ of chemical energy is released leaving the system as heat when C and O2 combine to form 1 kmol of CO2 Therefore a negative enthalpy of formation for a compound indicates that heat is released during the formation of that compound from its stable elements A positive value indicates heat is absorbed You will notice that two h f values are given for H2O in Table A26 one for liquid water and the other for water vapor This is because both phases of H2O are encountered at 25C and the effect of pressure on the enthalpy of formation is small Note that under equilibrium conditions water exists only as a liquid at 25C and 1 atm The difference between the two enthalpies of formation is equal to the hfg of water at 25C which is 24417 kJkg or 44000 kJkmol Another term commonly used in conjunction with the combustion of fuels is the heating value of the fuel which is defined as the amount of heat released when a fuel is burned completely in a steadyflow process and the products are returned to the state of the reactants In other words the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel That is Heating value h C kJkg fuel The heating value depends on the phase of the H2O in the products The heating value is called the higher heating value HHV when the H2O in the products is in the liquid form and it is called the lower heating value LHV when the H2O in the products is in the vapor form Fig 1518 The two heat ing values are related by HHV LHV m h fg H 2 O kJkg fuel 157 where m is the mass of H2O in the products per unit mass of fuel and hfg is the enthalpy of vaporization of water at the specified temperature Higher and lower heating values of common fuels are given in Table A27 The heating value or enthalpy of combustion of a fuel can be determined from a knowledge of the enthalpy of formation for the compounds involved This is illustrated with the following example FIGURE 1517 The enthalpy of formation of a compound represents the amount of energy absorbed or released as the component is formed from its stable elements during a steadyflow process at a specified state hfCO2 Q 393520 kJkmol CO2 1 kmol C 25C 1 atm 1 kmol CO2 25C 1 atm 1 kmol O2 25C 1 atm Combustion chamber FIGURE 1518 The higher heating value of a fuel is equal to the sum of the lower heating value of the fuel and the latent heat of vaporization of the H2O in the products Products liquid H2O mhfgH2O HHV LHV mhfgH2O Fuel 1 kg Air LHV Qout Products vapor H2O Combustion chamber Final PDF to printer 761 CHAPTER 15 cen22672ch15747790indd 761 110917 1154 AM EXAMPLE 155 Evaluation of the HHV and LHV of Propane Calculate the HHV and LHV of liquid propane fuel C 3 H 8 Compare your results with the values in Table A27 SOLUTION The higher and lower heating values of liquid propane are to be deter mined and compared to the listed values Assumptions 1 Combustion is complete 2 The combustion products contain CO 2 H 2 O and N 2 3 Combustion gases are ideal gases Properties The molar masses of C O 2 H 2 and air are 12 32 2 and 29 kgkmol respectively Table A1 Analysis The combustion of C 3 H 8 is illustrated in Fig 1519 The combustion reac tion with stoichiometric air is C 3 H 8 l 5 O 2 376 N 2 3 CO 2 4 H 2 O 18 8N 2 Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values The heat transfer for this process is equal to enthalpy of combustion Note that N 2 and O 2 are stable elements and thus their enthalpy of formation is zero Then q h C H prod H react N p h f p N r h f r N h f CO 2 N h f H 2 O N h f C 3 H 8 The h f of liquid propane is obtained by adding h fg of propane at 25C to h f of gas propane 103850 44097 335 118620 kJkmol For the HHV the water in the products is taken to be liquid Then h C 3 kmol 393520 kJkmol 4 kmol 285830 kJkmol 1 kmol 118620 kJkmol 2205260 kJkmol propane The HHV of the liquid propane is HHV h C M 2205260 kJkmol C 3 H 8 44097 kgkmol C 3 H 8 50010 kJkg C 3 H 8 The listed value from Table A27 is 50330 kJkg For the LHV the water in the products is taken to be vapor Then h C 3 kmol 393520 kJkmol 4 kmol 241820 kJkmol 1 kmol 118620 kJkmol 2029220 kJ kmol propane The LHV of the propane is then LHV h C M 2029220 kJ kmol C 3 H 8 44097 kg kmol C 3 H 8 46020 kJkg C 3 H 8 The listed value from Table A27 is 46340 kJkg The calculated and listed values are practically identical Discussion The higher heating value of liquid propane is 87 percent higher than its lower heating value Obtain the HHV and LHV of carbon monoxide CO from Table A27 Why are the two values equal to each other FIGURE 1519 Schematic for Example 155 CO2 H2O N2 hC Hprod Hreact C3H8 25C 1 atm 25C 1 atm Air 25C 1 atm Combustion chamber l Final PDF to printer 762 CHEMICAL REACTIONS cen22672ch15747790indd 762 110917 1154 AM When the exact composition of the fuel is known the enthalpy of combus tion of that fuel can be determined using enthalpy of formation data as shown above However for fuels that exhibit considerable variation in composition depending on the source such as coal natural gas and fuel oil it is more practical to determine their enthalpy of combustion experimentally by burn ing them directly in a bomb calorimeter at constant volume or in a steady flow device 154 FIRSTLAW ANALYSIS OF REACTING SYSTEMS The energy balance or the firstlaw relations developed in Chaps 4 and 5 are applicable to both reacting and nonreacting systems However chemi cally reacting systems involve changes in their chemical energy and thus it is more convenient to rewrite the energy balance relations so that the changes in chemical energies are explicitly expressed We do this first for steadyflow systems and then for closed systems SteadyFlow Systems Before writing the energy balance relation we need to express the enthalpy of a component in a form suitable for use for reacting systems That is we need to express the enthalpy such that it is relative to the standard reference state and the chemical energy term appears explicitly When expressed properly the enthalpy term should reduce to the enthalpy of formation hf at the standard reference state With this in mind we express the enthalpy of a component on a unitmole basis as Fig 1520 Enthalpy h f h h kJkmol where the term in the parentheses represents the sensible enthalpy relative to the standard reference state which is the difference between h the sensible enthalpy at the specified state and h the sensible enthalpy at the standard reference state of 25C and 1 atm This definition enables us to use enthalpy values from tables regardless of the reference state used in their construction When the changes in kinetic and potential energies are negligible the steadyflow energy balance relation Ėin Ėout can be expressed for a chemi cally reacting steadyflow system more explicitly as Q in W in n r h f h h r Rate of net energy transfer in by heat work and mass Q out W out n p h f h h p Rate of net energy transfer out by heat work and mass 158 where n p and n r represent the molar flow rates of the product p and the reactant r respectively In combustion analysis it is more convenient to work with quantities expressed per mole of fuel Such a relation is obtained by dividing each term of the equation above by the molal flow rate of the fuel yielding Q in W in N r h f h h r Energy transfer in per mole of fuel by heat work and mass Q out W out N p h f h h p Energy transfer out per mole of fuel by heat work and mass 159 FIGURE 1520 The enthalpy of a chemical component at a specified state is the sum of the enthalpy of the component at 25C 1 atm h f and the sensible enthalpy of the component relative to 25C 1 atm Enthalpy at 25C 1 atm Sensible enthalpy relative to 25C 1 atm H N hf h h Final PDF to printer 763 CHAPTER 15 cen22672ch15747790indd 763 110917 1154 AM where Nr and Np represent the number of moles of the reactant r and the product p respectively per mole of fuel Note that Nr 1 for the fuel and the other Nr and Np values can be picked directly from the balanced combustion equation Taking heat transfer to the system and work done by the system to be positive quantities the energy balance relation just discussed can be expressed more compactly as Q W N p h f h h p N r h f h h r 1510 or as Q W H prod H react kJkmol fuel 1511 where H prod N p h f h h p kJkmol fuel H react N r h f h h r kJkmol fuel If the enthalpy of combustion h C for a particular reaction is available the steadyflow energy equation per mole of fuel can be expressed as Q W h C N p h h p N r h h r kJkmol 1512 The energy balance relations above are sometimes written without the work term since most steadyflow combustion processes do not involve any work interactions A combustion chamber normally involves heat output but no heat input Then the energy balance for a typical steadyflow combustion process becomes Q out N r h f h h p Energy out by mass per mole of fuel N p h f h h p Energy out by mass per mole of fuel 1513 It expresses that the heat output during a combustion process is simply the difference between the energy of the reactants entering and the energy of the products leaving the combustion chamber Closed Systems The general closedsystem energy balance relation Ein Eout ΔEsystem can be expressed for a stationary chemically reacting closed system as Q in Q out W in W out U prod U react kJkmol fuel 1514 where Uprod represents the internal energy of the products and Ureact repre sents the internal energy of the reactants To avoid using another property the internal energy of formation u f we utilize the definition of enthalpy u h P v or u f u u h f h h P v and express the preceding equation as Fig 1521 Q W N p h f h h P v p N r h f h h P v r 1515 FIGURE 1521 An expression for the internal energy of a chemical component in terms of the enthalpy Nhf h h PV Nhf h h Pv U H PV Final PDF to printer 764 CHEMICAL REACTIONS cen22672ch15747790indd 764 110917 1154 AM where we have taken heat transfer to the system and work done by the sys tem to be positive quantities The P v terms are negligible for solids and liq uids and can be replaced by RuT for gases that behave as ideal gases Also if desired the h P v terms in Eq 1515 can be replaced by u The work term in Eq 1515 represents all forms of work including the boundary work It was shown in Chap 4 that ΔU Wb ΔH for nonreacting closed systems undergoing a quasiequilibrium P constant expansion or compression process This is also the case for chemically reacting systems There are several important considerations in the analysis of reacting sys tems For example we need to know whether the fuel is a solid a liquid or a gas since the enthalpy of formation h f of a fuel depends on the phase of the fuel We also need to know the state of the fuel when it enters the combus tion chamber in order to determine its enthalpy For entropy calculations it is especially important to know if the fuel and air enter the combustion chamber premixed or separately When the combustion products are cooled to low tem peratures we need to consider the possibility of condensation of some of the water vapor in the product gases FIGURE 1522 Schematic for Example 156 H2O CO2 CO O2 N2 Q C3H8 25C 005 kgmin 1500 K Air 7C Combustion chamber l EXAMPLE 156 FirstLaw Analysis of SteadyFlow Combustion Liquid propane C3H8 enters a combustion chamber at 25C at a rate of 005 kgmin where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7C as shown in Fig 1522 An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 90 percent of the carbon burns to CO2 with the remaining 10 percent forming CO If the exit temperature of the combustion gases is 1500 K determine a the mass flow rate of air and b the rate of heat transfer from the combustion chamber SOLUTION Liquid propane is burned steadily with excess air The mass flow rate of air and the rate of heat transfer are to be determined Assumptions 1 Steady operating conditions exist 2 Air and the combustion gases are ideal gases 3 Kinetic and potential energies are negligible Analysis We note that all the hydrogen in the fuel burns to H2O but 10 percent of the carbon burns incompletely and forms CO Also the fuel is burned with excess air and thus there is some free O2 in the product gases The theoretical amount of air is determined from the stoichiometric reaction to be C 3 H 8 l a th O 2 376 N 2 3 CO 2 4 H 2 O 376 a th N 2 O 2 balance a th 3 2 5 Then the balanced equation for the actual combustion process with 50 percent excess air and some CO in the products becomes C 3 H 8 l 75 O 2 376 N 2 27 CO 2 03CO 4H 2 O 265O 2 282 N 2 a The airfuel ratio for this combustion process is AF m air m fuel 75 476 kmol 29 kgkmol 3 kmol 12 kgkmol 4 kmol 2 kgkmol 2553 kg airkg fuel Final PDF to printer 765 CHAPTER 15 cen22672ch15747790indd 765 110917 1154 AM Thus m air AF m fuel 2353 kg airkg fuel 005 kg fuelmin 118 kg airmin b The heat transfer for this steadyflow combustion process is determined from the steadyflow energy balance Eout Ein applied on the combustion chamber per unit mole of the fuel Q out N p h f h h p N r h f h h r or Q out N r h f h h r N p h f h h p Assuming the air and the combustion products to be ideal gases we have h hT and we form the following minitable using data from the property tables Substance h f kJkmol h 280 K kJkmol h 298 K kJkmol h 1500 K kJkmol C 3 H 8 l 118910 O2 0 8150 8682 49292 N2 0 8141 8669 47073 H2Og 241820 9904 57999 CO2 393520 9364 71078 CO 110530 8669 47517 The h f of liquid propane is obtained by subtracting the h fg of propane at 25C from the h f of gas propane Substituting gives Q out 1 kmol C 3 H 8 118910 h 298 h 298 kJkmol C 3 H 8 75 kmol O 2 0 8150 8682 kJkmol O 2 282 kmol N 2 0 8141 8669 kJkmol N 2 2 7 kmol CO 2 393520 71078 9364 kJkmol CO 2 03 kmol CO 110530 47517 8669 kJkmol CO 4 kmol H 2 O 241820 57999 9904 kJkmol H 2 O 2 65 kmol O 2 0 49292 8682 kJkmol O 2 28 2 kmol N 2 0 47073 8669 kJkmol N 2 363880 kJkmol of C 3 H 8 Thus 363880 kJ of heat is transferred from the combustion chamber for each kmol 44 kg of propane This corresponds to 36388044 8270 kJ of heat loss per kilo gram of propane Then the rate of heat transfer for a mass flow rate of 005 kgmin for the propane becomes Q out m q out 005 kgmin8270 kJkg 4135 kJmin 689 kW Final PDF to printer 766 CHEMICAL REACTIONS cen22672ch15747790indd 766 110917 1154 AM FIGURE 1523 Schematic for Example 157 1 lbmol CH4 3 lbmol O2 77F 1 atm Before reaction CO2 H2O O2 1800 R P2 After reaction EXAMPLE 157 FirstLaw Analysis of Combustion in a Bomb The constantvolume tank shown in Fig 1523 contains 1 lbmol of methane CH4 gas and 3 lbmol of O2 at 77F and 1 atm The contents of the tank are ignited and the methane gas burns completely If the final temperature is 1800 R determine a the final pressure in the tank and b the heat transfer during this process SOLUTION Methane is burned in a rigid tank The final pressure in the tank and the heat transfer are to be determined Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO2 and all the hydrogen to H2O 2 The fuel the oxygen and the combustion gases are ideal gases 3 Kinetic and potential energies are negligible 4 There are no work interac tions involved Analysis The balanced combustion equation is CH 4 g 3 O 2 CO 2 2 H 2 O O 2 a At 1800 R water exists in the gas phase Using the idealgas relation for both the reactants and the products the final pressure in the tank is determined to be P react V N react R u T react P prod V N prod R u T prod P prod P react N prod N react T prod T react Substituting we get P prod 1 atm 4 lbmol 4 lbmol 1800 R 537 R 335 atm b Noting that the process involves no work interactions the heat transfer during this constantvolume combustion process can be determined from the energy balance Ein Eout ΔEsystem applied to the tank Q out N p h f h h P v p N r h f h h P v r Since both the reactants and the products are assumed to be ideal gases all the inter nal energy and enthalpies depend on temperature only and the P v terms in this equa tion can be replaced by RuT It yields Q out N r h f R u T r N p h f h 1800 R h 537 R R u T p since the reactants are at the standard reference temperature of 537 R From h f and idealgas tables in the Appendix Substance h f Btulbmol h 537 R Btulbmol h 1800 R Btulbmol CH4 32210 O2 0 37251 134858 CO2 169300 40275 183915 H2Og 104040 42580 154330 Final PDF to printer 767 CHAPTER 15 cen22672ch15747790indd 767 110917 1154 AM Substituting we have Q out 1 lbmol CH 4 32210 1986 537 Btulbmol CH 4 3 lbmol O 2 0 1986 537 Btulbmol O 2 1 lbmol CO 2 169300 183915 40275 1986 1800 Btulbmol CO 2 2 lbmol H 2 O 104040 154330 42580 1986 1800 Btulbmol H 2 O 1 lbmol O 2 0 134858 37251 1986 1800 Btulbmol O 2 308730 Btulbmol CH 4 Discussion On a mass basis the heat transfer from the tank would be 30873016 19300 Btulbm of methane 155 ADIABATIC FLAME TEMPERATURE In the absence of any work interactions and any changes in kinetic or potential energies the chemical energy released during a combustion process either is lost as heat to the surroundings or is used internally to raise the temper ature of the combustion products The smaller the heat loss the larger the temperature rise In the limiting case of no heat loss to the surroundings Q 0 the temperature of the products reaches a maximum which is called the adiabatic flame or adiabatic combustion temperature of the reaction Fig 1524 The adiabatic flame temperature of a steadyflow combustion process is determined from Eq 1511 by setting Q 0 and W 0 It yields H prod H react 1516 or N p h f h h p N r h f h h r 1517 Once the reactants and their states are specified the enthalpy of the reactants Hreact can be easily determined The calculation of the enthalpy of the prod ucts Hprod is not so straightforward however because the temperature of the products is not known prior to the calculations Therefore the determination of the adiabatic flame temperature requires the use of an iterative technique unless equations for the sensible enthalpy changes of the combustion products are available A temperature is assumed for the product gases and the Hprod is determined for this temperature If it is not equal to Hreact calculations are repeated with another temperature The adiabatic flame temperature is then determined from these two results by interpolation When the oxidant is air the product gases mostly consist of N2 and a good first guess for the adiabatic flame temperature is obtained by treating the product gases as N2 In combustion chambers the highest temperature to which a material can be exposed is limited by metallurgical considerations Therefore the adia batic flame temperature is an important consideration in the design of com bustion chambers gas turbines and nozzles The maximum temperatures that occur in these devices are considerably lower than the adiabatic flame temperature however since the combustion is usually incomplete some heat loss takes place and some combustion gases dissociate at high temperatures FIGURE 1524 The temperature of a combustion chamber becomes maximum when combustion is complete and no heat is lost to the surroundings Q 0 Fuel Products Insulation Tmax Air Combustion chamber Final PDF to printer 768 CHEMICAL REACTIONS cen22672ch15747790indd 768 110917 1154 AM Fig 1525 The maximum temperature in a combustion chamber can be controlled by adjusting the amount of excess air which serves as a coolant Note that the adiabatic flame temperature of a fuel is not unique Its value depends on 1 the state of the reactants 2 the degree of completion of the reaction and 3 the amount of air used For a specified fuel at a specified state burned with air at a specified state the adiabatic flame temperature attains its maximum value when complete combustion occurs with the theo retical amount of air FIGURE 1525 The maximum temperature encountered in a combustion chamber is lower than the theoretical adiabatic flame temperature Fuel Products Tprod Tmax Air Heat loss Incomplete combustion Dissociation EXAMPLE 158 Adiabatic Flame Temperature in Steady Combustion Liquid octane C8H18 enters the combustion chamber of a gas turbine steadily at 1 atm and 25C and it is burned with air that enters the combustion chamber at the same state as shown in Fig 1526 Determine the adiabatic flame temperature for a complete combustion with 100 percent theoretical air b complete combustion with 400 percent theoretical air and c incomplete combustion some CO in the products with 90 percent theoretical air SOLUTION Liquid octane is burned steadily The adiabatic flame temperature is to be determined for different cases Assumptions 1 This is a steadyflow combustion process 2 The combustion chamber is adiabatic 3 There are no work interactions 4 Air and the combustion gases are ideal gases 5 Changes in kinetic and potential energies are negligible Analysis a The balanced equation for the combustion process with the theoretical amount of air is C 8 H 18 l 125 O 2 376 N 2 8 CO 2 9 H 2 O 47N 2 The adiabatic flame temperature relation Hprod Hreact in this case reduces to N p h f h h p N r h fr N h f C 8 H 18 since all the reactants are at the standard reference state and h f 0 for O2 and N2 The h f and h values of various components at 298 K are Substance h f kJkmol h 298 K kJkmol C 3 H 18 l 249950 O2 0 8682 N2 0 8669 H2Og 241820 9904 CO2 393520 9364 Substituting we have 8 kmol CO 2 393520 h CO 2 9364 kJkmol CO 2 9 kmol H 2 O 241820 h H 2 O 9904 kJkmol H 2 O 47 kmol N 2 0 h N 2 8669 kJkmol N 2 1 kmol C 8 H 18 249950 kJkmol C 8 H 18 FIGURE 1526 Schematic for Example 158 CO2 H2O N2 O2 C8H18 25C 1 atm Tprod 1 atm Air 25C 1 atm Combustion chamber Final PDF to printer 769 CHAPTER 15 cen22672ch15747790indd 769 110917 1154 AM which yields 8 h CO 2 9 h H 2 O 47 h N 2 5646081 kJ It appears that we have one equation with three unknowns Actually we have only one unknownthe temperature of the products Tprodsince h hT for ideal gases Therefore we have to use an equation solver or a trialanderror approach to deter mine the temperature of the products A first guess is obtained by dividing the righthand side of the equation by the total number of moles which yields 56460818 9 47 88220 kJkmol This enthalpy value corresponds to about 2650 K for N2 2100 K for H2O and 1800 K for CO2 Noting that the majority of the moles are N2 we see that Tprod should be close to 2650 K but somewhat under it Therefore a good first guess is 2400 K At this temperature 8 h CO 2 9 h H 2 O 47 h N 2 8 125152 9 103508 47 79320 5660828 kJ This value is higher than 5646081 kJ Therefore the actual temperature is slightly under 2400 K Next we choose 2350 K It yields 8 122091 9 100846 47 77496 5526654 which is lower than 5646081 kJ Therefore the actual temperature of the products is between 2350 and 2400 K By interpolation it is found to be Tprod 2395 K b The balanced equation for the complete combustion process with 400 percent theoretical air is C 8 H 18 l 50 O 2 376 N 2 8 CO 2 9 H 2 O 375O 2 188 N 2 By following the procedure used in a the adiabatic flame temperature in this case is determined to be Tprod 962 K Notice that the temperature of the products decreases significantly as a result of using excess air c The balanced equation for the incomplete combustion process with 90 percent theoretical air is C 8 H 18 l 1125 O 2 376 N 2 55 CO 2 25CO 9H 2 O 423 N 2 Following the procedure used in a we find the adiabatic flame temperature in this case to be Tprod 2236 K Discussion Notice that the adiabatic flame temperature reaches its maximum value when complete combustion occurs with the theoretical amount of air part a It decreases as a result of incomplete combustion part c or using excess air part b 156 ENTROPY CHANGE OF REACTING SYSTEMS So far we have analyzed combustion processes from the conservation of mass and the conservation of energy points of view The thermodynamic analysis of a process is not complete however without the examination of the second law aspects Of particular interest are the exergy and exergy destruction both of which are related to entropy The entropy balance relations developed in Chap 7 are equally applica ble to both reacting and nonreacting systems provided that the entropies of Final PDF to printer 770 CHEMICAL REACTIONS cen22672ch15747790indd 770 110917 1154 AM individual constituents are evaluated properly using a common basis The entropy balance for any system including reacting systems undergoing any process can be expressed as S in S out Net entropy transfer by heat and mass S gen Entropy generation Δ S system Change in entropy kJK 1518 Using quantities per unit mole of fuel and taking the positive direction of heat transfer to be to the system the entropy balance relation can be expressed more explicitly for a closed or steadyflow reacting system as Fig 1527 Q k T k S gen S prod S react kJK 1519 where Tk is temperature at the boundary where Qk crosses it For an adiabatic process Q 0 the entropy transfer term drops out and Eq 1519 reduces to S genadiabatic S prod S react 0 1520 The total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes the system itself and its immediate surroundings where external irreversibilities might be occur ring When evaluating the entropy transfer between an extended system and the surroundings the boundary temperature of the extended system is simply taken to be the environment temperature as explained in Chap 7 The determination of the entropy change associated with a chemical reac tion seems to be straightforward except for one thing The entropy relations for the reactants and the products involve the entropies of the components not entropy changes which was the case for nonreacting systems Thus we are faced with the problem of finding a common base for the entropy of all sub stances as we did with enthalpy The search for such a common base led to the establishment of the third law of thermodynamics in the early part of the last century The third law was expressed in Chap 7 as follows The entropy of a pure crystalline substance at absolute zero temperature is zero Therefore the third law of thermodynamics provides an absolute base for the entropy values for all substances Entropy values relative to this base are called the absolute entropy The s values listed in Tables A18 through A25 for various gases such as N2 O2 CO CO2 H2 H2O OH and O are the idealgas absolute entropy values at the specified temperature and at a pressure of 1 atm The absolute entropy values for various fuels are listed in Table A26 together with the h f values at the standard reference state of 25C and 1 atm Equation 1520 is a general relation for the entropy change of a react ing system It requires the determination of the entropy of each individual component of the reactants and the products which in general is not very easy to do The entropy calculations can be simplified somewhat if the gas eous components of the reactants and the products are approximated as ideal gases However entropy calculations are never as easy as enthalpy or internal energy calculations since entropy is a function of both temperature and pres sure even for ideal gases When evaluating the entropy of a component of an idealgas mixture we should use the temperature and the partial pressure of the component Note FIGURE 1527 The entropy change associated with a chemical relation Surroundings Reaction chamber Ssys Products Sprod Reactants Sreact Final PDF to printer 771 CHAPTER 15 cen22672ch15747790indd 771 110917 1154 AM that the temperature of a component is the same as the temperature of the mixture and the partial pressure of a component is equal to the mixture pres sure multiplied by the mole fraction of the component Absolute entropy values at pressures other than P0 1 atm for any tempera ture T can be obtained from the idealgas entropy change relation written for an imaginary isothermal process between states T P0 and T P as illus trated in Fig 1528 s T P s T P 0 R u ln P P 0 1521 For the component i of an idealgas mixture this relation can be written as s i T P i s i T P 0 R u ln y i P m P 0 kJkmolK 1522 where P0 1 atm Pi is the partial pressure yi is the mole fraction of the com ponent and Pm is the total pressure of the mixture If a gas mixture is at a relatively high pressure or low temperature the devi ation from the idealgas behavior should be accounted for by incorporating more accurate equations of state or the generalized entropy charts 157 SECONDLAW ANALYSIS OF REACTING SYSTEMS Once the total entropy change or the entropy generation is evaluated the exergy destroyed Xdestroyed associated with a chemical reaction can be deter mined from X destroyed T 0 S gen kJ 1523 where T0 is the thermodynamic temperature of the surroundings When analyzing reacting systems we are more concerned with the changes in the exergy of reacting systems than with the values of exergy at various states Fig 1529 Recall from Chap 8 that the reversible work Wrev rep resents the maximum work that can be done during a process In the absence of any changes in kinetic and potential energies the reversible work relation for a steadyflow combustion process that involves heat transfer with only the surroundings at T0 can be obtained by replacing the enthalpy terms with h f h h yielding W rev N r h f h h T 0 s r N p h f h h T 0 s p 1524 An interesting situation arises when both the reactants and the products are at the temperature of the surroundings T0 In that case h T 0 s h T 0 s T 0 g 0 which is by definition the Gibbs function of a unit mole of a substance at temperature T0 The Wrev relation in this case can be written as W rev N r g 0r N p g 0p 1525 or W rev N r g f g T 0 g r N p g f g T 0 g p 1526 FIGURE 1528 At a specified temperature the absolute entropy of an ideal gas at pressures other than P0 1 atm can be determined by subtracting Ru ln PP0 from the tabulated value at 1 atm s Ru ln P0 T s P T P0 1 atm P sT P0 Tabulated sT P FIGURE 1529 The difference between the exergy of the reactants and of the products during a chemical reaction is the reversible work associated with that reaction T P State Reactants Reversible work Products Exergy Final PDF to printer 772 CHEMICAL REACTIONS cen22672ch15747790indd 772 110917 1154 AM where g f is the Gibbs function of formation g f 0 for stable elements like N2 and O2 at the standard reference state of 25C and 1 atm just like the enthalpy of formation and g T 0 g represents the value of the sensible Gibbs function of a substance at temperature T0 relative to the standard reference state For the very special case of Treact Tprod T0 25C ie the reactants the products and the surroundings are at 25C and the partial pressure Pi 1 atm for each component of the reactants and the products Eq 1526 reduces to W rev N r g fr N p g fp kJ 1527 We can conclude from this equation that the g f value the negative of the Gibbs function of formation at 25C and 1 atm of a compound represents the reversible work associated with the formation of that compound from its stable elements at 25C and 1 atm in an environment at 25C and 1 atm Fig 1530 The g f values of several substances are listed in Table A26 FIGURE 1530 The negative of the Gibbs function of formation of a compound at 25C 1 atm represents the reversible work associated with the formation of that compound from its stable elements at 25C 1 atm in an environment that is at 25C 1 atm 25C 1 atm C O2 CO2 Wrev gf CO2 394360 kJkmol Stable elements T0 25C Compound 25C 1 atm 25C 1 atm EXAMPLE 159 Reversible Work Associated with a Combustion Process One lbmol of carbon at 77F and 1 atm is burned steadily with 1 lbmol of oxygen at the same state as shown in Fig 1531 The CO2 formed during the process is then brought to 77F and 1 atm the conditions of the surroundings Assuming the combus tion is complete determine the reversible work for this process SOLUTION Carbon is burned steadily with pure oxygen The reversible work associated with this process is to be determined Assumptions 1 Combustion is complete 2 Steadyflow conditions exist during com bustion 3 Oxygen and the combustion gases are ideal gases 4 Changes in kinetic and potential energies are negligible Properties The Gibbs function of formation at 77F and 1 atm is 0 for C and O2 and 169680 Btulbmol for CO2 The enthalpy of formation is 0 for C and O2 and 169300 Btulbmol for CO2 The absolute entropy is 136 BtulbmolR for C 4900 Btu lbmolR for O2 and 5107 BtulbmolR for CO2 Table A26E Analysis The combustion equation is C O 2 CO 2 The C O2 and CO2 are at 77F and 1 atm which is the standard reference state and also the state of the surroundings Therefore the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products Eq 1527 W rev N r g fr N p g fp N C g fC 0 N O 2 g f O 2 0 N CO 2 g f CO 2 N CO 2 g f CO 2 1 lbmol 169680 Btulbmol 169680 Btu since the g f of stable elements at 77F and 1 atm is zero Therefore 169680 Btu of work could be done as 1 lbmol of C is burned with 1 lbmol of O2 at 77F and 1 atm in an environment at the same state The reversible work in this case represents the exergy of the reactants since the product the CO2 is at the state of the surroundings FIGURE 1531 Schematic for Example 159 C 77F 1 atm CO2 T0 77F O2 77F 1 atm 77F 1 atm P0 1 atm Combustion chamber Final PDF to printer 773 CHAPTER 15 cen22672ch15747790indd 773 110917 1154 AM Discussion We could also determine the reversible work without involving the Gibbs function by using Eq 1524 W rev N r h f h h T 0 s r N p h f h h T 0 s p N r h f T 0 s r N p h f T 0 s p N C h f T 0 s C N O 2 h f T 0 s O 2 N CO 2 h f T 0 s CO 2 Substituting the enthalpy of formation and absolute entropy values we obtain W rev 1 lbmol C 0 537 R136 Btulbmol R 1 lbmol O 2 0 537 R4900 Btulbmol R 1 lbmol CO 2 169300 Btulbmol 537 R 5107 Btulbmol R 169680 Btu which is identical to the result obtained before EXAMPLE 1510 SecondLaw Analysis of Adiabatic Combustion Methane CH4 gas enters a steadyflow adiabatic combustion chamber at 25C and 1 atm It is burned with 50 percent excess air that also enters at 25C and 1 atm as shown in Fig 1532 Assuming complete combustion determine a the temperature of the prod ucts b the entropy generation and c the reversible work and exergy destruction Assume that T0 298 K and the products leave the combustion chamber at 1 atm pressure SOLUTION Methane is burned with excess air in a steadyflow combustion cham ber The product temperature entropy generated reversible work and exergy destroyed are to be determined Assumptions 1 Steadyflow conditions exist during combustion 2 Air and the com bustion gases are ideal gases 3 Changes in kinetic and potential energies are negligible 4 The combustion chamber is adiabatic and thus there is no heat transfer 5 Combustion is complete Analysis a The balanced equation for the complete combustion process with 50 percent excess air is CH 4 g 3 O 2 376 N 2 CO 2 2H 2 O O 2 1128 N 2 Under steadyflow conditions the adiabatic flame temperature is determined from Hprod Hreact which reduces to N p h f h h p N r h fr N h f CH 4 since all the reactants are at the standard reference state and h f 0 for O2 and N2 Assuming idealgas behavior for air and for the products the h f and h values of vari ous components at 298 K can be listed as Substance h f kJkmol h 298 K kJkmol CH4g 74850 O2 0 8682 N2 0 8669 H2Og 241820 9904 CO2 393520 9364 FIGURE 1532 Schematic for Example 1510 CO2 H2O N2 O2 CH4 25C 1 atm Air 25C 1 atm Adiabatic combustion chamber T0 25C Final PDF to printer 774 CHEMICAL REACTIONS cen22672ch15747790indd 774 110917 1154 AM Substituting we have 1 kmol CO 2 393520 h CO 2 9364 kJkmol CO 2 2 kmol H 2 O 241820 h H 2 O 9904 kJkmol H 2 O 1128 kmol N 2 0 h N 2 8669 kJkmol N 2 1 kmol O 2 0 h O 2 8682 kJkmol O 2 1 kmol CH 4 74850 kJkmol CH 4 which yields h CO 2 2 h H 2 O h O 2 1128 h N 2 937950 kJ By trial and error the temperature of the products is found to be T prod 1789 K b Noting that combustion is adiabatic the entropy generation during this process is determined from Eq 1520 S gen S prod S react N p s p N r s r The CH4 is at 25C and 1 atm and thus its absolute entropy is s CH 4 18616 kJ kmolK Table A26 The entropy values listed in the idealgas tables are for 1 atm pressure Both the air and the product gases are at a total pressure of 1 atm but the entropies are to be calculated at the partial pressure of the components which is equal to Pi yiPtotal where yi is the mole fraction of component i From Eq 1522 S i N i s i T P i N i s iT P 0 R u ln y i P m The entropy calculations can be represented in tabular form as follows Ni yi s i T 1 atm Ru ln yiPm Ni s i CH4 1 100 18616 18616 O2 3 021 20504 1298 65406 N2 1128 079 19161 196 218347 Sreact 302369 CO2 1 00654 302517 22674 32519 H2O 2 01309 258957 16905 55172 O2 1 00654 264471 22674 28715 N2 1128 07382 247977 2524 282565 Sprod 398971 Thus S gen S prod S react 398971 302369 kJkmolK CH 4 9660 kJkmolK c The exergy destruction or irreversibility associated with this process is determined from Eq 1523 X destroyed T 0 S gen 298 K9660 kJkmolK 288 MJkmol CH 4 That is 288 MJ of work potential is wasted during this combustion process for each kmol of methane burned This example shows that even complete combustion pro cesses are highly irreversible Final PDF to printer 775 CHAPTER 15 cen22672ch15747790indd 775 110917 1154 AM This process involves no actual work Therefore the reversible work and exergy destroyed are identical W rev 288 MJkmol CH 4 That is 288 MJ of work could be done during this process but is not Instead the entire work potential is wasted EXAMPLE 1511 SecondLaw Analysis of Isothermal Combustion Methane CH4 gas enters a steadyflow combustion chamber at 25C and 1 atm and is burned with 50 percent excess air which also enters at 25C and 1 atm as shown in Fig 1533 After combustion the products are allowed to cool to 25C Assuming complete combustion determine a the heat transfer per kmol of CH4 b the entropy generation and c the reversible work and exergy destruction Assume that T0 298 K and that the products leave the combustion chamber at 1 atm pressure SOLUTION This is the same combustion process we discussed in Example 1510 except that the combustion products are brought to the state of the surroundings by transferring heat from them Thus the combustion equation remains the same CH 4 g 3 O 2 376 N 2 CO 2 2 H 2 O O 2 1128 N 2 At 25C part of the water will condense The amount of water vapor that remains in the products is determined from see Example 153 N v N gas P v P total 31698 kPa 101325 kPa 003128 and N v P v P total N gas 003128 1328 N v N v 043 kmol Therefore 157 kmol of the H2O formed is in the liquid form which is removed at 25C and 1 atm When one is evaluating the partial pressures of the components in the product gases the only water molecules that need to be considered are those that are in the vapor phase As before all the gaseous reactants and products are treated as ideal gases a Heat transfer during this steadyflow combustion process is determined from the steadyflow energy balance Eout Ein on the combustion chamber Q out N p h fp N r h fr since all the reactants and products are at the standard reference of 25C and the enthalpy of ideal gases depends on temperature only Solving for Qout and substituting the h f values we have Q out 1 kmol CH 4 74850 kJkmol CH 4 1 kmol CO 2 393520 kJkmol CO 2 043 kmol H 2 Og 241820 kJkmol H 2 Og 157 kmol H 2 Ol 285830 kJkmol H 2 Ol 871400 kJkmol CH 4 FIGURE 1533 Schematic for Example 1511 CO2 H2O N2 O2 CH4 25C 1 atm 25C 1 atm Air 25C 1 atm Combustion chamber T0 25C Final PDF to printer 776 CHEMICAL REACTIONS cen22672ch15747790indd 776 110917 1154 AM Fuels like methane are commonly burned to provide thermal energy at high temperatures for use in heat engines However a comparison of the reversible works obtained in the last two examples reveals that the exergy of the reactants 818 MJkmol CH4 decreases by 288 MJkmol as a result of the irreversible adiabatic combustion process alone That is the exergy of the hot combustion b The entropy of the reactants was evaluated in Example 1510 and was determined to be Sreact 302369 kJkmolK CH4 By following a similar approach the entropy of the products is determined to be Ni yi s i T 1 atm Ru ln yiPm Ni s i H 2 Ol 157 10000 6992 10977 H2O 043 00314 18883 2877 9357 CO2 1 00729 21380 2177 23557 O2 1 00729 20504 2177 22681 N2 1128 08228 19161 162 217963 Sprod 284535 Then the total entropy generation during this process is determined from an entropy balance applied on an extended system that includes the immediate surroundings of the combustion chamber S gen S prod S react Q out T surr 284535 302369 kJkmolK 871400 kJkmol 298 K 2746 kJkmolK CH 4 c The exergy destruction and reversible work associated with this process are deter mined from X destroyed T 0 S gen 298 K2746 kJkmolK 818 MJ kmol CH 4 and W rev X destroyed 818 MJ kmol CH 4 since this process involves no actual work Therefore 818 MJ of work could be done during this process but is not Instead the entire work potential is wasted The reversible work in this case represents the exergy of the reactants before the reaction starts since the products are in equilibrium with the surroundings that is they are at the dead state Discussion Note that for simplicity we calculated the entropy of the product gases before they actually entered the atmosphere and mixed with the atmospheric gases A more complete analysis would consider the composition of the atmosphere and the mixing of the product gases with the gases in the atmosphere forming a homogeneous mixture There is additional entropy generation during this mixing process and thus additional wasted work potential TOPIC OF SPECIAL INTEREST Fuel Cells This section can be skipped without a loss in continuity Final PDF to printer 777 CHAPTER 15 cen22672ch15747790indd 777 110917 1154 AM FIGURE 1535 The operation of a hydrogenoxygen fuel cell 2e O2 Load H2 O2 H2 2H 2e 2e Porous anode Porous cathode H2O gases at the end of the adiabatic combustion process is 818 288 530 MJ kmol CH4 In other words the work potential of the hot combustion gases is about 65 percent of the work potential of the reactants It seems that when methane is burned 35 percent of the work potential is lost before we even start using the thermal energy Fig 1534 Thus the second law of thermodynamics suggests that there should be a better way of converting the chemical energy to work The better way is of course the less irreversible way the best being the reversible case In chemi cal reactions the irreversibility is due to uncontrolled electron exchange between the reacting components The electron exchange can be controlled by replacing the combustion chamber with electrolytic cells like car batteries This is analogous to replacing unrestrained expansion of a gas in mechani cal systems with restrained expansion In the electrolytic cells the electrons are exchanged through conductor wires connected to a load and the chemical energy is directly converted to electric energy The energy conversion devices that work on this principle are called fuel cells Fuel cells are not heat engines and thus their efficiencies are not limited by the Carnot efficiency They con vert chemical energy to electric energy essentially in an isothermal manner A fuel cell functions like a battery except that it produces its own electricity by combining a fuel with oxygen in a cell electrochemically without combustion and discards the waste heat A fuel cell consists of two electrodes separated by an electrolyte such as a solid oxide phosphoric acid or molten carbonate The elec tric power generated by a single fuel cell is usually too small to be of any practical use Therefore fuel cells are usually stacked in practical applications This modu larity gives the fuel cells considerable flexibility in applications The same design can be used to generate a small amount of power for a remote switching station or a large amount of power to supply electricity to an entire town Therefore fuel cells are termed the microchip of the energy industry The operation of a hydrogenoxygen fuel cell is illustrated in Fig 1535 Hydrogen is ionized at the surface of the anode and hydrogen ions flow through the electrolyte to the cathode There is a potential difference between the anode and the cathode and free electrons flow from the anode to the cath ode through an external circuit such as a motor or a generator Hydrogen ions combine with oxygen and the free electrons at the surface of the cath ode forming water Therefore the fuel cell operates like an electrolysis system working in reverse In steady operation hydrogen and oxygen continuously enter the fuel cell as reactants and water leaves as the product Therefore the exhaust of the fuel cell is drinkablequality water The fuel cell was invented by William Groves in 1839 but it did not receive serious attention until the 1960s when fuel cells were used to produce electric ity and water for the Gemini and Apollo spacecraft Today they are used for the same purpose in the space shuttle missions Despite the irreversible effects such as internal resistance to electron flow fuel cells have a great potential for much higher conversion efficiencies At present fuel cells are available commercially but they are competitive only in some niche markets because of their high cost Fuel cells produce highquality electric power efficiently and quietly while gen erating low emissions using a variety of fuels such as hydrogen natural gas propane and biogas Recently many fuel cells have been installed to generate electricity For example a remote police station in Central Park in New York FIGURE 1534 The availability of methane decreases by 35 percent as a result of the irreversible combustion process Adiabatic combustion chamber 25C Reactants CH4 air Exergy 818 MJ 100 1789 K Products Exergy 530 MJ 65 Final PDF to printer 778 CHEMICAL REACTIONS cen22672ch15747790indd 778 110917 1154 AM City is powered by a 200kW phosphoric acid fuel cell that has an efficiency of 40 percent with negligible emissions it emits 1 ppm NOx and 5 ppm CO Hybrid power systems HPS that combine hightemperature fuel cells and gas turbines have the potential for very high efficiency in converting natural gas or even coal to electricity Also some car manufacturers are planning to introduce cars powered by fuelcell engines thus nearly doubling the effi ciency from about 30 percent for the gasoline engines to up to 60 percent for fuel cells Intense research and development programs by major car manu facturers are underway to make fuelcell cars economical and commercially available in the near future SUMMARY Any material that can be burned to release energy is called a fuel and a chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combus tion The oxidizer most often used in combustion processes is air The dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers Therefore 1 kmol O 2 376 kmol N 2 476 kmol air During a combustion process the components that exist before the reaction are called reactants and the components that exist after the reaction are called products Chemical equations are balanced on the basis of the conservation of mass principle which states that the total mass of each ele ment is conserved during a chemical reaction The ratio of the mass of air to the mass of fuel during a combustion process is called the airfuel ratio AF AF m air m fuel where m air NM air and m fuel N i M i fuel A combustion process is complete if all the carbon in the fuel burns to CO2 all the hydrogen burns to H2O and all the sulfur if any burns to SO2 The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air The theoretical air is also referred to as the chemically correct amount of air or 100 percent theoretical air The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel The air in excess of the stoichiometric amount is called the excess air The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air During a chemical reaction some chemical bonds are bro ken and others are formed Therefore a process that involves chemical reactions involves changes in chemical energies Because of the changed composition it is necessary to have a standard reference state for all substances which is chosen to be 25C 77F and 1 atm The difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction is called the enthalpy of reac tion hR For combustion processes the enthalpy of reaction is usually referred to as the enthalpy of combustion hC which represents the amount of heat released during a steadyflow combustion process when 1 kmol or 1 kg of fuel is burned completely at a specified temperature and pressure The enthalpy of a substance at a specified state due to its chemi cal composition is called the enthalpy of formation h f The enthalpy of formation of all stable elements is assigned a value of zero at the standard reference state of 25C and 1 atm The heating value of a fuel is defined as the amount of heat released when a fuel is burned completely in a steady flow process and the products are returned to the state of the reactants The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel Heating value h C kJkg fuel Taking heat transfer to the system and work done by the system to be positive quantities the conservation of energy relation for chemically reacting steadyflow systems can be expressed per unit mole of fuel as Q W N p h f h h p N r h f h h r where the superscript represents properties at the standard reference state of 25C and 1 atm For a closed system it becomes Q W N p h f h h P v p N r h f h h P v r Final PDF to printer 779 CHAPTER 15 cen22672ch15747790indd 779 110917 1154 AM The P v terms are negligible for solids and liquids and can be replaced by RuT for gases that behave as ideal gases In the absence of any heat loss to the surroundings Q 0 the temperature of the products will reach a maximum which is called the adiabatic flame temperature of the reaction The adiabatic flame temperature of a steadyflow combustion pro cess is determined from Hprod Hreact or N p h f h h p N r h f h h r Taking the positive direction of heat transfer to be to the system the entropy balance relation can be expressed for a closed system or steadyflow combustion chamber as Q k T k S gen S prod S react For an adiabatic process it reduces to S genadiabatic S prod S react 0 The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero The third law provides a common base for the entropy of all substances and the entropy values relative to this base are called the absolute entropy The idealgas tables list the absolute entropy values over a wide range of temperatures but at a fixed pressure of P0 1 atm Absolute entropy values at other pressures P for any temperature T are determined from s T P s T P 0 R u ln P P 0 For component i of an idealgas mixture this relation can be written as s i T P i s i T P 0 R u ln y i P m P 0 where Pi is the partial pressure yi is the mole fraction of the component and Pm is the total pressure of the mixture in atmospheres The exergy destruction and the reversible work associated with a chemical reaction are determined from X destroyed W rev W act T 0 S gen and Wrev Nr h f h h T0 s r Np h f h h T0 s p When both the reactants and the products are at the tem perature of the surroundings T0 the reversible work can be expressed in terms of the Gibbs functions as Wrev Nr g f g T 0 g r Np g f g T 0 g p REFERENCES AND SUGGESTED READINGS 1 S W Angrist Direct Energy Conversion 4th ed Boston Allyn and Bacon 1982 2 I Glassman Combustion New York Academic Press 1977 3 R Strehlow Fundamentals of Combustion Scranton PA International Textbook Co 1968 PROBLEMS Problems designated by a C are concept questions and stu dents are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Fuels and Combustion 151C What are the approximate chemical compositions of gasoline diesel fuel and natural gas 152C How does the presence of N2 in air affect the out come of a combustion process 153C Is the number of atoms of each element conserved dur ing a chemical reaction How about the total number of moles 154C What is the airfuel ratio How is it related to the fuelair ratio 155C Is the airfuel ratio expressed on a mole basis identi cal to the airfuel ratio expressed on a mass basis 156C How does the presence of moisture in air affect the outcome of a combustion process 157C What does the dewpoint temperature of the product gases represent How is it determined 158 Trace amounts of sulfur S in coal are burned in the presence of diatomic oxygen O2 to form sulfur dioxide SO2 Determine the minimum mass of oxygen required in the reac tants and the mass of sulfur dioxide in the products when 1 kg of sulfur is burned Final PDF to printer cen22672ch15747790indd 780 110917 1154 AM 780 CHEMICAL REACTIONS FIGURE P1525 25C Products C8H18 Air Combustion chamber P 1 atm 159E Methane CH4 is burned in the presence of diatomic oxygen The combustion products consist of water vapor and carbon dioxide gas Determine the mass of water vapor gener ated when 1 lbm of methane is burned Answer 225 lbm H2O lbm fuel Theoretical and Actual Combustion Processes 1510C Are complete combustion and theoretical combus tion identical If not how do they differ 1511C What does 100 percent theoretical air represent 1512C Consider a fuel that is burned with a 130 percent theoretical air and b 70 percent excess air In which case is the fuel burned with more air 1513C What are the causes of incomplete combustion 1514C Which is more likely to be found in the products of an incomplete combustion of a hydrocarbon fuel CO or OH Why 1515 Methane CH4 is burned with the stoichiometric amount of air during a combustion process Assuming com plete combustion determine the airfuel and fuelair ratios 1516 Acetylene C2H2 is burned with the stoichiometric amount of air during a combustion process Assuming com plete combustion determine the airfuel ratio on a mass and on a mole basis 1517 nButane fuel C4H10 is burned with the stoichiomet ric amount of air Determine the mass fraction of each product Also calculate the mass of carbon dioxide in the products and the mass of air required per unit of fuel mass burned 1518 nOctane C8H18 is burned with the stoichiometric amount of oxygen Calculate the mass fractions of each of the products and the mass of water in the products per unit mass of fuel burned 1519 Propane C3H8 is burned with 75 percent excess air during a combustion process Assuming complete combustion determine the airfuel ratio Answer 275 kg airkg fuel 1520 Propane fuel C3H8 is burned with 30 percent excess air Determine the mole fractions of each of the products Also calculate the mass of water in the products per unit mass of the fuel and the airfuel ratio 1521 In a combustion chamber ethane C2H6 is burned at a rate of 8 kgh with air that enters the combustion chamber at a rate of 176 kgh Determine the percentage of excess air used during this process Answer 37 percent 1522 Methyl alcohol CH3OH is burned with the stoichio metric amount of air Calculate the mole fractions of each of the products and the apparent molar mass of the product gas Also calculate the mass of water in the products per unit mass of fuel burned Answers 0116 CO2 0231 H2O 0653 N2 275 kgkmol 113 kg H2Okg fuel 1523E Ethylene C2H4 is burned with 175 percent theo retical air during a combustion process Assuming complete combustion and a total pressure of 145 psia determine a the airfuel ratio and b the dewpoint temperature of the prod ucts Answers a 259 lbm airlbm fuel b 105F 1524 Ethane C2H6 is burned with 20 percent excess air during a combustion process Assuming complete combustion and a total pressure of 100 kPa determine a the airfuel ratio and b the dewpoint temperature of the products FIGURE P1524 C2H6 CO2 H2O O2 N2 Air 20 excess Combustion chamber 100 kPa 1525 Octane C8H18 is burned with 250 percent theoretical air which enters the combustion chamber at 25C Assuming complete combustion and a total pressure of 1 atm determine a the airfuel ratio and b the dewpoint temperature of the products 1526 Butane C4H10 is burned in 200 percent theoretical air For complete combustion how many kmol of water must be sprayed into the combustion chamber per kmol of fuel if the products of combustion are to have a dewpoint temperature of 60C when the product pressure is 100 kPa 1527 A fuel mixture of 60 percent by mass methane CH4 and 40 percent by mass ethanol C2H6O is burned completely with theoretical air If the total flow rate of the fuel is 10 kgs determine the required flow rate of air Answer 139 kgs 1528 One kmol of ethane C2H6 is burned with an unknown amount of air during a combustion process An analysis of the combustion products reveals that the combustion is complete and there are 3 kmol of free O2 in the products Determine a the airfuel ratio and b the percentage of theoretical air used during this process 1529 A certain natural gas has the following volumetric analysis 65 percent CH4 8 percent H2 18 percent N2 3 percent O2 and 6 percent CO2 This gas is now burned completely with Final PDF to printer cen22672ch15747790indd 781 110917 1154 AM 781 CHAPTER 15 FIGURE P1536 CH3OH CO2 CO H2O O2 N2 Air 50 excess Combustion chamber the stoichiometric amount of dry air What is the airfuel ratio for this combustion process 1530 Repeat Prob 1529 by replacing the dry air with moist air that enters the combustion chamber at 25C 1 atm and 85 percent relative humidity 1531 A gaseous fuel with a volumetric analysis of 45 percent CH4 35 percent H2 and 20 percent N2 is burned to comple tion with 130 percent theoretical air Determine a the airfuel ratio and b the fraction of water vapor that would condense if the product gases were cooled to 25C at 1 atm Answers a 140 kg airkg fuel b 836 percent 1532 Reconsider Prob 1531 Using appropriate software study the effects of varying the per centages of CH4 H2 and N2 making up the fuel and the prod uct gas temperature in the range 5 to 85C 1533 Methane CH4 is burned with dry air The volumetric analysis of the products on a dry basis is 520 percent CO2 033 percent CO 1124 percent O2 and 8323 percent N2 Determine a the airfuel ratio and b the percentage of theo retical air used Answers a 345 kg airkg fuel b 200 percent 1534 The fuel mixer in a natural gas burner mixes meth ane CH4 with air to form a combustible mixture at the out let Determine the mass flow rates at the two inlets needed to produce 05 kgs of an ideal combustion mixture at the outlet 1535 nOctane C8H18 is burned with 60 percent excess air with 15 percent of the carbon in the fuel forming carbon monoxide Calculate the mole fractions of the products and the dewpoint temperature of the water vapor in the products when the products are at 1 atm pressure Answers 00678 CO2 00120 CO 00897 H2O 00808 O2 07498 N2 440C 1536 Methyl alcohol CH3OH is burned with 50 percent excess air The combustion is incomplete with 10 percent of the carbon in the fuel forming carbon monoxide Calculate the mole fraction of carbon monoxide and the apparent molecular weight of the products 1538 A coal from Utah which has an ultimate analysis by mass of 6140 percent C 579 percent H2 2531 percent O2 109 percent N2 141 percent S and 500 percent ash non combustibles is burned with the stoichiometric amount of air but the combustion is incomplete with 5 percent of the carbon in the fuel forming carbon monoxide Calculate the mass frac tion and the apparent molecular weight of the products and the mass of air required per unit mass of fuel burned 1539 Methyl alcohol CH3OH is burned with 100 percent excess air During the combustion process 60 percent of the carbon in the fuel is converted to CO2 and 40 percent is con verted to CO Write the balanced reaction equation and deter mine the airfuel ratio 1537 Determine the fuelair ratio when coal from Colorado which has an ultimate analysis by mass as 7961 percent C 466 percent H2 476 percent O2 183 percent N2 052 percent S and 862 percent ash noncombustibles is burned with 50 percent excess air Answer 00576 kg fuelkg air FIGURE P1537 CO2 H2O SO2 O2 N2 Coal Air 50 excess Combustion chamber FIGURE P1539 CO2 CO H2O O2 N2 CH3OH Air 100 excess Combustion chamber Enthalpy of Formation and Enthalpy of Combustion 1540C What is enthalpy of formation How does it differ from the enthalpy of combustion 1541C What is enthalpy of combustion How does it differ from the enthalpy of reaction 1542C When are the enthalpy of formation and the enthalpy of combustion identical 1543C Does the enthalpy of formation of a substance change with temperature 1544C What are the higher and the lower heating values of a fuel How do they differ How is the heating value of a fuel related to the enthalpy of combustion of that fuel 1545 Calculate the higher and lower heating values of a coal from Utah which has an ultimate analysis by mass of 6140 percent C 579 percent H2 2531 percent O2 109 percent N2 141 percent S and 500 percent ash noncombustibles The Final PDF to printer cen22672ch15747790indd 782 110917 1154 AM 782 CHEMICAL REACTIONS enthalpy of formation of SO2 is 297100 kJkmol Answers 30000 kJkg 28700 kJkg 1546 Determine the enthalpy of combustion of methane CH4 at 25C and 1 atm using the enthalpy of formation data from Table A26 Assume that the water in the products is in the liquid form Compare your result to the value listed in Table A27 Answer 890330 kJkmol 1547 Reconsider Prob 1546 Using appropriate software study the effect of temperature on the enthalpy of combustion Plot the enthalpy of combustion as a function of temperature over the range 25 to 600C 1548 Repeat Prob 1546 for gaseous ethane C2H6 1549 Repeat Prob 1546 for liquid octane C8H18 1550 Ethane C2H6 is burned at atmospheric pressure with the stoichiometric amount of air as the oxidizer Determine the heat rejected in kJkmol fuel when the products and reactants are at 25C and the water appears in the products as water vapor 1551 Reconsider Prob 1550 What minimum pressure of the products is needed to ensure that the water in the products will be in vapor form 1552 Calculate the HHV and LHV of gaseous noctane fuel C8H18 Compare your results with the values in Table A27 FirstLaw Analysis of Reacting Systems 1553C Consider a complete combustion process during which both the reactants and the products are maintained at the same state Combustion is achieved with a 100 percent theo retical air b 200 percent theoretical air and c the chemi cally correct amount of pure oxygen For which case will the amount of heat transfer be the highest Explain 1554C Consider a complete combustion process during which the reactants enter the combustion chamber at 20C and the products leave at 700C Combustion is achieved with a 100 percent theoretical air b 200 percent theoreti cal air and c the chemically correct amount of pure oxy gen For which case will the amount of heat transfer be the lowest Explain 1555C Derive an energy balance relation for a reacting closed system undergoing a quasiequilibrium constant pres sure expansion or compression process 1556 Liquid propane C3H8 enters a combustion cham ber at 25C at a rate of 12 kgmm where it is mixed and burned with 150 percent excess air that enters the combus tion chamber at 12C If the combustion is complete and the exit temperature of the combustion gases is 1200 K deter mine a the mass flow rate of air and b the rate of heat transfer from the combustion chamber Answers a 471 kg min b 5194 kJmin 1557E Liquid octane C8H18 at 77F is burned completely during a steadyflow combustion process with 180 percent theo retical air that enters the combustion chamber at 77F If the prod ucts leave at 2500 R determine a the airfuel ratio and b the heat transfer from the combustion chamber during this process 1558 Propane fuel C3H8 is burned in a space heater with 50 percent excess air The fuel and air enter this heater steadily at 1 atm and 17C while the combustion products leave at 1 atm and 97C Calculate the heat transferred in this heater in kJkmol fuel Answer 1953000 kJkmol fuel 1559 Propane fuel C3H8 is burned with an airfuel ratio of 25 in an atmospheric pressure heating furnace Determine the heat transfer per kilogram of fuel burned when the temperature of the products is such that liquid water just begins to form in the products 1560 Benzene gas C6H6 at 25C is burned during a steadyflow combustion process with 95 percent theoretical air that enters the combustion chamber at 25C All the hydrogen in the fuel burns to H2O but part of the carbon burns to CO If the products leave at 1000 K determine a the mole frac tion of the CO in the products and b the heat transfer from the combustion chamber during this process Answers a 21 percent b 2113 MJkmol C6H6 1561 Ethane gas C2H6 at 25C is burned in a steadyflow com bustion chamber at a rate of 5 kgh with the stoichiometric amount of air which is preheated to 500 K before entering the combustion chamber An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 95 percent of the carbon burns to CO2 the remaining 5 percent forming CO If the products leave the combustion chamber at 800 K determine the rate of heat transfer from the combustion chamber Answer 200 MJh FIGURE P1556 Qout C3H8 25C Products 1200 K Air 12C Combustion chamber FIGURE P1561 800 K Qout C2H6 25C Air 500 K Combustion chamber CO2 CO O H2O 2 N2 Final PDF to printer cen22672ch15747790indd 783 110917 1154 AM 783 CHAPTER 15 1562 A coal from Texas which has an ultimate analysis by mass of 3925 percent C 693 percent H2 4111 percent O2 072 percent N2 079 percent S and 1120 percent ash non combustibles is burned steadily with 40 percent excess air in a power plant boiler The coal and air enter this boiler at standard conditions and the products of combustion in the smokestack are at 127C Calculate the heat transfer in kJkg fuel in this boiler Include the effect of the sulfur in the energy analysis by noting that sulfur dioxide has an enthalpy of formation of 297100 kJkmol and an average specific heat at constant pressure of cp 417 kJkmolK 1563 Octane gas C8H18 at 25C is burned steadily with 80 percent excess air at 25C 1 atm and 40 percent relative humidity Assuming combustion is complete and the products leave the combustion chamber at 1000 K determine the heat transfer for this process per unit mass of octane 1564 Reconsider Prob 1563 Using appropriate software investigate the effect of the amount of excess air on the heat transfer for the combustion process Let the excess air vary from 0 to 200 percent Plot the heat transfer against excess air and discuss the results 1565 Liquid ethyl alcohol C2H5OHl at 25C is burned in a steadyflow combustion chamber with 40 percent excess air that also enters at 25C The products leave the combustion chamber at 600 K Assuming combustion is complete deter mine the required volume flow rate of the liquid ethyl alcohol to supply heat at a rate of 2000 kJs At 25C the density of liquid ethyl alcohol is 790 kgm3 the specific heat at a constant pressure is 11408 kJkmolK and the enthalpy of vaporization is 42340 kJkmol Answer 681 Lmin 1566 Gaseous propane C3H8 is burned in a steadyflow constantpressure process at 100 kPa with 200 percent theo retical air During the combustion process 90 percent of the carbon in the fuel is converted to CO2 and 10 percent is con verted to CO Determine a the balanced combustion equation b the dewpoint temperature of the products in C and c the heat transfer from the combustion chamber in kJ after 100 kmol of fuel are burned when the reactants enter the combustion chamber at 25C and the products are cooled to 25C h f kJkmol C3H8g 103850 CO2 393520 CO 110530 H2Og 241820 H2Ol 285830 1567 A gaseous fuel mixture that is 40 percent propane C3H8 and 60 percent methane CH4 by volume is mixed with the theoretical amount of dry air and burned in a steadyflow constantpressure process at 100 kPa Both the fuel and air enter the combustion chamber at 298 K and undergo a com plete combustion process The products leave the combustion chamber at 398 K Determine a the balanced combustion equation b the amount of water vapor condensed from the products and c the required airflow rate in kgh when the combustion process produces a heat transfer output of 97000 kJh h f kJkmol M kgkmol c p kJkmolK C3H8g 103850 44 CH4g 74850 16 CO2 393520 44 4116 CO 110530 28 2921 H2Og 241820 18 3428 H2Ol 285830 18 7524 O2 32 3014 N2 28 2927 Answer c 344 kgh 1568 A constantvolume tank contains a mixture of 120 g of methane CH4 gas and 600 g of O2 at 25C and 200 kPa The contents of the tank are now ignited and the methane gas burns completely If the final temperature is 1200 K determine a the final pressure in the tank and b the heat transfer dur ing this process 1569 Reconsider Prob 1568 Using appropriate software investigate the effect of the final tem perature on the final pressure and the heat transfer for the com bustion process Let the final temperature vary from 500 to 1500 K Plot the final pressure and heat transfer against the final temperature and discuss the results 1570 A closed combustion chamber is designed so that it maintains a constant pressure of 300 kPa during a combus tion process The combustion chamber has an initial volume of 05 m3 and contains a stoichiometric mixture of octane C8H18 gas and air at 25C The mixture is now ignited and the prod uct gases are observed to be at 1000 K at the end of the com bustion process Assuming complete combustion and treating both the reactants and the products as ideal gases determine the heat transfer from the combustion chamber during this pro cess Answer 3610 kJ 1571 To supply heated air to a house a highefficiency gas furnace burns gaseous propane C3H8 with a combustion efficiency of 96 percent Both the fuel and 140 percent theo retical air are supplied to the combustion chamber at 25C and 100 kPa and the combustion is complete Because this is a highefficiency furnace the product gases are cooled to 25C and 100 kPa before leaving the furnace To maintain the house at the desired temperature a heat transfer rate of 25000 kJh is required from the furnace Determine the volume of water condensed from the product gases per day Answer 686 Lday Final PDF to printer cen22672ch15747790indd 784 110917 1154 AM 784 CHEMICAL REACTIONS 1572E A constantvolume tank contains a mixture of 1 lbmol of benzene C6H6 gas and 60 percent excess air at 77F and 1 atm The contents of the tank are now ignited and all the hydrogen in the fuel burns to H2O but only 92 percent of the carbon burns to CO2 the remaining 8 percent forming CO If the final temperature in the tank is 2100 R determine the heat transfer from the combustion chamber during this pro cess Answer 757 105 Btu Adiabatic Flame Temperature 1573C A fuel is completely burned first with the stoichio metric amount of air and then with the stoichiometric amount of pure oxygen For which case will the adiabatic flame tem perature be higher 1574C A fuel at 25C is burned in a wellinsulated steady flow combustion chamber with air that is also at 25C Under what conditions will the adiabatic flame temperature of the combustion process be a maximum 1575E Hydrogen H2 at 40F is burned with 20 percent excess air that is also at 40F during an adiabatic steadyflow combustion process Assuming complete combustion find the exit temperature of the product gases 1576 What is the adiabatic flame temperature of methane CH4 when it is burned with 30 percent excess air 1577 Octane gas C8H18 at 25C is burned steadily with 30 percent excess air at 25C 1 atm and 60 percent relative humidity Assuming combustion is complete and adiabatic calculate the exit temperature of the product gases 1578 Reconsider Prob 1577 Using appropriate software investigate the effect of the relative humidity on the exit temperature of the product gases Plot the exit temperature of the product gases as a function of relative humidity for 0 ϕ 100 percent 1579 Acetylene gas C2H2 at 25C is burned during a steadyflow combustion process with 30 percent excess air at 27C It is observed that 75000 kJ of heat is being lost from the combustion chamber to the surroundings per kmol of acet ylene Assuming combustion is complete determine the exit temperature of the product gases Answer 2301 K 1580 Ethyl alcohol C2H5OHg is burned with 200 percent excess air in an adiabatic constantvolume container Ini tially air and ethyl alcohol are at 100 kPa and 25C Assum ing complete combustion determine the final temperature and pressure of the products of combustion Answers 1435 K 493 kPa 1581 Methane CH4 is burned with 300 percent excess air in an adiabatic constantvolume container Initially air and methane are at 1 atm and 25C Assuming complete combus tion determine the final pressure and temperature of the com bustion products Answers 394 kPa 1160 K 1582 A large railroad has experimented with burning pow dered coal in a gas turbine combustor Fifty percent excess air was introduced to the combustor at 1380 kPa and 127C while the powdered coal was injected at 25C The combustion was adiabatic and at constant pressure Based on a coal from Colo rado that has an ultimate analysis by mass of 7961 percent C 466 percent H2 476 percent O2 183 percent N2 052 percent S and 862 percent ash noncombustibles what is the esti mated temperature of the combustion products Neglect the effect of the sulfur in the energy balance FIGURE P1581 Air CH4 25C 100 kPa P T FIGURE P1582 50 excess air CO2 H2O SO2 O2 N2 Tprod Coal 25 C 127 C Combustion chamber 1380 kPa 1583 Reconsider Prob 1582 The combustion products are expanded in an isentropic turbine to 140 kPa Calculate the work produced by this turbine in kJkg fuel Entropy Change and SecondLaw Analysis of Reacting Systems 1584C Express the increase of entropy principle for chemi cally reacting systems 1585C How are the absolute entropy values of ideal gases at pressures different from 1 atm determined 1586C What does the Gibbs function of formation g f of a compound represent 1587 Liquid octane C8H18 enters a steadyflow combus tion chamber at 25C and 1 atm at a rate of 025 kgmin It is burned with 50 percent excess air that also enters at 25C and 1 atm After combustion the products are allowed to cool to 25C Assuming complete combustion and that all the H2O in the products is in liquid form determine a the heat trans fer rate from the combustion chamber b the entropy gen eration rate and c the exergy destruction rate Assume that Final PDF to printer cen22672ch15747790indd 785 110917 1154 AM 785 CHAPTER 15 T0 298 K and the products leave the combustion chamber at 1 atm pressure 1592 nOctane C8H18l is burned in the constantpressure combustor of an aircraft engine with 70 percent excess air Air enters this combustor at 600 kPa and 327C liquid fuel is injected at 25C and the products of combustion leave at 600 kPa and 1227C Determine the entropy generation and exergy destruction per unit mass of fuel during this combus tion process Take T0 25C 1593 A steadyflow combustion chamber is supplied with CO gas at 37C and 110 kPa at a rate of 04 m3min and air at 25C and 110 kPa at a rate of 15 kgmin Heat is transferred to a medium at 800 K and the combustion products leave the combustion chamber at 900 K Assuming the combustion is complete and T0 25C determine a the rate of heat trans fer from the combustion chamber and b the rate of exergy destruction Answers a 3567 kJmin b 1610 kJmin Review Problems 1594 Propane C3H8 fuel is burned with the stoichiomet ric amount of oxygen Determine the mass fractions of carbon dioxide and water in the products Also calculate the mass of water in the products per unit of fuel mass burned 1595 nOctane C8H18 is burned with 60 percent excess air in an automobile engine Assuming complete combustion and that the pressure in the exhaust system is 1 atm determine the minimum temperature of the combustion products before liq uid water will begin to form in the exhaust system 1596E A coal from Utah which has an ultimate analysis by mass of 6140 percent C 579 percent H2 2531 percent O2 109 percent N2 141 percent S and 500 percent ash noncombustibles is burned with 20 percent excess air in an atmospheric pressure boiler Calculate the mass of water in the products per unit mass of coal burned and the dewpoint tem perature of the water vapor in the products Answers 0549 lbm H2Olbm coal 108F 1597 A coal from Colorado which has an ultimate analysis by mass of 7961 percent C 466 percent H2 476 percent O2 183 percent N2 052 percent S and 862 percent ash noncom bustibles is burned in an industrial boiler with 10 percent excess air The temperature and pressure in the smokestack are 50C and 1 atm respectively Calculate the fraction of the water in the combustion products that is liquid and the fraction that is vapor 1598 A 1g sample of a certain fuel is burned in a bomb calorimeter that contains 2 kg of water in the presence of 100 g of air in the reaction chamber If the water temperature rises by 25C when equilibrium is established determine the heating value of the fuel in kJkg 1599E Hydrogen H2 is burned with 100 percent excess air that enters the combustion chamber at 90F 145 psia and 60 percent relative humidity Assuming complete combustion determine a the airfuel ratio and b the volume flow rate of air required to burn the hydrogen at a rate of 25 lbmh FIGURE P1587 Products 25C T0 298 K C8H18 25C 25C Air Combustion chamber 1 atm Qout l 1588 nOctane C8H18l is burned in an automobile engine with 200 percent excess air Air enters this engine at 1 atm and 25C Liquid fuel at 25C is mixed with this air before combustion The exhaust products leave the exhaust system at 1 atm and 77C What is the maximum amount of work in kJ kg fuel that can be produced by this engine Take T0 25C 1589 Reconsider Prob 1588 The automobile engine is to be converted to natural gas methane CH4 fuel Assum ing that all factors remain the same what is the maximum work that can be produced by the modified engine in kJkg fuel Answer 51050 kJkg fuel FIGURE P1589 200 excess air Products 77C Qout CH4 25C 25C 1 atm Combustion chamber 1590E Benzene gas C6H6 at 1 atm and 77F is burned during a steadyflow combustion process with 90 percent the oretical air that enters the combustion chamber at 77F and 1 atm All the hydrogen in the fuel burns to H2O but part of the carbon burns to CO Heat is lost to the surroundings at 77F and the products leave the combustion chamber at 1 atm and 1900 R Determine a the heat transfer from the combus tion chamber and b the exergy destruction 1591 Ethylene C2H4 gas enters an adiabatic combustion chamber at 25C and 1 atm and is burned with 20 percent excess air that enters at 25C and 1 atm The combustion is complete and the products leave the combustion chamber at 1 atm pres sure Assuming T0 25C determine a the temperature of the products b the entropy generation and c the exergy destruc tion Answers a 2270 K b 1311 kJkmolK c 390800 kJkmol Final PDF to printer cen22672ch15747790indd 786 110917 1154 AM 786 CHEMICAL REACTIONS 15100 nButane C4H10 is burned with the stoichiometric amount of air in a cook stove The products of combustion are at 1 atm pressure and 40C What fraction of the water in these products is liquid steadily with air at 25C and 1 atm What would your answer be if pure oxygen at 25C were used to burn the fuel instead of air 15108 Liquid propane C3H8l enters a combustion cham ber at 25C and 1 atm at a rate of 04 kgmin where it is mixed and burned with 150 percent excess air that enters the combus tion chamber at 25C The heat transfer from the combustion process is 53 kW Write the balanced combustion equation and determine a the mass flow rate of air b the average molar mass molecular weight of the product gases c the average specific heat at constant pressure of the product gases and d the temperature of the products of combustion Answers a 1563 kgmin b 2863 kgkmol c 3606 kJkmolK d 1282 K 15109 nOctane C8H18g is burned with the stoichiomet ric amount of air Determine the maximum work that can be produced in kJkg fuel when the air fuel and products are all at 25C and 1 atm Answer 45870 kJkg fuel 15110 Repeat Prob 15109 if 100 percent excess air is used for the combustion 15111E Methane CH4 is burned with 100 percent excess air with 10 percent of the carbon forming carbon monoxide Determine the maximum work that can be produced in Btulbm fuel when the air fuel and products are all at 77F and 1 atm 15112 A steam boiler heats liquid water at 200C to super heated steam at 4 MPa and 400C Methane fuel CH4 is burned at atmospheric pressure with 50 percent excess air The fuel and air enter the boiler at 25C and the products of combustion leave at 227C Calculate a the amount of steam generated per unit of fuel mass burned b the change in the exergy of the combus tion streams in kJkg fuel c the change in the exergy of the steam stream in kJkg steam and d the lost work potential in kJkg fuel Take T0 25C Answers a 1872 kg steamkg fuel b 49490 kJkg fuel c 1039 kJkg steam d 30040 kJkg fuel 15113 Repeat Prob 15112 using a coal from Utah that has an ultimate analysis by mass of 6140 percent C 579 percent H2 2531 percent O2 109 percent N2 141 percent S and 500 percent ash noncombustibles Neglect the effect of the sulfur in the energy and entropy balances 15114 Liquid octane C8H18 enters a steadyflow combustion chamber at 25C and 8 atm at a rate of 08 kgmin It is burned with 200 percent excess air that is compressed and preheated to 500 K and 8 atm before entering the combustion chamber After combustion the products enter an adiabatic turbine at 1300 K and 8 atm and leave at 950 K and 2 atm Assuming complete com bustion and T0 25C determine a the heat transfer rate from the combustion chamber b the power output of the turbine and c the reversible work and exergy destruction for the entire pro cess Answers a 770 kJmin b 263 kW c 514 kW 251 kW 15115 The furnace of a particular power plant can be considered to consist of two chambers an adiabatic combus tion chamber where the fuel is burned completely and adia batically and a heat exchanger where heat is transferred to a Carnot heat engine isothermally The combustion gases in the FIGURE P15100 Theoretical air CO2 H2O N2 40C C4H10 Combustion 1 atm chamber 15101 A gaseous fuel mixture of 60 percent propane C3H8 and 40 percent butane C4H10 on a volume basis is burned in air such that the airfuel ratio is 25 kg airkg fuel when the com bustion process is complete Determine a the moles of nitro gen in the air supplied to the combustion process in kmolkmol fuel b the moles of water formed in the combustion process in kmolkmol fuel and c the moles of oxygen in the product gases in kmolkmol fuel Answers a 338 b 440 c 338 15102 Calculate the higher and lower heating values of gaseous methane fuel CH4 Compare your results with the values in Table A27 15103 A steadyflow combustion chamber is supplied with CO gas at 37C and 110 kPa at a rate of 04 m3min and air at 25C and 110 kPa at a rate of 15 kgmin The combustion products leave the combustion chamber at 900 K Assuming combustion is complete determine the rate of heat transfer from the combustion chamber 15104 Methane gas CH4 at 25C is burned steadily with dry air that enters the combustion chamber at 17C The volu metric analysis of the products on a dry basis is 520 percent CO2 033 percent CO 1124 percent O2 and 8323 percent N2 Determine a the percentage of theoretical air used and b the heat transfer from the combustion chamber per kmol of CH4 if the combustion products leave at 700 K 15105 A 6m3 rigid tank initially contains a mixture of 1 kmol of hydrogen H2 gas and the stoichiometric amount of air at 25C The contents of the tank are ignited and all the hydrogen in the fuel burns to H2O If the combustion products are cooled to 25C determine a the fraction of the H2O that condenses and b the heat transfer from the combustion chamber during this process 15106 Propane gas C3H8 enters a steadyflow combustion chamber at 1 atm and 25C and is burned with air that enters the combustion chamber at the same state Determine the adia batic flame temperature for a complete combustion with 100 percent theoretical air b complete combustion with 200 percent theoretical air and c incomplete combustion some CO in the products with 90 percent theoretical air 15107 Determine the highest possible temperature that can be obtained when liquid gasoline assumed C8H18 at 25C is burned Final PDF to printer cen22672ch15747790indd 787 110917 1154 AM 787 CHAPTER 15 heat exchanger are well mixed so that the heat exchanger is at a uniform temperature at all times that is equal to the tempera ture of the exiting product gases Tp The work output of the Carnot heat engine can be expressed as w Q η c Q 1 T 0 T p where Q is the magnitude of the heat transfer to the heat engine and T0 is the temperature of the environment The work output of the Carnot engine will be zero either when Tp Taf which means the product gases will enter and exit the heat exchanger at the adiabatic flame temperature Taf and thus Q 0 or when Tp T0 which means the temperature of the product gases in the heat exchanger will be T0 and thus ηc 0 and will reach a maximum somewhere in between Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger show that the work output of the Carnot heat engine will be maximum when T p T af T 0 Also show that the maximum work output of the Carnot engine in this case becomes W max C T af 1 T 0 T af 2 where C is a constant whose value depends on the composition of the product gases and their specific heats 15116 Consider the combustion of a mixture of an alcohol CnHmOx and a hydrocarbon fuel CwHz with excess theoretical air and incomplete combustion according to the chemical reac tion as follows y 1 C n H m O x y 2 C w H z 1 B A th O 2 376N 2 D CO 2 E CO F H 2 O G O 2 J N 2 where y1 and y2 are the mole fractions of the fuel mixture Ath is the theoretical O2 required for this fuel and B is the excess amount of air in decimal form If a is the fraction of carbon in the fuel converted to carbon dioxide and b is the remaining fraction converted to carbon monoxide determine the coeffi cients Ath D E F G and J for a fixed B amount of excess air Write the coefficients D E F G and J as functions of y1 y2 n m x w z a b B and Ath in the simplest correct forms FIGURE P15115 Heat exchanger Tp const Q Surroundings T0 W Tp Fuel Air Adiabatic combustion chamber T0 FIGURE P15116 CO2 CO H2O O2 N2 CnHmOx CwHz Excess air Combustion chamber 15117 Develop an expression for the higher heating value of a gaseous alkane CnH2n2 in terms of n 15118 Using appropriate software write a general program to determine the adiabatic flame tem perature during the complete combustion of a hydrocarbon fuel CnHm at 25C in a steadyflow combustion chamber when the percent of excess air and its temperature are speci fied As a sample case determine the adiabatic flame tempera ture of liquid propane C3H8 as it is burned steadily with 120 percent excess air at 25C 15119 Using appropriate software determine the effect of the amount of air on the adiabatic flame temperature of liquid octane C8H18 Assume both the air and the octane are initially at 25C Determine the adiabatic flame temperature for 75 90 100 120 150 200 300 500 and 800 percent theoretical air Assume the hydrogen in the fuel always burns H2O and the carbon CO2 except when there is a deficiency of air In the latter case assume that part of the carbon forms CO Plot the adiabatic flame temperature against the percent theoretical air and discuss the results 15120 Using appropriate software determine the fuel among CH4g C2H2g C2H6g C3H8g C3H18l that gives the highest temperature when burned com pletely in an adiabatic constantvolume chamber with the theo retical amount of air Assume the reactants are at the standard reference state 15121 Using appropriate software determine the rate of heat transfer for the fuels CH4g C2H2g Final PDF to printer cen22672ch15747790indd 788 110917 1154 AM 788 CHEMICAL REACTIONS CH3OHg C3H8g and C3H18lwhen they are burned com pletely in a steadyflow combustion chamber with the theoreti cal amount of air Assume the reactants enter the combustion chamber at 298 K and the products leave at 1200 K 15122 Using appropriate software repeat Prob 15121 for a 50 b 100 and c 200 percent excess air 15123 Using appropriate software write a general program to determine the heat transfer during the complete combustion of a hydrocarbon fuel CnHm at 25C in a steadyflow combustion chamber when the percent of excess air and the temperatures of air and the products are specified As a sample case determine the heat transfer per unit mass of fuel as liquid propane C3H8 is burned steadily with 50 percent excess air at 25C and the combustion prod ucts leave the combustion chamber at 1800 K Fundamentals of Engineering FE Exam Problems 15124 A fuel is burned steadily in a combustion chamber The combustion temperature will be the highest except when a the fuel is preheated b the fuel is burned with a deficiency of air c the air is dry d the combustion chamber is well insulated e the combustion is complete 15125 A fuel is burned with 70 percent theoretical air This is equivalent to a 30 excess air b 70 excess air c 30 deficiency of air d 70 deficiency of air e stoichiometric amount of air 15126 Propane C3H8 is burned with 125 percent theoreti cal air The airfuel mass ratio for this combustion process is a 123 b 157 c 195 d 221 e 234 15127 Benzene gas C6H6 is burned with 90 percent theo retical air during a steadyflow combustion process The mole fraction of the CO in the products is a 17 b 23 c 36 d 44 e 143 15128 One kmol of methane CH4 is burned with an unknown amount of air during a combustion process If the combustion is complete and there is 1 kmol of free O2 in the products the airfuel mass ratio is a 346 b 257 c 172 d 143 e 119 15129 The higher heating value of a hydrocarbon fuel CnHm with m 8 is given to be 1560 MJkmol of fuel Then its lower heating value is a 1384 MJkmol b 1208 MJkmol c 1402 MJkmol d 1514 MJkmol e 1551 MJkmol 15130 Methane CH4 is burned completely with 80 per cent excess air during a steadyflow combustion process If both the reactants and the products are maintained at 25C and 1 atm and the water in the products exists in the liquid form the heat transfer from the combustion chamber per unit mass of methane is a 890 MJkg b 802 MJkg c 75 MJkg d 56 MJkg e 50 MJkg 15131 Acetylene gas C2H2 is burned completely during a steadyflow combustion process The fuel and the air enter the combustion chamber at 25C and the products leave at 1500 K If the enthalpy of the products relative to the standard reference state is 404 MJkmol of fuel the heat transfer from the combustion chamber is a 177 MJkmol b 227 MJkmol c 404 MJkmol d 631 MJkmol e 751 MJkmol 15132 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60C enters a dehumidifying section where all of the water vapor is condensed and removed from the mixture and the carbon dioxide leaves at 1 atm and 60C The entropy change of carbon dioxide in the dehumidifying section is a 28 kJkgK b 013 kJkgK c 0 d 013 kJkgK e 28 kJkgK 15133 A fuel is burned during a steadyflow combustion process Heat is lost to the surroundings at 300 K at a rate of 1120 kW The entropy of the reactants entering per unit time is 17 kWK and that of the products is 15 kWK The total rate of exergy destruction during this combustion process is a 520 kW b 600 kW c 1120 kW d 340 kW e 739 kW Design and Essay Problems 15134 Obtain the following information about a power plant that is closest to your town the net power output the type and amount of fuel the power consumed by the pumps fans and other auxiliary equipment stack gas losses and the rate of heat rejection at the condenser Using these data deter mine the rate of heat loss from the pipes and other compo nents and calculate the thermal efficiency of the plant 15135 A promising method of power generation by direct energy conversion is through the use of magnetohydrodynamic MHD generators Write an essay on the current status of MHD generators Explain their operating principles and how they differ from conventional power plants Discuss the prob lems that need to be overcome before MHD generators can become economical 15136 What is oxygenated fuel How would the heating value of oxygenated fuels compare to those of comparable hydrocarbon fuels on a unitmass basis Why is the use of oxygenated fuels mandated in some major cities in winter months 15137 Constantvolume vessels that contain flammable mixtures of hydrocarbon vapors and air at low pressures are often used Although the ignition of such mixtures is very unlikely as there is no source of ignition in the tank the Safety Final PDF to printer cen22672ch15747790indd 789 110917 1154 AM 789 CHAPTER 15 and Design Codes require that the tank withstand four times the pressure that may occur should an explosion take place in the tank For operating gauge pressures under 25 kPa deter mine the pressure for which these vessels must be designed in order to meet the requirements of the codes for a acetylene C2H2g b propane C3H8g and c noctane C8H18g Jus tify any assumptions that you make 15138 An electric utility uses a Pennsylvania coal that has an ultimate analysis by mass of 8436 percent C 189 per cent H2 440 percent O2 063 percent N2 089 percent S and 783 percent ash noncombustibles as fuel for its boilers The utility is changing from the Pennsylvania coal to an Illinois coal which has an ultimate analysis by mass of 6740 per cent C 531 percent H2 1511 percent O2 144 percent N2 236 percent S and 838 percent ash noncombustibles as fuel for its boilers With the Pennsylvania coal the boilers used 15 percent excess air Develop a schedule for the new coal show ing the heat released the smokestack dewpoint temperature adiabatic flame temperature and carbon dioxide production for various amounts of excess air Use this schedule to deter mine how to operate with the new coal as closely as possible to the conditions of the old coal Is there anything else that will have to be changed to use the new coal 15139 The safe disposal of hazardous waste material is a major environmental concern for industrialized societies and creates challenging problems for engineers The disposal methods commonly used include landfilling burying in the ground recycling and incineration or burning Incineration is frequently used as a practical means for the disposal of com bustible waste such as organic materials The EPA regulations require that the waste material be burned almost completely above a specified temperature without polluting the environ ment Maintaining the temperature above a certain level typi cally about 1100C necessitates the use of a fuel when the combustion of the waste material alone is not sufficient to obtain the minimum specified temperature A certain industrial process generates a liquid solution of etha nol and water as the waste product at a rate of 10 kgs The mass fraction of ethanol in the solution is 02 This solution is to be burned using methane CH4 in a steadyflow combustion cham ber Propose a combustion process that will accomplish this task with a minimal amount of methane State your assumptions Final PDF to printer cen22672ch15747790indd 790 110917 1154 AM Final PDF to printer cen22672ch16791822indd 791 110917 1223 PM 791 CHAPTER 16 C H E M I CA L A N D PHASE EQUI L I B R I U M I n Chap 15 we analyzed combustion processes under the assumption that combustion is complete when there is sufficient time and oxygen Often this is not the case however A chemical reaction may reach a state of equilibrium before reaching completion even when there is sufficient time and oxygen A system is said to be in equilibrium if no changes occur within the system when it is isolated from its surroundings An isolated system is in mechani cal equilibrium if no changes occur in pressure in thermal equilibrium if no changes occur in temperature in phase equilibrium if no transforma tions occur from one phase to another and in chemical equilibrium if no changes occur in the chemical composition of the system The conditions of mechanical and thermal equilibrium are straightforward but the conditions of chemical and phase equilibrium can be rather involved The equilibrium criterion for reacting systems is based on the second law of thermodynamics more specifically the increase of entropy principle For adiabatic systems chemical equilibrium is established when the entropy of the reacting system reaches a maximum Most reacting systems encountered in practice are not adiabatic however Therefore we need to develop an equi librium criterion applicable to any reacting system In this chapter we develop a general criterion for chemical equilibrium and apply it to reacting idealgas mixtures We then extend the analysis to simultaneous reactions Finally we discuss phase equilibrium for nonreacting systems OBJECTIVES The objectives of Chapter 16 are to Develop the equilibrium criterion for reacting systems based on the second law of thermodynamics Develop a general criterion for chemical equilibrium applicable to any reacting system based on minimizing the Gibbs function for the system Define and evaluate the chemical equilibrium constant Apply the general criterion for chemical equilibrium analysis to reacting idealgas mixtures Apply the general criterion for chemical equilibrium analysis to simultaneous reactions Relate the chemical equilibrium constant to the enthalpy of reaction Establish the phase equilibrium for nonreacting systems in terms of the specific Gibbs function of the phases of a pure substance Apply the Gibbs phase rule to determine the number of independent variables associated with a multicomponent multiphase system Apply Henrys law and Raoults law for gases dissolved in liquids Final PDF to printer 792 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 792 110917 1223 PM 161 CRITERION FOR CHEMICAL EQUILIBRIUM Consider a reaction chamber that contains a mixture of CO O2 and CO2 at a specified temperature and pressure Let us try to predict what will happen in this chamber Fig 161 Probably the first thing that comes to mind is a chemical reaction between CO and O2 to form more CO2 CO 1 2 O 2 CO 2 This reaction is certainly a possibility but it is not the only possibility It is also possible that some CO2 in the combustion chamber will dissociate into CO and O2 Yet a third possibility would be to have no reactions among the three components at all that is for the system to be in chemical equilibrium It appears that although we know the temperature pressure and composition thus the state of the system we are unable to predict whether the system is in chemical equilibrium In this chapter we develop the necessary tools to correct this Assume that the CO O2 and CO2 mixture mentioned above is in chemical equilibrium at the specified temperature and pressure The chemical composi tion of this mixture does not change unless the temperature or the pressure of the mixture is changed That is a reacting mixture in general has different equilibrium compositions at different pressures and temperatures Therefore when developing a general criterion for chemical equilibrium we consider a reacting system at a fixed temperature and pressure Taking the positive direction of heat transfer to be to the system the increase of entropy principle for a reacting or nonreacting system was expressed in Chap 7 as d S sys δQ T 161 A system and its surroundings form an adiabatic system and for such systems Eq 161 reduces to dSsys 0 That is a chemical reaction in an adiabatic chamber proceeds in the direction of increasing entropy When the entropy reaches a maximum the reaction stops Fig 162 Therefore entropy is a very useful property in the analysis of reacting adiabatic systems When a reacting system involves heat transfer the increase of entropy principle relation Eq 161 becomes impractical to use however since it requires a knowledge of heat transfer between the system and its surround ings A more practical approach would be to develop a relation for the equi librium criterion in terms of the properties of the reacting system only Such a relation is developed here Consider a reacting or nonreacting simple compressible system of fixed mass with only quasiequilibrium work modes at a specified temperature T and pressure P Fig 163 Combining the first and the secondlaw relations for this system gives δQ P dV dU dS δQ T dU P dV T dS 0 162 FIGURE 161 A reaction chamber that contains a mixture of CO2 CO and O2 at a specified temperature and pressure CO2 CO O2 O2 O2 CO2 CO2 CO CO FIGURE 162 Equilibrium criterion for a chemical reaction that takes place adiabatically 100 products Violation of second law S Equilibrium composition 100 reactants dS 0 dS 0 dS 0 FIGURE 163 A control mass undergoing a chemical reaction at a specified temperature and pressure dWb Reaction chamber Control mass T P δQ δWb Final PDF to printer 793 CHAPTER 16 cen22672ch16791822indd 793 110917 1223 PM The differential of the Gibbs function G H TS at constant temperature and pressure is dG TP dH T dS S dT dU P dV V dP T dS S d T dU P dV T dS 163 From Eqs 162 and 163 we have dGTP 0 Therefore a chemical reac tion at a specified temperature and pressure proceeds in the direction of a decreasing Gibbs function The reaction stops and chemical equilibrium is established when the Gibbs function attains a minimum value Fig 164 Therefore the criterion for chemical equilibrium can be expressed as dG TP 0 164 A chemical reaction at a specified temperature and pressure cannot proceed in the direction of the increasing Gibbs function since this will be a violation of the second law of thermodynamics Notice that if the temperature or the pres sure is changed the reacting system will assume a different equilibrium state which is the state of the minimum Gibbs function at the new temperature or pressure To obtain a relation for chemical equilibrium in terms of the properties of the individual components consider a mixture of four chemical compo nents A B C and D that exist in equilibrium at a specified temperature and pressure Let the number of moles of the respective components be NA NB NC and ND Now consider a reaction that occurs to an infinitesimal extent during which differential amounts of A and B reactants are converted to C and D products while the temperature and the pressure remain constant Fig 165 d N A A d N B B d N c C d N D D The equilibrium criterion Eq 164 requires that the change in the Gibbs function of the mixture during this process be equal to zero That is dG TP d G i TP g i d N i TP 0 165 or g C d N C g D d N D g A d N A g B d N B 0 166 where the g s are the molar Gibbs functions also called the chemical potentials at the specified temperature and pressure and the dNs are the dif ferential changes in the number of moles of the components To find a relation between the dNs we write the corresponding stoichio metric theoretical reaction ν A A ν B B ν C C ν D D 167 where the νs are the stoichiometric coefficients which are evaluated easily once the reaction is specified The stoichiometric reaction plays an impor tant role in the determination of the equilibrium composition of the reacting 0 0 FIGURE 164 Criterion for chemical equilibrium for a fixed mass at a specified temperature and pressure 100 products Violation of second law G Equilibrium composition 100 reactants dG 0 dG 0 dG 0 FIGURE 165 An infinitesimal reaction in a chamber at constant temperature and pressure Reaction chamber T P NA moles of A NB moles of B NC moles of C ND moles of D dNAA dNBB dNCC dNDD Final PDF to printer 794 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 794 110917 1223 PM mixtures because the changes in the number of moles of the components are proportional to the stoichiometric coefficients Fig 166 That is d N A ε ν A d N C ε ν C d N B ε ν B d N D ε ν D 168 where ε is the proportionality constant and represents the extent of a reaction A minus sign is added to the first two terms because the number of moles of the reactants A and B decreases as the reaction progresses For example if the reactants are C2H6 and O2 and the products are CO2 and H2O the reaction of 1 μmol 106 mol of C2H6 results in a 2μmol increase in CO2 a 3μmol increase in H2O and a 35μmol decrease in O2 in accordance with the stoichiometric equation C 2 H 6 35 O 2 2 CO 2 3 H 2 O That is the change in the number of moles of a component is onemillionth ε 106 of the stoichiometric coefficient of that component in this case Substituting the relations in Eq 168 into Eq 166 and canceling ε we obtain ν C g C ν D g D ν A g A ν B g B 0 169 This equation involves the stoichiometric coefficients and the molar Gibbs functions of the reactants and the products and it is known as the criterion for chemical equilibrium It is valid for any chemical reaction regardless of the phases involved Equation 169 is developed for a chemical reaction that involves two reac tants and two products for simplicity but it can easily be modified to handle chemical reactions with any number of reactants and products Next we ana lyze the equilibrium criterion for idealgas mixtures 162 THE EQUILIBRIUM CONSTANT FOR IDEALGAS MIXTURES Consider a mixture of ideal gases that exists in equilibrium at a specified temperature and pressure Like entropy the Gibbs function of an ideal gas depends on both the temperature and the pressure The Gibbs function values are usually listed versus temperature at a fixed reference pressure P0 which is taken to be 1 atm The variation of the Gibbs function of an ideal gas with pressure at a fixed temperature is determined by using the definition of the Gibbs function g h T s and the entropychange relation for isothermal processes Δ s R u ln P 2 P 1 It yields Δ g T Δ h 0 T Δ s T T Δ s T R u T ln P 2 P 1 Thus the Gibbs function of component i of an idealgas mixture at its partial pressure Pi and mixture temperature T can be expressed as g i T P i g i T R u T ln P i 1610 FIGURE 166 The changes in the number of moles of the components during a chemical reaction are proportional to the stoichiometric coefficients regardless of the extent of the reaction 01H2 H2 001H2 0001H2 02H 002H 0002H νH2 1 νH 2 2H Final PDF to printer 795 CHAPTER 16 cen22672ch16791822indd 795 110917 1223 PM where g i T represents the Gibbs function of component i at 1 atm pressure and temperature T and Pi represents the partial pressure of component i in atmospheres Substituting the Gibbs function expression for each component into Eq 169 we obtain ν C g C T R u T ln P C ν D g D T R u T ln P D ν A g A T R u T ln P A ν B g B T R u T ln P B 0 For convenience we define the standardstate Gibbs function change as ΔGT ν C g C T ν D g D T ν A g A T ν B g B T 0 1611 Substituting we get ΔGT R u T ν C ln P C ν D ln P D ν A ln P A ν B ln P B R u T ln P C ν C P D ν D P A ν A P B ν B 1612 Now we define the equilibrium constant KP for the chemical equilibrium of idealgas mixtures as K P P C ν C P D ν D P A ν A P B ν B 1613 Substituting into Eq 1612 and rearranging we obtain K P e Δ G T R u T 1614 Therefore the equilibrium constant KP of an idealgas mixture at a specified temperature can be determined from a knowledge of the standardstate Gibbs function change at the same temperature The KP values for several reactions are given in Table A28 Once the equilibrium constant is available it can be used to determine the equilibrium composition of reacting idealgas mixtures This is accomplished by expressing the partial pressures of the components in terms of their mole fractions P i y i P N i N total P where P is the total pressure and Ntotal is the total number of moles pres ent in the reaction chamber including any inert gases Replacing the par tial pressures in Eq 1613 by the above relation and rearranging we obtain Fig 167 K P N C ν C N D ν D N A ν A N B ν B P N total Δν 1615 where Δν ν C ν D ν A ν B Equation 1615 is written for a reaction involving two reactants and two prod ucts but it can be extended to reactions involving any number of reactants and products FIGURE 167 Three equivalent KP relations for reacting idealgas mixtures 1 In terms of partial pressures KP PC PD C D PA PB A B 3 In terms of the equilibrium composition KP NC ND NA NB A B Ntotal P 2 In terms of GT KP e GTRuT C D ν ν ν ν ν ν ν ν Δ Δ Δν Final PDF to printer 796 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 796 110917 1223 PM EXAMPLE 161 Equilibrium Constant of a Dissociation Process Using Eq 1614 and the Gibbs function data determine the equilibrium constant KP for the dissociation process N2 2N at 25C Compare your result to the KP value listed in Table A28 SOLUTION The equilibrium constant of the reaction N2 2N is listed in Table A28 at different temperatures It is to be verified using Gibbs function data Assumptions 1 The constituents of the mixture are ideal gases 2 The equilibrium mixture consists of N2 and N only Properties The equilibrium constant of this reaction at 298 K is ln KP 3675 Table A28 The Gibbs function of formation at 25C and 1 atm is 0 for N2 and 455510 kJkmol for N Table A26 Analysis In the absence of KP tables KP can be determined from the Gibbs function data and Eq 1614 K P e ΔG T R u T where from Eq 1611 ΔG T ν N g N T ν N 2 g N 2 T 2 455510 kJkmol 0 911020 kJkmol Substituting we find ln K P 911020 kJkmol 8314 kJkmolK 29815 K 3675 or K P 2 10 160 The calculated KP value is in agreement with the value listed in Table A28 The KP value for this reaction is practically zero indicating that this reaction will not occur at this temperature Discussion Note that this reaction involves one product N and one reactant N2 and the stoichiometric coefficients for this reaction are νN 2 and νN2 1 Also note that the Gibbs function of all stable elements such as N2 is assigned a value of zero at the stan dard reference state of 25C and 1 atm The Gibbs function values at other temperatures can be calculated from the enthalpy and absolute entropy data by using the definition of the Gibbs function g T h T T s T where h T h f h T h 298 K EXAMPLE 162 Producing Hydrogen by Heating Water Vapor to High Temperature As an alternative to electrolysis hydrogen gas can be produced thermally in accor dance with the dissociation process H 2 O H 2 1 2 O 2 by heating water vapor to very high temperatures Fig 168 Determine the percentage of water vapor that can be separated into hydrogen and oxygen when this reaction occurs at 4000 K and 10 kPa FIGURE 168 Schematic for Example 162 Initial composition 1 kmol H2O Equilibrium composition at 4000 K 10 kPa x H2O y H2 z O2 Final PDF to printer 797 CHAPTER 16 cen22672ch16791822indd 797 110917 1223 PM SOLUTION The reaction H2O H2 1 2 O2 is considered at a specified tem perature and pressure The percentage of water vapor that separates into hydrogen and oxygen is to be determined Assumptions 1 The equilibrium composition consists of H2O H2 and O2 only and dis sociation into H OH and O is negligible 2 The constituents of the mixture are ideal gases Analysis This is a dissociation process that is significant at very high temperatures only For simplicity we consider 1 kmol of H2O The stoichiometric and actual reactions in this case are as follows Stoichiometric H 2 O H 2 1 2 O 2 thus ν H 2 O 1 ν H 2 1 and ν O 2 05 Actual H 2 O x H 2 O reactants leftover y H 2 z O 2 products H balance 2 2x 2y y 1 x O balance 1 x 2z z 1 x 2 Total number of moles N total x y z 15 05x Pressure in atm P 10 kPa 009869 atm since 1 atm 101325 kPa The equilibrium constant for the reaction H2O H2 1 2 O2 at 4000 K is given in Table A28 to be ln KP 0542 and thus KP 05816 Assuming ideal gas behavior for all components in equilibrium composition the equilibrium constant relation in terms of mole numbers can be expressed in this case as K P N H 2 ν H 2 N O 2 ν O 2 N H 2 O ν H 2 O P N total ν H 2 ν O 2 ν H 2 O Substituting 05816 1 x 1 x 2 12 x 009869 15 05x 1 05 1 Using an equation solver or by trial and error the unknown x is determined to be x 0222 That is for each mole of H2O entering the reaction chamber there is only 0222 mole of H2O leaving Therefore the fraction of water vapor that dissociated into hydrogen and oxygen when heated to 4000 K is Fraction of dissociation 1 x 1 0222 0778 or 778 Therefore hydrogen can be produced at significant rates by heating water vapor to sufficiently high temperatures Discussion The dissociation of H2O into atomic H O and the compound OH can be significant at high temperatures and thus the first assumption is very simplistic This problem can be solved more realistically by considering all possible reactions that are likely to occur simultaneously as discussed later in this chapter A double arrow is used in equilibrium equations as an indication that a chemical reaction does not stop when chemical equilibrium is established rather it proceeds in both directions at the same rate That is at equilibrium the reactants are depleted at exactly the same rate as they are replenished from the products by the reverse reaction Final PDF to printer 798 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 798 110917 1223 PM 163 SOME REMARKS ABOUT THE KP OF IDEALGAS MIXTURES In Sec 162 we developed three equivalent expressions for the equilibrium constant KP of reacting idealgas mixtures Eq 1613 which expresses KP in terms of partial pressures Eq 1614 which expresses KP in terms of the standardstate Gibbs function change ΔGT and Eq 1615 which expresses KP in terms of the number of moles of the components All three relations are equivalent but sometimes one is more convenient to use than the others For example Eq 1615 is best suited for determining the equilibrium composition of a reacting idealgas mixture at a specified temperature and pressure On the basis of these relations we may draw the following conclu sions about the equilibrium constant KP of idealgas mixtures 1 The KP of a reaction depends on temperature only It is independent of the pressure of the equilibrium mixture and is not affected by the presence of inert gases This is because KP depends on ΔGT which depends on tem perature only and the ΔGT of inert gases is zero see Eq 1614 Thus at a specified temperature the following four reactions have the same KP value H 2 1 2 O 2 H 2 O at 1 atm H 2 1 2 O 2 H 2 O at 5 atm H 2 1 2 O 2 3 N 2 H 2 O 3 N 2 at 3 atm H 2 2 O 2 5 N 2 H 2 O 15 O 2 5 N 2 at 2 atm 2 The KP of the reverse reaction is 1KP This is easily seen from Eq 1613 For reverse reactions the products and reactants switch places and thus the terms in the numerator move to the denominator and vice versa Consequently the equilibrium constant of the reverse reaction becomes 1KP For example from Table A28 K P 01147 10 11 for H 2 1 2 O 2 H 2 O at 1000 K K P 8718 10 11 for H 2 O H 2 1 2 O 2 at 1000 K 3 The larger the KP the more complete the reaction This is also apparent from Fig 169 and Eq 1613 If the equilibrium composition consists largely of product gases the partial pressures of the products PC and PD are con siderably larger than the partial pressures of the reactants PA and PB which results in a large value of KP In the limiting case of a complete reaction no leftover reactants in the equilibrium mixture KP approaches infinity Con versely very small values of KP indicate that a reaction does not proceed to any appreciable degree Thus reactions with very small KP values at a speci fied temperature can be neglected A reaction with KP 1000 or ln KP 7 is usually assumed to proceed to completion and a reaction with KP 0001 or ln KP 7 is assumed not to occur at all For example ln KP 68 for the reaction N2 2N at 5000 K FIGURE 169 The larger the KP the more complete the reaction 4000 5000 1000 2000 3000 T K P 1 atm 6000 517 1018 265 106 2545 4147 0025 2677 7680 9770 000 016 1463 9963 KP mol H H2 2H Final PDF to printer 799 CHAPTER 16 cen22672ch16791822indd 799 110917 1223 PM Therefore the dissociation of N2 into monatomic nitrogen N can be disre garded at temperatures below 5000 K 4 The mixture pressure affects the equilibrium composition although it does not affect the equilibrium constant KP This can be seen from Eq 1615 which involves the term PΔν where Δν ΣνP ΣνR the difference between the number of moles of products and the number of moles of reactants in the stoichiometric reaction At a specified temperature the KP value of the reac tion and thus the righthand side of Eq 1615 remains constant Therefore the mole numbers of the reactants and the products must change to counteract any changes in the pressure term The direction of the change depends on the sign of Δν An increase in pressure at a specified temperature increases the number of moles of the reactants and decreases the number of moles of prod ucts if Δν is positive have the opposite effect if Δν is negative and have no effect if Δν is zero 5 The presence of inert gases affects the equilibrium composition although it does not affect the equilibrium constant KP This can be seen from Eq 1615 which involves the term 1NtotalΔν where Ntotal is the total number of moles of the idealgas mixture at equilibrium including inert gases The sign of Δν determines how the presence of inert gases influences the equilib rium composition Fig 1610 An increase in the number of moles of inert gases at a specified temperature and pressure decreases the number of moles of the reactants and increases the number of moles of products if Δν is posi tive have the opposite effect if Δν is negative and have no effect if Δν is zero 6 When the stoichiometric coefficients are doubled the value of KP is squared Therefore when one is using KP values from a table the stoichio metric coefficients the νs used in a reaction must be exactly the same ones appearing in the table from which the KP values are selected Multiplying all the coefficients of a stoichiometric equation does not affect the mass balance but it does affect the equilibrium constant calculations since the stoichiomet ric coefficients appear as exponents of partial pressures in Eq 1613 For example For H 2 1 2 O 2 H 2 O K P 1 P H 2 O P H 2 P O 2 12 But for 2H 2 O 2 2 H 2 O K P 2 P H 2 O 2 P H 2 2 P O 2 K P 1 2 7 Free electrons in the equilibrium composition can be treated as an ideal gas At high temperatures usually above 2500 K gas molecules start to dis sociate into unattached atoms such as H2 2H and at even higher tem peratures atoms start to lose electrons and ionize for example H H e 1616 The dissociation and ionization effects are more pronounced at low pressures Ionization occurs to an appreciable extent only at very high temperatures and the mixture of electrons ions and neutral atoms can be treated as an ideal gas Therefore the equilibrium composition of ionized gas mixtures can be FIGURE 1610 The presence of inert gases does not affect the equilibrium constant but it does affect the equilibrium composition 1 mol H2 a b 1 mol N2 1 mol H2 KP 00251 0158 mol H 0921 mol H2 KP 00251 1240 mol H 0380 mol H2 1 mol N2 Initial composition Equilibrium composition at 3000 K 1 atm Final PDF to printer 800 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 800 110917 1223 PM determined from Eq 1615 Fig 1611 This treatment may not be adequate in the presence of strong electric fields however since the electrons may be at a different temperature than the ions in this case 8 Equilibrium calculations provide information on the equilibrium compo sition of a reaction not on the reaction rate Sometimes it may even take years to achieve the indicated equilibrium composition For example the equilibrium constant of the reaction H2 1 2 O2 H2O at 298 K is about 1040 which suggests that a stoichiometric mixture of H2 and O2 at room tem perature should react to form H2O and the reaction should go to completion However the rate of this reaction is so slow that it practically does not occur But when the right catalyst is used the reaction goes to completion rather quickly to the predicted value FIGURE 1611 Equilibriumconstant relation for the ionization reaction of hydrogen where Ntotal NH NH Ne KP H H e Δν νH νe νH 1 1 1 1 NH Ne Ntotal NH H e P H Δν ν ν ν FIGURE 1612 Schematic for Example 163 2 kmol CO 3 kmol O2 Initial composition Equilibrium composition at 2600 K 304 kPa x CO2 y CO z O2 EXAMPLE 163 Equilibrium Composition at a Specified Temperature A mixture of 2 kmol of CO and 3 kmol of O2 is heated to 2600 K at a pressure of 304 kPa Determine the equilibrium composition assuming the mixture consists of CO2 CO and O2 Fig 1612 SOLUTION A reactive gas mixture is heated to a high temperature The equilib rium composition at that temperature is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO and O2 2 The con stituents of the mixture are ideal gases Analysis The stoichiometric and actual reactions in this case are as follows Stoichiometric CO 1 2 O 2 CO 2 thus ν CO 2 1 ν CO 1 and ν O 2 1 2 Actual 2CO 3O 2 x CO 2 products y CO z O 2 reactants leftover C balance 2 x y or y 2 x O balance 8 2x y 2z or z 3 x 2 Total number of moles N total x y z 5 x 2 Pressure P 304 kPa 30 atm The closest reaction listed in Table A28 is CO 2 CO 1 2 O 2 for which ln KP 2801 at 2600 K The reaction we have is the inverse of this and thus ln KP 2801 or KP 16461 in our case Assuming idealgas behavior for all components the equilibrium constant relation Eq 1615 becomes K P N CO 2 ν CO 2 N CO ν CO N O 2 ν O 2 P N total ν CO 2 ν CO ν O 2 Substituting we get 16461 x 2 x 3 x 2 12 3 5 x 2 12 Solving for x yields x 1906 Final PDF to printer 801 CHAPTER 16 cen22672ch16791822indd 801 110917 1223 PM Then y 2 x 0094 z 3 x 2 2047 Therefore the equilibrium composition of the mixture at 2600 K and 304 kPa is 1906 CO 2 00940CO 2 047O 2 Discussion In solving this problem we disregarded the dissociation of O2 into O according to the reaction O2 2O which is a real possibility at high temperatures This is because ln KP 7521 at 2600 K for this reaction which indicates that the amount of O2 that dissociates into O is negligible Besides we have not learned how to deal with simultaneous reactions yet We will do so in the next section FIGURE 1613 Schematic for Example 164 Initial composition 3 kmol CO 25 kmol O2 8 kmol N2 Equilibrium composition at 2600 K 5 atm x CO2 y CO z O2 8 N2 EXAMPLE 164 Effect of Inert Gases on Equilibrium Composition A mixture of 3 kmol of CO 25 kmol of O2 and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm Determine the equilibrium composition of the mixture Fig 1613 SOLUTION A gas mixture is heated to a high temperature The equilibrium com position at the specified temperature is to be determined Assumptions 1 The equilibrium composition consists of CO2 CO O2 and N2 2 The constituents of the mixture are ideal gases Analysis This problem is similar to Example 163 except that it involves an inert gas N2 At 2600 K some possible reactions are O2 2O ln KP 7521 N2 2N ln KP 28304 1 2 O2 1 2 N2 NO ln KP 2671 and CO 1 2 O2 CO2 ln KP 2801 or KP 16461 Based on these KP values we conclude that the O2 and N2 will not dissociate to any appreciable degree but a small amount will combine to form some oxides of nitrogen We disregard the oxides of nitrogen in this example but they should be considered in a more refined analysis We also conclude that most of the CO will combine with O2 to form CO2 Notice that despite the changes in pressure the num ber of moles of CO and O2 and the presence of an inert gas the KP value of the reaction is the same as that used in Example 163 The stoichiometric and actual reactions in this case are Stoichiometric CO 1 2 O 2 CO 2 thus ν CO 2 1 ν CO 1 and ν O 2 1 2 Actual 3CO 2 5O 2 8 N 2 x CO 2 products y CO z O 2 reactants leftover 8 N 2 insert C balance 3 x y or y 3 x O balance 8 2x y 2z or z 25 x 2 Total number of moles N total x y z 8 135 x 2 Assuming idealgas behavior for all components the equilibrium constant relation Eq 1615 becomes K P N CO 2 ν CO 2 N CO ν CO N O 2 ν O 2 P N total ν CO 2 ν CO ν O 2 Final PDF to printer 802 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 802 110917 1223 PM Substituting we get 16461 x 3 x 25 x 2 12 5 135 x 2 12 Solving for x yields x 2754 Then y 3 x 0246 z 25 x 2 1123 Therefore the equilibrium composition of the mixture at 2600 K and 5 atm is 2754 CO 2 0246CO 1 123O 2 8 N 2 Discussion Note that the inert gases do not affect the KP value or the KP relation for a reaction but they do affect the equilibrium composition 164 CHEMICAL EQUILIBRIUM FOR SIMULTANEOUS REACTIONS The reacting mixtures we have considered so far involved only one reaction and writing a KP relation for that reaction was sufficient to determine the equilibrium composition of the mixture However most practical chemical reactions involve two or more reactions that occur simultaneously which makes them more difficult to deal with In such cases it becomes necessary to apply the equilibrium criterion to all possible reactions that may occur in the reaction chamber When a chemical species appears in more than one reac tion the application of the equilibrium criterion together with the mass bal ance for each chemical species results in a system of simultaneous equations from which the equilibrium composition can be determined We have shown earlier that a reacting system at a specified tempera ture and pressure achieves chemical equilibrium when its Gibbs function reaches a minimum value that is dGTP 0 This is true regardless of the number of reactions that may be occurring When two or more reactions are involved this condition is satisfied only when dGTP 0 for each reaction Assuming idealgas behavior the KP of each reaction can be determined from Eq 1615 with Ntotal being the total number of moles present in the equilibrium mixture The determination of the equilibrium composition of a reacting mixture requires that we have as many equations as unknowns where the unknowns are the number of moles of each chemical species present in the equilibrium mixture The mass balance of each element involved provides one equation The rest of the equations must come from the KP relations written for each reaction Thus we conclude that the number of KP relations needed to deter mine the equilibrium composition of a reacting mixture is equal to the number of chemical species minus the number of elements present in equilibrium For an equilibrium mixture that consists of CO2 CO O2 and O for example Final PDF to printer 803 CHAPTER 16 cen22672ch16791822indd 803 110917 1223 PM two KP relations are needed to determine the equilibrium composition since it involves four chemical species and two elements Fig 1614 The determination of the equilibrium composition of a reacting mixture in the presence of two simultaneous reactions is presented here with an example FIGURE 1614 The number of KP relations needed to determine the equilibrium composition of a reacting mixture is the difference between the number of species and the number of elements Composition CO2 CO O2 O No of components 4 No of elements 2 No of KP relations needed 4 2 2 EXAMPLE 165 Equilibrium Composition for Simultaneous Reactions A mixture of 1 kmol of H2O and 2 kmol of O2 is heated to 4000 K at a pressure of 1 atm Determine the equilibrium composition of this mixture assuming that only H2O OH O2 and H2 are present Fig 1615 SOLUTION A gas mixture is heated to a specified temperature at a specified pres sure The equilibrium composition is to be determined Assumptions 1 The equilibrium composition consists of H2O OH O2 and H2 2 The constituents of the mixture are ideal gases Analysis The chemical reaction during this process can be expressed as H 2 O 2 O 2 x H 2 O y H 2 z O 2 wOH Mass balances for hydrogen and oxygen yield H balance 2 2x 2y w 1 O balance 5 x 2z w 2 The mass balances provide us with only two equations with four unknowns and thus we need to have two more equations to be obtained from the KP relations to deter mine the equilibrium composition of the mixture It appears that part of the H2O in the products is dissociated into H2 and OH during this process according to the stoichiometric reactions H 2 O H 2 1 2 O 2 reaction 1 H 2 O 1 2 H 2 OH reaction 2 The equilibrium constants for these two reactions at 4000 K are determined from Table A28 to be ln K P 1 0542 K P 1 05816 ln K P 2 0044 K P 2 09570 The KP relations for these two simultaneous reactions are K P 1 N H 2 ν H 2 N O 2 ν O 2 N H 2 O ν H 2 O P N total ν H 2 ν O 2 ν H 2 O K P 2 N H 2 ν H 2 N OH ν OH N H 2 O ν H 2 O P N total ν H 2 ν OH ν H 2 O where N total N H 2 O N H 2 N O 2 N OH x y z w FIGURE 1615 Schematic for Example 165 Equilibrium composition at 4000 K 1 atm x H2O y H2 z O2 w OH Initial composition 1 kmol H2O 2 kmol O2 Final PDF to printer 804 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 804 110917 1223 PM Substituting yields 05816 y z 12 x 1 x y z w 12 3 09570 w y 12 x 1 x y z w 12 4 Solving Eqs 1 2 3 and 4 simultaneously for the four unknowns x y z and w yields x 0271 y 0213 z 1849 w 1032 Therefore the equilibrium composition of 1 kmol H2O and 2 kmol O2 at 1 atm and 4000 K is 0271 H 2 O 0 213H 2 1849 O 2 1032OH Discussion We could also solve this problem by using the KP relation for the stoichio metric reaction O2 2O as one of the two equations Solving a system of simultaneous nonlinear equations is extremely tedious and timeconsuming if it is done by hand Thus it is often necessary to solve these kinds of problems by using an equation solver 165 VARIATION OF KP WITH TEMPERATURE It was shown in Sec 162 that the equilibrium constant KP of an ideal gas mixture depends on temperature only and it is related to the standardstate Gibbs function change ΔGT through the relation Eq 1614 ln K P ΔGT R u T In this section we develop a relation for the variation of KP with temperature in terms of other properties Substituting ΔGT ΔHT T ΔST into the preceding relation and differentiating with respect to temperature we get d ln K P dT ΔH T R u T 2 d ΔH T R u T dT d ΔST R u dT At constant pressure the second T ds relation T ds dh v dP reduces to T ds dh Also T dΔS dΔH since ΔS and ΔH consist of entropy and enthalpy terms of the reactants and the products Therefore the last two terms in the preceding relation cancel and it reduces to d ln K P dT ΔHT R u T 2 h R T R u T 2 1617 where h R T is the enthalpy of reaction at temperature T Notice that we dropped the superscript which indicates a constant pressure of 1 atm from ΔHT since the enthalpy of an ideal gas depends on temperature only and is independent of pressure Equation 1617 is an expression of the Final PDF to printer 805 CHAPTER 16 cen22672ch16791822indd 805 110917 1223 PM variation of KP with temperature in terms of h R T and it is known as the vant Hoff equation To integrate it we need to know how h R varies with T For small temperature intervals h R can be treated as a constant and Eq 1617 can be integrated to yield ln K P 2 K P 1 h R R u 1 T 1 1 T 2 1618 This equation has two important implications First it provides a means of calculating the h R of a reaction from a knowledge of KP which is eas ier to deter mine Second it shows that exothermic reactions h R 0 such as combustion processes are less complete at higher temperatures since KP decreases with temperature for such reactions Fig 1616 FIGURE 1616 Exothermic reactions are less complete at higher temperatures 4000 1000 2000 3000 T K KP Reaction C O2 CO2 478 1020 225 1010 780 106 141 105 EXAMPLE 166 The Enthalpy of Reaction of a Combustion Process Estimate the enthalpy of reaction h R for the combustion process of hydrogen H2 05O2 H2O at 2000 K using a enthalpy data and b KP data SOLUTION The h R at a specified temperature is to be determined using the enthalpy and Kp data Assumptions Both the reactants and the products are ideal gases Analysis a The h R of the combustion process of H2 at 2000 K is the amount of energy released as 1 kmol of H2 is burned in a steadyflow combustion chamber at a temperature of 2000 K It can be determined from Eq 156 h R N p h f h h p N r h f h h r N H 2 O h f h 2000 K h 298 K H 2 O N H 2 h f h 2000 K h 298 K H 2 N O 2 h f h 2000 K h 298 K O 2 Substituting yields h R 1 kmol H 2 O 241820 82593 9904 kJkmol H 2 O 1 kmol H 2 0 61400 8468 kJkmol H 2 05 kmol O 2 0 67881 8682 kJkmol O 2 251663 kJkmol b The h R value at 2000 K can be estimated by using KP values at 1800 and 2200 K the closest two temperatures to 2000 K for which KP data are available from Table A28 They are K P 1 18509 at T1 1800 K and K P 2 8696 at T2 2200 K By substituting these values into Eq 1618 the h R value is determined to be ln K P 2 K P 1 h R R u 1 T 1 1 T 2 ln 8696 18509 h R 8314 kJkmolK 1 1800 K 1 2200 K h R 251698 kJkmol Discussion Despite the large temperature difference between T1 and T2 400 K the two results are almost identical The agreement between the two results would be even better if a smaller temperature interval were used Final PDF to printer 806 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 806 110917 1223 PM 166 PHASE EQUILIBRIUM We showed at the beginning of this chapter that the equilibrium state of a sys tem at a specified temperature and pressure is the state of the minimum Gibbs function and the equilibrium criterion for a reacting or nonreacting system was expressed as Eq 164 dG TP 0 In the preceding sections we applied the equilibrium criterion to reacting sys tems In this section we apply it to nonreacting multiphase systems We know from experience that a wet Tshirt hanging in an open area even tually dries a small amount of water left in a glass evaporates and the after shave in an open bottle quickly disappears Fig 1617 These examples suggest that there is a driving force between the two phases of a substance that forces the mass to transform from one phase to another The magnitude of this force depends among other things on the relative concentrations of H2O in the two phases A wet Tshirt dries much more quickly in dry air than it does in humid air In fact it does not dry at all if the relative humidity of the environment is 100 percent In this case there is no transformation from the liquid phase to the vapor phase and the two phases are in phase equilibrium The conditions of phase equilibrium change however if the temperature or the pressure is changed Therefore we examine phase equilibrium at a speci fied temperature and pressure Phase Equilibrium for a SingleComponent System The equilibrium criterion for two phases of a pure substance such as water is easily developed by considering a mixture of saturated liquid and saturated vapor in equilibrium at a specified temperature and pressure such as that shown in Fig 1618 The total Gibbs function of this mixture is G m f g f m g g g where gf and gg are the Gibbs functions of the liquid and vapor phases per unit mass respectively Now imagine a disturbance during which a differential amount of liquid dmf evaporates at constant temperature and pressure The change in the total Gibbs function during this disturbance is dG TP g f d m f g g d m g since gf and gg remain constant at constant temperature and pressure At equi librium dGTP 0 Also from the conservation of mass dmg dmf Sub stituting we obtain dG TP g f g g d m f which must be equal to zero at equilibrium It yields g f g g 1619 Therefore the two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function Also at the triple point FIGURE 1617 Wet clothes hung in an open area eventually dry as a result of mass transfer from the liquid phase to the vapor phase C Squared StudiosGetty Images RF FIGURE 1618 A liquidvapor mixture in equilibrium at a constant temperature and pressure T P Vapor mg mf Liquid Final PDF to printer 807 CHAPTER 16 cen22672ch16791822indd 807 110917 1223 PM the state at which all three phases coexist in equilibrium the specific Gibbs functions of all three phases are equal to each other What happens if gf gg Obviously the two phases are not in equilibrium at that moment The second law requires that dGTP gf gg dmf 0 Thus dmf must be negative which means that some liquid must vaporize until gf gg Therefore the Gibbs function difference is the driving force for phase change just as the temperature difference is the driving force for heat transfer EXAMPLE 167 Phase Equilibrium for a Saturated Mixture Calculate the value of the Gibbs function for saturated refrigerant134a at 30F as a saturated liquid a saturated vapor and a mixture of liquid and vapor with a quality of 30 percent Demonstrate that phase equilibrium exists SOLUTION The values of the Gibbs function for saturated refrigerant134a at 30F as a saturated liquid a saturated vapor and a mixture of liquid and vapor are to be calculated Properties The properties of saturated water at 30F are h f 3008 Btulbm h fg 95608 Btulbm h g 9862 Btu lbm s f 000707 BtulbmR s fg 022250 Btu lbmR and s g 022957 BtulbmR Table A11E Analysis The Gibbs function for the liquid phase is g f h f T s f 3008 Btulbm 430 R 000707 BtulbmR 0032 Btulbm For the vapor phase g g h g T s g 9862 Btulbm 430 R 022957 kJkgK 0095 Btulbm For the saturated mixture with a quality of 30 percent h h f x h fg 3008 Btulbm 03095608 Btulbm 3169 Btulbm s s f x s fg 000707 BtulbmR 030022250 BtulbmR 007382 BtulbmR g h Ts 3169 Btulbm 430 R 007382 BtulbmR 0053 Btulbm Discussion The three results are in close agreement They would match exactly if more accurate property data were used Therefore the criterion for phase equilibrium is satisfied The Phase Rule Notice that a singlecomponent twophase system may exist in equilibrium at different temperatures or pressures However once the temperature is fixed the system is locked into an equilibrium state and all intensive properties of each phase except their relative amounts are fixed Therefore a single component twophase system has one independent property which may be taken to be the temperature or the pressure In general the number of independent variables associated with a multicomponent multiphase system is given by the Gibbs phase rule expressed as IV C PH 2 1620 Final PDF to printer 808 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 808 110917 1223 PM where IV the number of independent variables C the number of compo nents and PH the number of phases present in equilibrium For the single component C 1 twophase PH 2 system discussed above for example one independent intensive property needs to be specified IV 1 Fig 1619 At the triple point however PH 3 and thus IV 0 That is none of the properties of a pure substance at the triple point can be varied Also based on this rule a pure substance that exists in a single phase PH 1 has two inde pendent variables In other words two independent intensive properties need to be specified to fix the equilibrium state of a pure substance in a single phase Phase Equilibrium for a Multicomponent System Many multiphase systems encountered in practice involve two or more com ponents A multicomponent multiphase system at a specified temperature and pressure is in phase equilibrium when there is no driving force between the different phases of each component Thus for phase equilibrium the specific Gibbs function of each component must be the same in all phases Fig 1620 That is g f1 g g1 g s1 for component 1 g f2 g g2 g s2 for component 2 g fN g gN g sN for component N We could also derive these relations by using mathematical vigor instead of physical arguments Some components may exist in more than one solid phase at the specified temperature and pressure In this case the specific Gibbs function of each solid phase of a component must also be the same for phase equilibrium In this section we examine the phase equilibrium of twocomponent sys tems that involve two phases liquid and vapor in equilibrium For such sys tems C 2 PH 2 and thus IV 2 That is a twocomponent twophase system has two independent variables and such a system will not be in equi librium unless two independent intensive properties are fixed In general the two phases of a twocomponent system do not have the same composition in each phase That is the mole fraction of a component is differ ent in different phases This is illustrated in Fig 1621 for the twophase mix ture of oxygen and nitrogen at a pressure of 01 MPa On this diagram the vapor line represents the equilibrium composition of the vapor phase at vari ous temperatures and the liquid line does the same for the liquid phase At 84 K for example the mole fractions are 30 percent nitrogen and 70 percent oxygen in the liquid phase and 66 percent nitrogen and 34 percent oxygen in the vapor phase Notice that y f N 2 y f O 2 030 070 1 1621a y g N 2 y g O 2 066 034 1 1621b Therefore once the temperature and pressure two independent variables of a twocomponent twophase mixture are specified the equilibrium FIGURE 1619 According to the Gibbs phase rule a singlecomponent twophase system can have only one independent variable Water vapor T 100C 150C 200C Liquid water FIGURE 1620 A multicomponent multiphase system is in phase equilibrium when the specific Gibbs function of each component is the same in all phases T P NH3 H2O Vapor gfNH3 ggNH3 Liquid NH3 H2O gfH2O ggH2O FIGURE 1621 Equilibrium diagram for the two phase mixture of oxygen and nitrogen at 01 MPa 100 O2 Vapor T K 0 Liquid Vapor Liquid 10 20 30 40 50 60 70 80 90 0 N2 100 90 80 70 60 50 40 30 20 10 74 773 78 82 86 90 94 902 Final PDF to printer 809 CHAPTER 16 cen22672ch16791822indd 809 110917 1223 PM composition of each phase can be determined from the phase diagram which is based on experimental measurements It is interesting to note that temperature is a continuous function but mole fraction which is a dimensionless concentration in general is not The water and air temperatures at the free surface of a lake for example are always the same The mole fractions of air on the two sides of a waterair interface however are obviously very different in fact the mole fraction of air in water is close to zero Likewise the mole fractions of water on the two sides of a waterair interface are also different even when air is saturated Fig 1622 Therefore when specifying mole fractions in twophase mixtures we need to clearly specify the intended phase In most practical applications the two phases of a mixture are not in phase equilibrium since the establishment of phase equilibrium requires the dif fusion of species from higher concentration regions to lower concentration regions which may take a long time However phase equilibrium always exists at the interface of two phases of a species In the case of an airwater interface the mole fraction of water vapor in the air is easily determined from saturation data as shown in Example 168 The situation is similar at solidliquid interfaces Again at a given tem perature only a certain amount of solid can be dissolved in a liquid and the solubility of the solid in the liquid is determined from the requirement that thermodynamic equilibrium exists between the solid and the solution at the interface The solubility represents the maximum amount of solid that can be dissolved in a liquid at a specified temperature and is widely available in chem istry handbooks In Table 161 we present sample solubility data for sodium chloride NaCl and calcium bicarbonate CaHCO32 at various temperatures For example the solubility of salt NaCl in water at 310 K is 365 kg per 100 kg of water Therefore the mass fraction of salt in the saturated brine is simply mf saltliquid side m salt m 365 kg 100 365 kg 0267 or 267 percent whereas the mass fraction of salt in the pure solid salt is mf 10 Many processes involve the absorption of a gas into a liquid Most gases are weakly soluble in liquids such as air in water and for such dilute solutions the mole fractions of a species i in the gas and liquid phases at the interface are observed to be proportional to each other That is yigas side yiliquid side or Pigas side Pyiliquid side since yi PiP for idealgas mixtures This is known as the Henrys law and is expressed as y iliquid side P igas side H 1622 where H is the Henrys constant which is the product of the total pressure of the gas mixture and the proportionality constant For a given species it is a function of temperature only and is practically independent of pressure for pressures under about 5 atm Values of the Henrys constant for a number of aqueous solutions are given in Table 162 for various temperatures From this table and the equation above we make the following observations 1 The concentration of a gas dissolved in a liquid is inversely proportional to Henrys constant Therefore the larger the Henrys constant the smaller the concentration of dissolved gases in the liquid FIGURE 1622 Unlike temperature the mole fraction of species on the two sides of a liquid gas or solidgas or solid liquid interface are usually not the same yH2Oliquid side 1 Air Water yH2Ogas side Jump in concentration Concentration profile x TABLE 161 Solubility of two inorganic compounds in water at various temperatures in kg in 100 kg of water Solute Temperature K Salt NaCl Calcium bicarbonate CaHCO32 27315 357 1615 280 358 1630 290 359 1653 300 362 1675 310 365 1698 320 369 1720 330 372 1743 340 376 1765 350 382 1788 360 388 1810 370 395 1833 37315 398 1840 From Handbook of Chemistry McGrawHill 1961 Final PDF to printer 810 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 810 110917 1223 PM 2 The Henrys constant increases and thus the fraction of a dissolved gas in the liquid decreases with increasing temperature Therefore the dissolved gases in a liquid can be driven off by heating the liquid Fig 1623 3 The concentration of a gas dissolved in a liquid is proportional to the partial pressure of the gas Therefore the amount of gas dissolved in a liquid can be increased by increasing the pressure of the gas This can be used to advantage in the carbonation of soft drinks with CO2 gas Strictly speaking the result obtained from Eq 1622 for the mole fraction of dissolved gas is valid for the liquid layer just beneath the interface but not necessarily the entire liquid The latter will be the case only when thermody namic phase equilibrium is established throughout the entire liquid body We mentioned earlier that the use of Henrys law is limited to dilute gas liquid solutions that is liquids with a small amount of gas dissolved in them Then the question that arises naturally is what do we do when the gas is highly soluble in the liquid or solid such as ammonia in water In this case the linear relationship of Henrys law does not apply and the mole fraction of a gas dissolved in the liquid or solid is usually expressed as a function of the partial pressure of the gas in the gas phase and the temperature An approxi mate relation in this case for the mole fractions of a species on the liquid and gas sides of the interface is given by Raoults law as P igas side y igas side P total y iliquid side P isat T 1623 where PisatT is the saturation pressure of the species i at the interface tem perature and Ptotal is the total pressure on the gas phase side Tabular data are available in chemical handbooks for common solutions such as the ammonia water solution that is widely used in absorptionrefrigeration systems Gases may also dissolve in solids but the diffusion process in this case can be very complicated The dissolution of a gas may be independent of the struc ture of the solid or it may depend strongly on its porosity Some dissolution processes such as the dissolution of hydrogen in titanium similar to the dis solution of CO2 in water are reversible and thus maintaining the gas content in the solid requires constant contact of the solid with a reservoir of that gas Some other dissolution processes are irreversible For example oxygen gas dissolving in titanium forms TiO2 on the surface and the process does not reverse itself TABLE 162 Henrys constant H in bars for selected gases in water at low to moderate pressures for gas i H Pigas side yiwater side Solute 290 K 300 K 310 K 320 K 330 K 340 K H2S 440 560 700 830 980 1140 CO2 1280 1710 2170 2720 3220 O2 38000 45000 52000 57000 61000 65000 H2 67000 72000 75000 76000 77000 76000 CO 51000 60000 67000 74000 80000 84000 Air 62000 74000 84000 92000 99000 104000 N2 76000 89000 101000 110000 118000 124000 Table A21 from A F Mills Basic Heat and Mass Transfer Burr Ridge IL Richard D Irwin 1995 p 874 FIGURE 1623 Dissolved gases in a liquid can be driven off by heating the liquid yAgas side yAliquid side or yAgas side yAliquid side yAliquid side or PAgas side HyAliquid side Gas A Liquid B Gas A PAgas side P Final PDF to printer 811 CHAPTER 16 cen22672ch16791822indd 811 110917 1223 PM The molar density of the gas species i in the solid at the interface ρ isolid side is proportional to the partial pressure of the species i in the gas Pigas side on the gas side of the interface and is expressed as ρ isolid side S P igas side kmolm 3 1624 where S is the solubility Expressing the pressure in bars and noting that the unit of molar concentration is kmol of species i per m3 the unit of solubility is kmolm3bar Solubility data for selected gassolid combinations are given in Table 163 The product of solubility of a gas and the diffusion coefficient of the gas in a solid is referred to as the permeability which is a measure of the ability of the gas to penetrate a solid Permeability is inversely proportional to thickness and has the unit kmolsmbar Finally if a process involves the sublimation of a pure solid such as ice or the evaporation of a pure liquid such as water in a different medium such as air the mole or mass fraction of the substance in the liquid or solid phase is simply taken to be 10 and the partial pressure and thus the mole fraction of the sub stance in the gas phase can readily be determined from the saturation data of the substance at the specified temperature Also the assumption of thermodynamic equilibrium at the interface is very reasonable for pure solids pure liquids and solutions except when chemical reactions are occurring at the interface TABLE 163 Solubility of selected gases and solids for gas i S ρ isolid slide Pigas side S Gas Solid T K kmolm3bar O2 Rubber 298 000312 N2 Rubber 298 000156 CO2 Rubber 298 004015 He SiO2 298 000045 H2 Ni 358 000901 R M Barrer Diffusion in and through Solids New York Macmillan 1941 EXAMPLE 168 Mole Fraction of Water Vapor Just Over a Lake Determine the mole fraction of the water vapor at the surface of a lake whose tem perature is 15C and compare it to the mole fraction of water in the lake Fig 1624 Take the atmospheric pressure at lake level to be 92 kPa SOLUTION The mole fraction of water vapor at the surface of a lake is to be deter mined and to be compared to the mole fraction of water in the lake Assumptions 1 Both the air and water vapor are ideal gases 2 The amount of air dis solved in water is negligible Properties The saturation pressure of water at 15C is 17057 kPa Table A4 Analysis There exists phase equilibrium at the free surface of the lake and thus the air at the lake surface is always saturated at the interface temperature The air at the water surface is saturated Therefore the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 15C P v P sat 15C 17057 kPa The mole fraction of water vapor in the air at the surface of the lake is determined from Eq 1622 to be y v P v P 17057 kPa 92 kPa 00185 or 185 percent Water contains some dissolved air but the amount is negligible Therefore we can assume the entire lake to be liquid water Then its mole fraction becomes y waterliquid side 10 or 100 percent Discussion Note that the concentration of water on a molar basis is 100 percent just beneath the airwater interface and less than 2 percent just above it even though the air is assumed to be saturated so this is the highest value at 15C Therefore large discontinui ties can occur in the concentrations of a species across phase boundaries FIGURE 1624 Schematic for Example 168 yH2Oliquid side 10 Lake 15C Air 92 kPa yH2Oair side 00185 Saturated air Final PDF to printer 812 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 812 110917 1223 PM EXAMPLE 169 The Amount of Dissolved Air in Water Determine the mole fraction of air at the surface of a lake whose temperature is 17C Fig 1625 Take the atmospheric pressure at lake level to be 92 kPa SOLUTION The mole fraction of air in lake water is to be determined Assumptions 1 Both the air and vapor are ideal gases Properties The saturation pressure of water at 17C is 196 kPa Table A4 The Henrys constant for air dissolved in water at 290 K is H 62000 bar Table 162 Analysis This example is similar to the previous example Again the air at the water surface is saturated and thus the partial pressure of water vapor in the air at the lake sur face is the saturation pressure of water at 17C P v P sat 17 o C 196 kPa The partial pressure of dry air is P dry air P P v 92 196 9004 kPa 09004 bar Note that we could have ignored the vapor pressure since the amount of vapor in air is so small with little loss in accuracy an error of about 2 percent The mole fraction of air in the water is from Henrys law y dry airliquid side P dry airgas side H 09004 bar 62000 bar 145 10 5 Discussion This value is very small as expected Therefore the concentration of air in water just below the airwater interface is 145 moles per 100000 moles But obviously this is enough oxygen for fish and other creatures in the lake Note that the amount of air dissolved in water will decrease with increasing depth unless phase equilibrium exists throughout the entire lake FIGURE 1625 Schematic for Example 169 Air ydry airliquid side Lake 17C Pdry airgas side Saturated air FIGURE 1626 Schematic for Example 1610 Nickel plate H2 H2 Hydrogen gas 358 K 300 kPa EXAMPLE 1610 Diffusion of Hydrogen Gas into a Nickel Plate Consider a nickel plate that is placed into a tank filled with hydrogen gas at 358 K and 300 kPa Determine the molar and mass density of hydrogen in the nickel plate when phase equilibrium is established Fig 1626 SOLUTION A nickel plate is exposed to hydrogen gas The density of hydrogen in the plate is to be determined Properties 1 The molar mass of hydrogen H2 is M 2 kgkmol and the solubil ity of hydrogen in nickel at the specified temperature is given in Table 163 to be 000901 kmolm3bar Analysis Noting that 300 kPa 3 bar the molar density of hydrogen in the nickel plate is determined from Eq 1624 to be ρ H 2 solid side S P H 2 gas side 000901 kmolm 3 bar3 bar 0027 kmolm 3 It corresponds to a mass density of ρ H 2 solid side ρ H 2 solid side M H 2 0027 kmolm 3 2 kgkmol 0054 kgm 3 That is there will be 0027 kmol or 0054 kg of H2 gas in each m3 volume of nickel plate when phase equilibrium is established Final PDF to printer 813 CHAPTER 16 cen22672ch16791822indd 813 110917 1223 PM EXAMPLE 1611 Composition of Different Phases of a Mixture In absorption refrigeration systems a twophase equilibrium mixture of liquid ammo nia NH3 and water H2O is frequently used Consider one such mixture at 40C shown in Fig 1627 If the composition of the liquid phase is 70 percent NH3 and 30 percent H2O by mole numbers determine the composition of the vapor phase of this mixture SOLUTION A twophase mixture of ammonia and water at a specified temperature is considered The composition of the liquid phase is given and the composition of the vapor phase is to be determined Assumptions The mixture is ideal and thus Raoults law is applicable Properties The saturation pressures of H2O and NH3 at 40C are PH2Osat 73851 kPa and PNH3sat 155433 kPa Analysis The vapor pressures are determined from P H 2 Ogas side y H 2 Oliquid side P H 2 Osat T 030 73851 kPa 222 kPa P NH 3 gas side y NH 3 liquid side P NH 3 sat T 070 155433 kPa 108803 kPa The total pressure of the mixture is P total P H 2 O P NH 3 222 108803 109025 kPa Then the mole fractions in the gas phase are y H 2 Ogas side P H 2 Ogas side P total 222 kPa 109025 kPa 00020 y NH 3 gas side P NH 3 gas side P total 108803 kPa 109025 kPa 09980 Discussion Note that the gas phase consists almost entirely of ammonia making this mixture very suitable for absorption refrigeration FIGURE 1627 Schematic for Example 1611 Liquid yfH2O 030 yfNH3 070 40C ygH2O Vapor H2O NH3 ygNH3 SUMMARY An isolated system is said to be in chemical equilibrium if no changes occur in the chemical composition of the system The criterion for chemical equilibrium is based on the second law of thermodynamics and for a system at a specified tempera ture and pressure it can be expressed as dG TP 0 For the reaction ν A A ν B B ν c C ν D D where the νs are the stoichiometric coefficients the equilib rium criterion can be expressed in terms of the Gibbs func tions as ν C g C ν D g D ν A g A ν B g B 0 which is valid for any chemical reaction regardless of the phases involved For reacting systems that consist of ideal gases only the equilibrium constant KP can be expressed as K P e ΔG T R u T where the standardstate Gibbs function change ΔGT and the equilibrium constant KP are defined as ΔGT ν C g C T ν D g D T ν A g A T ν B g B T and K P P C ν C P D ν D P A ν A P B ν B Final PDF to printer 814 CHEMICAL AND PHASE EQUILIBRIUM cen22672ch16791822indd 814 110917 1223 PM Here Pis are the partial pressures of the components in atm The KP of idealgas mixtures can also be expressed in terms of the mole numbers of the components as K P N C ν C N D ν D N A ν A N B ν B P N total Δν where Δν νC νD νA νB P is the total pressure in atm and Ntotal is the total number of moles present in the reaction chamber including any inert gases The preceding equation is written for a reaction involving two reactants and two prod ucts but it can be extended to reactions involving any number of reactants and products The equilibrium constant KP of idealgas mixtures depends on temperature only It is independent of the pressure of the equilibrium mixture and it is not affected by the presence of inert gases The larger the KP the more complete the reaction Very small values of KP indicate that a reaction does not pro ceed to any appreciable degree A reaction with KP 1000 is usually assumed to proceed to completion and a reaction with KP 0001 is assumed not to occur at all The mixture pres sure affects the equilibrium composition although it does not affect the equilibrium constant KP The variation of KP with temperature is expressed in terms of other thermochemical properties through the vant Hoff equation d ln K P dT h R T R u T 2 where h R T is the enthalpy of reaction at temperature T For small temperature intervals it can be integrated to yield ln K P 2 K P 1 h R R u 1 T 1 1 T 2 This equation shows that combustion processes are less com plete at higher temperatures since KP decreases with tempera ture for exothermic reactions Two phases are said to be in phase equilibrium when there is no transformation from one phase to the other Two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function That is g f g g In general the number of independent variables associated with a multicomponent multiphase system is given by the Gibbs phase rule expressed as IV C PH 2 where IV the number of independent variables C the number of components and PH the number of phases pres ent in equilibrium A multicomponent multiphase system at a specified tem perature and pressure is in phase equilibrium when the spe cific Gibbs function of each component is the same in all phases For a gas i that is weakly soluble in a liquid such as air in water the mole fraction of the gas in the liquid yiliquid side is related to the partial pressure of the gas Pigas side by Henrys law y iliquid side P igas side H where H is Henrys constant When a gas is highly soluble in a liquid such as ammonia in water the mole fractions of the species of a twophase mixture in the liquid and gas phases are given approximately by Raoults law expressed as P igas side y igas side P total y iliquid side P isat T where Ptotal is the total pressure of the mixture PisatT is the saturation pressure of species i at the mixture temperature and yiliquid side and yigas side are the mole fractions of species i in the liquid and vapor phases respectively REFERENCES AND SUGGESTED READINGS 1 R M Barrer Diffusion in and through Solids New York Macmillan 1941 2 I Glassman Combustion New York Academic Press 1977 3 A M Kanury Introduction to Combustion Phenomena New York Gordon and Breach 1975 4 A F Mills Basic Heat and Mass Transfer Burr Ridge IL Richard D Irwin 1995 5 J M Smith and H C Van Ness Introduction to Chemical Engineering Thermodynamics 3rd ed New York John Wiley Sons 1986 Final PDF to printer cen22672ch16791822indd 815 110917 1223 PM 815 CHAPTER 16 PROBLEMS KP and the Equilibrium Composition of Ideal Gases 161C Why is the criterion for chemical equilibrium expressed in terms of the Gibbs function instead of entropy 162C Write three different KP relations for reacting ideal gas mixtures and state when each relation should be used 163C Is a wooden table in chemical equilibrium with the air 164C A reaction chamber contains a mixture of CO2 CO and O2 in equilibrium at a specified temperature and pressure How will a increasing the temperature at constant pressure and b increasing the pressure at constant temperature affect the number of moles of CO2 165C A reaction chamber contains a mixture of N2 and N in equilibrium at a specified temperature and pressure How will a increasing the temperature at constant pressure and b increasing the pressure at constant temperature affect the number of moles of N2 166C A reaction chamber contains a mixture of CO2 CO and O2 in equilibrium at a specified temperature and pressure Now some N2 is added to the mixture while the mixture tem perature and pressure are kept constant Will this affect the number of moles of O2 How 167C Which element is more likely to dissociate into its monatomic form at 3000 K H2 or N2 Why 168C Consider a mixture of NO O2 and N2 in equilibrium at a specified temperature and pressure Now the pressure is tripled a Will the equilibrium constant KP change b Will the number of moles of NO O2 and N2 change How 169C The equilibrium constant of the dissociation reaction H 2 2H at 3000 K and 1 atm is KP Express the equilibrium constants of the following reactions at 3000 K in terms of KP a H 2 2H at 2 atm b 2H H 2 at 1 atm c 2H 2 4H at 1 atm d H 2 2 N 2 2 H2N 2 at 2 atm e 6H 3 H 2 at 4 atm 1610C The equilibrium constant of the reaction CO 1 2 O 2 CO 2 at 1000 K and 1 atm is KP Express the equilibrium constant of the following reactions at 1000 K in terms of KP a CO 1 2 O 2 CO 2 at 3 atm b CO 2 CO 1 2 O 2 at 1 atm c CO O 2 CO 2 1 2 O 2 at 1 atm d CO 2O 2 5 N 2 CO 2 15 O 2 5 N 2 at 4 atm e 2CO O 2 2 CO 2 at 1 atm 1611C The equilibrium constant for C 1 2 O 2 CO reac tion at 100 kPa and 1600 K is KP Use this information to find the equilibrium constant for the following reactions at 1600 K a C 1 2 O 2 CO at 100 kPa b C 1 2 O 2 CO at 500 kPa c 2C O 2 2CO at 100 kPa d 2CO 2 C O 2 at 500 kPa 1612 Determine the temperature at which 10 percent of diatomic hydrogen H2 dissociates into monatomic hydrogen H at a pressure of 10 atm FIGURE P1614 10 CO2 60 H2O 30 CO 10 atm 800 K FIGURE P1612 1 kmol H2 Initial composition 10 atm 09H2 02H Equilibrium composition Problems designated by a C are concept questions and students are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software 1613E At what temperature will oxygen be 15 percent dis associated at a 3 psia and b 100 psia Answers a 5508 R b 6662 R 1614 A mixture of ideal gases consists of the following gases by mole fraction 10 percent CO2 60 percent H2O and 30 percent CO Determine the Gibbs function of the CO in this mixture when the mixture pressure is 10 atm and its tempera ture is 800 K 1615E Use the Gibbs function to determine the equilibrium constant of the H 2 O H 2 1 2 O 2 reaction at a 1440 R and b 3960 R How do these compare to the equilibrium con stants of Table A28 1616 An inventor claims she can produce hydrogen gas by the reversible reaction 2H 2 O 2 H 2 O 2 Determine the mole fractions of the hydrogen and oxygen produced when this Final PDF to printer cen22672ch16791822indd 816 110917 1223 PM 816 CHEMICAL AND PHASE EQUILIBRIUM reaction occurs at 4000 K and 10 kPa Answers 0560 H2 0280 O2 1617 Consider the reaction 2H 2 O 2 H 2 O 2 at 4000 K and 10 kPa Will the amount of hydrogen gas produced be increased when the reaction occurs at 100 kPa rather than 10 kPa 1618 Consider the reaction 2H 2 O 2 H 2 O 2 at 4000 K and 10 kPa How will the amount of hydrogen gas produced change if inert nitrogen is mixed with the water vapor such that the original mole fraction of the nitrogen is 20 percent 1619 Using the Gibbs function data determine the equilib rium constant KP for the reaction H 2 1 2 O 2 H 2 O at a 298 K and b 2000 K Compare your results with the KP val ues listed in Table A28 1620 Carbon dioxide is commonly produced through the reaction C O2 CO2 Determine the yield of carbon diox ide mole fraction when this is done in a reactor maintained at 1 atm and 3800 K The natural logarithm of the equilib rium constant for the reaction C O2 CO2 at 3800 K is 0461 Answer 0122 1621 The reaction N2 O2 2NO occurs in internal combustion engines Determine the equilibrium mole fraction of NO when the pressure is 101 kPa and the temperature is 1600 K 1622 Consider the disassociation reaction CO2 CO O at 1 atm and 2500 K Now 3 moles of nitrogen are added to the 1 mole of CO2 Determine the equilibrium composition of the products at the same temperature and pressure with the additional nitrogen Note First evaluate the KP of this reaction using the KP values of the reactions CO 2 CO 1 2 O 2 and 05O2 O 1623 Determine the equilibrium constant KP for the process CO 1 2 O 2 CO 2 at a 298 K and b 2000 K Compare your results with the values for KP listed in Table A28 1624 Study the effect of varying the percent excess air during the steadyflow combustion of hydrogen at a pressure of 1 atm At what temperature will 97 percent of H2 burn into H2O Assume the equilibrium mixture consists of H2O H2 O2 and N2 1625 Determine the equilibrium constant KP for the reaction CH4 2O2 CO2 2H2O at 25C Answer 196 X 10140 1626 Carbon dioxide CO2 is heated to 2400 K at a con stant pressure of 3 atm Determine the percentage of CO2 that will dissociate into CO and O2 during this process 1627 Using the Gibbs function data determine the equilib rium constant KP for the dissociation process CO 2 CO 1 2 O 2 at a 298 K and b 1800 K Compare your results with the KP values listed in Table A28 1628 Carbon monoxide is burned with 100 percent excess air during a steadyflow process at a pressure of 1 atm At what temperature will 97 percent of CO burn to CO2 Assume the equilibrium mixture consists of CO2 CO O2 and N2 Answer 2276 K 1629 Reconsider Prob 1628 Using appropriate software study the effect of varying the percent excess air during the steadyflow process from 0 to 200 per cent on the temperature at which 97 percent of CO burns into CO2 Plot the temperature against the percent excess air and discuss the results 1630E Repeat Prob 1628 using data in English units 1631 Air 79 percent N2 and 21 percent O2 is heated to 2000 K at a constant pressure of 2 atm Assuming the equilib rium mixture consists of N2 O2 and NO determine the equi librium composition at this state Is it realistic to assume that no monatomic oxygen or nitrogen will be present in the equi librium mixture Will the equilibrium composition change if the pressure is doubled at constant temperature 1632 Estimate KP for the following equilibrium reaction at 2500 K CO H2O CO2 H2 At 2000 K it is known that the enthalpy of reaction is 26176 kJkmol and KP is 02209 Compare your result with the value obtained from the definition of the equilibrium constant 1633 A gaseous mixture of 50 percent by mole fraction methane and 50 percent nitrogen is heated to 1000 K as its pressure is maintained at 1 atm Determine the equilibrium composition by mole fraction of the resulting mixture The natural logarithm of the equilibrium constant for the reaction C 2H2 CH4 at 1000 K is 2328 1634 A mixture of 3 mol of N2 1 mol of O2 and 01 mol of Ar is heated to 2400 K at a constant pressure of 10 atm Assuming the equilibrium mixture consists of N2 O2 Ar and NO determine the equilibrium composition Answers 00823NO 29589N2 09589O2 01Ar 1635 Determine the mole fraction of sodium that ionizes according to the reaction Na Na e at 2000 K and 15 atm KP 0668 for this reaction Answer 555 percent 1636 A mixture of ideal gases is blended in a rigid ves sel that is initially evacuated and is maintained at a constant temperature of 20C First nitrogen is added until the pressure is 110 kPa next carbon dioxide is added until the pressure is 230 kPa and finally NO is added until the pressure is 350 kPa Determine the Gibbs function of the N2 in this mixture Answer 200 kJkmol 1637E A steadyflow combustion chamber is supplied with CO gas at 560 R and 16 psia at a rate of 125 ft3min and with oxygen O2 at 537 R and 16 psia at a rate of 07 lbmmin The combustion products leave the combustion chamber at 3600 R and 16 psia If the combustion gases consist of CO2 CO and O2 determine a the equilibrium composition of the product Final PDF to printer cen22672ch16791822indd 817 110917 1223 PM 817 CHAPTER 16 gases and b the rate of heat transfer from the combustion chamber 1638 Liquid propane C3H8 enters a combustion chamber at 25C at a rate of 12 kgmin where it is mixed and burned with 150 percent excess air that enters the combustion cham ber at 12C If the combustion gases consist of CO2 H2O CO O2 and N2 that exit at 1200 K and 2 atm determine a the equilibrium composition of the product gases and b the rate of heat transfer from the combustion chamber Is it realistic to disregard the presence of NO in the product gases Answers a 3CO2 75O2 4H2O 47N2 b 5066 kJmin 1643C When determining the equilibrium composition of a mixture involving simultaneous reactions how would you determine the number of KP relations needed 1644 One mole of H2O is heated to 3400 K at a pressure of 1 atm Determine the equilibrium composition assuming that only H2O OH O2 and H2 are present Answers 0574H2O 0308H2 0095O2 0236OH 1645E Air 21 percent O2 79 percent N2 is heated to 5400 R at a pressure of 1 atm Determine the equilibrium composi tion assuming that only O2 N2 O and NO are present Is it realistic to assume that no N will be present in the final equi librium mixture FIGURE P1638 Combustion chamber 2 atm Air C3H8 12C 25C 1200 K H2O CO2 CO O2 N2 1639 Reconsider Prob 1638 Using appropriate software investigate if it is realistic to disregard the presence of NO in the product gases 1640 Oxygen O2 is heated during a steadyflow process at 1 atm from 298 to 3000 K at a rate of 05 kgmin Determine the rate of heat supply needed during this process assuming a some O2 dissociates into O and b no dissociation takes place FIGURE P1640 O2 O O2 298 K 3000 K Qin 1641 A constantvolume tank contains a mixture of 1 kmol H2 and 1 kmol O2 at 25C and 1 atm The contents are ignited Determine the final temperature and pressure in the tank when the combustion gases are H2O H2 and O2 Simultaneous Reactions 1642C What is the equilibrium criterion for systems that involve two or more simultaneous chemical reactions FIGURE P1645E O2 N2 O NO Reaction chamber Air 5400 R Qin 1646E Reconsider Prob 1645E Use appropriate software to obtain the equilibrium solution Compare your solution technique with that used in Prob 1645E 1647 A mixture of 2 mol of CO2 and 1 mol of O2 is heated to 2800 K at a pressure of 4 atm Determine the equilibrium composition of the mixture assuming that only CO2 CO O2 and O are present 1648 Water vapor H2O is heated during a steadyflow process at 1 atm from 298 to 3000 K at a rate of 02 kgmin Determine the rate of heat supply needed during this process assuming a some H2O dissociates into H2 O2 and OH and b no dissociation takes place Answers a 2056 kJmin b 1404 kJmin 1649 Reconsider Prob 1648 Using appropriate software study the effect of the pressure on the rate of heat supplied for the two cases Let the pressure vary from 1 to 10 atm For each of the two cases plot the rate of heat supplied as a function of pressure 1650 Ethyl alcohol C2H5OHg at 25C is burned in a steadyflow adiabatic combustion chamber with 40 percent excess air that also enters at 25C Determine the adiabatic flame temperature of the products at 1 atm assum ing the significant equilibrium reactions are CO 2 CO 1 2 O 2 and 1 2 N 2 1 2 O 2 NO Plot the adiabatic flame tempe rature and kmoles of CO2 CO and NO at equilibrium for val ues of percent excess air between 10 and 100 percent Final PDF to printer cen22672ch16791822indd 818 110917 1223 PM 818 CHEMICAL AND PHASE EQUILIBRIUM Variations of KP with Temperature 1651C What is the importance of the vant Hoff equation 1652C Will a fuel burn more completely at 2000 or 2500 K 1653 Estimate the enthalpy of reaction h R for the dissocia tion process O2 2O at 3100 K using a enthalpy data and b KP data Answers a 513614 kJkmol b 512808 kJkmol 1654E Estimate the enthalpy of reaction h R for the combus tion process of carbon monoxide at 3960 R using a enthalpy data and b KP data Answers a 119030 Btulbmol b 119041 Btulbmol 1655 Using the enthalpy of reaction h R data and the KP value at 3000 K estimate the KP value of the combustion pro cess H 2 1 2 O 2 H 2 O at 3200 K Answer 116 1656 Estimate the enthalpy of reaction h R for the dissocia tion process CO 2 CO 1 2 O 2 at 2200 K using a enthalpy data and b KP data 1657 Estimate the enthalpy of reaction for the equilib rium reaction CH4 2O2 CO2 2H2O at 2500 K using a enthalpy data and b KP data Obtain enthalpy and entropy properties from appropriate software Phase Equilibrium 1658C Consider a twophase mixture of ammonia and water in equilibrium Can this mixture exist in two phases at the same temperature but at a different pressure 1659C Consider a tank that contains a saturated liquid vapor mixture of water in equilibrium Some vapor is now allowed to escape the tank at constant temperature and pres sure Will this disturb the phase equilibrium and cause some of the liquid to evaporate 1660C Using the solubility data of a solid in a specified liq uid explain how you would determine the mole fraction of the solid in the liquid at the interface at a specified temperature 1661C Using solubility data of a gas in a solid explain how you would determine the molar concentration of the gas in the solid at the solidgas interface at a specified temperature 1662C Using the Henrys constant data for a gas dissolved in a liquid explain how you would determine the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature 1663E Water is sprayed into air at 80F and 143 psia and the falling water droplets are collected in a container on the floor Determine the mass and mole fractions of air dissolved in the water 1664 Show that a mixture of saturated liquid water and sat urated water vapor at 300 kPa satisfies the criterion for phase equilibrium 1665 Show that a mixture of saturated liquid water and saturated water vapor at 100C satisfies the criterion for phase equilibrium 1666 Calculate the value of the Gibbs function for saturated refrigerant134a at 0C as a saturated liquid saturated vapor and a mixture of liquid and vapor with a quality of 30 percent Demonstrate that phase equilibrium exists 1667 A liquidvapor mixture of refrigerant134a is at 10C with a quality of 40 percent Determine the value of the Gibbs function in kJkg when the two phases are in equilib rium Answer 225 kJkg FIGURE P1667 R134a 10C x 04 1668 At what temperature will the gaseous phase of an oxy gennitrogen mixture at 100 kPa have a nitrogen mole fraction of 30 percent What is the mass fraction of the oxygen in the liquid phase at this temperature 1669 Using the liquidvapor equilibrium diagram of an oxygennitrogen mixture at 100 kPa determine the tempera ture at which the composition of the liquid phase is 30 percent N2 and 70 percent O2 1670 An oxygennitrogen mixture consists of 30 kg of oxy gen and 40 kg of nitrogen This mixture is cooled to 84 K at 01 MPa pressure Determine the mass of the oxygen in the liquid and gaseous phase Answers 828 kg 217 kg 1671 Reconsider Prob 1670 What is the total mass of the liquid phase Answer 114 kg 1672 Consider a mixture of oxygen and nitrogen in the gas phase How many independent properties are needed to fix the state of the system 1673 A wall made of natural rubber separates O2 and N2 gases at 25C and 300 kPa Determine the molar concentra tions of O2 and N2 in the wall 1674 Consider a rubber plate that is in contact with nitro gen gas at 298 K and 250 kPa Determine the molar and mass density of nitrogen in the rubber at the interface 1675 An ammoniawater mixture is at 10C Determine the pressure of the ammonia vapor when the mole fraction of the ammonia in the liquid is a 20 percent and b 80 percent The saturation pressure of ammonia at 10C is 6153 kPa 1676 Consider a liquidvapor mixture of ammonia and water in equilibrium at 25C If the composition of the liquid Final PDF to printer cen22672ch16791822indd 819 110917 1223 PM 819 CHAPTER 16 FIGURE P1677 Condenser Weak solution Strong solution Evaporator 3 4 1 6 2 5 w Generator qg Absorber qa phase is 50 percent NH3 and 50 percent H2O by mole numbers determine the composition of the vapor phase of this mixture Saturation pressure of NH3 at 25C is 10035 kPa Answers 031 percent 9969 percent 1677 An ammoniawater absorption refrigeration unit operates its absorber at 0C and its generator at 46C The vapor mixture in the generator and absorber is to have an ammonia mole fraction of 96 percent Assuming ideal behav ior determine the operating pressure in the a generator and b absorber Also determine the mole fraction of the ammo nia in the c strong liquid mixture being pumped from the absorber and the d weak liquid solution being drained from the generator The saturation pressure of ammonia at 0C is 4306 kPa and at 46C it is 18302 kPa Answers a 223 kPa b 148 kPa c 0033 d 0117 vapor in the CO2 gas and b the mass of dissolved CO2 in a 300ml drink 1681E One lbmol of refrigerant134a is mixed with 1 lbmol of water in a closed container which is maintained at 147 psia and 77F Determine the mole fraction of the refrigerant134a in the liquid phase and the vapor phase Review Problems 1682 Determine the mole fraction of argon that ionizes according to the reaction Ar Ar e at 10000 K and 035 atm KP 000042 for this reaction 1683 Using the Gibbs function data determine the equilib rium constant KP for the dissociation process O2 2O at 2000 K Compare your result with the KP value listed in Table A28 Answer 44 107 1684 A mixture of 1 mol of H2 and 1 mol of Ar is heated at a constant pressure of 1 atm until 10 percent of H2 dissociates into monatomic hydrogen H Determine the final tempera ture of the mixture 1685 A mixture of 1 mol of H2O 2 mol of O2 and 5 mol of N2 is heated to 2200 K at a pressure of 5 atm Assuming the equilibrium mixture consists of H2O O2 N2 and H2 deter mine the equilibrium composition at this state Is it realistic to assume that no OH will be present in the equilibrium mixture 1686 Methane gas CH4 at 25C is burned with the stoichio metric amount of air at 25C during an adiabatic steadyflow combustion process at 1 atm Assuming the product gases con sist of CO2 H2O CO N2 and O2 determine a the equilibrium composition of the product gases and b the exit temperature 1687 Reconsider Prob 1686 Using appropriate software study the effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent Plot the exit tem perature against the percent excess air and discuss the results 1688 Solid carbon at 25C is burned with a stoichiometric amount of air which is at 1 atm pressure and 25C Determine the number of moles of CO2 formed per kmol of carbon when only CO2 CO O2 and N2 are present in the products and the products are at 1 atm and 967C 1689 Reconsider Prob 1688 Determine the amount of heat released per kilogram of carbon by the combus tion Answer 19670 kJkg carbon 1690 Propane gas is burned steadily at 1 atm pressure with 30 percent excess air a What is the equilibrium composition by mole fraction of the resulting products of combustion if the temperature is 1600 K and the products contain some NO b How much heat is released per kg of propane by this com bustion process Answers a 3CO2 4H2O 00302NO 1485O2 2419N2 b 14870 kJkg propane 1678 Rework Prob 1677 when the temperature in the absorber is increased to 6C and the temperature in the genera tor is reduced to 40C The saturation pressure of ammonia at 6C is 5348 kPa and at 40C it is 15567 kPa 1679 Consider a glass of water in a room at 27C and 92 kPa If the relative humidity in the room is 100 percent and the water and the air are in thermal and phase equilibrium deter mine a the mole fraction of the water vapor in the air and b the mole fraction of air in the water 1680 Consider a carbonated drink in a bottle at 27C and 115 kPa Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as water determine a the mole fraction of the water Final PDF to printer cen22672ch16791822indd 820 110917 1223 PM 820 CHEMICAL AND PHASE EQUILIBRIUM 1691 Methane gas is burned with 30 percent excess air This fuel enters a steadyflow combustor at 101 kPa and 25C and is mixed with the air The products of combustion leave this reactor at 101 kPa and 1600 K Determine the equilibrium composition of the products of combustion and the amount of heat released by this combustion in kJkmol methane 1695 Determine the equilibrium constant for the reaction CH4 2O2 CO2 2H2O when the reaction occurs at 100 kPa and 2000 K The natural logarithms of the equilibrium constants for the reactions C 2H2 CH4 and C O2 CO2 at 2000 K are 7847 and 23839 respectively 1696 Reconsider Prob 1695 What is the equilibrium mole fraction of the water vapor 1697 Estimate the enthalpy of reaction h R for the combustion process of hydrogen at 2400 K using a enthalpy data and b KP data Answers a 252377 kJkmol b 252047 kJkmol 1698 Reconsider Prob 1697 Using appropriate software investigate the effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K 1699 Using the enthalpy of reaction h R data and the KP value at 2800 K estimate the KP value of the dissociation pro cess O2 2O at 3000 K 16100 Consider a glass of water in a room at 25C and 100 kPa If the relative humidity in the room is 70 percent and the water and the air are in thermal equilibrium determine a the mole fraction of the water vapor in the room air b the mole fraction of the water vapor in the air adjacent to the water surface and c the mole fraction of air in the water near the surface 16101 Repeat Prob 16100 for a relative humidity of 25 percent 16102 A carbonated drink is fully charged with CO2 gas at 17C and 600 kPa such that the entire bulk of the drink is in thermodynamic equilibrium with the CO2water vapor mix ture Now consider a 2L soda bottle If the CO2 gas in that bottle were to be released and stored in a container at 20C and 100 kPa determine the volume of the container 16103 Tabulate the natural log of the equilibrium con stant as a function of temperature between 298 and 3000 K for the equilibrium reaction CO H2O CO2 H2 Compare your results to those obtained by combining the ln KP values for the two equilibrium reactions CO 2 CO 1 2 O 2 and H 2 O H 2 1 2 O 2 given in Table A28 16104 Ethyl alcohol C2H5OHg at 25C is burned in a steadyflow adiabatic combustion chamber with 90 percent excess air that also enters at 25C Determine the adiabatic flame temperature of the products at 1 atm assuming the only significant equilibrium reaction is CO 2 CO 1 2 O 2 Plot the adiabatic flame temperature as the percent excess air varies from 10 to 100 percent 16105 Show that as long as the extent of the reaction α for the dissociation reaction X2 2X is smaller than 1 α is given by α K P 4 K P FIGURE P1692E 40 excess air C8H18 Combustion chamber 600 psia CO2 H2O NO O2 N2 3600 R FIGURE P1691 Combustion chamber 1 atm 30 excess air CH4 25C 25C CO2 H2O NO O2 N2 1600 K Qout 1692E Gaseous octane is burned with 40 percent excess air in an automobile engine During combustion the pressure is 600 psia and the temperature reaches 3600 R Determine the equilibrium composition of the products of combustion 1693 A constantvolume tank contains a mixture of 1 mol of H2 and 05 mol of O2 at 25C and 1 atm The contents of the tank are ignited and the final temperature and pressure in the tank are 2800 K and 5 atm respectively If the combustion gases consist of H2O H2 and O2 determine a the equilib rium composition of the product gases and b the amount of heat transfer from the combustion chamber Is it realistic to assume that no OH will be present in the equilibrium mixture Answers a 0944H2O 0056H2 0028O2 b 132600 Jmol H2 1694 Ten kmol of methane gas are heated from 1 atm and 298 K to 1 atm and 1000 K Calculate the total amount of heat transfer required when a disassociation is neglected and b disassociation is considered The natural logarithm of the equilibrium constant for the reaction C 2H2 CH4 at 1000 K is 2328 For the solution of part a use empirical coefficients of Table A2c For the solution of part b use constant specific heats and take the constantvolume specific heats of methane hydrogen and carbon at 1000 K to be 633 217 and 0711 kJkmolK respectively The constantvolume specific heat of methane at 298 K is 278 kJkmolK Final PDF to printer cen22672ch16791822indd 821 110917 1223 PM 821 CHAPTER 16 16106 Show that when the three phases of a pure substance are in equilibrium the specific Gibbs function of each phase is the same 16107 Show that when the two phases of a twocomponent system are in equilibrium the specific Gibbs function of each phase of each component is the same 16108 Using Henrys law show that the dissolved gases in a liquid can be driven off by heating the liquid Fundamentals of Engineering FE Exam Problems 16109 Of the reactions given below the reaction whose equilibrium composition at a specified temperature is not affected by pressure is a H 2 1 2 O 2 H 2 O b CO 1 2 O 2 CO 2 c N 2 O 2 2NO d N 2 2N e all of the above 16110 Of the reactions given below the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and tem perature is a H 2 1 2 O 2 H 2 O b CO 1 2 O 2 CO 2 c N 2 O 2 2NO d N 2 2N e all of the above 16111 If the equilibrium constant for the reaction H 2 1 2 O 2 H 2 O is K the equilibrium constant for the reaction 2H2O 2H2 O2 at the same temperature is a 1K b 12K c 2K d K2 e 1K2 16112 If the equilibrium constant for the reaction CO 1 2 O 2 CO 2 is K the equilibrium constant for the reaction CO 2 3 N 2 CO 1 2 O 2 3N 2 at the same temperature is a 1K b 1K 3 c 4K d K e 1K2 16113 The equilibrium constant for the reaction H 2 1 2 O 2 H 2 O at 1 atm and 1500C is given to be K Of the reactions given below all at 1500C the reaction that has a different equilibrium constant is a H 2 1 2 O 2 H 2 O at 5 atm b 2H 2 O 2 2 H 2 O at 1 atm c H 2 O 2 H 2 O 1 2 O 2 at 2 atm d H 2 1 2 O 2 3N 2 H 2 O 3 N 2 at 5 atm e H 2 1 2 O 2 3N 2 H 2 O 3 N 2 at 1 atm 16114 Moist air is heated to a very high temperature If the equilibrium composition consists of H2O O2 N2 OH H2 and NO the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is a 1 b 2 c 3 d 4 e 5 16115 Propane C3H8 is burned with air and the combustion products consist of CO2 CO H2O O2 N2 OH H2 and NO The number of equilibrium constant relations needed to deter mine the equilibrium composition of the mixture is a 1 b 2 c 3 d 4 e 5 16116 Consider a gas mixture that consists of three com ponents The number of independent variables that need to be specified to fix the state of the mixture is a 1 b 2 c 3 d 4 e 5 16117 The value of Henrys constant for CO2 gas dissolved in water at 290 K is 128 MPa Consider water exposed to atmospheric air at 100 kPa that contains 3 percent CO2 by vol ume Under phase equilibrium conditions the mole fraction of CO2 gas dissolved in water at 290 K is a 23 104 b 30 104 c 080 104 d 22 104 e 56 104 16118 The solubility of nitrogen gas in rubber at 25C is 000156 kmolm3bar When phase equilibrium is established the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 300 kPa is a 0005 kgm3 b 0018 kgm3 c 0047 kgm3 d 013 kgm3 e 028 kgm3 Design and Essay Problems 16119 A gas turbine Brayton cycle at a natural gas pipe line pumping station uses natural gas methane as its fuel Air is drawn into the turbine at 101 kPa and 25C and the pressure ratio of the turbine is 8 The natural gas fuel is injected into the combustor such that the excess air is 40 percent Determine the net specific work produced by this engine and the engines overall thermal efficiency 16120 An engineer suggested that hightemperature dis association of water be used to produce a hydrogen fuel A reactorseparator has been designed that can accommodate temperatures as high as 4000 K and pressures as much as 5 atm Water enters this reactorseparator at 25C The separa tor separates the various constituents in the mixture into indi vidual streams whose temperature and pressure match those of the reactorseparator These streams are then cooled to 25C and stored in atmospheric pressure tanks with the exception of any remaining water which is returned to the reactor to repeat the process again Hydrogen gas from these tanks is later burned with a stoichiometric amount of air to provide heat for an electrical power plant The parameter that characterizes this system is the ratio of the heat released by burning the hydro gen to the amount of heat used to generate the hydrogen gas Select the operating pressure and temperature for the reactor separator that maximizes this ratio Can this ratio ever be big ger than unity 16121 An article that appeared in the Reno GazetteJournal on May 18 1992 quoted an inventor as saying that he had turned water into motor vehicle fuel in a breakthrough that Final PDF to printer cen22672ch16791822indd 822 110917 1223 PM 822 CHEMICAL AND PHASE EQUILIBRIUM would increase engine efficiency save gasoline and reduce smog There was also a picture of a car that the inventor modified to run on half water and half gasoline The inven tor claimed that sparks from catalytic poles in the converted engine would break down the water into oxygen and hydro gen which would be burned with the gasoline He added that hydrogen has a higher energy density than carbon and the high energy density enables one to get more power The inventor stated that the fuel efficiency of his car increased from 20 mpg miles per gallon to more than 50 mpg of gasoline as a result of conversion and noted that the conversion sharply reduced emissions of hydrocarbons carbon monoxide and other exhaust pollutants Evaluate the claims made by the inventor and write a report that is to be submitted to a group of investors who are consid ering financing this invention 16122 One means of producing liquid oxygen from atmo spheric air is to take advantage of the phaseequilibrium prop erties of oxygennitrogen mixtures This system is illustrated in Fig P16122 In this cascadedreactors system dry atmo spheric air is cooled in the first reactor until liquid is formed According to the phaseequilibrium properties this liquid will be richer in oxygen than in the vapor phase The vapor in the first reactor is discarded while the oxygenenriched liquid leaves the first reactor and is heated in a heat exchanger until it is again a vapor The vapor mixture enters the second reactor where it is again cooled until a liquid that is further enriched in oxygen is formed The vapor from the second reactor is routed back to the first reactor while the liquid is routed to another heat exchanger and another reactor to repeat the process once again The liquid formed in the third reactor is very rich in oxygen If all three reactors are operated at 1 atm pressure select the three temperatures that produce the greatest amount of 99 percent pure oxygen 16123 Automobiles are major emitters of air pollutants such as NOx CO and hydrocarbons HC Find out the legal limits of these pollutants in your area and estimate the total amount of each pollutant in kg that would be produced in your town if all the cars were emitting pollutants at the legal limit State your assumptions 16124 To protect the atmosphere it has been suggested that hydrogen be used as a fuel in aircraft that fly at high elevations This would avoid the formation of carbon dioxide and other carbonbased combustion products The combustion chamber of a Brayton cycle operates at about 400 kPa at these altitudes Assume that new materials that allow for a maximum tempera ture of 2600 K are available and the atmospheric composition at these altitudes is 21 percent oxygen and 79 percent nitrogen by volume The presence of NOx in the exhaust gases is criti cal at these altitudes and cannot exceed 01 percent by volume Excess air supply is used to control the maximum temperature of the combustion process Determine the quantity of excess air to be used so that neither the maximum temperature nor the maximum allowable NOx specification is exceeded What is the NOx mole fraction if the maximum temperature specifi cation governs If the NOx specification governs what is the temperature of the combustion gases FIGURE P16122 Oxygen enriched liquid Nitrogen enriched vapor Dry atmospheric air T3 T2 T1 Final PDF to printer cen22672ch17823880indd 823 110917 0138 PM 823 CHAPTER 17 C O M P R E S S I B L E FLOW F or the most part we have limited our consideration so far to flows for which density variations and thus compressibility effects are negligible In this chapter we lift this limitation and consider flows that involve significant changes in density Such flows are called compressible flows and they are often encountered in devices that involve the flow of gases at very high velocities Compressible flow combines fluid dynamics and thermody namics in that both are necessary to the development of the required theoreti cal background In this chapter we develop the general relations associated with onedimensional compressible flows for an ideal gas with constant spe cific heats We start this chapter by introducing the concepts of stagnation state speed of sound and Mach number for compressible flows The relationships between the static and stagnation fluid properties are developed for isentropic flows of ideal gases and they are expressed as functions of specificheat ratios and the Mach number The effects of area changes for onedimensional isentro pic subsonic and supersonic flows are discussed These effects are illustrated by considering the isentropic flow through converging and converging diverging nozzles The concept of shock waves and the variation of flow prop erties across normal and oblique shocks are discussed Finally we consider the effects of heat transfer on compressible flows and examine steam nozzles OBJECTIVES The objectives of Chapter 17 are to Develop the general relations for compressible flows encountered when gases flow at high speeds Introduce the concepts of stagnation state speed of sound and Mach number for a compressible fluid Develop the relationships between the static and stagnation fluid properties for isentropic flows of ideal gases Derive the relationships between the static and stagnation fluid properties as functions of specificheat ratios and Mach number Derive the effects of area changes for onedimensional isentropic subsonic and supersonic flows Solve problems of isentropic flow through converging and convergingdiverging nozzles Discuss the shock wave and the variation of flow properties across the shock wave Develop the concept of duct flow with heat transfer and negligible friction known as Rayleigh flow Examine the operation of steam nozzles commonly used in steam turbines Final PDF to printer 824 COMPRESSIBLE FLOW cen22672ch17823880indd 824 110917 0138 PM 171 STAGNATION PROPERTIES When analyzing control volumes we find it very convenient to combine the internal energy and the flow energy of a fluid into a single term enthalpy defined per unit mass as h u Pρ Whenever the kinetic and potential ener gies of the fluid are negligible as is often the case the enthalpy represents the total energy of a fluid For highspeed flows such as those encountered in jet engines Fig 171 the potential energy of the fluid is still negligible but the kinetic energy is not In such cases it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation or total enthalpy h0 defined per unit mass as h 0 h V 2 2 kJ kg 171 When the potential energy of the fluid is negligible the stagnation enthalpy represents the total energy of a flowing fluid stream per unit mass Thus it simplifies the thermodynamic analysis of highspeed flows Throughout this chapter the ordinary enthalpy h is referred to as the static enthalpy whenever necessary to distinguish it from the stagnation enthalpy Notice that the stagnation enthalpy is a combination property of a fluid just like the static enthalpy and these two enthalpies are identical when the kinetic energy of the fluid is negligible Consider the steady flow of a fluid through a duct such as a nozzle dif fuser or some other flow passage where the flow takes place adiabatically and with no shaft or electrical work as shown in Fig 172 Assuming the fluid experiences little or no change in its elevation and its potential energy the energy balance relation E in E out for this singlestream steadyflow device reduces to h 1 V 1 2 2 h 2 V 2 2 2 172 or h 01 h 02 173 That is in the absence of any heat and work interactions and any changes in potential energy the stagnation enthalpy of a fluid remains constant during a steadyflow process Flows through nozzles and diffusers usually satisfy these conditions and any increase in fluid velocity in these devices creates an equivalent decrease in the static enthalpy of the fluid If the fluid were brought to a complete stop then the velocity at state 2 would be zero and Eq 172 would become h 1 V 1 2 2 h 2 h 02 Thus the stagnation enthalpy represents the enthalpy of a fluid when it is brought to rest adiabatically During a stagnation process the kinetic energy of a fluid is converted to enthalpy internal energy flow energy which results in an increase in the fluid temperature and pressure The properties of a fluid at the stagnation state are called stagnation properties stagnation temperature stagnation FIGURE 172 Steady flow of a fluid through an adiabatic duct h02 h1 V1 h01 h01 h2 V2 Control volume FIGURE 171 Aircraft and jet engines involve high speeds and thus the kinetic energy term should always be considered when analyzing them aRoyaltyFreeCorbis b Reproduced by permission of United Technologies Corporation Pratt Whitney a Turbine Combustion chamber Exhaust nozzle Fan Compressors b Final PDF to printer 825 CHAPTER 17 cen22672ch17823880indd 825 110917 0138 PM pressure stagnation density etc The stagnation state and the stagnation properties are indicated by the subscript 0 The stagnation state is called the isentropic stagnation state when the stagnation process is reversible as well as adiabatic ie isentropic The entropy of a fluid remains constant during an isentropic stagnation process The actual irreversible and isentropic stagnation processes are shown on an hs diagram in Fig 173 Notice that the stagnation enthalpy of the fluid and the stagnation temperature if the fluid is an ideal gas is the same for both cases However the actual stagnation pressure is lower than the isentropic stagnation pressure since entropy increases during the actual stagnation pro cess as a result of fluid friction Many stagnation processes are approximated to be isentropic and isentropic stagnation properties are simply referred to as stagnation properties When the fluid is approximated as an ideal gas with constant specific heats its enthalpy can be replaced by cpT and Eq 171 is expressed as c p T 0 c p T V 2 2 or T 0 T V 2 2 c p 174 Here T0 is called the stagnation or total temperature and it represents the temperature an ideal gas attains when it is brought to rest adiabatically The term V 22cp corresponds to the temperature rise during such a process and is called the dynamic temperature For example the dynamic temperature of air flowing at 100 ms is 100 ms22 1005 kJkgK 50 K Therefore when air at 300 K and 100 ms is brought to rest adiabatically at the tip of a temperature probe for example its temperature rises to the stagnation value of 305 K Fig 174 Note that for lowspeed flows the stagnation and static or ordinary temperatures are practically the same But for highspeed flows the temperature measured by a stationary probe placed in the fluid the stag nation temperature may be significantly higher than the static temperature of the fluid The pressure a fluid attains when brought to rest isentropically is called the stagnation pressure P0 For ideal gases with constant specific heats P0 is related to the static pressure of the fluid by P 0 P T 0 T kk1 175 By noting that ρ 1v and using the isentropic relation P v k P 0 v 0 k the ratio of the stagnation density to static density is expressed as ρ 0 ρ T 0 T 1k1 176 When stagnation enthalpies are used there is no need to refer explicitly to kinetic energy Then the energy balance E in E out for a singlestream steadyflow device can be expressed as FIGURE 173 The actual state actual stagnation state and isentropic stagnation state of a fluid on an hs diagram s Actual state h Isentropic stagnation state P0 P0act Actual stagnation state h V 0 h P 2 2 FIGURE 174 The temperature of an ideal gas flowing at a velocity V rises by V22cp when it is brought to a complete stop Temperature rise during stagnation Air 100 ms 305 K 300 K Final PDF to printer 826 COMPRESSIBLE FLOW cen22672ch17823880indd 826 110917 0138 PM q in w in h 01 g z 1 q out w out h 02 g z 2 177 where h01 and h02 are the stagnation enthalpies at states 1 and 2 respectively When the fluid is an ideal gas with constant specific heats Eq 177 becomes q in q out w in w out c p T 02 T 01 g z 2 z 1 178 where T01 and T02 are the stagnation temperatures Notice that kinetic energy terms do not explicitly appear in Eqs 177 and 178 but the stagnation enthalpy terms account for their contribution EXAMPLE 171 Compression of HighSpeed Air in an Aircraft An aircraft is flying at a cruising speed of 250 ms at an altitude of 5000 m where the atmospheric pressure is 5405 kPa and the ambient air temperature is 2557 K The ambient air is first decelerated in a diffuser before it enters the compressor Fig 175 Approximating both the diffuser and the compressor to be isentropic determine a the stagnation pressure at the compressor inlet and b the required compressor work per unit mass if the stagnation pressure ratio of the compressor is 8 SOLUTION Highspeed air enters the diffuser and the compressor of an aircraft The stagnation pressure of the air and the compressor work input are to be determined Assumptions 1 Both the diffuser and the compressor are isentropic 2 Air is an ideal gas with constant specific heats at room temperature Properties The constantpressure specific heat cp and the specific heat ratio k of air at room temperature are c p 1005 kJkgK and k 14 Analysis a Under isentropic conditions the stagnation pressure at the compressor inlet diffuser exit can be determined from Eq 175 However first we need to find the stagnation temperature T01 at the compressor inlet Under the stated assumptions T01 is determined from Eq 174 to be T 01 T 1 V 1 2 2 c p 2557 K 250 ms 2 21005 kJkgK 1 kJkg 1000 m 2 s 2 2868 K Then from Eq 175 P 01 P 1 T 01 T 1 kk1 5405 kPa 2868 K 2557 K 14141 8077 kPa That is the temperature of air would increase by 311C and the pressure by 2672 kPa as air is decelerated from 250 ms to zero velocity These increases in the temperature and pressure of air are due to the conversion of the kinetic energy into enthalpy b To determine the compressor work we need to know the stagnation temperature of air at the compressor exit T02 The stagnation pressure ratio across the compressor P02 P01 is specified to be 8 Since the compression process is approximated as isentro pic T02 can be determined from the idealgas isentropic relation Eq 175 T 02 T 01 P 02 P 01 k1k 2868 K 8 14114 5195 K FIGURE 175 Schematic for Example 171 Compressor T1 P01 T01 P02 T02 2557 K V1 250 ms P1 5405 kPa Diffuser Aircraft engine Final PDF to printer 827 CHAPTER 17 cen22672ch17823880indd 827 110917 0138 PM 172 SPEED OF SOUND AND MACH NUMBER An important parameter in the study of compressible flow is the speed of sound or the sonic speed defined as the speed at which an infinitesimally small pressure wave travels through a medium The pressure wave may be caused by a small disturbance which creates a slight rise in local pressure To obtain a relation for the speed of sound in a medium consider a duct that is filled with a fluid at rest as shown in Fig 176 A piston fitted in the duct is now moved to the right with a constant incremental velocity dV creating a sonic wave The wave front moves to the right through the fluid at the speed of sound c and separates the moving fluid adjacent to the piston from the fluid still at rest The fluid to the left of the wave front experiences an incremental change in its thermodynamic properties while the fluid on the right of the wave front maintains its original thermodynamic properties as shown in Fig 176 To simplify the analysis consider a control volume that encloses the wave front and moves with it as shown in Fig 177 To an observer traveling with the wave front the fluid to the right appears to be moving toward the wave front with a speed of c and the fluid to the left to be moving away from the wave front with a speed of c dV Of course the observer sees the control volume that encloses the wave front and herself or himself as stationary and the observer is witnessing a steadyflow process The mass balance for this singlestream steadyflow process is expressed as m right m left or ρAc ρ dρAc dV By canceling the crosssectional or flow area A and neglecting the higher order terms this equation reduces to c dρ ρ dV 0 No heat or work crosses the boundaries of the control volume during this steadyflow process and the potential energy change can be neglected Then the steadyflow energy balance ein eout becomes Disregarding potential energy changes and heat transfer the compressor work per unit mass of air is determined from Eq 178 w in c p T 02 T 01 1005 kJkgK5195 K 2868 K 2339 kJkg Thus the work supplied to the compressor is 2339 kJkg Discussion Notice that using stagnation properties automatically accounts for any changes in the kinetic energy of a fluid stream FIGURE 176 Propagation of a small pressure wave along a duct x dV d ρ ρ ρ Moving wave front Piston Stationary fluid P dP h dh P h dV V x 0 P dP P P c FIGURE 177 Control volume moving with the small pressure wave along a duct dV d Control volume traveling with the wave front P dP h dh P h c c ρ ρ ρ Final PDF to printer 828 COMPRESSIBLE FLOW cen22672ch17823880indd 828 110917 0138 PM h c 2 2 h dh c dV 2 2 which yields dh c dV 0 where we have neglected the secondorder term dV2 The amplitude of the ordinary sonic wave is very small and does not cause any appreciable change in the pressure and temperature of the fluid Therefore the propagation of a sonic wave is not only adiabatic but also very nearly isentropic Then the ther modynamic relation T ds dh dPρ reduces to T ds dh dP p or dh dP ρ Combining the preceding equations yields the desired expression for the speed of sound as c 2 dP dρ at s constant or c 2 P ρ s 179 It is left as an exercise for the reader to show by using thermodynamic prop erty relations that Eq 179 can also be written as c 2 k P ρ T 1710 where k cp cv is the specific heat ratio of the fluid Note that the speed of sound in a fluid is a function of the thermodynamic properties of that fluid Fig 178 When the fluid is an ideal gas P ρRT the differentiation in Eq 1710 can be performed to yield c 2 k P ρ T k ρRT ρ T kRT or c kRT 1711 Noting that the gas constant R has a fixed value for a specified ideal gas and the specific heat ratio k of an ideal gas is at most a function of temperature we see that the speed of sound in a specified ideal gas is a function of tem perature alone Fig 179 0 FIGURE 178 The speed of sound in air increases with temperature At typical outside temperatures c is about 340 ms In round numbers therefore the sound of thunder from a lightning strike travels about 1 km in 3 seconds If you see the lightning and then hear the thunder less than 3 seconds later you know that the place where the lightning occurred is less than 1 km away Bear Dancer StudiosMark Dierker RF FIGURE 179 The speed of sound changes with temperature and varies with the fluid Air Helium 347 ms 634 ms 200 K 300 K 1000 K 284 ms 1861 ms 1019 ms 832 ms Final PDF to printer 829 CHAPTER 17 cen22672ch17823880indd 829 110917 0138 PM A second important parameter in the analysis of compressible fluid flow is the Mach number Ma named after the Austrian physicist Ernst Mach 18381916 It is the ratio of the actual speed of the fluid or an object in still fluid to the speed of sound in the same fluid at the same state Ma V c 1712 Note that the Mach number depends on the speed of sound which depends on the state of the fluid Therefore the Mach number of an aircraft cruis ing at constant velocity in still air may be different at different locations Fig 1710 Fluid flow regimes are often described in terms of the flow Mach number The flow is called sonic when Ma 1 subsonic when Ma 1 supersonic when Ma 1 hypersonic when Ma 1 and transonic when Ma 1 FIGURE 1711 Schematic for Example 172 Diffuser V 200 ms T 30C Air FIGURE 1710 The Mach number can be different at different temperatures even if the flight speed is the same Alamy RF V 320 ms Air 220 K Ma 108 V 320 ms Air 300 K Ma 092 EXAMPLE 172 Mach Number of Air Entering a Diffuser Air enters a diffuser shown in Fig 1711 with a speed of 200 ms Determine a the speed of sound and b the Mach number at the diffuser inlet when the air temperature is 30C SOLUTION Air enters a diffuser at high speed The speed of sound and the Mach number are to be determined at the diffuser inlet Assumption Air at the specified conditions behaves as an ideal gas Properties The gas constant of air is R 0287 kJkgK and its specific heat ratio at 30C is 14 Analysis We note that the speed of sound in a gas varies with temperature which is given to be 30C a The speed of sound in air at 30C is determined from Eq 1711 to be c kRT 14 0287 kJkgK 303 K 1000 m 2 s 2 1 kJkg 349 ms b Then the Mach number becomes Ma V c 200 ms 349 ms 0573 Discussion The flow at the diffuser inlet is subsonic since Ma 1 173 ONEDIMENSIONAL ISENTROPIC FLOW During fluid flow through many devices such as nozzles diffusers and tur bine blade passages flow quantities vary primarily in the flow direction only and the flow can be approximated as onedimensional isentropic flow with good accuracy Therefore it merits special consideration Before presenting a formal discussion of onedimensional isentropic flow we illustrate some important aspects of it with an example Final PDF to printer 830 COMPRESSIBLE FLOW cen22672ch17823880indd 830 110917 0138 PM FIGURE 1712 Schematic for Example 173 1400 Stagnation region 1400 kPa 200C CO2 1000 300 kgs 767 200 P kPa m EXAMPLE 173 Gas Flow Through a ConvergingDiverging Duct Carbon dioxide flows steadily through a varying crosssectional area duct such as a nozzle shown in Fig 1712 at a mass flow rate of 300 kgs The carbon diox ide enters the duct at a pressure of 1400 kPa and 200C with a low velocity and it expands in the nozzle to an exit pressure of 200 kPa The duct is designed so that the flow can be approximated as isentropic Determine the density velocity flow area and Mach number at each location along the duct that corresponds to an overall pres sure drop of 200 kPa SOLUTION Carbon dioxide enters a varying crosssectional area duct at specified conditions The flow properties are to be determined along the duct Assumptions 1 Carbon dioxide is an ideal gas with constant specific heats at room temperature 2 Flow through the duct is steady onedimensional and isentropic Properties For simplicity we use cp 0846 kJkgK and k 1289 throughout the calculations which are the constantpressure specific heat and specific heat ratio values of carbon dioxide at room temperature The gas constant of carbon dioxide is R 01889 kJkgK Analysis We note that the inlet temperature is nearly equal to the stagnation tempera ture since the inlet velocity is small The flow is isentropic and thus the stagnation tem perature and pressure throughout the duct remain constant Therefore T 0 T 1 200C 473 K and P 0 P 1 1400 kPa To illustrate the solution procedure we calculate the desired properties at the location where the pressure is 1200 kPa the first location that corresponds to a pressure drop of 200 kPa From Eq 175 T T 0 P P 0 k1k 473 K 1200 kPa 1400 kPa 1289 11289 457 K From Eq 174 V 2 c p T 0 T 2 0846 kJkgK473 K 457 K 1000 m 2 s 2 1 kJkg 1645 ms 165 ms From the idealgas relation ρ P RT 1200 kPa 01889 kPa m 3 kgK457 K 139 kgm 3 From the mass flow rate relation A m ρV 300 kgs 139 kgm 3 1645 ms 131 10 4 m 2 131 cm 2 Final PDF to printer 831 CHAPTER 17 cen22672ch17823880indd 831 110917 0138 PM From Eqs 1711 and 1712 c kRT 1289 01889 kJkgK457 K 1000 m 2 s 2 1 kJkg 3336 ms Ma V c 1645 ms 3336 ms 0493 The results for the other pressure steps are summarized in Table 171 and are plotted in Fig 1713 Discussion Note that as the pressure decreases the temperature and speed of sound decrease while the fluid velocity and Mach number increase in the flow direction The density decreases slowly at first and rapidly later as the fluid velocity increases FIGURE 1713 Variation of normalized fluid properties and crosssectional area along a duct as the pressure drops from 1400 to 200 kPa 200 400 600 800 1000 Ma A Ma ρ T V 1200 1400 P kPa T A V Flow direction ρ We note from Example 173 that the flow area decreases with decreas ing pressure down to a criticalpressure value where the Mach number is unity and then it begins to increase with further reductions in pressure The Mach number is unity at the location of smallest flow area called the throat Fig 1714 Note that the velocity of the fluid keeps increasing after pass ing the throat although the flow area increases rapidly in that region This increase in velocity past the throat is due to the rapid decrease in the fluid density The flow area of the duct considered in this example first decreases and then increases Such ducts are called convergingdiverging nozzles These nozzles are used to accelerate gases to supersonic speeds and should not be confused with Venturi nozzles which are used strictly for incompress ible flow The first use of such a nozzle occurred in 1893 in a steam turbine designed by a Swedish engineer Carl G B de Laval 18451913 and there fore convergingdiverging nozzles are often called Laval nozzles Variation of Fluid Velocity with Flow Area It is clear from Example 173 that the couplings among the velocity density and flow areas for isentropic duct flow are rather complex In the remain der of this section we investigate these couplings more thoroughly and we TABLE 171 Variation of fluid properties in flow direction in the duct described in Example 173 for m 3 kgs constant P kPa T K V ms ρ kgm3 c ms A cm2 Ma 1400 473 0 157 3394 0 1200 457 1645 139 3336 131 0493 1000 439 2407 121 3269 103 0736 800 417 3066 101 3188 964 0962 767 413 3172 982 3172 963 1000 600 391 3714 812 3087 100 1203 400 357 4419 593 2950 115 1498 200 306 5309 346 2729 163 1946 767 kPa is the critical pressure where the local Mach number is unity FIGURE 1714 The cross section of a nozzle at the smallest flow area is called the throat Fluid Convergingdiverging nozzle Throat Converging nozzle Throat Fluid Final PDF to printer 832 COMPRESSIBLE FLOW cen22672ch17823880indd 832 110917 0138 PM develop relations for the variation of statictostagnation property ratios with the Mach number for pressure temperature and density We begin our investigation by seeking relationships among the pressure tem perature density velocity flow area and Mach number for one dimensional isentropic flow Consider the mass balance for a steadyflow process m ρAV constant Differentiating and dividing the resultant equation by the mass flow rate we obtain dρ ρ dA A dV V 0 1713 Neglecting the potential energy the energy balance for an isentropic flow with no work interactions is expressed in differential form as Fig 1715 dP ρ V dV 0 1714 This relation is also the differential form of Bernoullis equation when changes in potential energy are negligible which is a form of Newtons sec ond law of motion for steadyflow control volumes Combining Eqs 1713 and 1714 gives dA A dP ρ 1 V 2 dρ dP 1715 Rearranging Eq 179 as ρPs 1c2 and substituting into Eq 1715 yield dA A dP ρ V 2 1 Ma 2 1716 This is an important relation for isentropic flow in ducts since it describes the variation of pressure with flow area We note that A ρ and V are positive quantities For subsonic flow Ma 1 the term 1 Ma2 is positive and thus dA and dP must have the same sign That is the pressure of the fluid must increase as the flow area of the duct increases and must decrease as the flow area of the duct decreases Thus at subsonic velocities the pressure decreases in converging ducts subsonic nozzles and increases in diverging ducts sub sonic diffusers In supersonic flow Ma 1 the term 1 Ma2 is negative and thus dA and dP must have opposite signs That is the pressure of the fluid must increase as the flow area of the duct decreases and must decrease as the flow area of the duct increases Thus at supersonic velocities the pressure decreases in diverging ducts supersonic nozzles and increases in converging ducts supersonic diffusers Another important relation for the isentropic flow of a fluid is obtained by substituting ρV dPdV from Eq 1714 into Eq 1716 dA A dV V 1 Ma 2 1717 FIGURE 1715 Derivation of the differential form of the energy equation for steady isentropic flow 0 isentropic dP h1 V 2 2 1 h2 V 2 2 2 or h V 2 2 constant Differentiate dh V dV 0 Also dh dP dh dP 1 Substitute dP V dV 0 T ds CONSERVATION OF ENERGY steady flow w 0 q 0 Δpe 0 v v ρ ρ Final PDF to printer 833 CHAPTER 17 cen22672ch17823880indd 833 110917 0138 PM This equation governs the shape of a nozzle or a diffuser in subsonic or super sonic isentropic flow Noting that A and V are positive quantities we conclude the following For subsonic flow Ma 1 dA dV 0 For supersonic flowMa 1 dA dV 0 For sonic flow Ma 1 dA dV 0 Thus the proper shape of a nozzle depends on the highest velocity desired relative to the sonic velocity To accelerate a fluid we must use a converging nozzle at subsonic velocities and a diverging nozzle at supersonic velocities The velocities encountered in most familiar applications are well below the sonic velocity and thus it is natural that we visualize a nozzle as a converging duct However the highest velocity we can achieve with a converging nozzle is the sonic velocity which occurs at the exit of the nozzle If we extend the converging nozzle by further decreasing the flow area in hopes of accelerat ing the fluid to supersonic velocities as shown in Fig 1716 we are up for disappointment Now the sonic velocity will occur at the exit of the converg ing extension instead of the exit of the original nozzle and the mass flow rate through the nozzle will decrease because of the reduced exit area Based on Eq 1716 which is an expression of the conservation of mass and energy principles we must add a diverging section to a converging noz zle to accelerate a fluid to supersonic velocities The result is a converging diverging nozzle The fluid first passes through a subsonic converging section where the Mach number increases as the flow area of the nozzle decreases and then reaches the value of unity at the nozzle throat The fluid continues to accelerate as it passes through a supersonic diverging section Noting that m ρAV for steady flow we see that the large decrease in density makes acceleration in the diverging section possible An example of this type of flow is the flow of hot combustion gases through a nozzle in a gas turbine The opposite process occurs in the engine inlet of a supersonic aircraft The fluid is decelerated by passing it first through a supersonic diffuser which has a flow area that decreases in the flow direction Ideally the flow reaches a Mach number of unity at the diffuser throat The fluid is further decelerated in a subsonic diffuser which has a flow area that increases in the flow direction as shown in Fig 1717 Property Relations for Isentropic Flow of Ideal Gases Next we develop relations between the static properties and stagnation prop erties of an ideal gas in terms of the specific heat ratio k and the Mach number Ma We assume the flow is isentropic and the gas has constant specific heats The temperature T of an ideal gas anywhere in the flow is related to the stagnation temperature T0 through Eq 174 T 0 T V 2 2 c p FIGURE 1716 We cannot attain supersonic velocities by extending the converging section of a converging nozzle Doing so will only move the sonic cross section farther downstream and decrease the mass flow rate P0 T0 Ma Ma Ma B 1 sonic A 1 B A P0 T0 A 1 sonic A Converging nozzle Converging nozzle Final PDF to printer 834 COMPRESSIBLE FLOW cen22672ch17823880indd 834 110917 0138 PM or T 0 T 1 V 2 2 c p T Noting that cp kRk 1 c2 kRT and Ma Vc we see that V 2 2 c p T V 2 2 kR k 1 T k 1 2 V 2 c 2 k 1 2 Ma 2 Substitution yields T 0 T 1 k 1 2 Ma 2 1718 which is the desired relation between T0 and T The ratio of the stagnation to static pressure is obtained by substituting Eq 1718 into Eq 175 P 0 P 1 k 1 2 Ma 2 kk1 1719 The ratio of the stagnation to static density is obtained by substituting Eq 1718 into Eq 176 ρ 0 ρ 1 k 1 2 Ma 2 1k1 1720 Numerical values of TT0 PP0 and ρρ0 are listed versus the Mach number in Table A32 for k 14 which are very useful for practical compressible flow calculations involving air The properties of a fluid at a location where the Mach number is unity the throat are called critical properties and the ratios in Eqs 1718 through FIGURE 1717 Variation of flow properties in subsonic and supersonic nozzles and diffusers Subsonic nozzle a Subsonic flow Ma 1 Supersonic diffuser Ma 1 Supersonic nozzle Ma 1 Subsonic diffuser Ma 1 b Supersonic flow P decreases V increases Ma increases T decreases ρ decreases P decreases V increases Ma increases T decreases ρ decreases P increases V decreases Ma decreases T increases ρ increases P increases V decreases Ma decreases T increases ρ increases Final PDF to printer 835 CHAPTER 17 cen22672ch17823880indd 835 110917 0138 PM 1720 are called critical ratios when Ma 1 Fig 1718 It is standard practice in the analysis of compressible flow to let the superscript asterisk represent the critical values Setting Ma 1 in Eqs 1718 through 1720 yields T T 0 2 k 1 1721 P P 0 2 k 1 kk1 1722 ρ ρ 0 2 k 1 1k1 1723 These ratios are evaluated for various values of k and are listed in Table 172 The critical properties of compressible flow should not be confused with the thermodynamic properties of substances at the critical point such as the criti cal temperature Tc and critical pressure Pc FIGURE 1718 When Mat 1 the properties at the nozzle throat are the critical properties Subsonic nozzle Supersonic nozzle T P T0 P0 ρ0 ρ ρ T P if Mat 1 Mat 1 Throat Throat T0 P0 ρ0 TABLE 172 The criticalpressure criticaltemperature and criticaldensity ratios for isentropic flow of some ideal gases Superheated steam k 13 Hot products of combustion k 133 Air k 14 Monatomic gases k 1667 P P 0 05457 05404 05283 04871 T T 0 08696 08584 08333 07499 ρ ρ 0 06276 06295 06340 06495 FIGURE 1719 Schematic for Example 174 T P 473 K 14 MPa CO2 T0 P0 EXAMPLE 174 Critical Temperature and Pressure in Gas Flow Calculate the critical pressure and temperature of carbon dioxide for the flow condi tions described in Example 173 Fig 1719 SOLUTION For the flow discussed in Example 173 the critical pressure and temperature are to be calculated Assumptions 1 The flow is steady adiabatic and onedimensional 2 Carbon dioxide is an ideal gas with constant specific heats Properties The specific heat ratio of carbon dioxide at room temperature is k 1289 Analysis The ratios of critical to stagnation temperature and pressure are determined to be T T 0 2 k 1 2 1289 1 08737 P P 0 2 k 1 kk1 2 1289 1 128912891 05477 Final PDF to printer 836 COMPRESSIBLE FLOW cen22672ch17823880indd 836 110917 0138 PM 174 ISENTROPIC FLOW THROUGH NOZZLES Converging or convergingdiverging nozzles are found in many engineering applications including steam and gas turbines aircraft and spacecraft propul sion systems and even industrial blasting nozzles and torch nozzles In this section we consider the effects of back pressure ie the pressure applied at the nozzle discharge region on the exit velocity the mass flow rate and the pressure distribution along the nozzle Converging Nozzles Consider the subsonic flow through a converging nozzle as shown in Fig 1720 The nozzle inlet is attached to a reservoir at pressure Pr and tem perature Tr The reservoir is sufficiently large so that the nozzle inlet velocity is negligible Since the fluid velocity in the reservoir is zero and the flow through the nozzle is approximated as isentropic the stagnation pressure and stagnation temperature of the fluid at any cross section through the nozzle are equal to the reservoir pressure and temperature respectively Now we begin to reduce the back pressure and observe the resulting effects on the pressure distribution along the length of the nozzle as shown in Fig 1720 If the back pressure Pb is equal to P1 which is equal to Pr there is no flow and the pressure distribution is uniform along the nozzle When the back pressure is reduced to P2 the exit plane pressure Pe also drops to P2 This causes the pressure along the nozzle to decrease in the flow direction When the back pressure is reduced to P3 P which is the pressure required to increase the fluid velocity to the speed of sound at the exit plane or throat the mass flow reaches a maximum value and the flow is said to be choked Further reduction of the back pressure to level P4 or below does not result in additional changes in the pressure distribution or anything else along the nozzle length Under steadyflow conditions the mass flow rate through the nozzle is con stant and is expressed as m ρAV P RT A Ma kRT PAMa k RT Solving for T from Eq 1718 and for P from Eq 1719 and substituting FIGURE 1720 The effect of back pressure on the pressure distribution along a converging nozzle x Lowest exit pressure PP0 Reservoir Pe x P V 0 0 1 P0 Pb P Pb P Pb 0 5 4 3 2 1 Pb P Pb P0 Pb Back pressure Pr P0 Tr r T0 Noting that the stagnation temperature and pressure are from Example 173 T0 473 K and P0 1400 kPa we see that the critical temperature and pressure in this case are T 08737 T 0 08737473 K 413 K P 05477 P 0 054771400 kPa 767 kPa Discussion Note that these values agree with those listed in the fifth row of Table 171 as expected Also property values other than these at the throat would indi cate that the flow is not critical and the Mach number is not unity Final PDF to printer 837 CHAPTER 17 cen22672ch17823880indd 837 110917 0138 PM m AMa P 0 k R T 0 1 k 1 Ma 2 2 k 1 2 k 1 1724 Thus the mass flow rate of a particular fluid through a nozzle is a function of the stagnation properties of the fluid the flow area and the Mach number Equation 1724 is valid at any cross section and thus m can be evaluated at any location along the length of the nozzle For a specified flow area A and stagnation properties T0 and P0 the maxi mum mass flow rate can be determined by differentiating Eq 1724 with respect to Ma and setting the result equal to zero It yields Ma 1 Since the only location in a nozzle where the Mach number can be unity is the location of minimum flow area the throat the mass flow rate through a nozzle is a maximum when Ma 1 at the throat Denoting this area by A we obtain an expression for the maximum mass flow rate by substituting Ma 1 in Eq 1724 m max A P 0 k R T 0 2 k 1 k 1 2 k 1 1725 Thus for a particular ideal gas the maximum mass flow rate through a nozzle with a given throat area is fixed by the stagnation pressure and temperature of the inlet flow The flow rate can be controlled by changing the stagna tion pressure or temperature and thus a converging nozzle can be used as a flowmeter The flow rate can also be controlled of course by varying the throat area This principle is very important for chemical processes medical devices flowmeters and anywhere the mass flux of a gas must be known and controlled A plot of m versus Pb P0 for a converging nozzle is shown in Fig 1721 Notice that the mass flow rate increases with decreasing Pb P0 reaches a maximum at Pb P and remains constant for Pb P0 values less than this critical ratio Also illustrated on this figure is the effect of back pressure on the nozzle exit pressure Pe We observe that P e P b P for P b P for P b P To summarize for all back pressures lower than the critical pressure P the pressure at the exit plane of the converging nozzle Pe is equal to P the Mach number at the exit plane is unity and the mass flow rate is the maxi mum or choked flow rate Because the velocity of the flow is sonic at the throat for the maximum flow rate a back pressure lower than the critical pres sure cannot be sensed in the nozzle upstream flow and does not affect the flow rate The effects of the stagnation temperature T0 and stagnation pressure P0 on the mass flow rate through a converging nozzle are illustrated in Fig 1722 where the mass flow rate is plotted against the statictostagnation pressure ratio at the throat Pt P0 An increase in P0 or a decrease of T0 will increase the mass flow rate through the converging nozzle a decrease in P0 or an increase in T0 will decrease it We could also conclude this by carefully observing Eqs 1724 and 1725 FIGURE 1721 The effect of back pressure Pb on the mass flow rate m and the exit pressure Pe of a converging nozzle P0 P 0 10 P0 Pb Pe 5 4 3 2 1 P0 10 P P0 max Pb 5 4 3 2 1 P0 10 P P0 m m FIGURE 1722 The variation of the mass flow rate through a nozzle with inlet stagnation properties 0 Pt Decrease in 10 P P0 m P 0 P0 P0 T0 increase in T0 or both Increase in P0 decrease in T0 or both Mat 1 Mat 1 Final PDF to printer 838 COMPRESSIBLE FLOW cen22672ch17823880indd 838 110917 0138 PM A relation for the variation of flow area A through the nozzle relative to throat area A can be obtained by combining Eqs 1724 and 1725 for the same mass flow rate and stagnation properties of a particular fluid This yields A A 1 Ma 2 k 1 1 k 1 2 Ma 2 k 1 2 k 1 1726 Table A32 gives values of AA as a function of the Mach number for air k 14 There is one value of AA for each value of the Mach number but there are two possible values of the Mach number for each value of AAone for subsonic flow and another for supersonic flow Another parameter sometimes used in the analysis of onedimensional isen tropic flow of ideal gases is Ma which is the ratio of the local velocity to the speed of sound at the throat Ma V c 1727 Equation 1727 can also be expressed as Ma V c c c Ma c c Ma kRT kR T Ma T T where Ma is the local Mach number T is the local temperature and T is the critical temperature Solving for T from Eq 1718 and for T from Eq 1721 and substituting we get Ma Ma k 1 2 k 1 Ma 2 1728 Values of Ma are also listed in Table A32 versus the Mach number for k 14 Fig 1723 Note that the parameter Ma differs from the Mach number Ma in that Ma is the local velocity nondimensionalized with respect to the sonic velocity at the throat whereas Ma is the local velocity nondi mensionalized with respect to the local sonic velocity Recall that the sonic velocity in a nozzle varies with temperature and thus with location FIGURE 1723 Various property ratios for isentropic flow through nozzles and diffusers are listed in Table A32 for k 14 air for convenience Ma 0 T0 090 A T P ρ 0 P ρ A Ma 100 110 09146 10000 10812 10089 10000 10079 05913 05283 04684 FIGURE 1724 Schematic for Example 175 Air nozzle 30 psia 630 R 450 fts Ma 1 EXAMPLE 175 Isentropic Flow of Air in a Nozzle Air enters a nozzle at 30 psia 630 R and a velocity of 450 fts Fig 1724 Approxi mating the flow as isentropic determine the pressure and temperature of air at a loca tion where the air velocity equals the speed of sound What is the ratio of the area at this location to the entrance area SOLUTION Air enters a nozzle at specified temperature pressure and velocity The exit pressure exit temperature and exittoinlet area ratio are to be determined for a Mach number of Ma 1 at the exit Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is approximated as steady onedimensional and isentropic Properties The properties of air are k 14 and cp 0240 BtulbmR Table A1E Final PDF to printer 839 CHAPTER 17 cen22672ch17823880indd 839 110917 0138 PM Analysis The properties of the fluid at the location where Ma 1 are the critical prop erties denoted by superscript We first determine the stagnation temperature and pres sure which remain constant throughout the nozzle since the flow is isentropic T 0 T i V i 2 2 c p 630 R 450 ft s 2 20240 Btu lbmR 1 Btu 1bm 25037 ft 2 s 2 6469 R P 0 P i T 0 T i kk 1 30 psia 6469 K 630 K 1414 1 329 psia From Table A32 or from Eqs 1718 and 1719 at Ma 1 we read T T 0 08333 P P 0 05283 Thus T 08333 T 0 083336469 R 539 R P 05283 P 0 05283329 psia 174 psia Also c i kR T i 14 006855 Btu lbmR 630 R 25037 ft 2 s 2 1 Btu lbm 1230 fts and Ma i V i c i 450 fts 1230 fts 03657 From Table A32 at this Mach number we read Ai A 17426 Thus the ratio of the throat area to the nozzle inlet area is A A i 1 17426 0574 Discussion If we solve this problem using the relations for compressible isentropic flow the results would be identical EXAMPLE 176 Air Loss from a Flat Tire Air in an automobile tire is maintained at a pressure of 220 kPa gage in an environ ment where the atmospheric pressure is 94 kPa The air in the tire is at the ambient temperature of 25C A 4mmdiameter leak develops in the tire as a result of an acci dent Fig 1725 Approximating the flow as isentropic determine the initial mass flow rate of air through the leak SOLUTION A leak develops in an automobile tire as a result of an accident The initial mass flow rate of air through the leak is to be determined Assumptions 1 Air is an ideal gas with constant specific heats 2 Flow of air through the hole is isentropic Properties The specific gas constant of air is R 0287 kPam3kgK The specific heat ratio of air at room temperature is k 14 FIGURE 1725 Schematic for Example 176 Air T 25C Pg 220 kPa Final PDF to printer 840 COMPRESSIBLE FLOW cen22672ch17823880indd 840 110917 0138 PM Analysis The absolute pressure in the tire is P P gage P atm 220 94 314 kPa The critical pressure is from Table 172 P 05283 P 0 05283 314 kPa 166 kPa 94 kPa Therefore the flow is choked and the velocity at the exit of the hole is the sonic speed Then the flow properties at the exit become ρ 0 P 0 R T 0 314 kPa 0287 kPa m 3 kgK298 K 3671 kgm 3 ρ ρ 2 k 1 1k1 3671 kgm 3 2 14 1 1141 2327 kgm 3 T 2 k 1 T 0 2 14 1 298 K 2483 K V c kR T 140287 kJkgK 1000 m 2 s 2 1 kJkg 2483 K 3159 ms Then the initial mass flow rate through the hole is m ρAV 2327 kgm 3 π 0004 m 2 4 3159 ms 000924 kgs 0554 kgmin Discussion The mass flow rate decreases with time as the pressure inside the tire drops ConvergingDiverging Nozzles When we think of nozzles we ordinarily think of flow passages whose crosssectional area decreases in the flow direction However the highest velocity to which a fluid can be accelerated in a converging nozzle is limited to the sonic velocity Ma 1 which occurs at the exit plane throat of the nozzle Accelerating a fluid to supersonic velocities Ma 1 can be accom plished only by attaching a diverging flow section to the subsonic nozzle at the throat The resulting combined flow section is a convergingdiverging nozzle which is standard equipment in supersonic aircraft and rocket propul sion Fig 1726 Forcing a fluid through a convergingdiverging nozzle is no guarantee that the fluid will be accelerated to a supersonic velocity In fact the fluid may find itself decelerating in the diverging section instead of accelerating if the back pressure is not in the right range The state of the nozzle flow is determined by the overall pressure ratio Pb P0 Therefore for given inlet conditions the flow through a convergingdiverging nozzle is governed by the back pressure Pb as will be explained Final PDF to printer 841 CHAPTER 17 cen22672ch17823880indd 841 110917 0138 PM FIGURE 1726 Convergingdiverging nozzles are commonly used in rocket engines to provide high thrust b NASA Nozzle Fuel Oxidizer Combustion chamber a b Consider the convergingdiverging nozzle shown in Fig 1727 A fluid enters the nozzle with a low velocity at stagnation pressure P0 When Pb P0 case A there is no flow through the nozzle This is expected since the flow in a nozzle is driven by the pressure difference between the nozzle inlet and the exit Now let us examine what happens as the back pressure is lowered 1 When P0 Pb PC the flow remains subsonic throughout the nozzle and the mass flow is less than that for choked flow The fluid velocity increases in the first converging section and reaches a maximum at the throat but Ma 1 However most of the gain in velocity is lost in the second diverging section of the nozzle which acts as a diffuser The pressure decreases in the converging section reaches a minimum at the throat and increases at the expense of velocity in the diverging section 2 When Pb PC the throat pressure becomes P and the fluid achieves sonic velocity at the throat But the diverging section of the nozzle still acts as a diffuser slowing the fluid to subsonic velocities The mass flow rate that was increasing with decreasing Pb also reaches its maximum value Recall that P is the lowest pressure that can be obtained at the throat and the sonic velocity is the highest velocity that can be achieved with a converging nozzle Thus lowering Pb further has no influence on the fluid flow in the converging part of the nozzle or the mass flow rate through the nozzle However it does influence the character of the flow in the diverging section 3 When PC Pb PE the fluid that achieved a sonic velocity at the throat continues accelerating to supersonic velocities in the diverging section as the pressure decreases This acceleration comes to a sudden stop however as a normal shock develops at a section between the throat and the exit plane which causes a sudden drop in velocity to subsonic levels and a sudden increase in pressure The fluid then continues to Final PDF to printer 842 COMPRESSIBLE FLOW cen22672ch17823880indd 842 110917 0138 PM decelerate further in the remaining part of the convergingdiverging nozzle Flow through the shock is highly irreversible and thus it cannot be approximated as isentropic The normal shock moves downstream away from the throat as Pb is decreased and it approaches the nozzle exit plane as Pb approaches PE When Pb PE the normal shock forms at the exit plane of the nozzle The flow is supersonic through the entire diverging section in this case and it can be approximated as isentropic However the fluid velocity drops to subsonic levels just before leaving the nozzle as it crosses the normal shock Normal shock waves are discussed in Sect 175 4 When PE Pb 0 the flow in the diverging section is supersonic and the fluid expands to PF at the nozzle exit with no normal shock forming within the nozzle Thus the flow through the nozzle can be approximated as isentropic When Pb PF no shocks occur within or outside the nozzle When Pb PF irreversible mixing and expansion waves occur downstream of the exit plane of the nozzle When Pb PF however the pressure of the fluid increases from PF to Pb irreversibly in the wake of the nozzle exit creating what are called oblique shocks FIGURE 1727 The effects of back pressure on the flow through a convergingdiverging nozzle 0 x Subsonic flow at nozzle exit shock in nozzle Exit P Supersonic flow at nozzle exit no shock in nozzle PA A Ma Throat Inlet B C D Subsonic flow at nozzle exit no shock Pe x Vi 0 Pb P0 E F G Shock in nozzle Sonic flow at throat Throat Exit Inlet 0 1 Sonic flow at throat Shock in nozzle E F G x Subsonic flow at nozzle exit shock in nozzle Supersonic flow at nozzle exit no shock in nozzle A B C D Subsonic flow at nozzle exit no shock P0 P Throat PB PC PD PE PG PF Pb Final PDF to printer 843 CHAPTER 17 cen22672ch17823880indd 843 110917 0138 PM EXAMPLE 177 Airflow Through a ConvergingDiverging Nozzle Air enters a convergingdiverging nozzle shown in Fig 1728 at 10 MPa and 800 K with negligible velocity The flow is steady onedimensional and isentropic with k 14 For an exit Mach number of Ma 2 and a throat area of 20 cm2 determine a the throat conditions b the exit plane conditions including the exit area and c the mass flow rate through the nozzle SOLUTION Air flows through a convergingdiverging nozzle The throat and the exit conditions and the mass flow rate are to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic Properties The specific heat ratio of air is given to be k 14 The gas constant of air is 0287 kJkgK Analysis The exit Mach number is given to be 2 Therefore the flow must be sonic at the throat and supersonic in the diverging section of the nozzle Since the inlet velocity is negligible the stagnation pressure and stagnation temperature are the same as the inlet temperature and pressure P0 10 MPa and T0 800 K Assuming idealgas behavior the stagnation density is ρ P 0 R T 0 1000 kPa 0287 kPa m 3 kgK800 K 4355 kgm 3 a At the throat of the nozzle Ma 1 and from Table A32 we read P P 0 05283 T T 0 08333 ρ ρ 0 06339 Thus P 05283 P 0 0528310 MPa 05283 MPa T 08333 T 0 08333800 K 6666 K ρ 06339 ρ 0 063394355 kgm 3 2761 kgm 3 Also V c kR T 140287 kJkgK6666 K 1000 m 2 s 2 1 kJkg 5175 ms b Since the flow is isentropic the properties at the exit plane can also be calculated by using data from Table A32 For Ma 2 we read P e P 0 01278 T e T 0 05556 ρ e ρ 0 02300 Ma e 16330 A e A 16875 Thus P e 01278 P 0 01278 10 MPa 01278 MPa T e 05556 T 0 05556 800 K 4445 K ρ e 02300 ρ 0 023004355 kgm 3 1002 kgm 3 A e 16875 A 1687520 cm 2 3375 cm 2 and V e Ma e c 163305175 ms 8451 ms FIGURE 1728 Schematic for Example 177 At 20 cm2 T0 800 K Mae 2 P0 10 MPa Vi 0 Final PDF to printer 844 COMPRESSIBLE FLOW cen22672ch17823880indd 844 110917 0138 PM 175 SHOCK WAVES AND EXPANSION WAVES We have seen that sound waves are caused by infinitesimally small pres sure disturbances and they travel through a medium at the speed of sound We have also seen in the present chapter that for some back pressure values abrupt changes in fluid properties occur in a very thin section of a converging diverging nozzle under supersonic flow conditions creating a shock wave It is of interest to study the conditions under which shock waves develop and how they affect the flow Normal Shocks First we consider shock waves that occur in a plane normal to the direction of flow called normal shock waves The flow process through the shock wave is highly irreversible and cannot be approximated as being isentropic Next we follow the footsteps of Pierre Laplace 17491827 G F Bernhard Riemann 18261866 William Rankine 18201872 Pierre Henry Hugo niot 18511887 Lord Rayleigh 18421919 and G I Taylor 18861975 and develop relationships for the flow properties before and after the shock We do this by applying the conservation of mass momentum and energy relations as well as some property relations to a stationary control volume that contains the shock as shown in Fig 1729 The normal shock waves are extremely thin so the entrance and exit flow areas for the control volume are approximately equal Fig 1730 We assume steady flow with no heat and work interactions and no potential energy changes Denoting the properties upstream of the shock by the sub script 1 and those downstream of the shock by 2 we have the following Conservation of mass ρ 1 A V 1 ρ 2 A V 2 1729 or ρ 1 V 1 ρ 2 V 2 The nozzle exit velocity could also be determined from Ve Maece where ce is the speed of sound at the exit conditions V e Ma e c e Ma e kR T e 2 14 0287 kJkgK 4445 K 1000 m 2 s 2 1 kJkg 8452 ms c Since the flow is steady the mass flow rate of the fluid is the same at all sections of the nozzle Thus it may be calculated by using properties at any cross section of the nozzle Using the properties at the throat we find that the mass flow rate is m ρAV 2761 kgm 3 20 10 4 m 2 5175 ms 286 kgs Discussion Note that this is the highest possible mass flow rate that can flow through this nozzle for the specified inlet conditions FIGURE 1729 Control volume for flow across a normal shock wave Control volume Flow Ma1 1 P V1 s Shock wave P V2 1 2 1 2 1 2 1 2 h h s Ma2 1 ρ ρ FIGURE 1730 Schlieren image of a normal shock in a Laval nozzle The Mach number in the nozzle just upstream to the left of the shock wave is about 13 Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the shock G S Settles Gas Dynamics Lab Penn State Uni versity Used with permission Final PDF to printer 845 CHAPTER 17 cen22672ch17823880indd 845 110917 0138 PM Conservation of energy h 1 V 1 2 2 h 2 V 2 2 2 1730 or h 01 h 02 1731 Linear momentum equation Rearranging Eq 1714 and integrating yield A P 1 P 2 m V 2 V 1 1732 Increase of entropy s 2 s 1 0 1733 We can combine the conservation of mass and energy relations into a single equation and plot it on an hs diagram using property relations The resultant curve is called the Fanno line and it is the locus of states that have the same value of stagnation enthalpy and mass flux mass flow per unit flow area Likewise combining the conservation of mass and momentum equations into a single equation and plotting it on the hs diagram yield a curve called the Rayleigh line Both these lines are shown on the hs diagram in Fig 1731 As proved later in Example 178 the points of maximum entropy on these lines points a and b correspond to Ma 1 The state on the upper part of each curve is subsonic and on the lower part supersonic The Fanno and Rayleigh lines intersect at two points points 1 and 2 which represent the two states at which all three conservation equations are satis fied One of these state 1 corresponds to the state before the shock and the other state 2 corresponds to the state after the shock Note that the flow is supersonic before the shock and subsonic afterward Therefore the flow must change from supersonic to subsonic if a shock is to occur The larger the Mach number before the shock the stronger the shock will be In the limiting case of Ma 1 the shock wave simply becomes a sound wave Notice from Fig 1731 that entropy increases s2 s1 This is expected since the flow through the shock is adiabatic but irreversible The conservation of energy principle Eq 1731 requires that the stag nation enthalpy remain constant across the shock h01 h02 For ideal gases h hT and thus T 01 T 02 1734 That is the stagnation temperature of an ideal gas also remains constant across the shock Note however that the stagnation pressure decreases across the shock because of the irreversibilities while the ordinary static tempera ture rises drastically because of the conversion of kinetic energy into enthalpy due to a large drop in fluid velocity Fig 1732 We now develop relations between various properties before and after the shock for an ideal gas with constant specific heats A relation for the ratio of the static temperatures T2T1 is obtained by applying Eq 1718 twice T 01 T 1 1 k 1 2 Ma 1 2 and T 02 T 2 1 k 1 2 Ma 2 2 Dividing the first equation by the second one and noting that T01 T02 we have T 2 T 1 1 Ma 1 2 k 1 2 1 Ma 2 2 k 1 2 1735 FIGURE 1731 The hs diagram for flow across a normal shock Ma 1 0 s SHOCK WAVE Subsonic flow h h a h01 1 2 h02 h02 h01 P02 P01 Ma 1 b V2 2 2 Ma 1 2 Supersonic flow Ma 1 h 1 s1 s2 V2 1 2 Fanno line Rayleigh line FIGURE 1732 Variation of flow properties across a normal shock in an ideal gas Normal shock P P0 V Ma T T0 ρ s increases decreases decreases decreases increases remains constant increases increases Final PDF to printer 846 COMPRESSIBLE FLOW cen22672ch17823880indd 846 110917 0138 PM From the idealgas equation of state ρ 1 P 1 R T 1 and ρ 2 P 2 R T 2 Substituting these into the conservation of mass relation ρ1V1 ρ2V2 and not ing that Ma Vc and c kRT we have T 2 T 1 P 2 V 2 P 1 V 1 P 2 Ma 2 c 2 P 1 Ma 1 c 1 P 2 Ma 2 T 2 P 1 Ma 1 T 1 P 2 P 1 2 Ma 2 Ma 1 2 1736 Combining Eqs 1735 and 1736 gives the pressure ratio across the shock Fanno line P 2 P 1 Ma 1 1 Ma 1 2 k 1 2 Ma 2 1 Ma 2 2 k 1 2 1737 Equation 1737 is a combination of the conservation of mass and energy equations thus it is also the equation of the Fanno line for an ideal gas with constant specific heats A similar relation for the Rayleigh line is obtained by combining the conservation of mass and momentum equations From Eq 1732 P 1 P 2 m A V 2 V 1 ρ 2 V 2 2 ρ 1 V 1 2 However ρ V 2 P RT Ma c 2 P RT Ma kRT 2 Pk Ma 2 Thus P 1 1 k Ma 1 2 P 2 1 k Ma 2 2 or Rayleigh line P 2 P 1 1 k Ma 1 2 1 k Ma 2 2 1738 Combining Eqs 1737 and 1738 yields Ma 2 2 Ma 1 2 2 k 1 2 Ma 1 2 k k 1 1 1739 This represents the intersections of the Fanno and Rayleigh lines and relates the Mach number upstream of the shock to that downstream of the shock The occurrence of shock waves is not limited to supersonic nozzles only This phenomenon is also observed at the engine inlet of supersonic aircraft where the air passes through a shock and decelerates to subsonic velocities before entering the diffuser of the engine Fig 1733 Explosions also pro duce powerful expanding spherical normal shocks which can be very destruc tive Fig 1734 FIGURE 1733 The air inlet of a supersonic fighter jet is designed such that a shock wave at the inlet decelerates the air to subsonic velocities increasing the pressure and temperature of the air before it enters the engine StockTrekGetty Images RF Final PDF to printer 847 CHAPTER 17 cen22672ch17823880indd 847 110917 0138 PM Various flow property ratios across the shock are listed in Table A33 for an ideal gas with k 14 Inspection of this table reveals that Ma2 the Mach number after the shock is always less than 1 and that the larger the supersonic Mach number before the shock the smaller the subsonic Mach number after the shock Also we see that the static pressure temperature and density all increase after the shock while the stagnation pressure decreases The entropy change across the shock is obtained by applying the entropy change equation for an ideal gas across the shock s 2 s 1 c p ln T 2 T 1 R ln P 2 P 1 1740 which can be expressed in terms of k R and Ma1 by using the relations devel oped earlier in this section A plot of nondimensional entropy change across the normal shock s2 s1R versus Ma1 is shown in Fig 1735 Since the flow across the shock is adiabatic and irreversible the second law of thermo dynamics requires that the entropy increase across the shock wave Thus a shock wave cannot exist for values of Ma1 less than unity where the entropy change would be negative For adiabatic flows shock waves can exist only for supersonic flows Ma1 1 FIGURE 1734 Schlieren image of the blast wave expanding spherical normal shock produced by the explosion of a firecracker The shock expanded radially outward in all directions at a supersonic speed that decreased with radius from the center of the explosion A microphone sensed the sudden change in pressure of the passing shock wave and triggered the microsecond flashlamp that exposed the photograph G S Settles Gas Dynamics Lab Penn State University Used with permission EXAMPLE 178 The Point of Maximum Entropy on the Fanno Line Show that the point of maximum entropy on the Fanno line point a of Fig 1731 for the adiabatic steady flow of a fluid in a duct corresponds to the sonic velocity Ma 1 SOLUTION It is to be shown that the point of maximum entropy on the Fanno line for steady adiabatic flow corresponds to sonic velocity FIGURE 1735 Entropy change across a normal shock 0 Impossible Subsonic flow before shock Ma1 Ma1 1 Supersonic flow before shock s2 s1 0 s2 s1 0 s2 s1R Final PDF to printer 848 COMPRESSIBLE FLOW cen22672ch17823880indd 848 110917 0138 PM Assumption The flow is steady adiabatic and onedimensional Analysis In the absence of any heat and work interactions and potential energy changes the steadyflow energy equation reduces to h V 2 2 constant Differentiating yields dh V dV 0 For a very thin shock with negligible change of duct area across the shock the steady flow continuity conservation of mass equation is expressed as ρV constant Differentiating we have ρ dV V dρ 0 Solving for dV gives dV V dρ ρ Combining this with the energy equation we have dh V 2 dρ ρ 0 which is the equation for the Fanno line in differential form At point a the point of maximum entropy ds 0 Then from the second T ds relation T ds dh v dP we have dh v dP dPρ Substituting yields dP ρ V 2 dρ ρ 0 at s constant Solving for V we have V P ρ s 12 which is the relation for the speed of sound Eq 179 Thus V c and the proof is complete EXAMPLE 179 Shock Wave in a ConvergingDiverging Nozzle If the air flowing through the convergingdiverging nozzle of Example 177 expe riences a normal shock wave at the nozzle exit plane Fig 1736 determine the following after the shock a the stagnation pressure static pressure static tempera ture and static density b the entropy change across the shock c the exit velocity and d the mass flow rate through the nozzle Approximate the flow as steady one dimensional and isentropic with k 14 from the nozzle inlet to the shock location SOLUTION Air flowing through a convergingdiverging nozzle experiences a normal shock at the exit The effect of the shock wave on various properties is to be determined FIGURE 1736 Schematic for Example 179 T1 4445 K Ma1 2 P01 10 MPa P1 01278 MPa ρ1 1002 kgm3 Shock wave 1 2 m 286 kgs Final PDF to printer 849 CHAPTER 17 cen22672ch17823880indd 849 110917 0138 PM Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 Flow through the nozzle is steady onedimensional and isentropic before the shock occurs 3 The shock wave occurs at the exit plane Properties The constantpressure specific heat and the specific heat ratio of air are cp 1005 kJkgK and k 14 The gas constant of air is 0287 kJkgK Analysis a The fluid properties at the exit of the nozzle just before the shock denoted by subscript 1 are those evaluated in Example 177 at the nozzle exit to be P 01 10 MPa P 1 01278 MPa T 1 4445 K ρ 1 1002 kgm 3 The fluid properties after the shock denoted by subscript 2 are related to those before the shock through the functions listed in Table A33 For Ma1 20 we read Ma 2 05774 P 02 P 01 07209 P 2 P 1 45000 T 2 T 1 16875 ρ 2 ρ 1 26667 Then the stagnation pressure P02 static pressure P2 static temperature T2 and static density ρ2 after the shock are P 02 07209 P 01 0720910 MPa 0721 MPa P 2 45000 P 1 4500001278 MPa 0575 MPa T 2 16875 T 1 168754445 K 750 K ρ 2 26667 ρ 1 266671002 kgm 3 267 kgm 3 b The entropy change across the shock is s 2 s 1 c p ln T 2 T 1 R ln P 2 P 1 1005 kJkgK ln 16875 0287 kJkgK ln 45000 00942 kJkgK Thus the entropy of the air increases as it passes through a normal shock which is highly irreversible c The air velocity after the shock is determined from V2 Ma2c2 where c2 is the speed of sound at the exit conditions after the shock V 2 Ma 2 c 2 Ma 2 kR T 2 05774 140287 kJkgK7501 K 1000 m 2 s 2 1 kJkg 317 ms d The mass flow rate through a convergingdiverging nozzle with sonic conditions at the throat is not affected by the presence of shock waves in the nozzle Therefore the mass flow rate in this case is the same as that determined in Example 177 m 286 kgs Discussion This result can easily be verified by using property values at the nozzle exit after the shock at all Mach numbers significantly greater than unity Final PDF to printer 850 COMPRESSIBLE FLOW cen22672ch17823880indd 850 110917 0138 PM Example 179 illustrates that the stagnation pressure and velocity decrease while the static pressure temperature density and entropy increase across the shock Fig 1737 The rise in the temperature of the fluid downstream of a shock wave is of major concern to the aerospace engineer because it creates heat transfer problems on the leading edges of wings and nose cones of space reentry vehicles and the recently proposed hypersonic space planes Overheating in fact led to the tragic loss of the space shuttle Columbia in February 2003 as it was reentering earths atmosphere Oblique Shocks Not all shock waves are normal shocks perpendicular to the flow direction For example when the space shuttle travels at supersonic speeds through the atmosphere it produces a complicated shock pattern consisting of inclined shock waves called oblique shocks Fig 1738 As you can see some por tions of an oblique shock are curved while other portions are straight First we consider straight oblique shocks like that produced when a uni form supersonic flow Ma1 1 impinges on a slender twodimensional wedge of halfangle δ Fig 1739 Since information about the wedge can not travel upstream in a supersonic flow the fluid knows nothing about the wedge until it hits the nose At that point since the fluid cannot flow through the wedge it turns suddenly through an angle called the turning angle or deflection angle θ The result is a straight oblique shock wave aligned at shock angle or wave angle β measured relative to the oncoming flow Fig 1740 To conserve mass β must obviously be greater than δ Since the Reynolds number of supersonic flows is typically large the boundary layer growing along the wedge is very thin and we ignore its effects The flow therefore turns by the same angle as the wedge namely deflection angle θ is equal to wedge halfangle δ If we take into account the displacement thick ness effect of the boundary layer the deflection angle θ of the oblique shock turns out to be slightly greater than wedge halfangle δ Like normal shocks the Mach number decreases across an oblique shock and oblique shocks are possible only if the upstream flow is supersonic How ever unlike normal shocks in which the downstream Mach number is always subsonic Ma2 downstream of an oblique shock can be subsonic sonic or super sonic depending on the upstream Mach number Ma1 and the turning angle FIGURE 1737 When a lion tamer cracks his whip a weak spherical shock wave forms near the tip and spreads out radially the pressure inside the expanding shock wave is higher than ambient air pressure and this is what causes the crack when the shock wave reaches the lions ear Joshua EtsHokinGetty Images RF FIGURE 1738 Schlieren image of a small model of the space shuttle orbiter being tested at Mach 3 in the supersonic wind tunnel of the Penn State Gas Dynamics Lab Several oblique shocks are seen in the air surrounding the spacecraft G S Settles Gas Dynamics Lab Penn State Uni versity Used with permission Final PDF to printer 851 CHAPTER 17 cen22672ch17823880indd 851 110917 0138 PM We analyze a straight oblique shock in Fig 1740 by decomposing the velocity vectors upstream and downstream of the shock into normal and tangential components and considering a small control volume around the shock Upstream of the shock all fluid properties velocity density pressure etc along the lower left face of the control volume are identical to those along the upper right face The same is true downstream of the shock There fore the mass flow rates entering and leaving those two faces cancel each other out and conservation of mass reduces to ρ 1 V 1 n A ρ 2 V 2 n A ρ 1 V 1 n ρ 2 V 2 n 1741 where A is the area of the control surface that is parallel to the shock Since A is the same on either side of the shock it has dropped out of Eq 1741 As you might expect the tangential component of velocity parallel to the oblique shock does not change across the shock that is V1 t V2 t This is easily proven by applying the tangential momentum equation to the control volume When we apply conservation of momentum in the direction normal to the oblique shock the only forces are pressure forces and we get P 1 A P 2 A ρ V 2 n A V 2 n ρ V 1 n A V 1 n P 1 P 2 ρ 2 V 2 n 2 ρ 1 V 1 n 2 1742 Finally since there is no work done by the control volume and no heat transfer into or out of the control volume stagnation enthalpy does not change across an oblique shock and conservation of energy yields h 01 h 02 h 0 h 1 1 2 V 1 n 2 1 2 V 1 t 2 h 2 1 2 V 2 n 2 1 2 V 2 t 2 But since V1 t V2 t this equation reduces to h 1 1 2 V 1 n 2 h 2 1 2 V 2 n 2 1743 Careful comparison reveals that the equations for conservation of mass momentum and energy Eqs 1741 through 1743 across an oblique shock are identical to those across a normal shock except that they are written in terms of the normal velocity component only Therefore the normal shock relations derived previously apply to oblique shocks as well but must be writ ten in terms of Mach numbers Ma1 n and Ma2 n normal to the oblique shock This is most easily visualized by rotating the velocity vectors in Fig 1740 by angle π2 β so that the oblique shock appears to be vertical Fig 1741 Trigonometry yields Ma 1 n Ma 1 sin β and Ma 2 n Ma 2 sin β θ 1744 where Ma1 n V1 n c1 and Ma2 n V2 n c2 From the point of view shown in Fig 1741 we see what looks like a normal shock but with some superposed tangential flow coming along for the ride Thus All the equations shock tables etc for normal shocks apply to oblique shocks as well provided that we use only the normal components of the Mach number In fact you may think of normal shocks as special oblique shocks in which shock angle β π2 or 90 We recognize immediately that an oblique shock FIGURE 1739 An oblique shock of shock angle β formed by a slender twodimensional wedge of halfangle δ The flow is turned by deflection angle θ downstream of the shock and the Mach number decreases Ma1 Ma1 Ma2 Oblique shock θ β δ FIGURE 1740 Velocity vectors through an oblique shock of shock angle β and deflection angle θ V2 n Oblique shock Control volume V1 n V1 V2 V1 t P1 P2 V2 t θ β Final PDF to printer 852 COMPRESSIBLE FLOW cen22672ch17823880indd 852 110917 0138 PM can exist only if Ma1 n 1 and Ma2 n 1 The normal shock equations appro priate for oblique shocks in an ideal gas are summarized in Fig 1742 in terms of Ma1 n For known shock angle β and known upstream Mach number Ma1 we use the first part of Eq 1744 to calculate Ma1 n and then use the normal shock tables or their corresponding equations to obtain Ma2 n If we also knew the deflection angle θ we could calculate Ma2 from the second part of Eq 1744 But in a typical application we know either β or θ but not both Fortunately a bit more algebra provides us with a relationship between θ β and Ma1 We begin by noting that tan β V1 n V1 t and tanβ θ V2 n V2 t Fig 1741 But since V1 t V2 t we combine these two expressions to yield V 2 n V 1 n tan β θ tan β 2 k 1 Ma 1 n 2 k 1 Ma 1 n 2 2 k 1 Ma 1 2 sin 2 β k 1 Ma 1 2 sin 2 β 1745 where we have also used Eq 1744 and the fourth equation of Fig 1742 We apply trigonometric identities for cos 2β and tanβ θ namely cos 2β cos 2 β sin 2 β and tan β θ tan β tan θ 1 tan β tan θ After some algebra Eq 1745 reduces to The θβMa relationship tan θ 2 cot β Ma 1 2 sin 2 β 1 Ma 1 2 k cos 2β 2 1746 Equation 1746 provides deflection angle θ as a unique function of shock angle β specific heat ratio k and upstream Mach number Ma1 For air k 14 we plot θ versus β for several values of Ma1 in Fig 1743 We note that this plot is often presented with the axes reversed β versus θ in com pressible flow textbooks since physically shock angle β is determined by deflection angle θ Much can be learned by studying Fig 1743 and we list some observations here Figure 1743 displays the full range of possible shock waves at a given freestream Mach number from the weakest to the strongest For any value of Mach number Ma1 greater than 1 the possible values of θ range from θ 0 at some value of β between 0 and 90 to a maximum value θ θmax at an intermediate value of β and then back to θ 0 at β 90 Straight oblique shocks for θ or β outside of this range cannot and do not exist At Ma1 15 for example straight oblique shocks cannot exist in air with shock angle β less than about 42 nor with deflection angle θ greater than about 12 If the wedge halfangle is greater than θmax the shock becomes curved and detaches from the nose of the wedge forming what is called a detached oblique shock or a bow wave Fig 1744 The shock angle β of the detached shock is 90 at the nose but β decreases as the shock curves downstream Detached shocks are much more complicated than simple straight oblique shocks to analyze In fact no simple solutions exist and prediction of detached shocks requires computational methods FIGURE 1742 Relationships across an oblique shock for an ideal gas in terms of the normal component of upstream Mach number Ma1 n P P k 1Ma 2 k 1Ma k 1 2k Ma k 1 T T 2 k 1Ma 2k Ma k 1 k 1 Ma V V k 1Ma 2 k 1Ma P P 2k Ma k 1 k 1 Ma k 1Ma 2 2k Ma k 1 h h T T ρ ρ FIGURE 1741 The same velocity vectors as in Fig 1740 but rotated by angle π2 β so that the oblique shock is vertical Normal Mach numbers Ma1 n and Ma2 n are also defined V1n P1 P2 V1 V1t Ma1n 1 Ma2n 1 Oblique shock V2n V2t V2 θ θ β β Final PDF to printer 853 CHAPTER 17 cen22672ch17823880indd 853 110917 0138 PM Similar oblique shock behavior is observed in axisymmetric flow over cones as in Fig 1745 although the θβMa relationship for axisymmetric flows differs from that of Eq 1746 When supersonic flow impinges on a blunt or bluff bodya body without a sharply pointed nose the wedge halfangle δ at the nose is 90 and an attached oblique shock cannot exist regardless of Mach number In fact a detached oblique shock occurs in front of all such blunt nosed bodies whether twodimensional axisymmetric or fully three dimensional For example a detached oblique shock is seen in front of the space shuttle model in Fig 1738 and in front of a sphere in Fig 1746 While θ is a unique function of Ma1 and β for a given value of k there are two possible values of β for θ θmax The dashed red line in Fig 1743 passes through the locus of θmax values dividing the shocks into weak oblique shocks the smaller value of β and strong oblique shocks the larger value of β At a given value of θ the weak shock is more common and is preferred by the flow unless the downstream pressure conditions are high enough for the formation of a strong shock For a given upstream Mach number Ma1 there is a unique value of θ for which the downstream Mach number Ma2 is exactly 1 The dashed green line in Fig 1743 passes through the locus of values where Ma2 1 To the left of this line Ma2 1 and to the right of this line Ma2 1 Downstream sonic conditions occur on the weak shock side of the plot with θ very close to θmax Thus the flow downstream of a strong oblique shock is always subsonic Ma2 1 The flow downstream of a weak oblique shock remains supersonic except for a narrow range of θ just below θmax where it is subsonic although it is still called a weak oblique shock As the upstream Mach number approaches infinity straight oblique shocks become possible for any β between 0 and 90 but the maximum possible turning angle for k 14 air is θmax 456 which occurs at β 678 Straight oblique shocks with turning angles above this value of θmax are not possible regardless of the Mach number FIGURE 1743 The dependence of straight oblique shock deflection angle θ on shock angle β for several values of upstream Mach number Ma1 Calculations are for an ideal gas with k 14 The dashed red line connects points of maximum deflection angle θ θmax Weak oblique shocks are to the left of this line while strong oblique shocks are to the right of this line The dashed green line connects points where the downstream Mach number is sonic Ma2 1 Supersonic downstream flow Ma2 1 is to the left of this line while subsonic downstream flow Ma2 1 is to the right of this line 0 10 20 30 40 β degrees θ degrees Ma2 1 Ma2 1 Ma1 θ θmax Weak 50 12 15 2 3 10 5 60 70 80 90 0 10 20 30 40 50 Strong Ma2 1 FIGURE 1744 A detached oblique shock occurs upstream of a twodimensional wedge of halfangle δ when δ is greater than the maximum possible deflection angle θ A shock of this kind is called a bow wave because of its resemblance to the water wave that forms at the bow of a ship Ma1 Detached oblique shock δ θmax Final PDF to printer 854 COMPRESSIBLE FLOW cen22672ch17823880indd 854 110917 0138 PM For a given value of upstream Mach number there are two shock angles where there is no turning of the flow θ 0 the strong case β 90 corresponds to a normal shock and the weak case β βmin represents the weakest possible oblique shock at that Mach number which is called a Mach wave Mach waves are caused for example by very small nonuniformities on the walls of a supersonic wind tunnel several can be seen in Figs 1738 and 1745 Mach waves have no effect on the flow since the shock is vanishingly weak In fact in the limit Mach waves are isentropic The shock angle for Mach waves is a unique function of the Mach number and is given the symbol μ not to be confused with the coefficient of viscosity Angle μ is called the Mach angle and is found by setting θ equal to zero in Eq 1746 solving for β μ and taking the smaller root We get Mach angle μ sin 1 1 Ma 1 1747 Since the specific heat ratio appears only in the denominator of Eq 1746 μ is independent of k Thus we can estimate the Mach number of any supersonic flow simply by measuring the Mach angle and applying Eq 1747 PrandtlMeyer Expansion Waves We now address situations where supersonic flow is turned in the opposite direction such as in the upper portion of a twodimensional wedge at an angle of attack greater than its halfangle δ Fig 1747 We refer to this type of flow as an expanding flow whereas a flow that produces an oblique shock may be called a compressing flow As previously the flow changes direction to conserve mass However unlike a compressing flow an expand ing flow does not result in a shock wave Rather a continuous expanding region called an expansion fan appears composed of an infinite number of Mach waves called PrandtlMeyer expansion waves In other words the flow does not turn suddenly as through a shock but graduallyeach suc cessive Mach wave turns the flow by an infinitesimal amount Since each individual expansion wave is nearly isentropic the flow across the entire expansion fan is also nearly isentropic The Mach number downstream of the expansion increases Ma2 Ma1 while pressure density and temper ature decrease just as they do in the supersonic expanding portion of a convergingdiverging nozzle FIGURE 1746 Color schlieren image of Mach 30 flow from left to right over a sphere A curved shock wave called a bow shock forms in front of the sphere and curves downstream G S Settles Gas Dynamics Lab Penn State University Used with permission Ma1 δ δ 20 δ 40 δ 60 a b c FIGURE 1745 Still frames from schlieren video graphy illustrating the detachment of an oblique shock from a cone with increasing cone halfangle δ in air at Mach 3 At a δ 20 and b δ 40 the oblique shock remains attached but by c δ 60 the oblique shock has detached forming a bow wave G S Settles Gas Dynamics Lab Penn State University Used with permission Final PDF to printer 855 CHAPTER 17 cen22672ch17823880indd 855 110917 0138 PM PrandtlMeyer expansion waves are inclined at the local Mach angle μ as sketched in Fig 1747 The Mach angle of the first expansion wave is easily determined as μ1 sin11Ma1 Similarly μ2 sin11Ma2 where we must be careful to measure the angle relative to the new direction of flow down stream of the expansion namely parallel to the upper wall of the wedge in Fig 1747 if we neglect the influence of the boundary layer along the wall But how do we determine Ma2 It turns out that the turning angle θ across the expansion fan can be calculated by integration making use of the isentropic flow relationships For an ideal gas the result is Anderson 2003 Turning angle across an expansion fan θ ν Ma 2 ν Ma 1 1748 where νMa is an angle called the PrandtlMeyer function not to be con fused with the kinematic viscosity νMa k 1 k 1 tan 1 k 1 k 1 Ma 2 1 tan 1 Ma 2 1 1749 Note that νMa is an angle and can be calculated in either degrees or radians Physically νMa is the angle through which the flow must expand starting with ν 0 at Ma 1 in order to reach a supersonic Mach number Ma 1 To find Ma2 for known values of Ma1 k and θ we calculate νMa1 from Eq 1749 νMa2 from Eq 1748 and then Ma2 from Eq 1749 noting that the last step involves solving an implicit equation for Ma2 Since there is no heat transfer or work and the flow can be approximated as isentropic through the expansion T0 and P0 remain constant and we use the isentropic flow rela tions derived previously to calculate other flow properties downstream of the expansion such as T2 ρ2 and P2 PrandtlMeyer expansion fans also occur in axisymmetric supersonic flows as in the corners and trailing edges of a conecylinder Fig 1748 Some very complex and to some of us beautiful interactions involving both shock waves and expansion waves occur in the supersonic jet produced by an overexpanded nozzle as in Fig 1749 When such patterns are visible in the exhaust of a jet engine pilots refer to it as a tiger tail Analysis of such flows is beyond the scope of the present text interested readers are referred to compressible flow textbooks such as Thompson 1972 Leipmann and Roshko 2001 and Anderson 2003 FIGURE 1747 An expansion fan in the upper portion of the flow formed by a two dimensional wedge at an angle of attack in a supersonic flow The flow is turned by angle θ and the Mach number increases across the expansion fan Mach angles upstream and downstream of the expansion fan are indicated Only three expansion waves are shown for simplicity but in fact there are an infinite number of them An oblique shock is also present in the bottom portion of this flow θ Ma1 1 μ1 μ2 Ma2 Expansion waves Oblique shock δ FIGURE 1748 a Mach 3 flow over an axisymmetric cone of 10 halfangle The boundary layer becomes turbulent shortly downstream of the nose generating Mach waves that are visible in the color schlieren image b A similar pattern is seen in this color schlieren image for Mach 3 flow over an 11 2D wedge Expansion waves are seen at the corners where the wedge flattens out a and b G S Settles Gas Dynamics Lab Penn State University Used with permission a b Final PDF to printer 856 COMPRESSIBLE FLOW cen22672ch17823880indd 856 110917 0138 PM EXAMPLE 1710 Estimation of the Mach Number from Mach Lines Estimate the Mach number of the freestream flow upstream of the space shuttle in Fig 1738 from the figure alone Compare with the known value of Mach number provided in the figure caption SOLUTION We are to estimate the Mach number from a figure and compare it to the known value Analysis Using a protractor we measure the angle of the Mach lines in the freestream flow µ 19 The Mach number is obtained from Eq 1747 μ sin 1 1 Ma 1 Ma 1 1 sin 19 ο Ma 1 307 Our estimated Mach number agrees with the experimental value of 30 01 Discussion The result is independent of the fluid properties FIGURE 1749 The complex interactions between shock waves and expansion waves in an overexpanded supersonic jet a The flow is visualized by a schlierenlike differential interferogram b Color schlieren image a Photo by H Oertel sen Reproduced by cour tesy of the FrenchGerman Research Institute of SaintLouis ISL Used with permission b G S Settles Gas Dynamics Lab Penn State University Used with permission a b Final PDF to printer 857 CHAPTER 17 cen22672ch17823880indd 857 110917 0138 PM EXAMPLE 1711 Oblique Shock Calculations Supersonic air at Ma1 20 and 750 kPa impinges on a twodimensional wedge of halfangle δ 10 Fig 1750 Calculate the two possible oblique shock angles βweak and βstrong that could be formed by this wedge For each case calculate the pres sure and Mach number downstream of the oblique shock compare and discuss SOLUTION We are to calculate the shock angle Mach number and pressure down stream of the weak and strong oblique shock formed by a twodimensional wedge Assumptions 1 The flow is steady 2 The boundary layer on the wedge is very thin Properties The fluid is air with k 14 Analysis Because of assumption 2 we approximate the oblique shock deflection angle as equal to the wedge halfangle that is θ δ 10 With Ma1 20 and θ 10 we solve Eq 1746 for the two possible values of oblique shock angle β βweak 393 and βstrong 837 From these values we use the first part of Eq 1744 to calculate upstream normal Mach number Ma1n Weak shock Ma 1 n Ma 1 sin β Ma 1 n 20 sin 393 1267 and Strong shock Ma 1 n Ma 1 sin β Ma1 n 20 sin 837 1988 We substitute these values of Ma1 n into the second equation of Fig 1742 to calculate the downstream normal Mach number Ma2 n For the weak shock Ma2 n 08032 and for the strong shock Ma2 n 05794 We also calculate the downstream pressure for each case using the third equation of Fig 1742 which gives Weak shock P 2 P 1 2k Ma 1 n 2 k 1 k 1 P 2 750 kPa 2 14 1267 2 14 1 14 1 128 kPa and Strong shock P 2 P 1 2k Ma 1 n 2 k 1 k 1 P 2 750 kPa 2 14 1988 2 14 1 14 1 333 kPa Finally we use the second part of Eq 1744 to calculate the downstream Mach number Weak shock Ma 2 Ma 2 n sin β θ 08032 sin 393 10 164 and Strong shock Ma 2 Ma 2 n sin β θ 05794 sin 837 10 0604 The changes in Mach number and pressure across the strong shock are much greater than the changes across the weak shock as expected Discussion Since Eq 1746 is implicit in β we solve it by an iterative approach or with an equation solver For both the weak and strong oblique shock cases Ma1 n is super sonic and Ma2 n is subsonic However Ma2 is supersonic across the weak oblique shock but subsonic across the strong oblique shock We could also use the normal shock tables in place of the equations but with loss of precision FIGURE 1750 Two possible oblique shock angles a βweak and b βstrong formed by a twodimensional wedge of halfangle δ 10 Ma1 Strong shock δ 10 βstrong a b Ma1 Weak shock δ 10 βweak Final PDF to printer 858 COMPRESSIBLE FLOW cen22672ch17823880indd 858 110917 0138 PM EXAMPLE 1712 PrandtlMeyer Expansion Wave Calculations Supersonic air at Ma1 20 and 230 kPa flows parallel to a flat wall that suddenly expands by δ 10 Fig 1751 Ignoring any effects caused by the boundary layer along the wall calculate downstream Mach number Ma2 and pressure P2 SOLUTION We are to calculate the Mach number and pressure downstream of a sudden expansion along a wall Assumptions 1 The flow is steady 2 The boundary layer on the wall is very thin Properties The fluid is air with k 14 Analysis Because of assumption 2 we approximate the total deflection angle as equal to the wall expansion angle ie θ δ 10 With Ma1 20 we solve Eq 1749 for the upstream PrandtlMeyer function νMa k 1 k 1 tan 1 k 1 k 1 Ma 2 1 tan 1 Ma 2 1 14 1 14 1 tan 1 14 1 14 1 20 2 1 tan 1 20 2 1 2638 Next we use Eq 1748 to calculate the downstream PrandtlMeyer function θ ν Ma 2 ν Ma 1 ν Ma 2 θ ν Ma 1 10 2638 3638 Ma2 is found by solving Eq 1749 which is implicitan equation solver is helpful We get Ma2 238 There are also compressible flow calculators on the Internet that solve these implicit equations along with both normal and oblique shock equations We use the isentropic relations to calculate the downstream pressure P 2 P 2 P 0 P 1 P 0 P 1 1 k 1 2 Ma 2 2 kk 1 1 k 1 2 Ma 1 2 kk 1 230 kPa 126 kPa Since this is an expansion Mach number increases and pressure decreases as expected Discussion We could also solve for downstream temperature density etc using the appropriate isentropic relations FIGURE 1751 An expansion fan caused by the sudden expansion of a wall with δ 10 Ma1 20 Ma2 δ 10 θ 176 DUCT FLOW WITH HEAT TRANSFER AND NEGLIGIBLE FRICTION RAYLEIGH FLOW So far we have limited our consideration mostly to isentropic flow also called reversible adiabatic flow since it involves no heat transfer and no irrevers ibilities such as friction Many compressible flow problems encountered in practice involve chemical reactions such as combustion nuclear reactions evaporation and condensation as well as heat gain or heat loss through the duct wall Such problems are difficult to analyze exactly since they may involve Final PDF to printer 859 CHAPTER 17 cen22672ch17823880indd 859 110917 0138 PM FIGURE 1752 Many practical compressible flow problems involve combustion which may be modeled as heat gain through the duct wall Fuel nozzles or spray bars Flame holders Air inlet FIGURE 1753 Control volume for flow in a constantarea duct with heat transfer and negligible friction P1 T1 ρ1 P2 T2 ρ2 V1 Control volume Q V2 significant changes in chemical composition during flow and the conversion of latent chemical and nuclear energies to thermal energy Fig 1752 The essential features of such complex flows can still be captured by a simple analysis by modeling the generation or absorption of thermal energy as heat transfer through the duct wall at the same rate and disregarding any changes in chemical composition This simplified problem is still too compli cated for an elementary treatment of the topic since the flow may involve fric tion variations in duct area and multidimensional effects In this section we limit our consideration to onedimensional flow in a duct of constant cross sectional area with negligible frictional effects Consider steady onedimensional flow of an ideal gas with constant spe cific heats through a constantarea duct with heat transfer but with negligible friction Such flows are referred to as Rayleigh flows after Lord Rayleigh 18421919 The conservation of mass momentum and energy equa tions entropy change and equation of state for the control volume shown in Fig 1753 are written as follows Mass equation Noting that the duct crosssectional area A is constant the relation m 1 m 2 or ρ1A1V1 ρ2A2V2 reduces to ρ 1 V 1 ρ 2 V 2 1750 xMomentum equation Noting that the frictional effects are negligible and thus there are no shear forces and assuming there are no external and body forces the momentum equation F out β m V in β m V in the flow or x direction becomes a balance between static pressure forces and momentum transfer Noting that the flows are high speed and turbulent and we are ignoring friction the momentum flux correction factor is approximately 1 β 1 and thus can be neglected Then P 1 A 1 P 2 A 2 m V 2 m V 1 P 1 P 2 ρ 2 V 2 V 2 ρ 1 V 1 V 1 or P 1 ρ 1 V 1 2 P 2 ρ 2 V 2 2 1751 Energy equation The control volume involves no shear shaft or other forms of work and the potential energy change is negligible If the rate of heat transfer is Q and the heat transfer per unit mass of fluid is q Q m the steadyflow energy balance E in E out becomes Q m h 1 V 1 2 2 m h 2 V 2 2 2 q h 1 V 1 2 2 h 2 V 2 2 2 1752 For an ideal gas with constant specific heats Δh cp ΔT and thus q c p T 2 T 1 V 2 2 V 1 2 2 1753 or q h 02 h 01 c p T 02 T 01 1754 Therefore the stagnation enthalpy h0 and stagnation temperature T0 change during Rayleigh flow both increase when heat is transferred to the fluid Final PDF to printer 860 COMPRESSIBLE FLOW cen22672ch17823880indd 860 110917 0138 PM and thus q is positive and both decrease when heat is transferred from the fluid and thus q is negative Entropy change In the absence of any irreversibilities such as friction the entropy of a system changes by heat transfer only it increases with heat gain and it decreases with heat loss Entropy is a property and thus a state function and the entropy change of an ideal gas with constant specific heats during a change of state from 1 to 2 is given by s 2 s 1 c p ln T 2 T 1 R ln P 2 P 1 1755 The entropy of a fluid may increase or decrease during Rayleigh flow depending on the direction of heat transfer Equation of state Noting that P ρRT the properties P ρ and T of an ideal gas at states 1 and 2 are related to each other by P 1 ρ 1 T 1 P 2 ρ 2 T 2 1756 Consider a gas with known properties R k and cp For a specified inlet state 1 the inlet properties P1 T1 ρ1 V1 and s1 are known The five exit proper ties P2 T2 ρ2 V2 and s2 can be determined from Eqs 1750 1751 1753 1755 and 1756 for any specified value of heat transfer q When the veloc ity and temperature are known the Mach number can be determined from Ma V c V kRT Obviously there is an infinite number of possible downstream states 2 cor responding to a given upstream state 1 A practical way of determining these downstream states is to assume various values of T2 and calculate all other properties as well as the heat transfer q for each assumed T2 from Eqs 1750 through 1756 Plotting the results on a Ts diagram gives a curve passing through the specified inlet state as shown in Fig 1754 The plot of Rayleigh flow on a Ts diagram is called the Rayleigh line and several important observations can be made from this plot and the results of the calculations 1 All the states that satisfy the conservation of mass momentum and energy equations as well as the property relations are on the Rayleigh line Therefore for a given initial state the fluid cannot exist at any downstream state outside the Rayleigh line on a Ts diagram In fact the Rayleigh line is the locus of all physically attainable downstream states corresponding to an initial state 2 Entropy increases with heat gain and thus we proceed to the right on the Rayleigh line as heat is transferred to the fluid The Mach number is Ma 1 at point a which is the point of maximum entropy see Example 1713 for proof The states on the upper arm of the Rayleigh line above point a are subsonic and the states on the lower arm below point a are supersonic Therefore a process proceeds to the right on the Rayleigh line with heat addition and to the left with heat rejection regardless of the initial value of the Mach number 3 Heating increases the Mach number for subsonic flow but decreases it for supersonic flow The flow Mach number approaches Ma 1 in both cases from 0 in subsonic flow and from in supersonic flow during heating FIGURE 1754 Ts diagram for flow in a constantarea duct with heat transfer and negligible friction Rayleigh flow Mab 1 k Maa 1 Ma 1 Ma 1 Cooling Ma 0 Cooling Ma Heating Ma 1 Heating Ma 1 Tmax smax s T a b Final PDF to printer 861 CHAPTER 17 cen22672ch17823880indd 861 110917 0138 PM 4 It is clear from the energy balance q cpT02 T01 that heating increases the stagnation temperature T0 for both subsonic and supersonic flows and cooling decreases it The maximum value of T0 occurs at Ma 1 This is also the case for the static temperature T except for the narrow Mach number range of 1 k Ma 1 in subsonic flow see Example 1713 Both temperature and the Mach number increase with heating in subsonic flow but T reaches a maximum Tmax at Ma 1 k which is 0845 for air and then decreases It may seem peculiar that the temperature of a fluid drops as heat is transferred to it But this is no more peculiar than the fluid velocity increasing in the diverging section of a convergingdiverging nozzle The cooling effect in this region is due to the large increase in the fluid velocity and the accompanying drop in temperature in accordance with the relation T0 T V2 2cp Note also that heat rejection in the region 1 k Ma 1 causes the fluid temperature to increase Fig 1755 5 The momentum equation P KV constant where K ρV constant from the continuity equation reveals that velocity and static pressure have opposite trends Therefore static pressure decreases with heat gain in subsonic flow since velocity and the Mach number increase but increases with heat gain in supersonic flow since velocity and the Mach number decrease 6 The continuity equation ρV constant indicates that density and velocity are inversely proportional Therefore density decreases with heat transfer to the fluid in subsonic flow since velocity and the Mach number increase but increases with heat gain in supersonic flow since velocity and the Mach number decrease 7 On the left half of Fig 1754 the lower arm of the Rayleigh line is steeper than the upper arm in terms of s as a function of T which indicates that the entropy change corresponding to a specified temperature change and thus a given amount of heat transfer is larger in supersonic flow The effects of heating and cooling on the properties of Rayleigh flow are listed in Table 173 Note that heating or cooling has opposite effects on most properties Also the stagnation pressure decreases during heating and increases during cooling regardless of whether the flow is subsonic or supersonic FIGURE 1755 During heating fluid temperature always increases if the Rayleigh flow is supersonic but the temperature may actually drop if the flow is subsonic T01 Supersonic flow Heating T02 T01 T1 T2 T1 T01 Subsonic flow Heating T02 T01 T1 T2 T1 or T2 T1 TABLE 173 The effects of heating and cooling on the properties of Rayleigh flow Heating Cooling Property Subsonic Supersonic Subsonic Supersonic Velocity V Increase Decrease Decrease Increase Mach number Ma Increase Decrease Decrease Increase Stagnation temperature T0 Increase Increase Decrease Decrease Temperature T Increase for Ma 1k12 Increase Decrease for Ma 1k12 Decrease Decrease for Ma 1k12 Increase for Ma 1k12 Density ρ Decrease Increase Increase Decrease Stagnation pressure P0 Decrease Decrease Increase Increase Pressure P Decrease Increase Increase Decrease Entropy s Increase Increase Decrease Decrease Final PDF to printer 862 COMPRESSIBLE FLOW cen22672ch17823880indd 862 110917 0138 PM EXAMPLE 1713 Extrema of Rayleigh Line Consider the Ts diagram of Rayleigh flow as shown in Fig 1756 Using the differ ential forms of the conservation equations and property relations show that the Mach number is Maa 1 at the point of maximum entropy point a and Ma b 1 k at the point of maximum temperature point b SOLUTION It is to be shown that Maa 1 at the point of maximum entropy and Ma b 1 k at the point of maximum temperature on the Rayleigh line Assumptions The assumptions associated with Rayleigh flow ie steady one dimensional flow of an ideal gas with constant properties through a constant cross sectional area duct with negligible frictional effects are valid Analysis The differential forms of the continuity ρV constant momentum rear ranged as P ρVV constant ideal gas P ρRT and enthalpy change Δh cp ΔT equations are expressed as ρV constant ρ dV V dρ 0 dρ ρ dV V 1 P ρV V constant dP ρV dV 0 dP dV ρV 2 P ρRT dP ρR dT RT dρ dP P dT T dρ ρ 3 The differential form of the entropy change relation Eq 1740 of an ideal gas with constant specific heats is ds c p dT T R dP P 4 Substituting Eq 3 into Eq 4 gives ds c p dT T R dT T dρ ρ c p R dT T R dρ ρ R k 1 dT T R dρ ρ 5 since c p R c v k c v R c v c v R k 1 Dividing both sides of Eq 5 by dT and combining with Eq 1 ds dT R Tk 1 R V dV dT 6 Dividing Eq 3 by dV and combining it with Eqs 1 and 2 give after rearranging dT dV T V V R 7 Substituting Eq 7 into Eq 6 and rearranging ds dT R Tk 1 R T V 2 R RkRT V 2 Tk 1RT V 2 8 FIGURE 1756 The Ts diagram of Rayleigh flow considered in Example 1713 smax Tmax Ma 1 0 Ma 1 T a a b b ds dT s 0 dT ds Final PDF to printer 863 CHAPTER 17 cen22672ch17823880indd 863 110917 0138 PM Setting dsdT 0 and solving the resulting equation RkRT V2 0 for V give the velocity at point a to be V a kR T a and Ma a V a c a kR T a kR T a 1 9 Therefore sonic conditions exist at point a and thus the Mach number is 1 Setting dTds dsdT1 0 and solving the resulting equation Tk 1 RT V2 0 for velocity at point b give V b R T b and Ma b V b c b R T b kR T b 1 k 10 Therefore the Mach number at point b is Ma b 1 k For air k 14 and thus Mab 0845 Discussion Note that in Rayleigh flow sonic conditions are reached as the entropy reaches its maximum value and maximum temperature occurs during subsonic flow EXAMPLE 1714 Effect of Heat Transfer on Flow Velocity Starting with the differential form of the energy equation show that the flow velocity increases with heat addition in subsonic Rayleigh flow but decreases in supersonic Rayleigh flow SOLUTION It is to be shown that flow velocity increases with heat addition in subsonic Rayleigh flow and that the opposite occurs in supersonic flow Assumptions 1 The assumptions associated with Rayleigh flow are valid 2 There are no work interactions and potential energy changes are negligible Analysis Consider heat transfer to the fluid in the differential amount of δq The dif ferential forms of the energy equations are expressed as δq d h 0 d h V 2 2 c p dT V dV 1 Dividing by cpT and factoring out dVV give δq c p T dT T V dV c p T dV V V dV dT T k 1 V 2 kRT 2 where we also used cp kRk 1 Noting that Ma2 V2c2 V2kRT and using Eq 7 for dTdV from Example 1713 give δq c p T dV V V T T V V R k 1 Ma 2 dV V 1 V 2 TR k Ma 2 Ma 2 3 Canceling the two middle terms in Eq 3 since V2TR k Ma2 and rearranging give the desired relation dV V δq c p T 1 1 Ma 2 4 Final PDF to printer 864 COMPRESSIBLE FLOW cen22672ch17823880indd 864 110917 0138 PM Property Relations for Rayleigh Flow It is often desirable to express the variations in properties in terms of the Mach number Ma Noting that Ma V c V kRT and thus V Ma kRT ρ V 2 ρkRT Ma 2 kP Ma 2 1757 since P ρRT Substituting into the momentum equation Eq 1751 gives P 1 k P 1 Ma 1 2 P 2 k P 2 Ma 2 2 which can be rearranged as P 2 P 1 1 k Ma 1 2 1 k Ma 2 2 1758 Again utilizing V Ma kRT the continuity equation ρ1V1 ρ2V2 is expressed as ρ 1 ρ 2 V 2 V 1 Ma 2 kR T 2 Ma 1 kR T 1 Ma 2 T 2 Ma 1 T 1 1759 Then the idealgas relation Eq 1756 becomes T 2 T 1 P 2 ρ 1 P 1 ρ 2 1 k Ma 1 2 1 k Ma 2 2 Ma 2 T 2 Ma 1 T 1 1760 Solving Eq 1760 for the temperature ratio T2T1 gives T 2 T 1 Ma 2 1 k Ma 1 2 Ma 1 1 k Ma 2 2 2 1761 Substituting this relation into Eq 1759 gives the density or velocity ratio as ρ 2 ρ 1 V 1 V 2 Ma 1 2 1 k Ma 2 2 Ma 2 2 1 k Ma 1 2 1762 Flow properties at sonic conditions are usually easy to determine and thus the critical state corresponding to Ma 1 serves as a convenient reference point in compressible flow Taking state 2 to be the sonic state Ma2 1 and superscript is used and state 1 to be any state no subscript the property relations in Eqs 1758 1761 and 1762 reduce to Fig 1758 P P 1 k 1 k Ma 2 T T Ma1 k 1 k Ma 2 2 V V ρ ρ 1 k Ma 2 1 k Ma 2 1763 In subsonic flow 1 Ma2 0 and thus heat transfer and velocity change have the same sign As a result heating the fluid δq 0 increases the flow velocity while cooling decreases it In supersonic flow however 1 Ma2 0 and heat transfer and velocity change have opposite signs As a result heating the fluid δq 0 decreases the flow velocity while cooling increases it Fig 1757 Discussion Note that heating the fluid has the opposite effect on flow velocity in sub sonic and supersonic Rayleigh flows FIGURE 1757 Heating increases the flow velocity in subsonic flow but decreases it in supersonic flow Supersonic flow V1 V2 V1 Subsonic flow δq δq V1 V2 V1 FIGURE 1758 Summary of relations for Rayleigh flow P P 1 k 1 kMa T T k 1Ma 2 k 1Ma 1 kMa V V 1 kMa 1 kMa ρ ρ P P k 1 1 kMa 2 k 1Ma k 1 T T Ma1 k 1 kMa Final PDF to printer 865 CHAPTER 17 cen22672ch17823880indd 865 110917 0138 PM Similar relations can be obtained for dimensionless stagnation temperature and stagnation pressure as follows T 0 T 0 T 0 T T T T T 0 1 k 1 2 Ma 2 Ma1 k 1 k Ma 2 2 1 k 1 2 1 1764 which simplifies to T 0 T 0 k 1 Ma 2 2 k 1 Ma 2 1 k Ma 2 2 1765 Also P 0 P 0 P 0 P P P P P 0 1 k 1 2 Ma 2 kk1 1 k 1 k Ma 2 1 k 1 2 kk1 1766 which simplifies to P 0 P 0 k 1 1 k Ma 2 2 k 1 Ma 2 k 1 kk1 1767 The five relations in Eqs 1763 1765 and 1767 enable us to calculate the dimensionless pressure temperature density velocity stagnation temper ature and stagnation pressure for Rayleigh flow of an ideal gas with a speci fied k for any given Mach number Representative results are given in tabular and graphical form in Table A34 for k 14 Choked Rayleigh Flow It is clear from the earlier discussions that subsonic Rayleigh flow in a duct may accelerate to sonic velocity Ma 1 with heating What happens if we continue to heat the fluid Does the fluid continue to accelerate to supersonic velocities An examination of the Rayleigh line indicates that the fluid at the critical state of Ma 1 cannot be accelerated to supersonic velocities by heating Therefore the flow is choked This is analogous to not being able to accelerate a fluid to supersonic velocities in a converging nozzle by simply extending the converging flow section If we keep heating the fluid we will simply move the critical state further downstream and reduce the flow rate since fluid density at the critical state will now be lower Therefore for a given inlet state the corresponding critical state fixes the maximum possible heat transfer for steady flow Fig 1759 That is q max h 0 h 01 c p T 0 T 01 1768 Further heat transfer causes choking and thus causes the inlet state to change eg inlet velocity will decrease and the flow no longer follows the same Rayleigh line Cooling the subsonic Rayleigh flow reduces the velocity and the Mach number approaches zero as the temperature approaches absolute zero Note that the stagnation temperature T0 is maximum at the critical state of Ma 1 In supersonic Rayleigh flow heating decreases the flow velocity Further heating simply increases the temperature and moves the critical state farther downstream resulting in a reduction in the mass flow rate of the fluid It may seem like supersonic Rayleigh flow can be cooled indefinitely but it turns FIGURE 1759 For a given inlet state the maximum possible heat transfer occurs when sonic conditions are reached at the exit state T01 Rayleigh flow Choked flow qmax T02 T01 T1 T2 T Final PDF to printer 866 COMPRESSIBLE FLOW cen22672ch17823880indd 866 110917 0138 PM out that there is a limit Taking the limit of Eq 1765 as the Mach number approaches infinity gives lim Ma T 0 T 0 1 1 k 2 1769 which yields T 0 T 0 049 for k 14 Therefore if the critical stagnation tem perature is 1000 K air cannot be cooled below T0 490 K in Rayleigh flow Physically this means that the flow velocity reaches infinity by the time the tem perature reaches 490 Ka physical impossibility When supersonic flow cannot be sustained the flow undergoes a normal shock wave and becomes subsonic EXAMPLE 1715 Rayleigh Flow in a Tubular Combustor A combustion chamber consists of tubular combustors of 15cm diameter Compressed air enters the tubes at 550 K 480 kPa and 80 ms Fig 1760 Fuel with a heating value of 42000 kJkg is injected into the air and is burned with an airfuel mass ratio of 40 Approximating combustion as a heat transfer process to air determine the tem perature pressure velocity and Mach number at the exit of the combustion chamber SOLUTION Fuel is burned in a tubular combustion chamber with compressed air The exit temperature pressure velocity and Mach number are to be determined Assumptions 1 The assumptions associated with Rayleigh flow ie steady one dimensional flow of an ideal gas with constant properties through a constant crosssec tional area duct with negligible frictional effects are valid 2 Combustion is complete and it is treated as a heat addition process with no change in the chemical composition of the flow 3 The increase in mass flow rate due to fuel injection is disregarded Properties We take the properties of air to be k 14 cp 1005 kJkgK and R 0287 kJkgK Analysis The inlet density and mass flow rate of air are ρ 1 P 1 R T 1 480 kPa 0287 kJkgK550 K 3041 kgm 3 m air ρ 1 A 1 V 1 3041 kgm 3 π 015 m 2 4 80 ms 4299 kgs The mass flow rate of fuel and the rate of heat transfer are m fuel m air AF 4299 kgs 40 01075 kgs Q m fuel HV 01075 kgs 42000 kJkg 4514 kW q Q m air 4514 kJs 4299 kgs 1050 kJkg The stagnation temperature and Mach number at the inlet are T 01 T 1 V 1 2 2 c p 550 K 80 ms 2 21005 kJkgK 1 kJkg 1000 m 2 s 2 5532 K c 1 kR T 1 14 0287 kJkgK550 K 1000 m 2 s 2 1 kJkg 4701 ms Ma 1 V 1 c 1 80 ms 4701 ms 01702 FIGURE 1760 Schematic of the combustor tube analyzed in Example 1715 Combustor tube P1 480 kPa P2 T2 V2 T1 550 K V1 80 ms Q Final PDF to printer 867 CHAPTER 17 cen22672ch17823880indd 867 110917 0138 PM The exit stagnation temperature is from the energy equation q cpT02 T01 T 02 T 01 q c p 5532 K 1050 kJkg 1005 kJkgK 1598 K The maximum value of stagnation temperature T 0 occurs at Ma 1 and its value can be determined from Table A34 or from Eq 1765 At Ma1 01702 we read T 0 T 0 01291 Therefore T 0 T 01 01291 5532 K 01291 4284 K The stagnation temperature ratio at the exit state and the Mach number corresponding to it are from Table A34 T 02 T 0 1598 K 4284 K 03730 Ma 2 03142 0314 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are Table A34 Ma 1 01702 T 1 T 01541 P 1 P 23065 V 1 V 00668 Ma 2 03142 T 2 T 04389 P 2 P 21086 V 2 V 02082 Then the exit temperature pressure and velocity are determined to be T 2 T 1 T 2 T T 1 T 04389 01541 2848 T 2 2848 T 1 2848 550 K 1566 K P 2 P 1 P 2 P P 1 P 21086 23065 09142 P 2 09142 P 1 09142 480 kPa 439 kPa V 2 V 1 V 2 V V 1 V 02082 00668 3117 V 2 3117 V 1 3117 80 ms 249 ms Discussion Note that the temperature and velocity increase and pressure decreases during this subsonic Rayleigh flow with heating as expected This problem can also be solved using appropriate relations instead of tabulated values which can likewise be coded for convenient computer solutions 177 STEAM NOZZLES We have seen in Chap 3 that water vapor at moderate or high pressures devi ates considerably from idealgas behavior and thus most of the relations developed in this chapter are not applicable to the flow of steam through the nozzles or blade passages encountered in steam turbines Given that the steam properties such as enthalpy are functions of pressure as well as temperature and that no simple property relations exist an accurate analysis of steam flow through the nozzles is no easy matter Often it becomes necessary to use steam tables an hs diagram or a computer program for the properties of steam Final PDF to printer 868 COMPRESSIBLE FLOW cen22672ch17823880indd 868 110917 0138 PM A further complication in the expansion of steam through nozzles occurs as the steam expands into the saturation region as shown in Fig 1761 As the steam expands in the nozzle its pressure and temperature drop and ordinarily one would expect the steam to start condensing when it strikes the saturation line However this is not always the case Owing to the high speeds the residence time of the steam in the nozzle is small and there may not be sufficient time for the necessary heat transfer and the formation of liquid droplets Consequently the condensation of the steam may be delayed for a little while This phenomenon is known as supersaturation and the steam that exists in the wet region without containing any liquid is called supersaturated steam Supersaturation states are nonequilibrium or meta stable states During the expansion process the steam reaches a temperature lower than that normally required for the condensation process to begin Once the tem perature drops a sufficient amount below the saturation temperature corre sponding to the local pressure groups of steam moisture droplets of sufficient size are formed and condensation occurs rapidly The locus of points where condensation takes place regardless of the initial temperature and pressure at the nozzle entrance is called the Wilson line The Wilson line lies between the 4 and 5 percent moisture curves in the saturation region on the hs dia gram for steam and it is often approximated by the 4 percent moisture line Therefore steam flowing through a highvelocity nozzle is assumed to begin condensation when the 4 percent moisture line is crossed The criticalpressure ratio PP0 for steam depends on the nozzle inlet state as well as on whether the steam is superheated or saturated at the nozzle inlet However the idealgas relation for the criticalpressure ratio Eq 1722 gives reasonably good results over a wide range of inlet states As indicated in Table 172 the specific heat ratio of superheated steam is approximated as k 13 Then the criticalpressure ratio becomes P P 0 2 k 1 kk1 0546 When steam enters the nozzle as a saturated vapor instead of superheated vapor a common occurrence in the lower stages of a steam turbine the criticalpressure ratio is taken to be 0576 which corresponds to a specific heat ratio of k 114 FIGURE 1761 The hs diagram for the isentropic expansion of steam in a nozzle s h P1 1 P2 Saturation line 2 Wilson line x 096 EXAMPLE 1716 Steam Flow Through a Converging Diverging Nozzle Steam enters a convergingdiverging nozzle at 2 MPa and 400C with a negligible velocity and a mass flow rate of 25 kgs and it exits at a pressure of 300 kPa The flow is isentropic between the nozzle entrance and throat and the overall nozzle effi ciency is 93 percent Determine a the throat and exit areas and b the Mach number at the throat and the nozzle exit SOLUTION Steam enters a convergingdiverging nozzle with a low velocity The throat and exit areas and the Mach number are to be determined Final PDF to printer 869 CHAPTER 17 cen22672ch17823880indd 869 110917 0138 PM FIGURE 1762 Schematic and hs diagram for Example 1716 s h 1 P2 300 kPa 2 2s t Pt P1 P01 2 MPa T1 400C P1 2 MPa V1 0 Steam Throat m 25 kgs ηN 93 Assumptions 1 Flow through the nozzle is onedimensional 2 The flow is isentropic between the inlet and the throat and is adiabatic and irreversible between the throat and the exit 3 The inlet velocity is negligible Analysis We denote the entrance throat and exit states by 1 t and 2 respectively as shown in Fig 1762 a Since the inlet velocity is negligible the inlet stagnation and static states are identical The ratio of the exittoinlet stagnation pressure is P 2 P 01 300 kPa 2000 kPa 015 It is much smaller than the criticalpressure ratio which is taken to be PP01 0546 since the steam is superheated at the nozzle inlet Therefore the flow surely is super sonic at the exit Then the velocity at the throat is the sonic velocity and the throat pressure is P t 0546 P 01 05462 MPa 109 MPa At the inlet P 1 P 01 2 MPa T 1 T 01 400C h 1 h 01 32484 kJkg s 1 s t s 2s 71292 kJkgK Also at the throat P t 109 MPa s t 71292 kJkgK h t 30768 kJkg v t 024196 m 3 kg Then the throat velocity is determined from Eq 173 to be V t 2 h 01 h t 232484 30768 kJkg 1000 m 2 s 2 1 kJkg 5858 ms The flow area at the throat is determined from the mass flow rate relation A t m v t V t 25 kgs02420 m 3 kg 5858 ms 1033 10 4 m 2 1033 cm 2 At state 2s P 2s P 2 300 kPa s 2s s 1 71292 kJkgK h 2s 27836 kJkg The enthalpy of the steam at the actual exit state is see Chap 7 η N h 01 h 2 h 01 h 2s 093 32484 h 2 32484 27836 h 2 28161 kJkg Therefore P 2 300 kPa h 2 28161 kJkg v 2 067723 m 3 kg s 2 72019 kJkgK Final PDF to printer 870 COMPRESSIBLE FLOW cen22672ch17823880indd 870 110917 0138 PM Then the exit velocity and the exit area become V 2 2 h 01 h 2 232484 28161 kJkg 1000 m 2 s 2 1 kJkg 9298 ms A 2 m v 2 V 2 25 kgs067723 m 3 kg 9298 ms 1821 10 4 m 2 1821 cm 2 b The velocity of sound and the Mach numbers at the throat and the exit of the nozzle are determined by replacing differential quantities with differences c P ρ s 12 ΔP Δ 1 v s 12 The velocity of sound at the throat is determined by evaluating the specific volume at st 71292 kJkgK and at pressures of 1115 and 1065 MPa Pt 25 kPa c 1115 1065 kPa 1 023776 1024633 kgm 3 1000 m 2 s 2 1 kPa m 3 kg 5846 ms The Mach number at the throat is determined from Eq 1712 to be Ma V c 5858 ms 5846 ms 1002 Thus the flow at the throat is sonic as expected The slight deviation of the Mach number from unity is due to replacing the derivatives with differences The velocity of sound and the Mach number at the nozzle exit are determined by evaluating the specific volume at s2 72019 kJkgK and at pressures of 325 and 275 kPa P2 25 kPa c 325 275 kPa 1063596 1072245 kgm 3 1000 m 2 s 2 1 kPa m 3 kg 5154 ms and Ma V c 9298 ms 5154 ms 1804 Thus the flow of steam at the nozzle exit is supersonic SUMMARY In this chapter the effects of compressibility on gas flow are examined When dealing with compressible flow it is conve nient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation or total enthalpy h0 defined as h 0 h V 2 2 The properties of a fluid at the stagnation state are called stag nation properties and are indicated by the subscript zero The stagnation temperature of an ideal gas with constant specific heats is T 0 T V 2 2 c p Final PDF to printer cen22672ch17823880indd 871 110917 0138 PM 871 CHAPTER 17 which represents the temperature an ideal gas would attain if it is brought to rest adiabatically The stagnation proper ties of an ideal gas are related to the static properties of the fluid by P 0 P T 0 T kk1 and ρ 0 ρ T 0 T 1k1 The speed at which an infinitesimally small pressure wave travels through a medium is the speed of sound For an ideal gas it is expressed as c P ρ s kRT The Mach number is the ratio of the actual velocity of the fluid to the speed of sound at the same state Ma V c The flow is called sonic when Ma 1 subsonic when Ma 1 supersonic when Ma 1 hypersonic when Ma 1 and transonic when Ma 1 Nozzles whose flow area decreases in the flow direction are called converging nozzles Nozzles whose flow area first decreases and then increases are called convergingdiverging nozzles The location of the smallest flow area of a nozzle is called the throat The highest velocity to which a fluid can be accelerated in a converging nozzle is the sonic velocity Accelerating a fluid to supersonic velocities is possible only in convergingdiverging nozzles In all supersonic converging diverging nozzles the flow velocity at the throat is the speed of sound The ratios of the stagnationtostatic properties for ideal gases with constant specific heats can be expressed in terms of the Mach number as T 0 T 1 k 1 2 Ma 2 P 0 P 1 k 1 2 Ma 2 kk1 and ρ 0 ρ 1 k 1 2 Ma 2 1k1 When Ma 1 the resulting statictostagnation property ratios for the temperature pressure and density are called critical ratios and are denoted by the superscript asterisk T T 0 2 k 1 P P 0 2 k 1 kk1 and ρ ρ 0 2 k 1 1k1 The pressure outside the exit plane of a nozzle is called the back pressure For all back pressures lower than P the pres sure at the exit plane of the converging nozzle is equal to P the Mach number at the exit plane is unity and the mass flow rate is the maximum or choked flow rate In some range of back pressure the fluid that achieved a sonic velocity at the throat of a convergingdiverging nozzle and is accelerating to supersonic velocities in the diverging section experiences a normal shock which causes a sudden rise in pressure and temperature and a sudden drop in velocity to subsonic levels Flow through the shock is highly irrevers ible and thus it cannot be approximated as isentropic The properties of an ideal gas with constant specific heats before subscript 1 and after subscript 2 a shock are related by T 01 T 02 Ma 2 k 1 Ma 1 2 2 2kMa 1 2 k 1 T 2 T 1 2 Ma 1 2 k 1 2 Ma 2 2 k 1 and P 2 P 1 1 k Ma 1 2 1 k Ma 2 2 2k Ma 1 2 k 1 k 1 These equations also hold across an oblique shock provided that the component of the Mach number normal to the oblique shock is used in place of the Mach number Steady onedimensional flow of an ideal gas with constant specific heats through a constantarea duct with heat transfer and negligible friction is referred to as Rayleigh flow The property relations and curves for Rayleigh flow are given in Table A34 Heat transfer during Rayleigh flow can be determined from q c p T 02 T 01 c p T 2 T 1 V 2 2 V 1 2 2 Final PDF to printer cen22672ch17823880indd 872 110917 0138 PM 872 COMPRESSIBLE FLOW REFERENCES AND SUGGESTED READINGS 1 J D Anderson Modern Compressible Flow with Histori cal Perspective 3rd ed New York McGrawHill 2003 2 Y A Çengel and J M Cimbala Fluid Mechanics Fundamentals and Applications 4th ed New York McGrawHill Education 2018 3 H Cohen G F C Rogers and H I H Saravanamuttoo Gas Turbine Theory 3rd ed New York Wiley 1987 4 H Liepmann and A Roshko Elements of Gas Dynamics Mineola NY Dover Publications 2001 5 C E Mackey responsible NACA officer and curator Equations Tables and Charts for Compressible Flow NACA Report 1135 httpnacalarcnasagov reports1953nacareport1135 6 A H Shapiro The Dynamics and Thermodynamics of Compressible Fluid Flow Vol 1 New York Ronald Press Company 1953 7 P A Thompson CompressibleFluid Dynamics New York McGrawHill 1972 8 United Technologies Corporation The Aircraft Gas Turbine and Its Operation 1982 9 M Van Dyke An Album of Fluid Motion Stanford CA The Parabolic Press 1982 PROBLEMS Stagnation Properties 171C A highspeed aircraft is cruising in still air How does the temperature of air at the nose of the aircraft differ from the temperature of air at some distance from the aircraft 172C What is dynamic temperature 173C In airconditioning applications the temperature of air is measured by inserting a probe into the flow stream Thus the probe actually measures the stagnation temperature Does this cause any significant error 174C How and why is the stagnation enthalpy h0 defined How does it differ from ordinary static enthalpy 175 Air flows through a device such that the stagnation pressure is 04 MPa the stagnation temperature is 400C and the velocity is 520 ms Determine the static pressure and tem perature of the air at this state Answers 545 K 0184 MPa 176E Steam flows through a device with a stagnation pressure of 120 psia a stagnation temperature of 700F and a velocity of 900 fts Assuming idealgas behavior determine the static pressure and temperature of the steam at this state 177 Calculate the stagnation temperature and pressure for the following substances flowing through a duct a helium at 025 MPa 50C and 240 ms b nitrogen at 015 MPa 50C and 300 ms and c steam at 01 MPa 350C and 480 ms 178 Determine the stagnation temperature and stagnation pressure of air that is flowing at 36 kPa 238 K and 325 ms Answers 291 K 724 kPa 179 Air enters a compressor with a stagnation pressure of 100 kPa and a stagnation temperature of 35C and it is com pressed to a stagnation pressure of 900 kPa Assuming the com pression process to be isentropic determine the power input to the compressor for a mass flow rate of 004 kgs Answer 108 kW 1710 Products of combustion enter a gas turbine with a stag nation pressure of 090 MPa and a stagnation temperature of 840C and they expand to a stagnation pressure of 100 kPa Taking k 133 and R 0287 kJkgK for the products of com bustion and assuming the expansion process to be isentropic determine the power output of the turbine per unit mass flow Speed of Sound and Mach Number 1711C What is sound How is it generated How does it travel Can sound waves travel in a vacuum 1712C In which medium does a sound wave travel faster in cool air or in warm air 1713C In which medium will sound travel fastest for a given temperature air helium or argon 1714C In which medium does a sound wave travel faster in air at 20C and 1 atm or in air at 20C and 5 atm 1715C Is the sonic velocity in a specified medium a fixed quantity or does it change as the properties of the medium change Explain 1716C Does the Mach number of a gas flowing at a con stant velocity remain constant Explain 1717C Is it realistic to approximate that the propagation of sound waves is an isentropic process Explain Problems designated by a C are concept questions and stu dents are encouraged to answer them all Problems designated by an E are in English units and the SI users can ignore them Problems with the icon are comprehensive in nature and are intended to be solved with appropriate software Final PDF to printer cen22672ch17823880indd 873 110917 0138 PM 873 CHAPTER 17 1718 Carbon dioxide enters an adiabatic nozzle at 800 K with a velocity of 50 ms and leaves at 400 K Assuming con stant specific heats at room temperature determine the Mach number a at the inlet and b at the exit of the nozzle Assess the accuracy of the constantspecificheat approximation Answers a 0113 b 264 1719 Nitrogen enters a steadyflow heat exchanger at 150 kPa 10C and 100 ms and it receives heat in the amount of 120 kJkg as it flows through it Nitrogen leaves the heat exchanger at 100 kPa with a velocity of 200 ms Determine the Mach number of the nitrogen at the inlet and the exit of the heat exchanger 1720 Assuming ideal gas behavior determine the speed of sound in refrigerant134a at 09 MPa and 60C 1721 Determine the speed of sound in air at a 300 K and b 800 K Also determine the Mach number of an aircraft moving in air at a velocity of 330 ms for both cases 1722E Steam flows through a device with a pressure of 120 psia a temperature of 700F and a velocity of 900 fts Determine the Mach number of the steam at this state by assuming idealgas behavior with k 13 Answer 0441 1723E Reconsider Prob 1722E Using appropriate software compare the Mach number of steam flow over the temperature range 350 to 700F Plot the Mach number as a function of temperature 1724 Air expands isentropically from 22 MPa and 77C to 04 MPa Calculate the ratio of the initial to the final speed of sound Answer 128 1725 Repeat Prob 1724 for helium gas 1726 The Airbus A340 passenger plane has a maximum takeoff weight of about 260000 kg a length of 64 m a wing span of 60 m a maximum cruising speed of 945 kmh a seat ing capacity of 271 passengers a maximum cruising altitude of 14000 m and a maximum range of 12000 km The air tem perature at the cruising altitude is about 60C Determine the Mach number of this plane for the stated limiting conditions 1727 The isentropic process for an ideal gas is expressed as Pvk constant Using this process equation and the definition of the speed of sound Eq 179 obtain the expression for the speed of sound for an ideal gas Eq 1711 OneDimensional Isentropic Flow 1728C Is it possible to accelerate a gas to a supersonic velocity in a converging nozzle Explain 1729C A gas initially at a subsonic velocity enters an adia batic diverging duct Discuss how this affects a the velocity b the temperature c the pressure and d the density of the fluid 1730C A gas at a specified stagnation temperature and pressure is accelerated to Ma 2 in a convergingdiverging nozzle and to Ma 3 in another nozzle What can you say about the pressures at the throats of these two nozzles 1731C A gas initially at a supersonic velocity enters an adi abatic converging duct Discuss how this affects a the veloc ity b the temperature c the pressure and d the density of the fluid 1732C A gas initially at a supersonic velocity enters an adi abatic diverging duct Discuss how this affects a the velocity b the temperature c the pressure and d the density of the fluid 1733C Consider a converging nozzle with sonic speed at the exit plane Now the nozzle exit area is reduced while the nozzle inlet conditions are maintained constant What will hap pen to a the exit velocity and b the mass flow rate through the nozzle 1734C A gas initially at a subsonic velocity enters an adia batic converging duct Discuss how this affects a the veloc ity b the temperature c the pressure and d the density of the fluid 1735 Helium enters a convergingdiverging nozzle at 07 MPa 800 K and 100 ms What are the lowest temperature and pressure that can be obtained at the throat of the nozzle 1736 Consider a large commercial airplane cruising at a speed of 1050 kmh in air at an altitude of 10 km where the standard air temperature is 50C Determine if the speed of this airplane is subsonic or supersonic 1737 Calculate the critical temperature pressure and den sity of a air at 200 kPa 100C and 325 ms and b helium at 200 kPa 60C and 300 ms 1738E Air at 25 psia 320F and Mach number Ma 07 flows through a duct Calculate the velocity and the stag nation pressure temperature and density of air Answers 958 fts 856 R 347 psia 0109 lbmft3 1739 Air enters a convergingdiverging nozzle at a pres sure of 1200 kPa with negligible velocity What is the low est pressure that can be obtained at the throat of the nozzle Answer 634 kPa 1740 An aircraft is designed to cruise at Mach number Ma 11 at 12000 m where the atmospheric temperature is 23615 K Determine the stagnation temperature on the leading edge of the wing 1741 Quiescent carbon dioxide at 900 kPa and 500 K is accelerated isentropically to a Mach number of 06 Determine the temperature and pressure of the carbon dioxide after accel eration Answers 475 K 718 kPa 1742 In March 2004 NASA successfully launched an experimental supersoniccombustion ramjet engine called a scramjet that reached a recordsetting Mach number of 7 Taking the air temperature to be 20C determine the speed of this engine Answer 8040 kmh Final PDF to printer cen22672ch17823880indd 874 110917 0138 PM 874 COMPRESSIBLE FLOW 1743E Reconsider Prob 1742 Determine the speed of this engine in miles per hour corresponding to a Mach number of 7 in air at a temperature of 0F Isentropic Flow through Nozzles 1744C Is it possible to accelerate a fluid to supersonic velocities with a velocity other than the sonic velocity at the throat Explain 1745C What would happen if we tried to further accelerate a supersonic fluid with a diverging diffuser 1746C How does the parameter Ma differ from the Mach number Ma 1747C What would happen if we attempted to decelerate a supersonic fluid with a diverging diffuser 1748C Consider subsonic flow in a converging nozzle with specified conditions at the nozzle inlet and critical pressure at the nozzle exit What is the effect of dropping the back pressure well below the critical pressure on a the exit velocity b the exit pressure and c the mass flow rate through the nozzle 1749C Consider a converging nozzle and a converging diverging nozzle having the same throat areas For the same inlet conditions how would you compare the mass flow rates through these two nozzles 1750C Consider gas flow through a converging nozzle with specified inlet conditions We know that the highest velocity the fluid can have at the nozzle exit is the sonic velocity at which point the mass flow rate through the nozzle is a maximum If it were possible to achieve hypersonic velocities at the nozzle exit how would it affect the mass flow rate through the nozzle 1751C Consider subsonic flow in a converging nozzle with fixed inlet conditions What is the effect of dropping the back pressure to the critical pressure on a the exit velocity b the exit pressure and c the mass flow rate through the nozzle 1752C Consider the isentropic flow of a fluid through a convergingdiverging nozzle with a subsonic velocity at the throat How does the diverging section affect a the velocity b the pressure and c the mass flow rate of the fluid 1753 Nitrogen enters a convergingdiverging nozzle at 700 kPa and 400 K with a negligible velocity Determine the critical velocity pressure temperature and density in the nozzle 1754 Air enters a convergingdiverging nozzle at 12 MPa with a negligible velocity Approximating the flow as isentro pic determine the back pressure that would result in an exit Mach number of 18 Answer 209 kPa 1755 An ideal gas flows through a passage that first converges and then diverges during an adiabatic reversible steadyflow process For subsonic flow at the inlet sketch the variation of pressure velocity and Mach number along the length of the nozzle when the Mach number at the minimum flow area is equal to unity 1756 Repeat Prob 1755 for supersonic flow at the inlet 1757 For an ideal gas obtain an expression for the ratio of the speed of sound where Ma 1 to the speed of sound based on the stagnation temperature cc0 1758 Explain why the maximum flow rate per unit area for a given ideal gas depends only on P 0 T 0 For an ideal gas with k 14 and R 0287 kJkgK find the constant a such that m A a P 0 T 0 1759 An ideal gas with k 14 is flowing through a noz zle such that the Mach number is 16 where the flow area is 45 cm2 Approximating the flow as isentropic determine the flow area at the location where the Mach number is 08 1760 Repeat Prob 1759 for an ideal gas with k 133 1761E Air enters a convergingdiverging nozzle of a super sonic wind tunnel at 150 psia and 100F with a low velocity The flow area of the test section is equal to the exit area of the nozzle which is 5 ft2 Calculate the pressure temperature velocity and mass flow rate in the test section for a Mach number Ma 2 Explain why the air must be very dry for this application Answers 191 psia 311 R 1729 fts 1435 lbms 1762 Air enters a nozzle at 05 MPa 420 K and a velocity of 110 ms Approximating the flow as isentropic determine the pressure and temperature of air at a location where the air velocity equals the speed of sound What is the ratio of the area at this location to the entrance area Answers 355 K 278 kPa 0428 1763 Repeat Prob 1762 assuming the entrance velocity is negligible 1764 Air at 900 kPa and 400 K enters a converging nozzle with a negligible velocity The throat area of the nozzle is 10 cm2 Approximating the flow as isen tropic calculate and plot the exit pressure the exit velocity and the mass flow rate versus the back pressure Pb for 09 Pb 01 MPa 1765 Reconsider Prob 1764 Using appropriate software solve the problem for the inlet condi tions of 08 MPa and 1200 K Shock Waves and Expansion Waves 1766C Are the isentropic relations of ideal gases applicable for flows across a normal shock waves b oblique shock waves and c PrandtlMeyer expansion waves 1767C What do the states on the Fanno line and the Ray leigh line represent What do the intersection points of these two curves represent 1768C It is claimed that an oblique shock can be analyzed like a normal shock provided that the normal component of velocity normal to the shock surface is used in the analysis Do you agree with this claim Final PDF to printer cen22672ch17823880indd 875 110917 0138 PM 875 CHAPTER 17 1769C How does the normal shock affect a the fluid velocity b the static temperature c the stagnation tempera ture d the static pressure and e the stagnation pressure 1770C How do oblique shocks occur How do oblique shocks differ from normal shocks 1771C For an oblique shock to occur does the upstream flow have to be supersonic Does the flow downstream of an oblique shock have to be subsonic 1772C Can the Mach number of a fluid be greater than 1 after a normal shock wave Explain 1773C Can a shock wave develop in the converging section of a convergingdiverging nozzle Explain 1774C Consider supersonic airflow approaching the nose of a twodimensional wedge and experiencing an oblique shock Under what conditions does an oblique shock detach from the nose of the wedge and form a bow wave What is the numerical value of the shock angle of the detached shock at the nose 1775C Consider supersonic flow impinging on the rounded nose of an aircraft Is the oblique shock that forms in front of the nose an attached or a detached shock Explain 1776 Air enters a normal shock at 26 kPa 230 K and 815 ms Calculate the stagnation pressure and Mach number upstream of the shock as well as pressure temperature veloc ity Mach number and stagnation pressure downstream of the shock 1777 Reconsider Prob 1776 Calculate the entropy change of air across the normal shock wave Answer 0242 kJkgK 1778 Air enters a convergingdiverging nozzle with low velocity at 24 MPa and 120C If the exit area of the noz zle is 35 times the throat area what must the back pressure be to produce a normal shock at the exit plane of the nozzle Answer 0793 MPa 1779 Reconsider Prob 1778 What must the back pressure be for a normal shock to occur at a location where the cross sectional area is twice the throat area 1780E Air flowing steadily in a nozzle experiences a nor mal shock at a Mach number of Ma 25 If the pressure and temperature of air are 100 psia and 4405 R respectively upstream of the shock calculate the pressure temperature velocity Mach number and stagnation pressure downstream of the shock Compare these results to those for helium under going a normal shock under the same conditions 1781E Reconsider Prob 1780E Using appropriate software study the effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 Ma1 35 In addi tion to the required information calculate the entropy change of the air and helium across the normal shock Tabulate the results in a parametric table 1782 Air enters a convergingdiverging nozzle of a super sonic wind tunnel at 1 MPa and 300 K with a low velocity If a normal shock wave occurs at the exit plane of the nozzle at Ma 24 determine the pressure temperature Mach num ber velocity and stagnation pressure after the shock wave Answers 448 kPa 284 K 0523 177 ms 540 kPa 1783 Using appropriate software calculate and plot the entropy change of air across the normal shock for upstream Mach numbers between 05 and 15 in increments of 01 Explain why normal shock waves can occur only for upstream Mach numbers greater than Ma 1 1784 Consider supersonic airflow approaching the nose of a twodimensional wedge at a Mach number of 3 Using Fig 1743 determine the minimum shock angle and the maxi mum deflection angle a straight oblique shock can have 1785 Air flowing at 32 kPa 240 K and Ma1 36 is forced to undergo an expansion turn of 15 Determine the Mach number pressure and temperature of air after the expansion Answers 481 665 kPa 153 K 1786 Consider the supersonic flow of air at upstream con ditions of 70 kPa and 260 K and a Mach number of 24 over a twodimensional wedge of halfangle 10 If the axis of the wedge is tilted 25 with respect to the upstream airflow deter mine the downstream Mach number pressure and temperature above the wedge Answers 311 238 kPa 191 K FIGURE P1786 Ma1 24 Ma2 25 10 1787 Reconsider Prob 1786 Determine the downstream Mach number pressure and temperature below the wedge for a strong oblique shock for an upstream Mach number of 5 1788E Air at 12 psia 30F and a Mach number of 20 is forced to turn upward by a ramp that makes an 8 angle off the flow direction As a result a weak oblique shock forms Deter mine the wave angle Mach number pressure and temperature after the shock 1789 Air flowing at 40 kPa 210 K and a Mach number of 34 impinges on a twodimensional wedge of halfangle 8 Determine the two possible oblique shock angles βweak and βstrong that could be formed by this wedge For each case calculate the pressure and Mach number downstream of the oblique shock Final PDF to printer cen22672ch17823880indd 876 110917 0138 PM 876 COMPRESSIBLE FLOW 1790 Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma 26 If the pressure and tem perature of air are 58 kPa and 270 K respectively upstream of the shock calculate the pressure temperature velocity Mach number and stagnation pressure downstream of the shock Compare these results to those for helium undergoing a normal shock under the same conditions 1791 Reconsider Prob 1790 Calculate the entropy changes of air and helium across the normal shock 1792 For an ideal gas flowing through a normal shock develop a relation for V2V1 in terms of k Ma1 and Ma2 Duct Flow with Heat Transfer and Negligible Friction Rayleigh Flow 1793C What is the characteristic aspect of Rayleigh flow What are the main assumptions associated with Rayleigh flow 1794C What is the effect of heating the fluid on the flow velocity in subsonic Rayleigh flow Answer the same ques tions for supersonic Rayleigh flow 1795C On a Ts diagram of Rayleigh flow what do the points on the Rayleigh line represent 1796C What is the effect of heat gain and heat loss on the entropy of the fluid during Rayleigh flow 1797C Consider subsonic Rayleigh flow of air with a Mach number of 092 Heat is now transferred to the fluid and the Mach number increases to 095 Does the temperature T of the fluid increase decrease or remain constant during this pro cess How about the stagnation temperature T0 1798C Consider subsonic Rayleigh flow that is accelerated to sonic velocity Ma 1 at the duct exit by heating If the fluid continues to be heated will the flow at duct exit be super sonic subsonic or remain sonic 1799 Argon gas enters a constant crosssectionalarea duct at Ma1 02 P1 320 kPa and T1 400 K at a rate of 085 kgs Disregarding frictional losses determine the highest rate of heat transfer to the argon without reducing the mass flow rate 17100 Air is heated as it flows subsonically through a duct When the amount of heat transfer reaches 67 kJkg the flow is observed to be choked and the velocity and the static pres sure are measured to be 680 ms and 270 kPa Disregarding frictional losses determine the velocity static temperature and static pressure at the duct inlet 17101 Compressed air from the compressor of a gas turbine enters the combustion chamber at T1 700 K P1 600 kPa and Ma1 02 at a rate of 03 kgs Via combustion heat is transferred to the air at a rate of 150 kJs as it flows through the duct with negligible friction Determine the Mach number at the duct exit and the drop in stagnation pressure P01 P02 during this process Answers 0271 127 kPa 17102 Repeat Prob 17101 for a heat transfer rate of 300 kJs 17103E Air flows with negligible friction through a 6in diameter duct at a rate of 9 lbms The temperature and pres sure at the inlet are T1 800 R and P1 30 psia and the Mach number at the exit is Ma2 1 Determine the rate of heat trans fer and the pressure drop for this section of the duct 17104 Air enters an approximately frictionless duct with V1 70 ms T1 600 K and P1 350 kPa Letting the exit temperature T2 vary from 600 to 5000 K evaluate the entropy change at intervals of 200 K and plot the Rayleigh line on a Ts diagram 17105 Air enters a rectangular duct at T1 300 K P1 420 kPa and Ma1 2 Heat is transferred to the air in the amount of 55 kJkg as it flows through the duct Disregarding frictional losses determine the temperature and Mach number at the duct exit Answers 386 K 164 FIGURE P17105 Air 55 kJkg P1 420 kPa T1 300 K Ma1 2 17106 Repeat Prob 17105 assuming air is cooled in the amount of 55 kJkg 17107 Consider supersonic flow of air through a 7cmdiameter duct with negligible friction Air enters the duct at Ma1 18 P01 140 kPa and T01 600 K and it is decelerated by heating Determine the highest temperature that air can be heated by heat addition while the mass flow rate remains constant 17108E Air is heated as it flows through a 6 in 6 in square duct with negligible friction At the inlet air is at T1 700 R P1 80 psia and V1 260 fts Determine the rate at which heat must be transferred to the air to choke the flow at the duct exit and the entropy change of air during this process Steam Nozzles 17109C What is supersaturation Under what conditions does it occur 17110 Steam enters a converging nozzle at 50 MPa and 400C with a negligible velocity and it exits at 30 MPa For a nozzle exit area of 75 cm2 determine the exit velocity mass flow rate and exit Mach number if the nozzle a is isentropic and b has an efficiency of 94 percent Answers a 529 ms 461 kgs 0917 b 512 ms 443 kgs 0885 17111 Steam enters a convergingdiverging nozzle at 1 MPa and 500C with a negligible velocity at a mass flow rate of 25 kgs and it exits at a pressure of 200 kPa Assuming Final PDF to printer cen22672ch17823880indd 877 110917 0138 PM 877 CHAPTER 17 the flow through the nozzle to be isentropic determine the exit area and the exit Mach number Answers 315 cm2 174 17112 Repeat Prob 17111 for a nozzle efficiency of 85 percent Review Problems 17113 A subsonic airplane is flying at a 5000m altitude where the atmospheric conditions are 54 kPa and 256 K A Pitot static probe measures the difference between the static and stagnation pressures to be 16 kPa Calculate the speed of the airplane and the flight Mach number Answers 199 ms 0620 17114 The thrust developed by the engine of a Boeing 777 is about 380 kN Assuming choked flow in the nozzles deter mine the mass flow rate of air through the nozzle Take the ambient conditions to be 215 K and 35 kPa 17115 A stationary temperature probe inserted into a duct where air is flowing at 190 ms reads 85C What is the actual temperature of the air Answer 670C 17116 Nitrogen enters a steadyflow heat exchanger at 150 kPa 10C and 100 ms and it receives heat in the amount of 150 kJkg as it flows through it The nitrogen leaves the heat exchanger at 100 kPa with a velocity of 200 ms Determine the stagnation pressure and temperature of the nitrogen at the inlet and exit states 17117 Plot the mass flow parameter m R T 0 A P 0 versus the Mach number for k 12 14 and 16 in the range of 0 Ma 1 17118 Obtain Eq 1710 by starting with Eq 179 and using the cyclic rule and the thermodynamic property relations c p T s T p and c v T s T v 17119 For ideal gases undergoing isentropic flows obtain expressions for PP TT and ρρ as functions of k and Ma 17120 Using Eqs 174 1713 and 1714 verify that for the steady flow of ideal gases dT0T dAA 1 Ma2 dVV Explain the effect of heating and area changes on the velocity of an ideal gas in steady flow for a subsonic flow and b supersonic flow 17121 Find the expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave Ma1 17122 Derive an expression for the speed of sound based on van der Waals equation of state P RTv b av2 Using this relation determine the speed of sound in carbon dioxide at 80C and 320 kPa and compare your result to that obtained by assuming idealgas behavior The van der Waals constants for carbon dioxide are a 3643 kPam6kmol2 and b 00427 m3kmol 17123 Helium enters a nozzle at 05 MPa 600 K and a velocity of 120 ms Assuming isentropic flow determine the pressure and temperature of helium at a location where the velocity equals the speed of sound What is the ratio of the area at this location to the entrance area 17124 Repeat Prob 17123 assuming the entrance velocity is negligible 17125 Nitrogen enters a duct with varying flow area at 400 K 100 kPa and a Mach number of 03 Assuming a steady isentro pic flow determine the temperature pressure and Mach number at a location where the flow area has been reduced by 20 percent 17126 Repeat Prob 17125 for an inlet Mach number of 05 17127 Nitrogen enters a convergingdiverging nozzle at 620 kPa and 310 K with a negligible velocity and it experi ences a normal shock at a location where the Mach number is Ma 30 Calculate the pressure temperature velocity Mach number and stagnation pressure downstream of the shock Compare these results to those of air undergoing a normal shock at the same conditions 17128 An aircraft flies with a Mach number Ma1 09 at an altitude of 7000 m where the pressure is 411 kPa and the temper ature is 2427 K The diffuser at the engine inlet has an exit Mach number of Ma2 03 For a mass flow rate of 38 kgs determine the static pressure rise across the diffuser and the exit area 17129 Consider an equimolar mixture of oxygen and nitro gen Determine the critical temperature pressure and density for stagnation temperature and pressure of 550 K and 350 kPa 17130E Helium expands in a nozzle from 220 psia 740 R and negligible velocity to 15 psia Calculate the throat and exit areas for a mass flow rate of 02 lbms assuming the nozzle is isentropic Why must this nozzle be convergingdiverging 17131 Helium expands in a nozzle from 08 MPa 500 K and negligible velocity to 01 MPa Calculate the throat and exit areas for a mass flow rate of 034 kgs assuming the nozzle is isentropic Why must this nozzle be converging diverging Answers 596 cm2 897 cm2 17132 Air is heated as it flows subsonically through a 10 cm 10 cm square duct The properties of air at the inlet are maintained at Ma1 06 P1 350 kPa and T1 420 K at all times Disregarding frictional losses determine the highest rate of heat transfer to the air in the duct without affecting the inlet conditions Answer 716 kW FIGURE P17132 P1 350 kPa T1 420 K Ma1 06 Qmax Final PDF to printer cen22672ch17823880indd 878 110917 0138 PM 878 COMPRESSIBLE FLOW 17133 Repeat Prob 17132 for helium 17134 Air is accelerated as it is heated in a duct with negli gible friction Air enters at V1 100 ms T1 400 K and P1 35 kPa and then exits at a Mach number of Ma2 08 Deter mine the heat transfer to the air in kJkg Also determine the maximum amount of heat transfer without reducing the mass flow rate of air 17135 Air at sonic conditions and at static temperature and pressure of 340 K and 250 kPa respectively is to be acceler ated to a Mach number of 16 by cooling it as it flows through a channel with constant crosssectional area Disregarding fric tional effects determine the required heat transfer from the air in kJkg Answer 475 kJkg 17136 Air is cooled as it flows through a 30cmdiameter duct The inlet conditions are Ma1 12 T01 350 K and P01 240 kPa and the exit Mach number is Ma2 20 Disre garding frictional effects determine the rate of cooling of air 17137 Saturated steam enters a convergingdiverging noz zle at 175 MPa 10 percent moisture and negligible veloc ity and it exits at 12 MPa For a nozzle exit area of 25 cm2 determine the throat area exit velocity mass flow rate and exit Mach number if the nozzle a is isentropic and b has an efficiency of 92 percent 17138 Air flows through a convergingdiverging nozzle in which the exit area is 2896 times the throat area Upstream of the nozzle entrance the velocity is negligibly small and the pressure and temperature are 20 MPa and 150C respectively Calculate the back pressure just outside the nozzle such that a normal shock sits right at the nozzle exit plane 17139 Using appropriate software and the relations in Table A32 calculate the onedimensional compressible flow functions for an ideal gas with k 1667 and present your results by duplicating Table A32 17140 Using appropriate software and the relations in Table A33 calculate the onedimensional normal shock functions for an ideal gas with k 1667 and present your results by duplicating Table A33 17141 Using appropriate software determine the shape of a convergingdiverging nozzle for air for a mass flow rate of 3 kgs and inlet stagnation conditions of 1400 kPa and 200C Approximate the flow as isentropic Repeat the calculations for 50kPa increments of pressure drop to an exit pressure of 100 kPa Plot the nozzle to scale Also calculate and plot the Mach number along the nozzle 17142 Steam at 60 MPa and 700 K enters a converg ing nozzle with a negligible velocity The noz zle throat area is 8 cm2 Approximating the flow as isentropic plot the exit pressure the exit velocity and the mass flow rate through the nozzle versus the back pressure Pb for 60 Pb 30 MPa Treat the steam as an ideal gas with k 13 cp 1872 kJkgK and R 0462 kJkgK 17143 Using appropriate software and the relations given in Table A32 calculate the one dimensional isentropic compressibleflow functions by varying the upstream Mach number from 1 to 10 in increments of 05 for air with k 14 17144 Repeat Prob 17143 for methane with k 13 Fundamentals of Engineering FE Exam Problems 17145 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane Now the nozzle exit diameter is reduced by half while the nozzle inlet tem perature and pressure are maintained the same The nozzle exit velocity will a remain the same b double c quadruple d go down by half e go down by onefourth 17146 An aircraft is cruising in still air at 5C at a veloc ity of 400 ms The air temperature at the nose of the aircraft where stagnation occurs is a 5C b 25C c 55C d 80C e 85C 17147 Air is flowing in a wind tunnel at 25C 95 kPa and 250 ms The stagnation pressure at the location of a probe inserted into the flow section is a 184 kPa b 96 kPa c 161 kPa d 122 kPa e 135 kPa 17148 Air is flowing in a wind tunnel at 12C and 66 kPa at a velocity of 190 ms The Mach number of the flow is a 056 b 065 c 073 d 087 e 17 17149 An aircraft is reported to be cruising in still air at 20C and 40 kPa at a Mach number of 086 The velocity of the aircraft is a 91 ms b 220 ms c 186 ms d 274 ms e 378 ms 17150 Air is approaching a convergingdiverging nozzle with a low velocity at 12C and 200 kPa and it leaves the noz zle at a supersonic velocity The velocity of air at the throat of the nozzle is a 338 ms b 309 ms c 280 ms d 256 ms e 95 ms 17151 Argon gas is approaching a convergingdiverging nozzle with a low velocity at 20C and 150 kPa and it leaves the nozzle at a supersonic velocity If the crosssectional area of the throat is 0015 m2 the mass flow rate of argon through the nozzle is a 047 kgs b 17 kgs c 26 kgs d 66 kgs e 102 kgs Final PDF to printer cen22672ch17823880indd 879 110917 0138 PM 879 CHAPTER 17 17152 Carbon dioxide enters a convergingdiverging noz zle at 60 ms 310C and 300 kPa and it leaves the nozzle at a supersonic velocity The velocity of carbon dioxide at the throat of the nozzle is a 125 ms b 225 ms c 312 ms d 353 ms e 377 ms 17153 Consider gas flow through a convergingdiverging nozzle Of the five following statements select the one that is incorrect a The fluid velocity at the throat can never exceed the speed of sound b If the fluid velocity at the throat is below the speed of sound the diverging section will act like a diffuser c If the fluid enters the diverging section with a Mach num ber greater than one the flow at the nozzle exit will be supersonic d There will be no flow through the nozzle if the back pres sure equals the stagnation pressure e The fluid velocity decreases the entropy increases and stagnation enthalpy remains constant during flow through a normal shock 17154 Combustion gases with k 133 enter a converging nozzle at stagnation temperature and pressure of 350C and 400 kPa and are discharged into the atmospheric air at 20C and 100 kPa The lowest pressure that will occur within the nozzle is a 13 kPa b 100 kPa c 216 kPa d 290 kPa e 315 kPa Design and Essay Problems 17155 Find out if there is a supersonic wind tunnel on your campus If there is obtain the dimensions of the wind tun nel and the temperatures and pressures as well as the Mach number at several locations during operation For what typical experiments is the wind tunnel used 17156 Assuming you have a thermometer and a device to mea sure the speed of sound in a gas explain how you can determine the mole fraction of helium in a mixture of helium gas and air 17157 Design a 1mlong cylindrical wind tunnel whose diameter is 25 cm operating at a Mach number of 18 Atmo spheric air enters the wind tunnel through a converging diverging nozzle where it is accelerated to supersonic velocities Air leaves the tunnel through a convergingdiverging diffuser where it is decelerated to a very low velocity before entering the fan section Disregard any irreversibilities Specify the tem peratures and pressures at several locations as well as the mass flow rate of air at steadyflow conditions Why is it often neces sary to dehumidify the air before it enters the wind tunnel FIGURE P17157 P0 Ma 18 D 25 cm T0 17158 In your own words write a summary of the differ ences between incompressible flow subsonic flow and super sonic flow Final PDF to printer cen22672ch17823880indd 880 110917 0138 PM Final PDF to printer 881 cen22672app01881930indd 881 110617 0932 AM 1 APPENDIX P R O P E RT Y TA B L E S AN D C H A RTS S I U N ITS Table A1 Molar mass gas constant and criticalpoint properties 882 Table A2 Idealgas specific heats of various common gases 883 Table A3 Properties of common liquids solids and foods 886 Table A4 Saturated waterTemperature table 888 Table A5 Saturated waterPressure table 890 Table A6 Superheated water 892 Table A7 Compressed liquid water 896 Table A8 Saturated icewater vapor 897 Figure A9 Ts diagram for water 898 Figure A10 Mollier diagram for water 899 Table A11 Saturated refrigerant134aTemperature table 900 Table A12 Saturated refrigerant134aPressure table 902 Table A13 Superheated refrigerant134a 903 Figure A14 Ph diagram for refrigerant134a 905 Figure A15 NelsonObert generalized compressibility charts 906 Table A16 Properties of the atmosphere at high altitude 907 Table A17 Idealgas properties of air 908 Table A18 Idealgas properties of nitrogen N2 910 Table A19 Idealgas properties of oxygen O2 912 Table A20 Idealgas properties of carbon dioxide CO2 914 Table A21 Idealgas properties of carbon monoxide CO 916 Table A22 Idealgas properties of hydrogen H2 918 Table A23 Idealgas properties of water vapor H2O 919 Table A24 Idealgas properties of monatomic oxygen O 921 Table A25 Idealgas properties of hydroxyl OH 921 Table A26 Enthalpy of formation Gibbs function of formation and absolute entropy at 25C 1 atm 922 Table A27 Properties of some common fuels and hydrocarbons 923 Table A28 Natural logarithms of the equilibrium constant Kp 924 Figure A29 Generalized enthalpy departure chart 925 Figure A30 Generalized entropy departure chart 926 Figure A31 Psychrometric chart at 1 atm total pressure 927 Table A32 Onedimensional isentropic compressibleflow functions for an ideal gas with k 14 928 Table A33 Onedimensional normalshock functions for an ideal gas with k 14 929 Table A34 Rayleigh flow functions for an ideal gas with k 14 930 Final PDF to printer 882 PROPERTY TABLES AND CHARTS cen22672app01881930indd 882 110617 0932 AM The unit kJkgK is equivalent to kPam3kgK The gas constant is calculated from R RuM where Ru 831447 kJkmolK and M is the molar mass Source of Data K A Kobe and R E Lynn Jr Chemical Review 52 1953 pp 117236 and ASHRAE Handbook of Fundamentals Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 pp 164 and 361 TABLE A1 Molar mass gas constant and criticalpoint properties Substance Formula Molar mass M kgkmol Gas constant R kJkgK Criticalpoint properties Temperature K Pressure MPa Volume m3kmol Air 2897 02870 1325 377 00883 Ammonia NH3 1703 04882 4055 1128 00724 Argon Ar 39948 02081 151 486 00749 Benzene C6H6 78115 01064 562 492 02603 Bromine Br2 159808 00520 584 1034 01355 nButane C4H10 58124 01430 4252 380 02547 Carbon dioxide CO2 4401 01889 3042 739 00943 Carbon monoxide CO 28011 02968 133 350 00930 Carbon tetrachloride CCl4 15382 005405 5564 456 02759 Chlorine Cl2 70906 01173 417 771 01242 Chloroform CHCl3 11938 006964 5366 547 02403 Dichlorodifluoromethane R12 CCl2F2 12091 006876 3847 401 02179 Dichlorofluoromethane R21 CHCl2F 10292 008078 4517 517 01973 Ethane C2H6 30070 02765 3055 448 01480 Ethyl alcohol C2H5OH 4607 01805 516 638 01673 Ethylene C2H4 28054 02964 2824 512 01242 Helium He 4003 20769 53 023 00578 nHexane C6H14 86179 009647 5079 303 03677 Hydrogen normal H2 2016 41240 333 130 00649 Krypton Kr 8380 009921 2094 550 00924 Methane CH4 16043 05182 1911 464 00993 Methyl alcohol CH3OH 32042 02595 5132 795 01180 Methyl chloride CH3Cl 50488 01647 4163 668 01430 Neon Ne 20183 04119 445 273 00417 Nitrogen N2 28013 02968 1262 339 00899 Nitrous oxide N2O 44013 01889 3097 727 00961 Oxygen O2 31999 02598 1548 508 00780 Propane C3H8 44097 01885 370 426 01998 Propylene C3H6 42081 01976 365 462 01810 Sulfur dioxide SO2 64063 01298 4307 788 01217 Tetrafluoroethane R134a CF3CH2F 10203 008149 3742 4059 01993 Trichlorofluoromethane R11 CCl3F 13737 006052 4712 438 02478 Water H2O 18015 04615 6471 2206 00560 Xenon Xe 13130 006332 2898 588 01186 Final PDF to printer 883 APPENDIX 1 cen22672app01881930indd 883 110617 0932 AM TABLE A2 Idealgas specific heats of various common gases a At 300 K Gas Formula Gas constant R kJkgK cp kJkgK cv kJkgK k Air 02870 1005 0718 1400 Argon Ar 02081 05203 03122 1667 Butane C4H10 01433 17164 15734 1091 Carbon dioxide CO2 01889 0846 0657 1289 Carbon monoxide CO 02968 1040 0744 1400 Ethane C2H6 02765 17662 14897 1186 Ethylene C2H4 02964 15482 12518 1237 Helium He 20769 51926 31156 1667 Hydrogen H2 41240 14307 10183 1405 Methane CH4 05182 22537 17354 1299 Neon Ne 04119 10299 06179 1667 Nitrogen N2 02968 1039 0743 1400 Octane C8H18 00729 17113 16385 1044 Oxygen O2 02598 0918 0658 1395 Propane C3H8 01885 16794 14909 1126 Steam H2O 04615 18723 14108 1327 Note The unit kJkgK is equivalent to kJkgºC Source of Data B G Kyle Chemical and Process Thermodynamics 3rd ed Upper Saddle River NJ Prentice Hall 2000 Final PDF to printer 884 PROPERTY TABLES AND CHARTS cen22672app01881930indd 884 110617 0932 AM TABLE A2 Idealgas specific heats of various common gases Continued b At various temperatures Temperature K cp kJkgK cv kJkgK k cp kJkgK cv kJkgK k cp kJkgK cv kJkgK k Air Carbon dioxide CO2 Carbon monoxide CO 250 1003 0716 1401 0791 0602 1314 1039 0743 1400 300 1005 0718 1400 0846 0657 1288 1040 0744 1399 350 1008 0721 1398 0895 0706 1268 1043 0746 1398 400 1013 0726 1395 0939 0750 1252 1047 0751 1395 450 1020 0733 1391 0978 0790 1239 1054 0757 1392 500 1029 0742 1387 1014 0825 1229 1063 0767 1387 550 1040 0753 1381 1046 0857 1220 1075 0778 1382 600 1051 0764 1376 1075 0886 1213 1087 0790 1376 650 1063 0776 1370 1102 0913 1207 1100 0803 1370 700 1075 0788 1364 1126 0937 1202 1113 0816 1364 750 1087 0800 1359 1148 0959 1197 1126 0829 1358 800 1099 0812 1354 1169 0980 1193 1139 0842 1353 900 1121 0834 1344 1204 1015 1186 1163 0866 1343 1000 1142 0855 1336 1234 1045 1181 1185 0888 1335 Hydrogen H2 Nitrogen N2 Oxygen O2 250 14051 9927 1416 1039 0742 1400 0913 0653 1398 300 14307 10183 1405 1039 0743 1400 0918 0658 1395 350 14427 10302 1400 1041 0744 1399 0928 0668 1389 400 14476 10352 1398 1044 0747 1397 0941 0681 1382 450 14501 10377 1398 1049 0752 1395 0956 0696 1373 500 14513 10389 1397 1056 0759 1391 0972 0712 1365 550 14530 10405 1396 1065 0768 1387 0988 0728 1358 600 14546 10422 1396 1075 0778 1382 1003 0743 1350 650 14571 10447 1395 1086 0789 1376 1017 0758 1343 700 14604 10480 1394 1098 0801 1371 1031 0771 1337 750 14645 10521 1392 1110 0813 1365 1043 0783 1332 800 14695 10570 1390 1121 0825 1360 1054 0794 1327 900 14822 10698 1385 1145 0849 1349 1074 0814 1319 1000 14983 10859 1380 1167 0870 1341 1090 0830 1313 Source of Data Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 p 783 Table A4M Originally published in Tables of Thermal Properties of Gases NBS Circular 564 1955 Final PDF to printer 885 APPENDIX 1 cen22672app01881930indd 885 110617 0932 AM Source of Data B G Kyle Chemical and Process Thermodynamics Englewood Cliffs NJ PrenticeHall 1984 TABLE A2 Idealgas specific heats of various common gases Concluded c As a function of temperature c p a bT c T 2 d T 3 T in K cp in kJkmolK Substance Formula a b c d Temperature range K error Max Avg Nitrogen N2 2890 01571 102 08081 105 2873 109 2731800 059 034 Oxygen O2 2548 1520 102 07155 105 1312 109 2731800 119 028 Air 2811 01967 102 04802 105 1966 109 2731800 072 033 Hydrogen H2 2911 01916 102 04003 105 08704 109 2731800 101 026 Carbon monoxide CO 2816 01675 102 05372 105 2222 109 2731800 089 037 Carbon dioxide CO2 2226 5981 102 3501 105 7469 109 2731800 067 022 Water vapor H2O 3224 01923 102 1055 105 3595 109 2731800 053 024 Nitric oxide NO 2934 009395 102 09747 105 4187 109 2731500 097 036 Nitrous oxide N2O 2411 58632 102 3562 105 1058 109 2731500 059 026 Nitrogen dioxide NO2 229 5715 102 352 105 787 109 2731500 046 018 Ammonia NH3 27568 25630 102 099072 105 66909 109 2731500 091 036 Sulfur S 2721 2218 102 1628 105 3986 109 2731800 099 038 Sulfur dioxide SO2 2578 5795 102 3812 105 8612 109 2731800 045 024 Sulfur trioxide SO3 1640 1458 102 1120 105 3242 109 2731300 029 013 Acetylene C2H2 218 92143 102 6527 105 1821 109 2731500 146 059 Benzene C6H6 3622 48475 102 3157 105 7762 109 2731500 034 020 Methanol CH4O 190 9152 102 122 105 8039 109 2731000 018 008 Ethanol C2H6O 199 2096 102 1038 105 2005 109 2731500 040 022 Hydrogen chloride HCl 3033 07620 102 1327 105 4338 109 2731500 022 008 Methane CH4 1989 5024 102 1269 105 1101 109 2731500 133 057 Ethane C2H6 6900 1727 102 6406 105 7285 109 2731500 083 028 Propane C3H8 404 3048 102 1572 105 3174 109 2731500 040 012 nButane C4H10 396 3715 102 1834 105 3500 109 2731500 054 024 iButane C4H10 7913 4160 102 2301 105 4991 109 2731500 025 013 nPentane C5H12 6774 4543 102 2246 105 4229 109 2731500 056 021 nHexane C6H14 6938 5522 102 2865 105 5769 109 2731500 072 020 Ethylene C2H4 395 1564 102 8344 105 1767 109 2731500 054 013 Propylene C3H6 315 2383 102 1218 105 2462 109 2731500 073 017 Final PDF to printer 886 PROPERTY TABLES AND CHARTS cen22672app01881930indd 886 110617 0932 AM TABLE A3 Properties of common liquids solids and foods a Liquids Substance Boiling data at 1 atm Freezing data Liquid properties Normal boiling point C Latent heat of vaporization hfg kJkg Freezing point C Latent heat of fusion hifkJkg Temperature C Density ρ kgm3 Specific heat cp kJkgK Ammonia 333 1357 777 3224 333 682 443 20 665 452 0 639 460 25 602 480 Argon 1859 1616 1893 28 1856 1394 114 Benzene 802 394 55 126 20 879 172 Brine 20 sodium chloride by mass 1039 174 20 1150 311 nButane 05 3852 1385 803 05 601 231 Carbon dioxide 784 2305 at 0C 566 0 298 059 Ethanol 782 8383 1142 109 25 783 246 Ethyl alcohol 786 855 156 108 20 789 284 Ethylene glycol 1981 8001 108 1811 20 1109 284 Glycerine 1799 974 189 2006 20 1261 232 Helium 2689 228 2689 1462 228 Hydrogen 2528 4457 2592 595 2528 707 100 Isobutane 117 3671 160 1057 117 5938 228 Kerosene 204293 251 249 20 820 200 Mercury 3567 2947 389 114 25 13560 0139 Methane 1615 5104 1822 584 1615 423 349 100 301 579 Methanol 645 1100 977 992 25 787 255 Nitrogen 1958 1986 210 253 1958 809 206 160 596 297 Octane 1248 3063 575 1807 20 703 210 Oil light 25 910 180 Oxygen 183 2127 2188 137 183 1141 171 Petroleum 230384 20 640 20 Propane 421 4278 1877 800 421 581 225 0 529 253 50 449 313 Refrigerant134a 261 2170 966 50 1443 123 261 1374 127 0 1295 134 25 1207 143 Water 100 2257 00 3337 0 1000 422 25 997 418 50 988 418 75 975 419 100 958 422 Sublimation temperature At pressures below the triplepoint pressure of 518 kPa carbon dioxide exists as a solid or gas Also the freezingpoint temperature of carbon dioxide is the triplepoint temperature of 565C Final PDF to printer 887 APPENDIX 1 cen22672app01881930indd 887 110617 0932 AM TABLE A3 Properties of common liquids solids and foods Concluded b Solids values are for room temperature unless indicated otherwise Substance Density ρ kgm3 Specific heat cp kJkgK Substance Density ρ kgm3 Specific heat cp kJkgK Metals Nonmetals Aluminum Asphalt 2110 0920 200 K 0797 Brick common 1922 079 250 K 0859 Brick fireclay 500C 2300 0960 300 K 2700 0902 Concrete 2300 0653 350 K 0929 Clay 1000 0920 400 K 0949 Diamond 2420 0616 450 K 0973 Glass window 2700 0800 500 K 0997 Glass pyrex 2230 0840 Bronze 76 Cu 2 Zn 2 Al 8280 0400 Graphite 2500 0711 Granite 2700 1017 Brass yellow 65 Cu 35 Zn 8310 0400 Gypsum or plaster board 800 109 Ice Copper 200 K 156 173C 0254 220 K 171 100C 0342 240 K 186 50C 0367 260 K 201 0C 0381 273 K 921 211 27C 8900 0386 Limestone 1650 0909 100C 0393 Marble 2600 0880 200C 0403 Plywood Douglas Fir 545 121 Iron 7840 045 Rubber soft 1100 1840 Lead 11310 0128 Rubber hard 1150 2009 Magnesium 1730 1000 Sand 1520 0800 Nickel 8890 0440 Stone 1500 0800 Silver 10470 0235 Woods hard maple oak etc 721 126 Steel mild 7830 0500 Woods soft fir pine etc 513 138 Tungsten 19400 0130 c Foods Food Water content mass Freezing point C Specific heat kJkgK Latent heat of fusion kJkg Food Water content mass Freezing point C Specific heat kJkgK Latent heat of fusion kJkg Above freezing Below freezing Above freezing Below freezing Apples 84 11 365 190 281 Lettuce 95 02 402 204 317 Bananas 75 08 335 178 251 Milk whole 88 06 379 195 294 Beef round 67 308 168 224 Oranges 87 08 375 194 291 Broccoli 90 06 386 197 301 Potatoes 78 06 345 182 261 Butter 16 104 53 Salmon fish 64 22 298 165 214 Cheese swiss 39 100 215 133 130 Shrimp 83 22 362 189 277 Cherries 80 18 352 185 267 Spinach 93 03 396 201 311 Chicken 74 28 332 177 247 Strawberries 90 08 386 197 301 Corn sweet 74 06 332 177 247 Tomatoes ripe 94 05 399 202 314 Eggs whole 74 06 332 177 247 Turkey 64 298 165 214 Ice cream 63 56 295 163 210 Watermelon 93 04 396 201 311 Source of Data Values are obtained from various handbooks and other sources or are calculated Water content and freezingpoint data of foods are from ASHRAE Handbook of Fundamentals SI version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 Chapter 30 Table1 Freezing point is the temperature at which freezing starts for fruits and vegetables and the average freezing temperature for other foods Final PDF to printer 888 PROPERTY TABLES AND CHARTS cen22672app01881930indd 888 110617 0932 AM TABLE A4 Saturated waterTemperature table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Temp T C Sat Press Psat kPa Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 001 06117 0001000 20600 0000 23749 23749 0001 25009 25009 00000 91556 91556 5 08725 0001000 14703 21019 23608 23818 21020 24891 25101 00763 89487 90249 10 12281 0001000 10632 42020 23466 23887 42022 24772 25192 01511 87488 88999 15 17057 0001001 77885 62980 23325 23955 62982 24654 25283 02245 85559 87803 20 23392 0001002 57762 83913 23184 24023 83915 24535 25374 02965 83696 86661 25 31698 0001003 43340 10483 23043 24091 10483 24417 25465 03672 81895 85567 30 42469 0001004 32879 12573 22902 24159 12574 24298 25556 04368 80152 84520 35 56291 0001006 25205 14663 22760 24227 14664 24179 25646 05051 78466 83517 40 73851 0001008 19515 16753 22619 24294 16753 24060 25735 05724 76832 82556 45 95953 0001010 15251 18843 22477 24361 18844 23940 25824 06386 75247 81633 50 12352 0001012 12026 20933 22334 24427 20934 23820 25913 07038 73710 80748 55 15763 0001015 95639 23024 22191 24493 23026 23698 26001 07680 72218 79898 60 19947 0001017 76670 25116 22047 24559 25118 23577 26088 08313 70769 79082 65 25043 0001020 61935 27209 21903 24624 27212 23454 26175 08937 69360 78296 70 31202 0001023 50396 29304 21758 24689 29307 23330 26261 09551 67989 77540 75 38597 0001026 41291 31399 21613 24753 31403 23206 26346 10158 66655 76812 80 47416 0001029 34053 33497 21466 24816 33502 23080 26430 10756 65355 76111 85 57868 0001032 28261 35596 21319 24878 35602 22953 26514 11346 64089 75435 90 70183 0001036 23593 37697 21170 24940 37704 22825 26596 11929 62853 74782 95 84609 0001040 19808 39800 21020 25001 39809 22696 26676 12504 61647 74151 100 10142 0001043 16720 41906 20870 25060 41917 22564 26756 13072 60470 73542 105 12090 0001047 14186 44015 20718 25119 44028 22431 26834 13634 59319 72952 110 14338 0001052 12094 46127 20564 25177 46142 22297 26911 14188 58193 72382 115 16918 0001056 10360 48242 20409 25233 48259 22160 26986 14737 57092 71829 120 19867 0001060 089133 50360 20253 25289 50381 22021 27060 15279 56013 71292 125 23223 0001065 077012 52483 20095 25343 52507 21881 27131 15816 54956 70771 130 27028 0001070 066808 54610 19934 25395 54638 21737 27201 16346 53919 70265 135 31322 0001075 058179 56741 19773 25447 56775 21591 27269 16872 52901 69773 140 36153 0001080 050850 58877 19609 25496 58916 21443 27335 17392 51901 69294 145 41568 0001085 044600 61019 19442 25544 61064 21292 27398 17908 50919 68827 150 47616 0001091 039248 63166 19274 25591 63218 21138 27459 18418 49953 68371 155 54349 0001096 034648 65319 19103 25635 65379 20980 27518 18924 49002 67927 160 61823 0001102 030680 67479 18930 25678 67547 20820 27575 19426 48066 67492 165 70093 0001108 027244 69646 18754 25719 69724 20656 27628 19923 47143 67067 170 79218 0001114 024260 71820 18575 25757 71908 20488 27679 20417 46233 66650 175 89260 0001121 021659 74002 18394 25794 74102 20317 27727 20906 45335 66242 180 10028 0001127 019384 76192 18209 25828 76305 20142 27772 21392 44448 65841 185 11235 0001134 017390 78391 18021 25860 78519 19962 27814 21875 43572 65447 190 12552 0001141 015636 80600 17830 25890 80743 19779 27853 22355 42705 65059 195 13988 0001149 014089 82818 17636 25917 82978 19590 27888 22831 41847 64678 200 15549 0001157 012721 85046 17437 25942 85226 19398 27920 23305 40997 64302 Final PDF to printer 889 APPENDIX 1 cen22672app01881930indd 889 110617 0932 AM TABLE A4 Saturated waterTemperature table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Temp T C Sat Press Psat kPa Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 205 17243 0001164 011508 87286 17235 25964 87487 19200 27948 23776 40154 63930 210 19077 0001173 010429 89538 17029 25983 89761 18997 27973 24245 39318 63563 215 21059 0001181 0094680 91802 16819 25999 92050 18788 27993 24712 38489 63200 220 23196 0001190 0086094 94079 16605 26013 94355 18574 28010 25176 37664 62840 225 25497 0001199 0078405 96370 16386 26023 96676 18354 28022 25639 36844 62483 230 27971 0001209 0071505 98676 16161 26029 99014 18128 28029 26100 36028 62128 235 30626 0001219 0065300 10100 15932 26032 10137 17895 28032 26560 35216 61775 240 33470 0001229 0059707 10334 15698 26031 10375 17655 28030 27018 34405 61424 245 36512 0001240 0054656 10569 15457 26027 10615 17408 28022 27476 33596 61072 250 39762 0001252 0050085 10807 15211 26018 10857 17153 28010 27933 32788 60721 255 43229 0001263 0045941 11047 14958 26005 11101 16890 27991 28390 31979 60369 260 46923 0001276 0042175 11288 14699 25987 11348 16618 27966 28847 31169 60017 265 50853 0001289 0038748 11533 14432 25965 11598 16337 27935 29304 30358 59662 270 55030 0001303 0035622 11779 14157 25937 11851 16046 27897 29762 29542 59305 275 59464 0001317 0032767 12029 13874 25903 12107 15745 27852 30221 28723 58944 280 64166 0001333 0030153 12282 13582 25864 12367 15432 27799 30681 27898 58579 285 69146 0001349 0027756 12537 13281 25818 12631 15107 27737 31144 27066 58210 290 74418 0001366 0025554 12797 12969 25765 12898 14769 27667 31608 26225 57834 295 79990 0001384 0023528 13060 12645 25705 13171 14416 27587 32076 25374 57450 300 85879 0001404 0021659 13327 12309 25636 13448 14048 27496 32548 24511 57059 305 92094 0001425 0019932 13600 11959 25558 13731 13663 27394 33024 23633 56657 310 98650 0001447 0018333 13877 11593 25471 14020 13259 27279 33506 22737 56243 315 10556 0001472 0016849 14161 11211 25372 14316 12834 27150 33994 21821 55816 320 11284 0001499 0015470 14451 10809 25260 14620 12385 27006 34491 20881 55372 325 12051 0001528 0014183 14750 10385 25134 14934 11910 26843 34998 19911 54908 330 12858 0001560 0012979 15057 9935 24992 15258 11403 26660 35516 18906 54422 335 13707 0001597 0011848 15375 9455 24830 15594 10860 26454 36050 17857 53907 340 14601 0001638 0010783 15707 8938 24645 15946 10274 26220 36602 16756 53358 345 15541 0001685 0009772 16055 8377 24432 16317 9634 25951 37179 15585 52765 350 16529 0001741 0008806 16424 7759 24183 16712 8927 25639 37788 14326 52114 355 17570 0001808 0007872 16822 7064 23886 17140 8129 25269 38442 12942 51384 360 18666 0001895 0006950 17262 6257 23519 17615 7201 24816 39165 11373 50537 365 19822 0002015 0006009 17772 5264 23036 18172 6055 24227 40004 09489 49493 370 21044 0002217 0004953 18445 3856 22301 18912 4431 23343 41119 06890 48009 37395 22064 0003106 0003106 20157 0 20157 20843 0 20843 44070 0 44070 Source of Data Tables A4 through A8 are generated using the Engineering Equation Solver EES software developed by S A Klein and F L Alvarado The routine used in calculations is the highly accurate SteamIAPWS which incorporates the 1995 Formulation for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use issued by The International Association for the Properties of Water and Steam IAPWS This formulation replaces the 1984 formulation of Haar Gallagher and Kell NBSNRC Steam Tables Hemisphere Publishing Co 1984 which is also available in EES as the routine STEAM The new formulation is based on the correlations of Saul and Wagner J Phys Chem Ref Data 16 893 1987 with modifications to adjust to the International Temperature Scale of 1990 The modifications are described by Wagner and Pruss J Phys Chem Ref Data 22 783 1993 The properties of ice are based on Hyland and Wexler Formulations for the Thermodynamic Properties of the Saturated Phases of H2O from 17315 K to 47315 K ASHRAE Trans Part 2A Paper 2793 1983 Concluded Final PDF to printer 890 PROPERTY TABLES AND CHARTS cen22672app01881930indd 890 110617 0932 AM TABLE A5 Saturated waterPressure table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Press P kPa Sat temp Tsat C Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 10 697 0001000 12919 29302 23552 23845 29303 24844 25137 01059 88690 89749 15 1302 0001001 87964 54686 23381 23928 54688 24701 25247 01956 86314 88270 20 1750 0001001 66990 73431 23255 23989 73433 24595 25329 02606 84621 87227 25 2108 0001002 54242 88422 23154 24038 88424 24510 25394 03118 83302 86421 30 2408 0001003 45654 10098 23069 24079 10098 24439 25448 03543 82222 85765 40 2896 0001004 34791 12139 22931 24145 12139 24323 25537 04224 80510 84734 50 3287 0001005 28185 13775 22821 24198 13775 24230 25607 04762 79176 83938 75 4029 0001008 19233 16874 22611 24298 16875 24053 25740 05763 76738 82501 10 4581 0001010 14670 19179 22454 24372 19181 23921 25839 06492 74996 81488 15 5397 0001014 10020 22593 22221 24480 22594 23723 25983 07549 72522 80071 20 6006 0001017 76481 25140 22046 24560 25142 23575 26089 08320 70752 79073 25 6496 0001020 62034 27193 21904 24624 27196 23455 26175 08932 69370 78302 30 6909 0001022 52287 28924 21785 24677 28927 23353 26246 09441 68234 77675 40 7586 0001026 39933 31758 21588 24763 31762 23184 26361 10261 66430 76691 50 8132 0001030 32403 34049 21427 24832 34054 23047 26452 10912 65019 75931 75 9176 0001037 22172 38436 21118 24961 38444 22780 26624 12132 62426 74558 100 9961 0001043 16941 41740 20882 25056 41751 22575 26750 13028 60562 73589 101325 9997 0001043 16734 41895 20870 25060 41906 22565 26756 13069 60476 73545 125 10597 0001048 13750 44423 20688 25130 44436 22406 26849 13741 59100 72841 150 11135 0001053 11594 46697 20523 25192 46713 22260 26931 14337 57894 72231 175 11604 0001057 10037 48682 20377 25245 48701 22131 27002 14850 56865 71716 200 12021 0001061 088578 50450 20246 25291 50471 22016 27063 15302 55968 71270 225 12397 0001064 079329 52047 20127 25332 52071 21910 27117 15706 55171 70877 250 12741 0001067 071873 53508 20018 25368 53535 21812 27165 16072 54453 70525 275 13058 0001070 065732 54857 19916 25401 54886 21720 27209 16408 53800 70207 300 13352 0001073 060582 56111 19821 25432 56143 21635 27249 16717 53200 69917 325 13627 0001076 056199 57284 19731 25459 57319 21554 27286 17005 52645 69650 350 13886 0001079 052422 58389 19646 25485 58426 21477 27320 17274 52128 69402 375 14130 0001081 049133 59432 19566 25509 59473 21404 27351 17526 51645 69171 400 14361 0001084 046242 60422 19489 25531 60466 21334 27381 17765 51191 68955 450 14790 0001088 041392 62265 19345 25571 62314 21203 27434 18205 50356 68561 500 15183 0001093 037483 63954 19212 25607 64009 21080 27481 18604 49603 68207 550 15546 0001097 034261 65516 19088 25639 65577 20966 27524 18970 48916 67886 600 15883 0001101 031560 66972 18971 25668 67038 20858 27562 19308 48285 67593 650 16198 0001104 029260 68337 18861 25694 68408 20755 27596 19623 47699 67322 700 16495 0001108 027278 69623 18756 25718 69700 20658 27628 19918 47153 67071 750 16775 0001111 025552 70840 18656 25740 70924 20564 27657 20195 46642 66837 Final PDF to printer 891 APPENDIX 1 cen22672app01881930indd 891 110617 0932 AM TABLE A5 Saturated waterPressure table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Press P kPa Sat temp Tsat C Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 800 17041 0001115 024035 71997 18561 25760 72087 20475 27683 20457 46160 66616 850 17294 0001118 022690 73100 18469 25779 73195 20388 27708 20705 45705 66409 900 17535 0001121 021489 74155 18381 25796 74256 20305 27730 20941 45273 66213 950 17766 0001124 020411 75167 18296 25813 75274 20224 27752 21166 44862 66027 1000 17988 0001127 019436 76139 18214 25828 76251 20146 27771 21381 44470 65850 1100 18406 0001133 017745 77978 18057 25855 78103 19996 27807 21785 43735 65520 1200 18796 0001138 016326 79696 17909 25878 79833 19854 27838 22159 43058 65217 1300 19160 0001144 015119 81310 17768 25899 81459 19719 27865 22508 42428 64936 1400 19504 0001149 014078 82835 17634 25918 82996 19589 27889 22835 41840 64675 1500 19829 0001154 013171 84282 17506 25934 84455 19464 27910 23143 41287 64430 1750 20572 0001166 011344 87612 17206 25967 87816 19171 27952 23844 40033 63877 2000 21238 0001177 0099587 90612 16930 25991 90847 18898 27983 24467 38923 63390 2250 21841 0001187 0088717 93354 16673 26009 93621 18643 28005 25029 37926 62954 2500 22395 0001197 0079952 95887 16432 26021 96187 18401 28019 25542 37016 62558 3000 23385 0001217 0066667 10046 15985 26032 10083 17949 28032 26454 35402 61856 3500 24256 0001235 0057061 10454 15576 26030 10497 17530 28027 27253 33991 61244 4000 25035 0001252 0049779 10824 15193 26017 10874 17135 28008 27966 32731 60696 5000 26394 0001286 0039448 11481 14489 25970 11545 16397 27942 29207 30530 59737 6000 27559 0001319 0032449 12058 13841 25899 12138 15709 27846 30275 28627 58902 7000 28583 0001352 0027378 12580 13230 25810 12675 15052 27726 31220 26927 58148 8000 29501 0001384 0023525 13060 12645 25705 13171 14416 27587 32077 25373 57450 9000 30335 0001418 0020489 13509 12076 25585 13637 13793 27429 32866 23925 56791 10000 31100 0001452 0018028 13933 11518 25452 14078 13176 27255 33603 22556 56159 11000 31808 0001488 0015988 14339 10966 25304 14502 12561 27063 34299 21245 55544 12000 32468 0001526 0014264 14730 10413 25143 14913 11941 26854 34964 19975 54939 13000 33085 0001566 0012781 15110 9855 24966 15314 11313 26627 35606 18730 54336 14000 33667 0001610 0011487 15484 9287 24771 15710 10670 26379 36232 17497 53728 15000 34216 0001657 0010341 15855 8703 24557 16103 10005 26108 36848 16261 53108 16000 34736 0001710 0009312 16226 8094 24320 16499 9311 25810 37461 15005 52466 17000 35229 0001770 0008374 16602 7451 24054 16903 8574 25477 38082 13709 51791 18000 35699 0001840 0007504 16991 6759 23750 17322 7778 25100 38720 12343 51064 19000 36147 0001926 0006677 17403 5989 23392 17768 6892 24660 39396 10860 50256 20000 36575 0002038 0005862 17858 5090 22948 18266 5855 24121 40146 09164 49310 21000 36983 0002207 0004994 18416 3919 22335 18880 4504 23384 41071 07005 48076 22000 37371 0002703 0003644 19517 1408 20924 20111 1615 21726 42942 02496 45439 22064 37395 0003106 0003106 20157 0 20157 20843 0 20843 44070 0 44070 Concluded Final PDF to printer 892 PROPERTY TABLES AND CHARTS cen22672app01881930indd 892 110617 0932 AM TABLE A6 Superheated water T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 001 MPa 4581C P 005 MPa 8132C P 010 MPa 9961C Sat 14670 24372 25839 81488 32403 24832 26452 75931 16941 25056 26750 73589 50 14867 24433 25920 81741 100 17196 25155 26875 84489 34187 25115 26824 76953 16959 25062 26758 73611 150 19513 25879 27830 86893 38897 25857 27802 79413 19367 25829 27766 76148 200 21826 26614 28796 89049 43562 26600 28778 81592 21724 26582 28755 78356 250 24136 27361 29775 91015 48206 27351 29762 83568 24062 27339 29745 80346 300 26446 28123 30767 92827 52841 28116 30758 85387 26389 28107 30745 82172 400 31063 29693 32800 96094 62094 29689 32793 88659 31027 29683 32786 85452 500 35680 31329 34897 98998 71338 31326 34893 91566 35655 31322 34887 88362 600 40296 33033 37063 101631 80577 33031 37060 94201 40279 33028 37056 90999 700 44911 34808 39299 104056 89813 34806 39297 96626 44900 34804 39294 93424 800 49527 36654 41606 106312 99047 36652 41604 98883 49519 36650 41602 95682 900 54143 38569 43983 108429 108280 38568 43982 101000 54137 38567 43980 97800 1000 58758 40553 46428 110429 117513 40552 46427 103000 58755 40550 46426 99800 1100 63373 42600 48938 112326 126745 42599 48937 104897 63372 42598 48936 101698 1200 67989 44709 51508 114132 135977 44708 51507 106704 67988 44707 51506 103504 1300 72604 46874 54134 115857 145209 46873 54133 108429 72605 46872 54133 105229 P 020 MPa 12021C P 030 MPa 13352C P 040 MPa 14361C Sat 088578 25291 27063 71270 060582 25432 27249 69917 046242 25531 27381 68955 150 095986 25771 27691 72810 063402 25710 27612 70792 047088 25644 27528 69306 200 108049 26546 28707 75081 071643 26510 28659 73132 053434 26472 28609 71723 250 119890 27314 29712 77100 079645 27289 29679 75180 059520 27264 29645 73804 300 131623 28088 30721 78941 087535 28070 30696 77037 065489 28051 30671 75677 400 154934 29672 32770 82236 103155 29660 32755 80347 077265 29649 32739 79003 500 178142 31314 34877 85153 118672 31306 34866 83271 088936 31298 34855 81933 600 201302 33022 37048 87793 134139 33016 37040 85915 100558 33010 37033 84580 700 224434 34799 39288 90221 149580 34795 39282 88345 112152 34790 39276 87012 800 247550 36647 41598 92479 165004 36643 41593 90605 123730 36639 41589 89274 900 270656 38563 43977 94598 180417 38560 43973 92725 135298 38557 43969 91394 1000 293755 40548 46423 96599 195824 40545 46420 94726 146859 40543 46417 93396 1100 316848 42596 48933 98497 211226 42594 48931 96624 158414 42592 48929 95295 1200 339938 44705 51504 100304 226624 44703 51502 98431 169966 44702 51500 97102 1300 363026 46871 54131 102029 242019 46869 54130 100157 181516 46867 54128 98828 P 050 MPa 15183C P 060 MPa 15883C P 080 MPa 17041C Sat 037483 25607 27481 68207 031560 25668 27562 67593 024035 25760 27683 66616 200 042503 26433 28558 70610 035212 26394 28506 69683 026088 26311 28398 68177 250 047443 27238 29610 72725 039390 27212 29576 71833 029321 27159 29504 70402 300 052261 28033 30646 74614 043442 28014 30620 73740 032416 27975 30569 72345 350 057015 28830 31681 76346 047428 28816 31661 75481 035442 28786 31622 74107 400 061731 29637 32724 77956 051374 29625 32708 77097 038429 29602 32677 75735 500 071095 31290 34845 80893 059200 31282 34834 80041 044332 31266 34813 78692 600 080409 33004 37025 83544 066976 32998 37017 82695 050186 32987 37001 81354 700 089696 34786 39270 85978 074725 34781 39264 85132 056011 34772 39253 83794 800 098966 36636 41584 88240 082457 36632 41579 87395 061820 36625 41570 86061 900 108227 38554 43966 90362 090179 38551 43962 89518 067619 38545 43955 88185 1000 117480 40540 46414 92364 097893 40538 46411 91521 073411 40533 46405 90189 1100 126728 42590 48926 94263 105603 42588 48924 93420 079197 42583 48919 92090 1200 135972 44700 51498 96071 113309 44698 51496 95229 084980 44694 51493 93898 1300 145214 46866 54126 97797 121012 46864 54125 96955 090761 46861 54122 95625 The temperature in parentheses is the saturation temperature at the specified pressure Properties of saturated vapor at the specified pressure Final PDF to printer 893 APPENDIX 1 cen22672app01881930indd 893 110617 0932 AM TABLE A6 Superheated water T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 100 MPa 17988C P 120 MPa 18796C P 140 MPa 19504C Sat 019437 25828 27771 65850 016326 25878 27838 65217 014078 25918 27889 64675 200 020602 26223 28283 66956 016934 26129 28161 65909 014303 26027 28030 64975 250 023275 27104 29431 69265 019241 27047 29356 68313 016356 26989 29279 67488 300 025799 27937 30516 71246 021386 27897 30463 70335 018233 27857 30409 69553 350 028250 28757 31582 73029 023455 28727 31542 72139 020029 28697 31501 71379 400 030661 29579 32645 74670 025482 29555 32613 73793 021782 29531 32581 73046 500 035411 31250 34791 77642 029464 31234 34770 76779 025216 31218 34748 76047 600 040111 32975 36986 80311 033395 32963 36970 79456 028597 32951 36955 78730 700 044783 34763 39241 82755 037297 34753 39229 81904 031951 34744 39217 81183 800 049438 36617 41561 85024 041184 36610 41552 84176 035288 36603 41543 83458 900 054083 38539 43948 87150 045059 38533 43940 86303 038614 38527 43933 85587 1000 058721 40527 46400 89155 048928 40522 46394 88310 041933 40517 46388 87595 1100 063354 42579 48914 91057 052792 42575 48910 90212 045247 42570 48905 89497 1200 067983 44690 51489 92866 056652 44687 51485 92022 048558 44683 51481 91308 1300 072610 46858 54119 94593 060509 46855 54116 93750 051866 46851 54113 93036 P 160 MPa 20137C P 180 MPa 20711C P 200 MPa 21238C Sat 012374 25948 27928 64200 011037 25973 27959 63775 009959 25991 27983 63390 225 013293 26451 28578 65537 011678 26370 28472 64825 010381 26285 28361 64160 250 014190 26929 29199 66753 012502 26867 29117 66088 011150 26803 29033 65475 300 015866 27816 30354 68864 014025 27774 30299 68246 012551 27732 30242 67684 350 017459 28666 31460 70713 015460 28636 31419 70120 013860 28605 31377 69583 400 019007 29508 32549 72394 016849 29483 32516 71814 015122 29459 32484 71292 500 022029 31201 34726 75410 019551 31185 34704 74845 017568 31169 34683 74337 600 024999 32939 36939 78101 022200 32927 36923 77543 019962 32915 36907 77043 700 027941 34735 39205 80558 024822 34726 39194 80005 022326 34717 39182 79509 800 030865 36595 41534 82834 027426 36588 41524 82284 024674 36580 41515 81791 900 033780 38521 43926 84965 030020 38515 43919 84417 027012 38509 43911 83925 1000 036687 40512 46382 86974 032606 40507 46376 86427 029342 40502 46371 85936 1100 039589 42566 48900 88878 035188 42562 48896 88331 031667 42557 48891 87842 1200 042488 44679 51477 90689 037766 44676 51473 90143 033989 44672 51470 89654 1300 045383 46848 54109 92418 040341 46845 54106 91872 036308 46842 54103 91384 P 250 MPa 22395C P 300 MPa 23385C P 350 MPa 24256C Sat 007995 26021 28019 62558 006667 26032 28032 61856 005706 26030 28027 61244 225 008026 26048 28055 62629 250 008705 26633 28809 64107 007063 26447 28565 62893 005876 26240 28297 61764 300 009894 27622 30096 266459 008118 27508 29943 65412 006845 27388 29784 64484 350 010979 28525 31270 68424 009056 28444 31161 67450 007680 28360 31049 66601 400 012012 29398 32401 70170 009938 29336 32317 69235 008456 29272 32232 68428 450 013015 30262 33516 71768 010789 30212 33449 70856 009198 30161 33381 70074 500 013999 31128 34628 73254 011620 31086 34572 72359 009919 31045 34517 71593 600 015931 32885 36868 75979 013245 32855 36828 75103 011325 32825 36789 74357 700 017835 34693 39152 78455 014841 34670 39122 77590 012702 34647 39093 76855 800 019722 36562 41492 80744 016420 36543 41469 79885 014061 36525 41446 79156 900 021597 38494 43893 82882 017988 38479 43875 82028 015410 38464 43857 81304 1000 023466 40490 46356 84897 019549 40477 46342 84045 016751 40464 46327 83324 1100 025330 42547 48879 86804 021105 42536 48867 85955 018087 42525 48856 85236 1200 027190 44663 51460 88618 022658 44653 51451 87771 019420 44644 51441 87053 1300 029048 46834 54095 90349 024207 46826 54088 89502 020750 46818 54080 88786 Continued Final PDF to printer 894 PROPERTY TABLES AND CHARTS cen22672app01881930indd 894 110617 0932 AM TABLE A6 Superheated water T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 40 MPa 25035C P 45 MPa 25744C P 50 MPa 26394C Sat 004978 26017 28008 60696 004406 25997 27980 60198 003945 25970 27942 59737 275 005461 26689 28873 62312 004733 26514 28644 61429 004144 26323 28395 60571 300 005887 27262 29617 63639 005138 27130 29442 62854 004535 26990 29257 62111 350 006647 28274 30933 65843 005842 28186 30815 65153 005197 28095 30693 64516 400 007343 29208 32145 67714 006477 29142 32057 67071 005784 29075 31967 66483 450 008004 30110 33312 69386 007076 30058 33242 68770 006332 30006 33172 68210 500 008644 31003 34460 70922 007652 30960 34404 70323 006858 30918 34347 69781 600 009886 32794 36749 73706 008766 32764 36709 73127 007870 32733 36669 72605 700 011098 34624 39063 76214 009850 34600 39033 75647 008852 34577 39003 75136 800 012292 36506 41423 78523 010916 36488 41400 77962 009816 36469 41377 77458 900 013476 38448 43839 80675 011972 38433 43821 80118 010769 38418 43802 79619 1000 014653 40451 46312 82698 013020 40439 46298 82144 011715 40426 46283 81648 1100 015824 42514 48844 84612 014064 42504 48832 84060 012655 42493 48821 83566 1200 016992 44635 51432 86430 015103 44626 51422 85880 013592 44616 51413 85388 1300 018157 46809 54072 88164 016140 46801 54065 87616 014527 46793 54057 87124 P 60 MPa 27559C P 70 MPa 28583C P 80 MPa 29501C Sat 003245 25899 27846 58902 0027378 25810 27726 58148 0023525 25705 27587 57450 300 003619 26684 28856 60703 0029492 26335 28399 59337 0024279 25923 27865 57937 350 004225 27904 30439 63357 0035262 27701 30169 62305 0029975 27483 29881 61321 400 004742 28937 31783 65432 0039958 28795 31592 64502 0034344 28646 31394 63658 450 005217 29899 33029 67219 0044187 29790 32883 66353 0038194 29678 32733 65579 500 005667 30831 34231 68826 0048157 30743 34114 68000 0041767 30654 33995 67266 550 006102 31752 35413 70308 0051966 31679 35316 69507 0045172 31605 35218 68800 600 006527 32672 36588 71693 0055665 32610 36506 70910 0048463 32547 36424 70221 700 007355 34530 38943 74247 0062850 34483 38883 73487 0054829 34436 38822 72822 800 008165 36432 41331 76582 0069856 36395 41285 75836 0061011 36357 41238 75185 900 008964 38388 43766 78751 0076750 38357 43730 78014 0067082 38327 43693 77372 1000 009756 40401 46254 80786 0083571 40375 46225 80055 0073079 40350 46196 79419 1100 010543 42471 48797 82709 0090341 42450 48774 81982 0079025 42428 48750 81350 1200 011326 44598 51394 84534 0097075 44579 51374 83810 0084934 44561 51355 83181 1300 012107 46777 54041 86273 0103781 46761 54026 85551 0090817 46745 54010 84925 P 90 MPa 30335C P 100 MPa 31100C P 125 MPa 32781C Sat 0020489 25585 27429 56791 0018028 25452 27255 56159 0013496 25056 26743 54638 325 0023284 26476 28571 58738 0019877 26116 28103 57596 350 0025816 27250 29573 60380 0022440 26996 29240 59460 0016138 26249 28266 57130 400 0029960 28492 31188 62876 0026436 28331 30975 62141 0020030 27896 30400 60433 450 0033524 29563 32580 64872 0029782 29445 32424 64219 0023019 29137 32015 62749 500 0036793 30563 33874 66603 0032811 30470 33751 65995 0025630 30232 33436 64651 550 0039885 31530 35120 68164 0035655 31454 35020 67585 0028033 31261 34765 66317 600 0042861 32484 36341 69605 0038378 32420 36258 69045 0030306 32258 36046 67828 650 0045755 33434 37552 70954 0041018 33380 37481 70408 0032491 33241 37302 69227 700 0048589 34388 38761 72229 0043597 34340 38700 71693 0034612 34220 38546 70540 800 0054132 36320 41192 74606 0048629 36282 41145 74085 0038724 36188 41028 72967 900 0059562 38296 43657 76802 0053547 38265 43620 76290 0042720 38189 43529 75195 1000 0064919 40324 46167 78855 0058391 40299 46138 78349 0046641 40235 46065 77269 1100 0070224 42407 48727 80791 0063183 42385 48703 80289 0050510 42331 48645 79220 1200 0075492 44542 51336 82625 0067938 44524 51317 82126 0054342 44477 51270 81065 1300 0080733 46729 53995 84371 0072667 46713 53980 83874 0058147 46673 53941 82819 Continued Final PDF to printer 895 APPENDIX 1 cen22672app01881930indd 895 110617 0932 AM TABLE A6 Superheated water T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 150 MPa 34216C P 175 MPa 35467C P 200 MPa 36575C Sat 0010341 24557 26108 53108 0007932 23907 25295 51435 0005862 22948 24121 49310 350 0011481 25209 26931 54438 400 0015671 27406 29757 58819 0012463 26843 29024 57211 0009950 26179 28169 55526 450 0018477 28808 31579 61434 0015204 28454 31114 60212 0012721 28073 30617 59043 500 0020828 29984 33108 63480 0017385 29724 32767 62424 0014793 29453 32412 61446 550 0022945 31062 34504 65230 0019305 30858 34236 64266 0016571 30647 33962 63390 600 0024921 32093 35831 66796 0021073 31925 35613 65890 0018185 31753 35390 65075 650 0026804 33101 37121 68233 0022742 32958 36938 67366 0019695 32814 36753 66593 700 0028621 34098 38391 69573 0024342 33975 38235 68735 0021134 33851 38078 67991 800 0032121 36093 40911 72037 0027405 35997 40793 71237 0023870 35901 40675 70531 900 0035503 38112 43437 74288 0030348 38035 43346 73511 0026484 37957 43254 72829 1000 0038808 40171 45992 76378 0033215 40107 45920 75616 0029020 40043 45847 74950 1100 0042062 42277 48586 78339 0036029 42223 48528 77588 0031504 42169 48470 76933 1200 0045279 44431 51223 80192 0038806 44385 51176 79449 0033952 44338 51129 78802 1300 0048469 46633 53903 81952 0041556 46592 53865 81215 0036371 46552 53827 80574 P 250 MPa P 300 MPa P 350 MPa 375 0001978 17999 18494 40345 0001792 17381 17919 39313 0001701 17028 17624 38724 400 0006005 24285 25787 51400 0002798 20689 21528 44758 0002105 19149 19886 42144 425 0007886 26078 28050 54708 0005299 24529 26118 51473 0003434 22533 23735 47751 450 0009176 27212 29506 56759 0006737 26189 28210 54422 0004957 24975 26710 51946 500 0011143 28873 31659 59643 0008691 28240 30848 57956 0006933 27553 29979 56331 550 0012736 30208 33392 61816 0010175 29745 32797 60403 0008348 29258 32180 59093 600 0014140 31400 34935 63637 0011445 31034 34468 62373 0009523 30656 33990 61229 650 0015430 32519 36377 65243 0012590 32217 35994 64074 0010565 31909 35607 63030 700 0016643 33599 37760 66702 0013654 33343 37439 65599 0011523 33083 37116 64623 800 0018922 35707 40438 69322 0015628 35512 40200 68301 0013278 35316 39963 67409 900 0021075 37802 43071 71668 0017473 37646 42888 70695 0014904 37490 42706 69853 1000 0023150 39915 45702 73821 0019240 39786 45558 72880 0016450 39658 45415 72069 1100 0025172 42061 48354 75825 0020954 41952 48239 74906 0017942 41844 48124 74118 1200 0027157 44246 51035 77710 0022630 44153 50942 76807 0019398 44061 50850 76034 1300 0029115 46472 53751 79494 0024279 46392 53676 78602 0020827 46312 53602 77841 P 400 MPa P 500 MPa P 600 MPa 375 0001641 16770 17426 38290 0001560 16386 17166 37642 0001503 16097 16999 37149 400 0001911 18550 19314 41145 0001731 17878 18744 40029 0001633 17452 18432 39317 425 0002538 20975 21990 45044 0002009 19603 20607 42746 0001816 18929 20018 41630 450 0003692 23642 25118 49449 0002487 21603 22847 45896 0002086 20551 21802 44140 500 0005623 26816 29065 54744 0003890 25281 27226 51762 0002952 23932 25703 49356 550 0006985 28751 31544 57857 0005118 27695 30254 55563 0003955 26646 29019 53517 600 0008089 30268 33504 60170 0006108 29471 32526 58245 0004833 28668 31568 56527 650 0009053 31595 35216 62078 0006957 30956 34435 60373 0005591 30313 33668 58867 700 0009930 32820 36792 63740 0007717 32287 36146 62179 0006265 31754 35513 60814 800 0011521 35118 39726 66613 0009073 34722 39258 65225 0007456 34326 38800 64033 900 0012980 37333 42525 69107 0010296 37020 42168 67819 0008519 36709 41821 66725 1000 0014360 39529 45273 71355 0011441 39274 44994 70131 0009504 39020 44722 69099 1100 0015686 41737 48011 73425 0012534 41522 47789 72244 0010439 41309 47573 71255 1200 0016976 43969 50759 75357 0013590 43786 50581 74207 0011339 43605 50408 73248 1300 0018239 46233 53528 77175 0014620 46075 53385 76048 0012213 45918 53245 75111 Concluded Final PDF to printer 896 PROPERTY TABLES AND CHARTS cen22672app01881930indd 896 110617 0932 AM TABLE A7 Compressed liquid water T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 5 MPa 26394C P 10 MPa 31100C P 15 MPa 34216C Sat 00012862 11481 11545 29207 00014522 13933 14079 33603 00016572 15855 16103 36848 0 00009977 004 503 00001 00009952 012 1007 00003 00009928 018 1507 00004 20 00009996 8361 8861 02954 00009973 8331 9328 02943 00009951 8301 9793 02932 40 00010057 16692 17195 05705 00010035 16633 17637 05685 00010013 16575 18077 05666 60 00010149 25029 25536 08287 00010127 24943 25955 08260 00010105 24858 26374 08234 80 00010267 33382 33896 10723 00010244 33269 34294 10691 00010221 33159 34692 10659 100 00010410 41765 42285 13034 00010385 41623 42662 12996 00010361 41485 43039 12958 120 00010576 50191 50719 15236 00010549 50018 51073 15191 00010522 49850 51428 15148 140 00010769 58680 59218 17344 00010738 58472 59545 17293 00010708 58269 59875 17243 160 00010988 67255 67804 19374 00010954 67006 68101 19316 00010920 66763 68401 19259 180 00011240 75947 76509 21338 00011200 75648 76768 21271 00011160 75358 77032 21206 200 00011531 84792 85368 23251 00011482 84432 85580 23174 00011435 84084 85800 23100 220 00011868 93839 94432 25127 00011809 93401 94582 25037 00011752 92981 94743 24951 240 00012268 10316 10377 26983 00012192 10262 10383 26876 00012121 10210 10392 26774 260 00012755 11285 11349 28841 00012653 11216 11343 28710 00012560 11151 11340 28586 280 00013226 12218 12350 30565 00013096 12134 12330 30410 300 00013980 13294 13433 32488 00013783 13176 13383 32279 320 00014733 14319 14540 34263 340 00016311 15679 15924 36555 P 20 MPa 36575C P 30 MPa P 50 MPa Sat 00020378 17858 18266 40146 0 00009904 023 2003 00005 00009857 029 2986 00003 00009767 029 4913 00010 20 00009929 8271 10257 02921 00009886 8211 11177 02897 00009805 8093 12995 02845 40 00009992 16517 18516 05646 00009951 16405 19390 05607 00009872 16190 21125 05528 60 00010084 24775 26792 08208 00010042 24614 27626 08156 00009962 24308 29288 08055 80 00010199 33050 35090 10627 00010155 32840 35886 10564 00010072 32442 37478 10442 100 00010337 41350 43417 12920 00010290 41087 44174 12847 00010201 40594 45694 12705 120 00010496 49685 51784 15105 00010445 49366 52500 15020 00010349 48769 53943 14859 140 00010679 58071 60207 17194 00010623 57690 60876 17098 00010517 56977 62236 16916 160 00010886 66528 68705 19203 00010823 66074 69321 19094 00010704 65233 70585 18889 180 00011122 75078 77302 21143 00011049 74540 77855 21020 00010914 73549 79006 20790 200 00011390 83749 86027 23027 00011304 83111 86502 22888 00011149 81945 87519 22628 220 00011697 92577 94916 24867 00011595 91815 95293 24707 00011412 90439 96145 24414 240 00012053 10161 10402 26676 00011927 10069 10427 26491 00011708 99055 10491 26156 260 00012472 11090 11340 28469 00012314 10978 11347 28250 00012044 10782 11384 27864 280 00012978 12056 12315 30265 00012770 11915 12298 30001 00012430 11677 12299 29547 300 00013611 13072 13344 32091 00013322 12889 13289 31761 00012879 12596 13240 31218 320 00014450 14166 14455 33996 00014014 13917 14337 33558 00013409 13543 14214 32888 340 00015693 15402 15716 36086 00014932 15024 15471 35438 00014049 14529 15231 34575 360 00018248 17036 17401 38787 00016276 16268 16756 37499 00014848 15565 16307 36301 380 00018729 17820 18382 40026 00015884 16671 17465 38102 Final PDF to printer 897 APPENDIX 1 cen22672app01881930indd 897 110617 0932 AM TABLE A8 Saturated icewater vapor Temp T C Sat press Psat kPa Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Sat ice vi Sat vapor vg Sat ice ui Subl uig Sat vapor ug Sat ice hi Subl hig Sat vapor hg Sat ice si Subl sig Sat vapor sg 001 061169 0001091 20599 33340 27079 23745 33340 28339 25005 12202 10374 9154 0 061115 0001091 20617 33343 27079 23745 33343 28339 25005 12204 10375 9154 2 051772 0001091 24162 33763 27094 23718 33763 28345 24968 12358 10453 9218 4 043748 0001090 28384 34180 27108 23690 34180 28350 24932 12513 10533 9282 6 036873 0001090 33427 34594 27122 23662 34593 28354 24895 12667 10613 9347 8 030998 0001090 39466 35004 27135 23635 35004 28358 24858 12821 10695 9413 10 025990 0001089 46717 35412 27148 23607 35412 28362 24821 12976 10778 9480 12 021732 0001089 55447 35817 27161 23579 35817 28366 24784 13130 10862 9549 14 018121 0001088 65988 36218 27173 23552 36218 28369 24747 13284 10947 9618 16 015068 0001088 78751 36617 27186 23524 36617 28372 24710 13439 11033 9689 18 012492 0001088 94251 37013 27197 23496 37013 28375 24673 13593 11121 9761 20 010326 0001087 11313 37406 27209 23468 37406 28377 24636 13748 11209 9835 22 008510 0001087 13620 37795 27220 23441 37795 28379 24599 13903 11300 9909 24 006991 0001087 16447 38182 27231 23413 38182 28381 24562 14057 11391 9985 26 005725 0001087 19922 38566 27242 23385 38566 28382 24525 14212 11484 10063 28 004673 0001086 24210 38947 27252 23357 38947 28383 24488 14367 11578 10141 30 003802 0001086 29517 39325 27262 23329 39325 28384 24451 14521 11673 10221 32 003082 0001086 36109 39700 27272 23302 39700 28384 24414 14676 11770 10303 34 002490 0001085 44324 40072 27281 23274 40072 28385 24377 14831 11869 10386 36 002004 0001085 54601 40440 27290 23246 40440 28384 24340 14986 11969 10470 38 001608 0001085 67505 40807 27299 23218 40807 28384 24303 15141 12071 10557 40 001285 0001084 83767 41170 27307 23190 41170 28383 24266 15296 12174 10644 Final PDF to printer 898 PROPERTY TABLES AND CHARTS cen22672app01881930indd 898 110617 0932 AM Qu alit y 90 3 2 0 0 3 8 0 0 4 0 0 0 4 5 0 0 3 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1200 1100 1000 800 700 600 500 400 300 200 100 100 9 8 7 6 5 4 Entropy kJkgK 3 2 1 0 10 9 8 7 6 5 4 3 2 1 0 900 Temperature C 2800 h 2650 kJkg 2650 2550 2900 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 4100 4200 4300 4400 4500 4600 4700 4800 4900 5000 5500 5000 100 kgm3 30 kgm3 10 kgm3 3 kgm3 1 kgm3 03 kgm3 01 kgm3 003 kgm3 Density 001 kgm3 20000 15000 10000 8000 6000 5000 1500 1000 800 600 500 400 350 300 250 200 150 100 80 60 40 80 70 60 50 40 30 20 30 20 15 10 8 6 4 3 2 15 10 0806 04 03 02 015 01 008 006 004 003 002 0015 0008 0006 0004 0003 0002 2000 3000 4000 2800 h 2600 kJkg 2650 2550 2900 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 4100 4200 4300 4400 4500 4600 4700 4800 4900 5000 5500 5000 300 kgm3 100 kgm3 30 kgm3 10 kgm3 3 kgm3 1 kgm3 03 kgm3 01 kgm3 003 kgm3 Density 001 kgm3 20000 15000 10000 8000 6000 5000 1500 1000 800 600 500 400 350 300 250 200 150 100 80 60 40 80 70 60 50 40 30 20 30 20 15 10 8 6 4 3 2 15 10 0806 04 03 02 015 01 008 006 004 003 002 0015 0008 0006 0004 0003 0002 2000 3000 4000 h 4 0 0 kJ k g 0 0 6 0 0 2 0 0 8 1 0 0 0 1 2 0 0 1 4 0 0 16 00 18 00 20 00 22 00 24 00 h 260 0 kJ kg Saturated liquid 4 5 0 0 4 0 0 0 3 8 0 0 Saturated vapor Quality 10 3 6 0 0 3 4 0 0 3 2 0 0 2 6 0 0 2 4 0 0 2 0 0 0 1 8 0 0 1 6 0 0 14 0 0 12 00 10 00 h 2 2 0 0 k J k g 3 0 0 0 Density kgm3 h 4 2 0 0 k J k g 3 0 0 0 3 0 0 0 3 0 0 0 P 30000 bar h 5000 kJkg FIGURE A9 Ts diagram for water Source of Data From NBSNRC Steam Tables1 by Lester Haar John S Gallagher and George S Kell RoutledgeTaylor Francis Books Inc 1984 Final PDF to printer 899 APPENDIX 1 cen22672app01881930indd 899 110617 0932 AM Enthalpy kJkg 4000 3000 2000 1000 5000 4000 3000 2000 1000 2 3 4 5 6 Entropy kJkgK 7 8 9 10 2 3 4 5 6 7 8 9 10 P 30000 bar 25000 T 300C 400C 500C 600C 700C 800C 900C 1000C 1100C T 1100C 20000 15000 10000 8000 6000 5000 4000 3000 2000 1500 1000 800 600 400 300 200 150 100 80 60 50 30 20 15 10 8 6 5 3 2 15 10 08 98 96 94 92 88 86 84 82 80 78 76 06 04 03 02 015 01 008 006 004 005 003 002 0015 001 P 0008 bar 4 40 500 Den sit y 0 01 k gm3 01 k gm3 1 kg m3 10 kgm3 100 kgm 3 Density 1000 kgm3 Qualit y 90 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 P 30000 bar 25000 T 300C 400C 500C 600C 700C 800C 900C 1000C 1100C T 1100C 20000 15000 10000 8000 6000 5000 4000 3000 2000 1500 800 1000 600 400 300 200 150 100 80 60 50 30 20 15 10 8 6 5 3 2 15 10 08 98 96 94 92 88 86 84 82 80 78 76 06 04 03 02 015 01 008 006 004 005 003 002 0015 001 P 0008 bar 4 40 500 Den sit y 0 01 k gm3 01 k gm3 1 kg m3 10 kgm3 100 kgm 3 Density 1000 kgm3 Qualit y 90 74 72 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 5000 FIGURE A10 Mollier diagram for water Source of Data From NBSNRC Steam Tables1 by Lester Haar John S Gallagher and George S Kell RoutledgeTaylor Francis Books Inc 1984 Final PDF to printer 900 PROPERTY TABLES AND CHARTS cen22672app01881930indd 900 110617 0932 AM TABLE A11 Saturated refrigerant134aTemperature table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Temp T C Sat press Psat kPa Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 40 5125 00007053 036064 0036 20742 20738 000 22586 22586 000000 096869 096869 38 5686 00007082 032718 2472 20606 20853 2512 22462 22713 001071 095516 096588 36 6295 00007111 029740 4987 20469 20968 5032 22337 22840 002137 094182 096319 34 6956 00007141 027082 7509 20332 21083 7559 22210 22966 003196 092867 096063 32 7671 00007171 024706 1004 20194 21197 1009 22083 23093 004249 091569 095819 30 8443 00007201 022577 1258 20055 21312 1264 21955 23219 005297 090289 095586 28 9276 00007232 020666 1512 19915 21427 1519 21825 23344 006339 089024 095364 26 10173 00007264 018947 1767 19775 21542 1775 21695 23470 007376 087776 095152 24 11137 00007296 017398 2023 19634 21657 2031 21563 23594 008408 086542 094950 22 12172 00007328 015999 2280 19492 21771 2289 21430 23719 009435 085323 094758 20 13282 00007361 014735 2537 19349 21886 2547 21296 23843 010456 084119 094575 18 14469 00007394 013589 2796 19205 22000 2807 21160 23967 011473 082927 094401 16 15738 00007428 012550 3055 19060 22115 3067 21023 24090 012486 081749 094234 14 17093 00007463 011605 3315 18914 22229 3328 20884 24212 013493 080583 094076 12 18537 00007498 010744 3576 18766 22342 3590 20744 24334 014497 079429 093925 10 20074 00007533 0099600 3838 18618 22456 3853 20602 24455 015496 078286 093782 8 21708 00007570 0092438 4101 18469 22569 4117 20459 24576 016491 077154 093645 6 23444 00007607 0085888 4364 18318 22682 4382 20314 24695 017482 076033 093514 4 25285 00007644 0079889 4629 18166 22794 4648 20166 24814 018469 074921 093390 2 27236 00007683 0074388 4894 18012 22907 4915 20017 24933 019452 073819 093271 0 29301 00007722 0069335 5161 17858 23018 5183 19867 25050 020432 072726 093158 2 31484 00007761 0064690 5428 17701 23130 5453 19714 25166 021408 071641 093050 4 33790 00007802 0060412 5697 17544 23240 5723 19558 25282 022381 070565 092946 6 36223 00007843 0056469 5966 17384 23351 5995 19401 25396 023351 069496 092847 8 38788 00007886 0052829 6237 17223 23460 6268 19242 25509 024318 068435 092752 10 41489 00007929 0049466 6509 17061 23569 6542 19080 25622 025282 067380 092661 12 44331 00007973 0046354 6782 16896 23678 6817 18916 25733 026243 066331 092574 14 47319 00008018 0043471 7056 16730 23786 7094 18749 25843 027201 065289 092490 16 50458 00008064 0040798 7331 16562 23893 7372 18580 25951 028157 064252 092409 18 53752 00008112 0038317 7607 16392 23999 7651 18408 26059 029111 063219 092330 20 57207 00008160 0036012 7885 16219 24104 7932 18233 26164 030062 062192 092254 22 60827 00008209 0033867 8164 16045 24209 8214 18055 26269 031012 061168 092180 24 64618 00008260 0031869 8444 15868 24313 8498 17874 26372 031959 060148 092107 26 68584 00008312 0030008 8726 15689 24415 8783 17690 26473 032905 059131 092036 28 72731 00008366 0028271 9009 15508 24517 9070 17503 26573 033849 058117 091967 30 77064 00008421 0026648 9293 15324 24617 9358 17313 26671 034792 057105 091897 32 81589 00008477 0025131 9579 15137 24717 9649 17119 26767 035734 056095 091829 34 86311 00008535 0023712 9867 14948 24815 9941 16921 26861 036675 055086 091760 36 91235 00008595 0022383 10156 14755 24911 10234 16719 26953 037615 054077 091692 38 96368 00008657 0021137 10447 14560 25007 10530 16513 27044 038554 053068 091622 40 10171 00008720 0019968 10739 14361 25100 10828 16303 27131 039493 052059 091552 42 10728 00008786 0018870 11034 14159 25192 11128 16089 27217 040432 051048 091480 44 11307 00008854 0017837 11330 13953 25283 11430 15870 27300 041371 050036 091407 Final PDF to printer 901 APPENDIX 1 cen22672app01881930indd 901 110617 0932 AM TABLE A11 Saturated refrigerant134aTemperature table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Temp T C Sat press Psat kPa Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 46 11910 00008924 0016866 11628 13743 25371 11734 15646 27380 042311 049020 091331 48 12536 00008997 0015951 11928 13530 25458 12041 15417 27457 043251 048001 091252 52 13862 00009151 0014276 12535 13089 25624 12662 14941 27603 045136 045948 091084 56 15291 00009317 0012782 13152 12629 25781 13294 14441 27735 047028 043870 090898 60 16828 00009498 0011434 13779 12145 25923 13938 13909 27847 048930 041746 090676 65 18910 00009751 0009959 14580 11506 26086 14764 13205 27969 051330 039048 090379 70 21182 00010037 0008650 15403 10817 26220 15615 12437 28052 053763 036239 090002 75 23658 00010373 0007486 16255 10062 26317 16501 11587 28088 056252 033279 089531 80 26353 00010774 0006439 17143 9222 26366 17427 10635 28063 058812 030113 088925 85 29282 00011273 0005484 18081 8264 26345 18411 9539 27951 061487 026632 088120 90 32469 00011938 0004591 19094 7119 26213 19482 8222 27704 064354 022638 086991 95 35941 00012945 0003713 20249 5625 25873 20714 6494 27208 067605 017638 085243 100 39751 00015269 0002657 21873 2972 24846 22480 3422 25902 072224 009169 081393 Source of Data Tables A11 through A13 are generated using the Engineering Equation Solver EES software developed by S A Klein and F L Alvarado The routine used in calculations is the R134a which is based on the fundamental equation of state developed by R TillnerRoth and HD Baehr An International Standard Formulation for the Thermodynamic Properties of 1112Tetrafluoroethane HFC134a for temperatures from 170 K to 455 K and pressures up to 70 MPa J Phys Chem Ref Data Vol 23 No 5 1994 The enthalpy and entropy values of saturated liquid are set to zero at 40C and 40F Concluded Final PDF to printer 902 PROPERTY TABLES AND CHARTS cen22672app01881930indd 902 110617 0932 AM TABLE A12 Saturated refrigerant134aPressure table Specific volume m3kg Internal energy kJkg Enthalpy kJkg Entropy kJkgK Press P kPa Sat temp Tsat C Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 60 3695 00007097 031108 3795 20534 20913 3837 22396 22780 001633 094812 096445 70 3387 00007143 026921 7672 20323 21090 7722 22202 22974 003264 092783 096047 80 3113 00007184 023749 1114 20133 21248 1120 22027 23147 004707 091009 095716 90 2865 00007222 021261 1430 19960 21390 1436 21867 23304 006003 089431 095434 100 2637 00007258 019255 1719 19801 21521 1727 21719 23446 007182 088008 095191 120 2232 00007323 016216 2238 19515 21753 2247 21452 23699 009269 085520 094789 140 1877 00007381 014020 2696 19260 21956 2706 21213 23919 011080 083387 094467 160 1560 00007435 012355 3106 19031 22137 3118 20996 24114 012686 081517 094202 180 1273 00007485 011049 3481 18820 22301 3494 20795 24290 014131 079848 093979 200 1009 00007532 0099951 3826 18625 22451 3841 20609 24450 015449 078339 093788 240 538 00007618 0083983 4446 18271 22717 4464 20268 24732 017786 075689 093475 280 125 00007697 0072434 4995 17954 22949 5016 19961 24977 019822 073406 093228 320 246 00007771 0063681 5490 17665 23155 5514 19678 25193 021631 071395 093026 360 582 00007840 0056809 5942 17399 23341 5970 19415 25386 023265 069591 092856 400 891 00007905 0051266 6361 17149 23510 6392 19168 25561 024757 067954 092711 450 1246 00007983 0045677 6844 16858 23703 6880 18878 25758 026462 066093 092555 500 1571 00008058 0041168 7292 16586 23877 7332 18604 25936 028021 064399 092420 550 1873 00008129 0037452 7709 16329 24038 7754 18344 26098 029460 062842 092302 600 2155 00008198 0034335 8101 16084 24186 8150 18095 26246 030799 061398 092196 650 2420 00008265 0031680 8472 15851 24323 8526 17856 26382 032052 060048 092100 700 2669 00008331 0029392 8824 15627 24451 8882 17626 26508 033232 058780 092012 750 2906 00008395 0027398 9159 15411 24570 9222 17403 26625 034348 057582 091930 800 3131 00008457 0025645 9480 15202 24682 9548 17186 26734 035408 056445 091853 850 3345 00008519 0024091 9788 15000 24788 9861 16975 26836 036417 055362 091779 900 3551 00008580 0022703 10084 14803 24888 10162 16769 26931 037383 054326 091709 950 3748 00008640 0021456 10370 14611 24982 10452 16568 27020 038307 053333 091641 1000 3937 00008700 0020329 10647 14424 25071 10734 16370 27104 039196 052378 091574 1200 4629 00008935 0016728 11672 13712 25384 11779 15612 27392 042449 048870 091320 1400 5240 00009167 0014119 12596 13044 25640 12725 14892 27617 045325 045742 091067 1600 5788 00009400 0012134 13445 12405 25850 13596 14196 27792 047921 042881 090802 1800 6287 00009639 0010568 14236 11785 26021 14409 13514 27923 050304 040213 090517 2000 6745 00009887 0009297 14981 11175 26156 15178 12836 28015 052519 037684 090204 2500 7754 00010567 0006941 16702 9647 26349 16966 11118 28084 057542 031701 089243 3000 8616 00011410 0005272 18309 8017 26326 18651 9257 27908 062133 025759 087893 Final PDF to printer 903 APPENDIX 1 cen22672app01881930indd 903 110617 0932 AM TABLE A13 Superheated refrigerant134a T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 006 MPa Tsat 3695C P 010 MPa Tsat 2637C P 014 MPa Tsat 1877C Sat 031108 20913 22780 09645 019255 21521 23446 09519 014020 21956 23919 09447 20 033608 22062 24078 10175 019841 21968 23952 09721 10 035048 22757 24860 10478 020743 22677 24751 10031 014605 22593 24637 09724 0 036476 23467 25656 10775 021630 23397 25560 10333 015263 23325 25461 10032 10 037893 24194 26468 11067 022506 24132 26382 10628 015908 24068 26295 10331 20 039302 24937 27295 11354 023373 24881 27218 10919 016544 24824 27140 10625 30 040705 25697 28139 11637 024233 25646 28069 11204 017172 25595 27999 10913 40 042102 26473 28999 11916 025088 26427 28936 11485 017794 26380 28872 11196 50 043495 27266 29875 12192 025937 27224 29817 11762 018412 27181 29759 11475 60 044883 28075 30768 12464 026783 28036 30715 12036 019025 27997 30661 11750 70 046269 28901 31677 12732 027626 28865 31628 12306 019635 28829 31578 12021 80 047651 29743 32602 12998 028465 29710 32557 12573 020242 29677 32511 12289 90 049032 30602 33543 13261 029303 30571 33501 12836 020847 30540 33459 12554 100 050410 31476 34501 13521 030138 31448 34461 13097 021449 31419 34422 12815 P 018 MPa Tsat 1273C P 020 MPa Tsat 1009C P 024 MPa Tsat 538C Sat 011049 22301 24290 09398 009995 22451 24450 09379 008398 22717 24732 09348 10 011189 22504 24518 09485 009991 22457 24456 09381 0 011722 23249 25359 09799 010481 23211 25307 09699 008617 23130 25198 09520 10 012240 24002 26205 10103 010955 23969 26160 10005 009026 23900 26066 09832 20 012748 24766 27060 10400 011418 24736 27020 10304 009423 24676 26938 10134 30 013248 25543 27927 10691 011874 25516 27891 10596 009812 25463 27817 10429 40 013741 26333 28807 10976 012322 26309 28774 10882 010193 26261 28707 10718 50 014230 27138 29700 11257 012766 27116 29670 11164 010570 27073 29609 11002 60 014715 27958 30607 11533 013206 27938 30579 11441 010942 27898 30524 11281 70 015196 28793 31528 11806 013641 28775 31503 11714 011310 28738 31453 11555 80 015673 29643 32465 12075 014074 29627 32441 11984 011675 29593 32395 11826 90 016149 30509 33416 12340 014504 30493 33394 12250 012038 30462 33351 12093 100 016622 31390 34382 12603 014933 31375 34362 12513 012398 31346 34322 12356 P 028 MPa Tsat 125C P 032 MPa Tsat 246C P 040 MPa Tsat 891C Sat 007243 22949 24977 09323 006368 23155 25193 09303 0051266 23510 25561 09271 0 007282 23046 25085 09362 10 007646 23829 25970 09681 006609 23756 25870 09545 0051506 23599 25659 09306 20 007997 24615 26854 09987 006925 24551 26767 09856 0054213 24419 26588 09628 30 008338 25408 27742 10285 007231 25352 27666 10158 0056796 25237 27509 09937 40 008672 26212 28640 10577 007530 26162 28572 10452 0059292 26060 28432 10237 50 009000 27028 29548 10862 007823 26983 29487 10739 0061724 26892 29361 10529 60 009324 27858 30469 11143 008111 27817 30412 11022 0064104 27734 30298 10814 70 009644 28701 31401 11419 008395 28664 31350 11299 0066443 28588 31245 11095 80 009961 29559 32348 11690 008675 29524 32300 11572 0068747 29454 32204 11370 90 010275 30430 33307 11958 008953 30399 33264 11841 0071023 30334 33175 11641 100 010587 31317 34281 12223 009229 31287 34241 12106 0073274 31228 34159 11908 110 010897 32218 35269 12484 009503 32191 35231 12368 0075504 32135 35155 12172 120 011205 33134 36272 12742 009775 33108 36236 12627 0077717 33056 36165 12432 130 011512 34065 37288 12998 010045 34041 37255 12883 0079913 33992 37189 12689 140 011818 35011 38320 13251 010314 34988 38289 13136 0082096 34942 38226 12943 Final PDF to printer 904 PROPERTY TABLES AND CHARTS cen22672app01881930indd 904 110617 0932 AM TABLE A13 Superheated refrigerant134a T C v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK v m3kg u kJkg h kJkg s kJkgK P 050 MPa Tsat 1571C P 060 MPa Tsat 2155C P 070 MPa Tsat 2669C Sat 0041168 23877 25936 09242 0034335 24186 26246 09220 0029392 24451 26508 09201 20 0042115 24242 26348 09384 30 0044338 25086 27303 09704 0035984 24924 27083 09500 0029966 24749 26847 09314 40 0046456 25927 28250 10011 0037865 25788 28060 09817 0031696 25641 27859 09642 50 0048499 26773 29198 10309 0039659 26650 29030 10122 0033322 26522 28854 09955 60 0050485 27627 30151 10600 0041389 27517 30000 10417 0034875 27403 29844 10257 70 0052427 28491 31112 10884 0043069 28391 30975 10706 0036373 28288 30834 10550 80 0054331 29365 32082 11163 0044710 29274 31957 10988 0037829 29181 31829 10835 90 0056205 30252 33063 11436 0046318 30169 32948 11265 0039250 30084 32831 11115 100 0058053 31152 34055 11706 0047900 31075 33949 11536 0040642 30996 33841 11389 110 0059880 32065 35059 11971 0049458 31993 34961 11804 0042010 31921 34861 11659 120 0061687 32991 36075 12233 0050997 32924 35984 12068 0043358 32857 35892 11925 130 0063479 33931 37105 12492 0052519 33869 37020 12328 0044688 33806 36934 12186 140 0065256 34885 38147 12747 0054027 34826 38068 12585 0046004 34767 37988 12445 150 0067021 35852 39204 13000 0055522 35798 39129 12838 0047306 35742 39054 12700 160 0068775 36834 40273 13250 0057006 36783 40203 13089 0048597 36731 40132 12952 P 080 MPa Tsat 3131C P 090 MPa Tsat 3551C P 100 MPa Tsat 3937C Sat 0025645 24682 26734 09185 0022686 24882 26925 09169 0020319 25071 27104 09157 40 0027035 25484 27646 09481 0023375 25315 27419 09328 0020406 25132 27173 09180 50 0028547 26387 28671 09803 0024809 26246 28479 09661 0021796 26096 28276 09526 60 0029973 27285 29682 10111 0026146 27162 29515 09977 0023068 27033 29340 09851 70 0031340 28183 30690 10409 0027413 28074 30541 10280 0024261 27961 30387 10160 80 0032659 29086 31699 10699 0028630 28988 31565 10574 0025398 28887 31427 10459 90 0033941 29997 32712 10982 0029806 29908 32590 10861 0026492 29817 32466 10749 100 0035193 30917 33732 11259 0030951 30835 33621 11141 0027552 30752 33508 11032 110 0036420 31847 34761 11531 0032068 31772 34658 11415 0028584 31696 34554 11309 120 0037625 32789 35799 11798 0033164 32719 35704 11684 0029592 32649 35608 11580 130 0038813 33742 36847 12062 0034241 33678 36759 11949 0030581 33612 36670 11847 140 0039985 34708 37907 12321 0035302 34648 37825 12211 0031554 34587 37742 12110 150 0041143 35686 38978 12577 0036349 35630 38901 12468 0032512 35573 38824 12369 160 0042290 36678 40061 12830 0037384 36625 39989 12722 0033457 36571 39917 12624 170 0043427 37683 41157 13081 0038408 37633 41089 12973 0034392 37582 41022 12876 180 0044554 38701 42265 13328 0039423 38654 42202 13221 0035317 38606 42138 13125 P 120 MPa Tsat 4629C P 140 MPa Tsat 5240C P 160 MPa Tsat 5788C Sat 0016728 25384 27392 09132 0014119 25640 27617 09107 0012134 25850 27792 09080 50 0017201 25764 27828 09268 60 0018404 26757 28966 09615 0015005 26446 28547 09389 0012372 26091 28071 09164 70 0019502 27723 30063 09939 0016060 27462 29710 09733 0013430 27178 29327 09536 80 0020529 28677 31140 10249 0017023 28451 30834 10056 0014362 28211 30509 09875 90 0021506 29628 32209 10547 0017923 29428 31937 10364 0015215 29219 31653 10195 100 0022442 30581 33274 10836 0018778 30401 33030 10661 0016014 30216 32778 10501 110 0023348 31540 34341 11119 0019597 31376 34119 10949 0016773 31209 33893 10795 120 0024228 32505 35412 11395 0020388 32355 35209 11230 0017500 32203 35003 11081 130 0025086 33479 36490 11665 0021155 33341 36302 11504 0018201 33202 36114 11360 140 0025927 34463 37574 11931 0021904 34334 37401 11773 0018882 34206 37227 11633 150 0026753 35457 38668 12192 0022636 35337 38507 12038 0019545 35219 38346 11901 160 0027566 36463 39771 12450 0023355 36351 39620 12298 0020194 36240 39471 12164 170 0028367 37480 40884 12704 0024061 37375 40743 12554 0020830 37271 40604 12422 180 0029158 38510 42009 12955 0024757 38412 41878 12808 0021456 38313 41746 12677 Concluded Final PDF to printer 905 APPENDIX 1 cen22672app01881930indd 905 110617 0932 AM 20 10 4 1 01 004 002 001 100 150 200 250 300 350 400 Enthalpy kJkg Pressure MPa 450 500 550 600 650 700 750 100 150 200 250 300 350 400 450 500 550 600 650 700 750 02 04 2 20 10 4 1 01 004 002 001 02 R134a T 0C 20 20 40 60 40 60 80 100 1500 064 01 02 03 04 06 07 08 09 40 20 0 20 40 60 80 100 120 140 T 160C 180 200 248 264 272 s 256 220 240 260 280 300 saturated vapor X 05 072 080 088 saturated liquids 096 104 112 120 128 136 144 152 160 168 176 184 192 200 208 224 232 240 248 s 216 kJkgK 1450 1400 1350 1300 1250 1200 kgm3 1150 1100 1050 1000 950 900 850 800 700 600 500 400 300 160 120 90 70 60 50 40 32 24 16 12 8 6 Densityh 200 kgm3 4 32 24 16 12 08 06 04 03 R134a T 0C 20 20 40 60 40 60 80 100 1500 064 01 02 03 04 06 07 08 09 40 20 0 20 40 60 80 100 120 140 T 160C 180 200 248 264 272 s 256 220 240 260 280 300 saturated vapor X 05 072 080 088 saturated liquids 096 104 112 120 128 136 144 152 160 168 176 184 192 200 208 224 232 240 248 s 216 kJkgK 1450 1400 1350 1300 1250 1200 kgm3 1150 1100 1050 1000 950 900 850 800 700 600 500 400 300 160 120 90 70 60 50 40 32 24 16 12 8 6 Density 200 kgm3 4 32 24 16 12 08 06 04 03 04 2 FIGURE A14 Ph diagram for refrigerant134a Note The reference point used for the chart is different than that used in the R134a tables Therefore problems should be solved using all property data either from the tables or from the chart but not from both Source of Data American Society of Heating Refrigerating and AirConditioning Engineers Inc Atlanta GA Final PDF to printer 906 PROPERTY TABLES AND CHARTS cen22672app01881930indd 906 110617 0932 AM 100 REDUCED PRESSURE PR Pv RT COMPRESSIBILITY FACTOR Z 03 04 05 06 07 08 09 10 00 01 02 03 04 05 06 07 08 09 10 a Low pressures 0 PR 10 065 060 TR 10 105 110 115 120 130 140 160 200 300 095 090 085 080 07 5 070 TR 500 065 060 TR 10 105 110 115 120 130 140 160 200 300 095 090 085 080 07 5 070 TR 500 180 160 150 140 130 120 110 100 090 080 vR 07 200 220 240 260 300 350 400 500 60 0 80 0 180 160 150 140 130 120 110 100 090 080 vR 07 200 220 240 260 300 350 400 500 60 0 80 0 200 160 140 120 105 110 100 095 090 085 080 075 000 005 010 090 095 Z PR 070 065 060 30 20 15 12 10 vr 070 065 060 30 20 15 12 10 vR NELSON OBERT GENERALIZED COMPRESSIBILITY CHARTS CHART No 1 PSEUDO REDUCED VOLUME P Pcr REDUCED PRESSURE PR TR 25 Z 100 TR 15 TR 300 REDUCED TEMPERATURE T TR Tcr v RTcr Pcr vR 1953 NOTE DEVIATION 10 NELSON OBERT GENERALIZED COMPRESSIBILITY CHARTS CHART No 2 PSEUDO REDUCED VOLUME P Pcr REDUCED PRESSURE PR REDUCED TEMPERATURE T Tcr TR vR 1953 TR 100 105 110 115 120 130 140 150 160 180 200 350 250 TR 500 TR 100 105 110 115 120 130 140 150 160 180 200 350 250 TR 500 vR 020 025 030 035 040 045 vR 0 50 060 070 08 0 09 0 10 0 1 20 1 40 1 60 3 0 0 2 0 0 vR 020 025 030 035 040 045 vR 0 50 060 070 08 0 09 0 10 0 1 20 1 40 1 60 3 0 0 3 0 0 5 0 0 2 0 0 REDUCED PRESSURE PR COMPRESSIBILITY FACTOR Z b Intermediate pressures 0 PR 7 00 05 15 25 35 45 55 65 10 20 30 40 50 60 70 020 030 040 050 060 070 080 090 100 110 Pv RT v RTcr Pcr NELSON OBERT GENERALIZED COMPRESSIBILITY CHARTS CHART No 2 PSEUDO REDUCED VOLUME P Pcr REDUCED PRESSURE PR REDUCED TEMPERATURE T TR Tcr v RTcr Pcr vR 1953 FIGURE A15 NelsonObert generalized compressibility chart Used with permission of Dr Edward E Obert University of Wisconsin Final PDF to printer 907 APPENDIX 1 cen22672app01881930indd 907 110617 0932 AM TABLE A16 Properties of the atmosphere at high altitude Altitude m Temperature C Pressure kPa Gravity g ms2 Speed of sound ms Density kgm3 Viscosity μ kgms Thermal conductivity WmK 0 1500 10133 9807 3403 1225 1789 105 00253 200 1370 9895 9806 3395 1202 1783 105 00252 400 1240 9661 9805 3388 1179 1777 105 00252 600 1110 9432 9805 3380 1156 1771 105 00251 800 980 9208 9804 3372 1134 1764 105 00250 1000 850 8988 9804 3364 1112 1758 105 00249 1200 720 8772 9803 3357 1090 1752 105 00248 1400 590 8560 9802 3349 1069 1745 105 00247 1600 460 8353 9802 3341 1048 1739 105 00245 1800 330 8149 9801 3333 1027 1732 105 00244 2000 200 7950 9800 3325 1007 1726 105 00243 2200 070 7755 9800 3317 0987 1720 105 00242 2400 059 7563 9799 3310 0967 1713 105 00241 2600 189 7376 9799 3302 0947 1707 105 00240 2800 319 7192 9798 3294 0928 1700 105 00239 3000 449 7012 9797 3286 0909 1694 105 00238 3200 579 6836 9797 3278 0891 1687 105 00237 3400 709 6663 9796 3270 0872 1681 105 00236 3600 839 6494 9796 3262 0854 1674 105 00235 3800 969 6328 9795 3254 0837 1668 105 00234 4000 1098 6166 9794 3246 0819 1661 105 00233 4200 123 6007 9794 3238 0802 1655 105 00232 4400 136 5852 9793 3230 0785 1648 105 00231 4600 149 5700 9793 3222 0769 1642 105 00230 4800 162 5551 9792 3214 0752 1635 105 00229 5000 175 5405 9791 3205 0736 1628 105 00228 5200 188 5262 9791 3197 0721 1622 105 00227 5400 201 5123 9790 3189 0705 1615 105 00226 5600 214 4986 9789 3181 0690 1608 105 00224 5800 227 4852 9785 3173 0675 1602 105 00223 6000 240 4722 9788 3165 0660 1595 105 00222 6200 253 4594 9788 3156 0646 1588 105 00221 6400 266 4469 9787 3148 0631 1582 105 00220 6600 279 4347 9786 3140 0617 1575 105 00219 6800 292 4227 9785 3131 0604 1568 105 00218 7000 305 4111 9785 3123 0590 1561 105 00217 8000 369 3565 9782 3081 0526 1527 105 00212 9000 434 3080 9779 3038 0467 1493 105 00206 10000 499 2650 9776 2995 0414 1458 105 00201 12000 565 1940 9770 2951 0312 1422 105 00195 14000 565 1417 9764 2951 0228 1422 105 00195 16000 565 1053 9758 2951 0166 1422 105 00195 18000 565 757 9751 2951 0122 1422 105 00195 Source of Data US Standard Atmosphere Supplements US Government Printing Office 1966 Based on yearround mean conditions at 45 latitude and varies with the time of the year and the weather patterns The conditions at sea level z 0 are taken to be P 101325 kPa T 15C ρ 12250 kgm3 g 980665 m2s Final PDF to printer 908 PROPERTY TABLES AND CHARTS cen22672app01881930indd 908 110617 0932 AM TABLE A17 Idealgas properties of air T K h kJkg Pr u kJkg vr s kJkgK T K h kJkg Pr u kJkg vr s kJkgK 200 19997 03363 14256 17070 129559 580 58604 1438 41955 1157 237348 210 20997 03987 14969 15120 134444 590 59652 1531 42715 1106 239140 220 21997 04690 15682 13460 139105 600 60702 1628 43478 1058 240902 230 23002 05477 16400 12050 143557 610 61753 1730 44242 1012 242644 240 24002 06355 17113 10840 147824 620 62807 1836 45009 9692 244356 250 25005 07329 17828 9790 151917 630 63863 1984 45778 9284 246048 260 26009 08405 18545 8878 155848 640 64922 2064 46550 8899 247716 270 27011 09590 19260 8080 159634 650 65984 2186 47325 8534 249364 280 28013 10889 19975 7380 163279 660 67047 2313 48101 8189 250985 285 28514 11584 20333 7061 165055 670 68114 2446 48881 7861 252589 290 29016 12311 20691 6761 166802 680 69182 2585 49662 7550 254175 295 29517 13068 21049 6479 168515 690 70252 2729 50445 7256 255731 298 29818 13543 21264 6319 169528 700 71327 2880 51233 6976 257277 300 30019 13860 21407 6212 170203 710 72404 3038 52023 6707 258810 305 30522 14686 21767 5960 171865 720 73482 3202 52814 6453 260319 310 31024 15546 22125 5723 173498 730 74562 3372 53607 6213 261803 315 31527 16442 22485 5498 175106 740 75644 3550 54402 5982 263280 320 32029 17375 22842 5286 176690 750 76729 3735 55199 5763 264737 325 32531 18345 23202 5084 178249 760 77818 3927 56001 5554 266176 330 33034 19352 23561 4894 179783 780 80003 4335 57612 5164 269013 340 34042 2149 24282 4541 182790 800 82195 4775 59230 4808 271787 350 35049 2379 25002 4222 185708 820 84398 5259 60859 4484 274504 360 36058 2626 25724 3934 188543 840 86608 5760 62495 4185 277170 370 37067 2892 26446 3672 191313 860 88827 6309 64140 3912 279783 380 38077 3176 27169 3434 194001 880 91056 6898 65795 3661 282344 390 39088 3481 27893 3215 196633 900 93293 7529 67458 3431 284856 400 40098 3806 28616 3016 199194 920 95538 8205 69128 3218 287324 410 41112 4153 29343 2833 201699 940 97792 8928 70808 3022 289748 420 42126 4522 30069 2666 204142 960 100055 9700 72502 2840 292128 430 43143 4915 30799 2511 206533 980 102325 1052 74198 2673 294468 440 44161 5332 31530 2368 208870 1000 104604 1140 75894 2517 296770 450 45180 5775 32262 2236 211161 1020 106889 1234 77610 2372 299034 460 46202 6245 32997 2114 213407 1040 109185 1333 79336 2329 301260 470 47224 6742 33732 2001 215604 1060 111486 1439 81062 2114 303449 480 48249 7268 34470 1895 217760 1080 113789 1552 82788 1998 305608 490 49274 7824 35208 1797 219876 1100 116107 1671 84533 18896 307732 500 50302 8411 35949 1706 221952 1120 118428 1797 86279 17886 309825 510 51332 9031 36692 1621 223993 1140 120757 1931 88035 16946 311883 520 52363 9684 37436 1541 225997 1160 123092 2072 89791 16064 313916 530 53398 1037 38184 1467 227967 1180 125434 2222 91557 15241 315916 540 54435 1110 38934 1397 229906 1200 127779 2380 93333 14470 317888 550 55474 1186 39686 1331 231809 1220 130131 2547 95109 13747 319834 560 56517 1266 40442 1270 233685 1240 132493 2723 96895 13069 321751 570 57559 1350 41197 1212 235531 Final PDF to printer 909 APPENDIX 1 cen22672app01881930indd 909 110617 0932 AM TABLE A17 Idealgas properties of air T K h kJkg Pr u kJkg vr s kJkgK T K h kJkg Pr u kJkg vr s kJkgK 1260 134855 2908 98690 12435 323638 1600 175757 7912 129830 5804 352364 1280 137224 3104 100476 11835 325510 1620 178200 8341 131696 5574 353879 1300 139597 3309 102282 11275 327345 1640 180646 8789 133572 5355 355381 1320 141976 3525 104088 10747 329160 1660 183096 9256 135448 5147 356867 1340 144360 3753 105894 10247 330959 1680 185550 9742 137324 4949 358335 1360 146749 3991 107710 9780 332724 1700 18801 1025 13927 4761 35979 1380 149144 4242 109526 9337 334474 1750 19416 1161 14398 4328 36336 1400 151542 4505 111352 8919 336200 1800 20033 1310 14872 3994 36684 1420 153944 4780 113177 8526 337901 1850 20653 1475 15349 3601 37023 1440 156351 5069 115013 8153 339586 1900 21274 1655 15826 3295 37354 1460 158763 5371 116849 7801 341247 1950 21897 1852 16306 3022 37677 1480 161179 5688 118695 7468 342892 2000 22521 2068 16787 2776 37994 1500 163597 6019 120541 7152 344516 2050 23146 2303 17268 2555 38303 1520 166023 6365 122387 6854 346120 2100 23777 2559 17753 2356 38605 1540 168451 6728 124243 6569 347712 2150 24403 2837 18238 2175 38901 1560 170882 7105 126099 6301 349276 2200 25032 3138 18724 2012 39191 1580 173317 7500 127965 6046 350829 2250 25664 3464 19213 1864 39474 Note The properties Pr relative pressure and vr relative specific volume are dimensionless quantities used in the analysis of isentropic processes and should not be confused with the properties pressure and specific volume Source of Data Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 pp 78586 table A5 Originally published in J H Keenan and J Kaye Gas Tables New York John Wiley Sons 1948 Concluded Final PDF to printer 910 PROPERTY TABLES AND CHARTS cen22672app01881930indd 910 110617 0932 AM TABLE A18 Idealgas properties of nitrogen N2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 600 17563 12574 212066 220 6391 4562 182639 610 17864 12792 212564 230 6683 4770 183938 620 18166 13011 213055 240 6975 4979 185180 630 18468 13230 213541 250 7266 5188 186370 640 18772 13450 214018 260 7558 5396 187514 650 19075 13671 214489 270 7849 5604 188614 660 19380 13892 214954 280 8141 5813 189673 670 19685 14114 215413 290 8432 6021 190695 680 19991 14337 215866 298 8669 6190 191502 690 20297 14560 216314 300 8723 6229 191682 700 20604 14784 216756 310 9014 6437 192638 710 20912 15008 217192 320 9306 6645 193562 720 21220 15234 217624 330 9597 6853 194459 730 21529 15460 218059 340 9888 7061 195328 740 21839 15686 218472 350 10180 7270 196173 750 22149 15913 218889 360 10471 7478 196995 760 22460 16141 219301 370 10763 7687 197794 770 22772 16370 219709 380 11055 7895 198572 780 23085 16599 220113 390 11347 8104 199331 790 23398 16830 220512 400 11640 8314 200071 800 23714 17061 220907 410 11932 8523 200794 810 24027 17292 221298 420 12225 8733 201499 820 24342 17524 221684 430 12518 8943 202189 830 24658 17757 222067 440 12811 9153 202863 840 24974 17990 222447 450 13105 9363 203523 850 25292 18224 222822 460 13399 9574 204170 860 25610 18459 223194 470 13693 9786 204803 870 25928 18695 223562 480 13988 9997 205424 880 26248 18931 223927 490 14285 10210 206033 890 26568 19168 224288 500 14581 10423 206630 900 26890 19407 224647 510 14876 10635 207216 910 27210 19644 225002 520 15172 10848 207792 920 27532 19883 225353 530 15469 11062 208358 930 27854 20122 225701 540 15766 11277 208914 940 28178 20362 226047 550 16064 11492 209461 950 28501 20603 226389 560 16363 11707 209999 960 28826 20844 226728 570 16662 11923 210528 970 29151 21086 227064 580 16962 12139 211049 980 29476 21328 227398 590 17262 12356 211562 990 29803 21571 227728 Final PDF to printer 911 APPENDIX 1 cen22672app01881930indd 911 110617 0932 AM Source of Data Tables A18 through A25 are adapted from Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 pp 78798 Originally published in JANAF Thermochemical Tables NSRDSNBS37 1971 TABLE A18 Idealgas properties of nitrogen N2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 1000 30129 21815 228057 1760 56227 41594 247396 1020 30784 22304 228706 1780 56938 42139 247798 1040 31442 22795 229344 1800 57651 42685 248195 1060 32101 23288 229973 1820 58363 43231 248589 1080 32762 23782 230591 1840 59075 43777 248979 1100 33426 24280 231199 1860 59790 44324 249365 1120 34092 24780 231799 1880 60504 44873 249748 1140 34760 25282 232391 1900 61220 45423 250128 1160 35430 25786 232973 1920 61936 45973 250502 1180 36104 26291 233549 1940 62654 46524 250874 1200 36777 26799 234115 1960 63381 47075 251242 1220 37452 27308 234673 1980 64090 47627 251607 1240 38129 27819 235223 2000 64810 48181 251969 1260 38807 28331 235766 2050 66612 49567 252858 1280 39488 28845 236302 2100 68417 50957 253726 1300 40170 29361 236831 2150 70226 52351 254578 1320 40853 29378 237353 2200 72040 53749 255412 1340 41539 30398 237867 2250 73856 55149 256227 1360 42227 30919 238376 2300 75676 56553 257027 1380 42915 31441 238878 2350 77496 57958 257810 1400 43605 31964 239375 2400 79320 59366 258580 1420 44295 32489 239865 2450 81149 60779 259332 1440 44988 33014 240350 2500 82981 62195 260073 1460 45682 33543 240827 2550 84814 63613 260799 1480 46377 34071 241301 2600 86650 65033 261512 1500 47073 34601 241768 2650 88488 66455 262213 1520 47771 35133 242228 2700 90328 67880 262902 1540 48470 35665 242685 2750 92171 69306 263577 1560 49168 36197 243137 2800 94014 70734 264241 1580 49869 36732 243585 2850 95859 72163 264895 1600 50571 37268 244028 2900 97705 73593 265538 1620 51275 37806 244464 2950 99556 75028 266170 1640 51980 38344 244896 3000 101407 76464 266793 1660 52686 38884 245324 3050 103260 77902 267404 1680 53393 39424 245747 3100 105115 79341 268007 1700 54099 39965 246166 3150 106972 80782 268601 1720 54807 40507 246580 3200 108830 82224 269186 1740 55516 41049 246990 3250 110690 83668 269763 Concluded Final PDF to printer 912 PROPERTY TABLES AND CHARTS cen22672app01881930indd 912 110617 0932 AM TABLE A19 Idealgas properties of oxygen O2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 600 17929 12940 226346 220 6404 4575 196171 610 18250 13178 226877 230 6694 4782 197461 620 18572 13417 227400 240 6984 4989 198696 630 18895 13657 227918 250 7275 5197 199885 640 19219 13898 228429 260 7566 5405 201027 650 19544 14140 228932 270 7858 5613 202128 660 19870 14383 229430 280 8150 5822 203191 670 20197 14626 229920 290 8443 6032 204218 680 20524 14871 230405 298 8682 6203 205033 690 20854 15116 230885 300 8736 6242 205213 700 21184 15364 231358 310 9030 6453 206177 710 21514 15611 231827 320 9325 6664 207112 720 21845 15859 232291 330 9620 6877 208020 730 22177 16107 232748 340 9916 7090 208904 740 22510 16357 233201 350 10213 7303 209765 750 22844 16607 233649 360 10511 7518 210604 760 23178 16859 234091 370 10809 7733 211423 770 23513 17111 234528 380 11109 7949 212222 780 23850 17364 234960 390 11409 8166 213002 790 24186 17618 235387 400 11711 8384 213765 800 24523 17872 235810 410 12012 8603 214510 810 24861 18126 236230 420 12314 8822 215241 820 25199 18382 236644 430 12618 9043 215955 830 25537 18637 237055 440 12923 9264 216656 840 25877 18893 237462 450 13228 9487 217342 850 26218 19150 237864 460 13525 9710 218016 860 26559 19408 238264 470 13842 9935 218676 870 26899 19666 238660 480 14151 10160 219326 880 27242 19925 239051 490 14460 10386 219963 890 27584 20185 239439 500 14770 10614 220589 900 27928 20445 239823 510 15082 10842 221206 910 28272 20706 240203 520 15395 11071 221812 920 28616 20967 240580 530 15708 11301 222409 930 28960 21228 240953 540 16022 11533 222997 940 29306 21491 241323 550 16338 11765 223576 950 29652 21754 241689 560 16654 11998 224146 960 29999 22017 242052 570 16971 12232 224708 970 30345 22280 242411 580 17290 12467 225262 980 30692 22544 242768 590 17609 12703 225808 990 31041 22809 242120 Final PDF to printer 913 APPENDIX 1 cen22672app01881930indd 913 110617 0932 AM TABLE A19 Idealgas properties of oxygen O2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 1000 31389 23075 243471 1760 58880 44247 263861 1020 32088 23607 244164 1780 59624 44825 264283 1040 32789 24142 244844 1800 60371 45405 264701 1060 33490 24677 245513 1820 61118 45986 265113 1080 34194 25214 246171 1840 61866 46568 265521 1100 34899 25753 246818 1860 62616 47151 265925 1120 35606 26294 247454 1880 63365 47734 266326 1140 36314 26836 248081 1900 64116 48319 266722 1160 37023 27379 248698 1920 64868 48904 267115 1180 37734 27923 249307 1940 65620 49490 267505 1200 38447 28469 249906 1960 66374 50078 267891 1220 39162 29018 250497 1980 67127 50665 268275 1240 39877 29568 251079 2000 67881 51253 268655 1260 40594 30118 251653 2050 69772 52727 269588 1280 41312 30670 252219 2100 71668 54208 270504 1300 42033 31224 252776 2150 73573 55697 271399 1320 42753 31778 253325 2200 75484 57192 272278 1340 43475 32334 253868 2250 77397 58690 273136 1360 44198 32891 254404 2300 79316 60193 273891 1380 44923 33449 254932 2350 81243 61704 274809 1400 45648 34008 255454 2400 83174 63219 275625 1420 46374 34567 255968 2450 85112 64742 276424 1440 47102 35129 256475 2500 87057 66271 277207 1460 47831 35692 256978 2550 89004 67802 277979 1480 48561 36256 257474 2600 90956 69339 278738 1500 49292 36821 257965 2650 92916 70883 279485 1520 50024 37387 258450 2700 94881 72433 280219 1540 50756 37952 258928 2750 96852 73987 280942 1560 51490 38520 259402 2800 98826 75546 281654 1580 52224 39088 259870 2850 100808 77112 282357 1600 52961 39658 260333 2900 102793 78682 283048 1620 53696 40227 260791 2950 104785 80258 283728 1640 54434 40799 261242 3000 106780 81837 284399 1660 55172 41370 261690 3050 108778 83419 285060 1680 55912 41944 262132 3100 110784 85009 285713 1700 56652 42517 262571 3150 112795 86601 286355 1720 57394 43093 263005 3200 114809 88203 286989 1740 58136 43669 263435 3250 116827 89804 287614 Concluded Final PDF to printer 914 PROPERTY TABLES AND CHARTS cen22672app01881930indd 914 110617 0932 AM TABLE A20 Idealgas properties of carbon dioxide CO2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 600 22280 17291 243199 220 6601 4772 202966 610 22754 17683 243983 230 6938 5026 204464 620 23231 18076 244758 240 7280 5285 205920 630 23709 18471 245524 250 7627 5548 207337 640 24190 18869 246282 260 7979 5817 208717 650 24674 19270 247032 270 8335 6091 210062 660 25160 19672 247773 280 8697 6369 211376 670 25648 20078 248507 290 9063 6651 212660 680 26138 20484 249233 298 9364 6885 213685 690 26631 20894 249952 300 9431 6939 213915 700 27125 21305 250663 310 9807 7230 215146 710 27622 21719 251368 320 10186 7526 216351 720 28121 22134 252065 330 10570 7826 217534 730 28622 22522 252755 340 10959 8131 218694 740 29124 22972 253439 350 11351 8439 219831 750 29629 23393 254117 360 11748 8752 220948 760 30135 23817 254787 370 12148 9068 222044 770 30644 24242 255452 380 12552 9392 223122 780 31154 24669 256110 390 12960 9718 224182 790 31665 25097 256762 400 13372 10046 225225 800 32179 25527 257408 410 13787 10378 226250 810 32694 25959 258048 420 14206 10714 227258 820 33212 26394 258682 430 14628 11053 228252 830 33730 26829 259311 440 15054 11393 229230 840 34251 27267 259934 450 15483 11742 230194 850 34773 27706 260551 460 15916 12091 231144 860 35296 28125 261164 470 16351 12444 232080 870 35821 28588 261770 480 16791 12800 233004 880 36347 29031 262371 490 17232 13158 233916 890 36876 29476 262968 500 17678 13521 234814 900 37405 29922 263559 510 18126 13885 235700 910 37935 30369 264146 520 18576 14253 236575 920 38467 30818 264728 530 19029 14622 237439 930 39000 31268 265304 540 19485 14996 238292 940 39535 31719 265877 550 19945 15372 239135 950 40070 32171 266444 560 20407 15751 239962 960 40607 32625 267007 570 20870 16131 240789 970 41145 33081 267566 580 21337 16515 241602 980 41685 33537 268119 590 21807 16902 242405 990 42226 33995 268670 Final PDF to printer 915 APPENDIX 1 cen22672app01881930indd 915 110617 0932 AM TABLE A20 Idealgas properties of carbon dioxide CO2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 1000 42769 34455 269215 1760 86420 71787 301543 1020 43859 35378 270293 1780 87612 72812 302217 1040 44953 36306 271354 1800 88806 73840 302884 1060 46051 37238 272400 1820 90000 74868 303544 1080 47153 38174 273430 1840 91196 75897 304198 1100 48258 39112 274445 1860 92394 76929 304845 1120 49369 40057 275444 1880 93593 77962 305487 1140 50484 41006 276430 1900 94793 78996 306122 1160 51602 41957 277403 1920 95995 80031 306751 1180 52724 42913 278361 1940 97197 81067 307374 1200 53848 43871 297307 1960 98401 82105 307992 1220 54977 44834 280238 1980 99606 83144 308604 1240 56108 45799 281158 2000 100804 84185 309210 1260 57244 46768 282066 2050 103835 86791 310701 1280 58381 47739 282962 2100 106864 89404 312160 1300 59522 48713 283847 2150 109898 92023 313589 1320 60666 49691 284722 2200 112939 94648 314988 1340 61813 50672 285586 2250 115984 97277 316356 1360 62963 51656 286439 2300 119035 99912 317695 1380 64116 52643 287283 2350 122091 102552 319011 1400 65271 53631 288106 2400 125152 105197 320302 1420 66427 54621 288934 2450 128219 107849 321566 1440 67586 55614 289743 2500 131290 110504 322808 1460 68748 56609 290542 2550 134368 113166 324026 1480 66911 57606 291333 2600 137449 115832 325222 1500 71078 58606 292114 2650 140533 118500 326396 1520 72246 59609 292888 2700 143620 121172 327549 1540 73417 60613 292654 2750 146713 123849 328684 1560 74590 61620 294411 2800 149808 126528 329800 1580 76767 62630 295161 2850 152908 129212 330896 1600 76944 63741 295901 2900 156009 131898 331975 1620 78123 64653 296632 2950 159117 134589 333037 1640 79303 65668 297356 3000 162226 137283 334084 1660 80486 66592 298072 3050 165341 139982 335114 1680 81670 67702 298781 3100 168456 142681 336126 1700 82856 68721 299482 3150 171576 145385 337124 1720 84043 69742 300177 3200 174695 148089 338109 1740 85231 70764 300863 3250 177822 150801 339069 Concluded Final PDF to printer 916 PROPERTY TABLES AND CHARTS cen22672app01881930indd 916 110617 0932 AM TABLE A21 Idealgas properties of carbon monoxide CO T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 600 17611 12622 218204 220 6391 4562 188683 610 17915 12843 218708 230 6683 4771 189980 620 18221 13066 219205 240 6975 4979 191221 630 18527 13289 219695 250 7266 5188 192411 640 18833 13512 220179 260 7558 5396 193554 650 19141 13736 220656 270 7849 5604 194654 660 19449 13962 221127 280 8140 5812 195713 670 19758 14187 221592 290 8432 6020 196735 680 20068 14414 222052 298 8669 6190 197543 690 20378 14641 222505 300 8723 6229 197723 700 20690 14870 222953 310 9014 6437 198678 710 21002 15099 223396 320 9306 6645 199603 720 21315 15328 223833 330 9597 6854 200500 730 21628 15558 224265 340 9889 7062 201371 740 21943 15789 224692 350 10181 7271 202217 750 22258 16022 225115 360 10473 7480 203040 760 22573 16255 225533 370 10765 7689 203842 770 22890 16488 225947 380 11058 7899 204622 780 23208 16723 226357 390 11351 8108 205383 790 23526 16957 226762 400 11644 8319 206125 800 23844 17193 227162 410 11938 8529 206850 810 24164 17429 227559 420 12232 8740 207549 820 24483 17665 227952 430 12526 8951 208252 830 24803 17902 228339 440 12821 9163 208929 840 25124 18140 228724 450 13116 9375 209593 850 25446 18379 229106 460 13412 9587 210243 860 25768 18617 229482 470 13708 9800 210880 870 26091 18858 229856 480 14005 10014 211504 880 26415 19099 230227 490 14302 10228 212117 890 26740 19341 230593 500 14600 10443 212719 900 27066 19583 230957 510 14898 10658 213310 910 27392 19826 231317 520 15197 10874 213890 920 27719 20070 231674 530 15497 11090 214460 930 28046 20314 232028 540 15797 11307 215020 940 28375 20559 232379 550 16097 11524 215572 950 28703 20805 232727 560 16399 11743 216115 960 29033 21051 233072 570 16701 11961 216649 970 29362 21298 233413 580 17003 12181 217175 980 29693 21545 233752 590 17307 12401 217693 990 30024 21793 234088 Final PDF to printer 917 APPENDIX 1 cen22672app01881930indd 917 110617 0932 AM TABLE A21 Idealgas properties of carbon monoxide CO T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 1000 30355 22041 234421 1760 56756 42123 253991 1020 31020 22540 235079 1780 57473 42673 254398 1040 31688 23041 235728 1800 58191 43225 254797 1060 32357 23544 236364 1820 58910 43778 255194 1080 33029 24049 236992 1840 59629 44331 255587 1100 33702 24557 237609 1860 60351 44886 255976 1120 34377 25065 238217 1880 61072 45441 256361 1140 35054 25575 238817 1900 61794 45997 256743 1160 35733 26088 239407 1920 62516 46552 257122 1180 36406 26602 239989 1940 63238 47108 257497 1200 37095 27118 240663 1960 63961 47665 257868 1220 37780 27637 241128 1980 64684 48221 258236 1240 38466 28426 241686 2000 65408 48780 258600 1260 39154 28678 242236 2050 67224 50179 259494 1280 39844 29201 242780 2100 69044 51584 260370 1300 40534 29725 243316 2150 70864 52988 261226 1320 41226 30251 243844 2200 72688 54396 262065 1340 41919 30778 244366 2250 74516 55809 262887 1360 42613 31306 244880 2300 76345 57222 263692 1380 43309 31836 245388 2350 78178 58640 264480 1400 44007 32367 245889 2400 80015 60060 265253 1420 44707 32900 246385 2450 81852 61482 266012 1440 45408 33434 246876 2500 83692 62906 266755 1460 46110 33971 247360 2550 85537 64335 267485 1480 46813 34508 247839 2600 87383 65766 268202 1500 47517 35046 248312 2650 89230 67197 268905 1520 48222 35584 248778 2700 91077 68628 269596 1540 48928 36124 249240 2750 92930 70066 270285 1560 49635 36665 249695 2800 94784 71504 270943 1580 50344 37207 250147 2850 96639 72945 271602 1600 51053 37750 250592 2900 98495 74383 272249 1620 51763 38293 251033 2950 100352 75825 272884 1640 52472 38837 251470 3000 102210 77267 273508 1660 53184 39382 251901 3050 104073 78715 274123 1680 53895 39927 252329 3100 105939 80164 274730 1700 54609 40474 252751 3150 107802 81612 275326 1720 55323 41023 253169 3200 109667 83061 275914 1740 56039 41572 253582 3250 111534 84513 276494 Concluded Final PDF to printer 918 PROPERTY TABLES AND CHARTS cen22672app01881930indd 918 110617 0932 AM TABLE A22 Idealgas properties of hydrogen H2 T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 1440 42808 30835 177410 260 7370 5209 126636 1480 44091 31786 178291 270 7657 5412 127719 1520 45384 32746 179153 280 7945 5617 128765 1560 46683 33713 179995 290 8233 5822 129775 1600 47990 34687 180820 298 8468 5989 130574 1640 49303 35668 181632 300 8522 6027 130754 1680 50622 36654 182428 320 9100 6440 132621 1720 51947 37646 183208 340 9680 6853 134378 1760 53279 38645 183973 360 10262 7268 136039 1800 54618 39652 184724 380 10843 7684 137612 1840 55962 40663 185463 400 11426 8100 139106 1880 57311 41680 186190 420 12010 8518 140529 1920 58668 42705 186904 440 12594 8936 141888 1960 60031 43735 187607 460 13179 9355 143187 2000 61400 44771 188297 480 13764 9773 144432 2050 63119 46074 189148 500 14350 10193 145628 2100 64847 47386 189979 520 14935 10611 146775 2150 66584 48708 190796 560 16107 11451 148945 2200 68328 50037 191598 600 17280 12291 150968 2250 70080 51373 192385 640 18453 13133 152863 2300 71839 52716 193159 680 19630 13976 154645 2350 73608 54069 193921 720 20807 14821 156328 2400 75383 55429 194669 760 21988 15669 157923 2450 77168 56798 195403 800 23171 16520 159440 2500 78960 58175 196125 840 24359 17375 160891 2550 80755 59554 196837 880 25551 18235 162277 2600 82558 60941 197539 920 26747 19098 163607 2650 84368 62335 198229 960 27948 19966 164884 2700 86186 63737 198907 1000 29154 20839 166114 2750 88008 65144 199575 1040 30364 21717 167300 2800 89838 66558 200234 1080 31580 22601 168449 2850 91671 67976 200885 1120 32802 23490 169560 2900 93512 69401 201527 1160 34028 24384 170636 2950 95358 70831 202157 1200 35262 25284 171682 3000 97211 72268 202778 1240 36502 26192 172698 3050 99065 73707 203391 1280 37749 27106 173687 3100 100926 75152 203995 1320 39002 28027 174652 3150 102793 76604 204592 1360 40263 28955 175593 3200 104667 78061 205181 1400 41530 29889 176510 3250 106545 79523 205765 Final PDF to printer 919 APPENDIX 1 cen22672app01881930indd 919 110617 0932 AM TABLE A23 Idealgas properties of water vapor H2O T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 600 20402 15413 212920 220 7295 5466 178576 610 20765 15693 213529 230 7628 5715 180054 620 21130 15975 214122 240 7961 5965 181471 630 21495 16257 214707 250 8294 6215 182831 640 21862 16541 215285 260 8627 6466 184139 650 22230 16826 215856 270 8961 6716 185399 660 22600 17112 216419 280 9296 6968 186616 670 22970 17399 216976 290 9631 7219 187791 680 23342 17688 217527 298 9904 7425 188720 690 23714 17978 218071 300 9966 7472 188928 700 24088 18268 218610 310 10302 7725 190030 710 24464 18561 219142 320 10639 7978 191098 720 24840 18854 219668 330 10976 8232 192136 730 25218 19148 220189 340 11314 8487 193144 740 25597 19444 220707 350 11652 8742 194125 750 25977 19741 221215 360 11992 8998 195081 760 26358 20039 221720 370 12331 9255 196012 770 26741 20339 222221 380 12672 9513 196920 780 27125 20639 222717 390 13014 9771 197807 790 27510 20941 223207 400 13356 10030 198673 800 27896 21245 223693 410 13699 10290 199521 810 28284 21549 224174 420 14043 10551 200350 820 28672 21855 224651 430 14388 10813 201160 830 29062 22162 225123 440 14734 11075 201955 840 29454 22470 225592 450 15080 11339 202734 850 29846 22779 226057 460 15428 11603 203497 860 30240 23090 226517 470 15777 11869 204247 870 30635 23402 226973 480 16126 12135 204982 880 31032 23715 227426 490 16477 12403 205705 890 31429 24029 227875 500 16828 12671 206413 900 31828 24345 228321 510 17181 12940 207112 910 32228 24662 228763 520 17534 13211 207799 920 32629 24980 229202 530 17889 13482 208475 930 33032 25300 229637 540 18245 13755 209139 940 33436 25621 230070 550 18601 14028 209795 950 33841 25943 230499 560 18959 14303 210440 960 34247 26265 230924 570 19318 14579 211075 970 34653 26588 231347 580 19678 14856 211702 980 35061 26913 231767 590 20039 15134 212320 990 35472 27240 232184 Final PDF to printer 920 PROPERTY TABLES AND CHARTS cen22672app01881930indd 920 110617 0932 AM TABLE A23 Idealgas properties of water vapor H2O T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 1000 35882 27568 232597 1760 70535 55902 258151 1020 36709 28228 233415 1780 71523 56723 258708 1040 37542 28895 234223 1800 72513 57547 259262 1060 38380 29567 235020 1820 73507 58375 259811 1080 39223 30243 235806 1840 74506 59207 260357 1100 40071 30925 236584 1860 75506 60042 260898 1120 40923 31611 237352 1880 76511 60880 261436 1140 41780 32301 238110 1900 77517 61720 261969 1160 42642 32997 238859 1920 78527 62564 262497 1180 43509 33698 239600 1940 79540 63411 263022 1200 44380 34403 240333 1960 80555 64259 263542 1220 45256 35112 241057 1980 81573 65111 264059 1240 46137 35827 241773 2000 82593 65965 264571 1260 47022 36546 242482 2050 85156 68111 265838 1280 47912 37270 243183 2100 87735 70275 267081 1300 48807 38000 243877 2150 90330 72454 268301 1320 49707 38732 244564 2200 92940 74649 269500 1340 50612 39470 245243 2250 95562 76855 270679 1360 51521 40213 245915 2300 98199 79076 271839 1380 52434 40960 246582 2350 100846 81308 272978 1400 53351 41711 247241 2400 103508 83553 274098 1420 54273 42466 247895 2450 106183 85811 275201 1440 55198 43226 248543 2500 108868 88082 276286 1460 56128 43989 249185 2550 111565 90364 277354 1480 57062 44756 249820 2600 114273 92656 278407 1500 57999 45528 250450 2650 116991 94958 279441 1520 58942 46304 251074 2700 119717 97269 280462 1540 59888 47084 251693 2750 122453 99588 281464 1560 60838 47868 252305 2800 125198 101917 282453 1580 61792 48655 252912 2850 127952 104256 283429 1600 62748 49445 253513 2900 130717 106605 284390 1620 63709 50240 254111 2950 133486 108959 285338 1640 64675 51039 254703 3000 136264 111321 286273 1660 65643 51841 255290 3050 139051 113692 287194 1680 66614 52646 255873 3100 141846 116072 288102 1700 67589 53455 256450 3150 144648 118458 288999 1720 68567 54267 257022 3200 147457 120851 289884 1740 69550 55083 257589 3250 150272 123250 290756 Continued Final PDF to printer 921 APPENDIX 1 cen22672app01881930indd 921 110617 0932 AM TABLE A24 Idealgas properties of monatomic oxygen O T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 2400 50894 30940 204932 298 6852 4373 160944 2450 51936 31566 205362 300 6892 4398 161079 2500 52979 32193 205783 500 11197 7040 172088 2550 54021 32820 206196 1000 21713 13398 186678 2600 55064 33447 206601 1500 32150 19679 195143 2650 56108 34075 206999 1600 34234 20931 196488 2700 57152 34703 207389 1700 36317 22183 197751 2750 58196 35332 207772 1800 38400 23434 198941 2800 59241 35961 208148 1900 40482 24685 200067 2850 60286 36590 208518 2000 42564 25935 201135 2900 61332 37220 208882 2050 43605 26560 201649 2950 62378 37851 209240 2100 44646 27186 202151 3000 63425 38482 209592 2150 45687 27811 202641 3100 65520 39746 210279 2200 46728 28436 203119 3200 67619 41013 210945 2250 47769 29062 203588 3300 69720 42283 211592 2300 48811 29688 204045 3400 71824 43556 212220 2350 49852 30314 204493 3500 73932 44832 212831 TABLE A25 Idealgas properties of hydroxyl OH T K h kJkmol u kJkmol s kJkmolK T K h kJkmol u kJkmol s kJkmolK 0 0 0 0 2400 77015 57061 248628 298 9188 6709 183594 2450 78801 58431 249364 300 9244 6749 183779 2500 80592 59806 250088 500 15181 11024 198955 2550 82388 61186 250799 1000 30123 21809 219624 2600 84189 62572 251499 1500 46046 33575 232506 2650 85995 63962 252187 1600 49358 36055 234642 2700 87806 65358 252864 1700 52706 38571 236672 2750 89622 66757 253530 1800 56089 41123 238606 2800 91442 68162 254186 1900 59505 43708 240453 2850 93266 69570 254832 2000 62952 46323 242221 2900 95095 70983 255468 2050 64687 47642 243077 2950 96927 72400 256094 2100 66428 48968 243917 3000 98763 73820 256712 2150 68177 50301 244740 3100 102447 76673 257919 2200 69932 51641 245547 3200 106145 79539 259093 2250 71694 52987 246338 3300 109855 82418 260235 2300 73462 54339 247116 3400 113578 85309 261347 2350 75236 55697 247879 3500 117312 88212 262429 Final PDF to printer 922 PROPERTY TABLES AND CHARTS cen22672app01881930indd 922 110617 0932 AM TABLE A26 Enthalpy of formation Gibbs function of formation and absolute entropy at 25C 1 atm Substance Formula h f kJkmol g f kJkmol s kJkmolK Carbon Cs 0 0 574 Hydrogen H2g 0 0 13068 Nitrogen N2g 0 0 19161 Oxygen O2g 0 0 20504 Carbon monoxide COg 110530 137150 19765 Carbon dioxide CO2g 393520 394360 21380 Water vapor H2Og 241820 228590 18883 Water H2Ol 285830 237180 6992 Hydrogen peroxide H2O2g 136310 105600 23263 Ammonia NH3g 46190 16590 19233 Methane CH4g 74850 50790 18616 Acetylene C2H2g 226730 209170 20085 Ethylene C2H4g 52280 68120 21983 Ethane C2H6g 84680 32890 22949 Propylene C3H6g 20410 62720 26694 Propane C3H8g 103850 23490 26991 nButane C4H10g 126150 15710 31012 nOctane C8H18g 208450 16530 46673 nOctane C8H18l 249950 6610 36079 nDodecane C12H26g 291010 50150 62283 Benzene C6H6g 82930 129660 26920 Methyl alcohol CH3OHg 200670 162000 23970 Methyl alcohol CH3OHl 238660 166360 12680 Ethyl alcohol C2H5OHg 235310 168570 28259 Ethyl alcohol C2H5OHl 277690 174890 16070 Oxygen Og 249190 231770 16106 Hydrogen Hg 218000 203290 11472 Nitrogen Ng 472650 455510 15330 Hydroxyl OHg 39460 34280 18370 Source of Data From JANAF Thermochemical Tables Midland MI Dow Chemical Co 1971 Selected Values of Chemical Thermodynamic Properties NBS Technical Note 2703 1968 and API Research Project 44 Carnegie Press 1953 Final PDF to printer 923 APPENDIX 1 cen22672app01881930indd 923 110617 0932 AM TABLE A27 Properties of some common fuels and hydrocarbons Fuel phase Formula Molar mass kgkmol Density1 kgL Enthalpy of vaporization2 kJkg Specific heat1 cp kJkgK Higher heating value3 kJkg Lower heating value3 kJkg Carbon s C 12011 2 0708 32800 32800 Hydrogen g H2 2016 144 141800 120000 Carbon monoxide g CO 28013 105 10100 10100 Methane g CH4 16043 509 220 55530 50050 Methanol l CH4O 32042 0790 1168 253 22660 19920 Acetylene g C2H2 26038 169 49970 48280 Ethane g C2H6 30070 172 175 51900 47520 Ethanol l C2H6O 46069 0790 919 244 29670 26810 Propane l C3H8 44097 0500 335 277 50330 46340 Butane l C4H10 58123 0579 362 242 49150 45370 1Pentene l C5H10 70134 0641 363 220 47760 44630 Isopentane l C5H12 72150 0626 232 48570 44910 Benzene l C6H6 78114 0877 433 172 41800 40100 Hexene l C6H12 84161 0673 392 184 47500 44400 Hexane l C6H14 86177 0660 366 227 48310 44740 Toluene l C7H8 92141 0867 412 171 42400 40500 Heptane l C7H16 100204 0684 365 224 48100 44600 Octane l C8H18 114231 0703 363 223 47890 44430 Decane l C10H22 142285 0730 361 221 47640 44240 Gasoline l CnH187n 100110 072078 350 24 47300 44000 Light diesel l CnH18n 170 078084 270 22 46100 43200 Heavy diesel l CnH17n 200 082088 230 19 45500 42800 Natural gas g CnH38nN01n 18 2 50000 45000 1At 1 atm and 20C 2At 25C for liquid fuels and 1 atm and normal boiling temperature for gaseous fuels 3At 25C Multiply by molar mass to obtain heating values in kJkmol Final PDF to printer 924 PROPERTY TABLES AND CHARTS cen22672app01881930indd 924 110617 0932 AM TABLE A28 Natural logarithms of the equilibrium constant Kp The equilibrium constant Kp for the reaction νAA νBB νCC νDD is defined as K p P C ν C P D ν D P A ν A P B ν B Temp K H2 2H O2 2O N2 2N H2O H2 ½O2 H2O ½H2 OH CO2 CO ½O2 ½N2 ½O2 NO 298 164005 186975 367480 92208 106208 103762 35052 500 92827 105630 213372 52691 60281 57616 20295 1000 39803 45150 99127 23163 26034 23529 9388 1200 30874 35005 80011 18182 20283 17871 7569 1400 24463 27742 66329 14609 16099 13842 6270 1600 19637 22285 56055 11921 13066 10830 5294 1800 15866 18030 48051 9826 10657 8497 4536 2000 12840 14622 41645 8145 8728 6635 3931 2200 10353 11827 36391 6768 7148 5120 3433 2400 8276 9497 32011 5619 5832 3860 3019 2600 6517 7521 28304 4648 4719 2801 2671 2800 5002 5826 25117 3812 3763 1894 2372 3000 3685 4357 22359 3086 2937 1111 2114 3200 2534 3072 19937 2451 2212 0429 1888 3400 1516 1935 17800 1891 1576 0169 1690 3600 0609 0926 15898 1392 1088 0701 1513 3800 0202 0019 14199 0945 0501 1176 1356 4000 0934 0796 12660 0542 0044 1599 1216 4500 2486 2513 9414 0312 0920 2490 0921 5000 3725 3895 6807 0996 1689 3197 0686 5500 4743 5023 4666 1560 2318 3771 0497 6000 5590 5963 2865 2032 2843 4245 0341 Source of Data Gordon J Van Wylen and Richard E Sonntag Fundamentals of Classical Thermodynamics EnglishSI Version 3rd ed New York John Wiley Sons 1986 p 723 table A14 Based on thermodynamic data given in JANAF Thermochemical Tables Midland MI Thermal Research Laboratory The Dow Chemical Company 1971 Final PDF to printer 925 APPENDIX 1 cen22672app01881930indd 925 110617 0932 AM FIGURE A29 Generalized enthalpy departure chart Source of Data Redrawn from Gordon van Wylen and Richard Sontag Fundamentals of Classical Thermodynamics SI version 2d ed Wiley New York 1976 TR 090 0 10 20 30 40 50 60 70 05 15 25 35 45 55 65 75 01 02 03 04 05 10 20 30 4050 10 20 30 hideal h RuTcr Enthalpy departure Zh hideal h RTcr hideal h RuTcr Enthalpy departure Zh hideal h RTcr Reduced pressure PR 075 080 085 090 092 094 096 098 050 055 060 065 070 075 080 085 090 092 094 096 098 100 100 102 104 106 108 110 110 115 125 120 120 140 160 180 170 150 150 130 130 280 260 240 220 200 200 190 300 400 090 095 TR Saturated vapor Saturated liquid 050 055 060 065 070 075 080 085 090 092 094 096 098 100 110 075 080 085 090 092 094 096 098 100 102 104 106 108 110 115 125 120 120 140 160 180 170 150 150 130 130 280 260 240 220 200 200 190 300 400 090 095 TR PR 080 100 110 120 140 160 200 Saturated vapour 080 TR 090 100 110 120 140 160 200 0 005 010 015 020 025 030 0000 0100 0200 0300 0400 0500 Saturated vapor Saturated vapor Saturated liquid Final PDF to printer 926 PROPERTY TABLES AND CHARTS cen22672app01881930indd 926 110617 0932 AM 0 10 20 30 40 50 60 70 80 90 100 01 02 03 0405 10 20 30 4050 10 20 30 sideal s Ru Entropy departure Zs sideal s R Entropy departure sideal s Ru sideal s R Reduced pressure PR 050 055 060 065 070 075 075 080 080 085 085 090 090 092 092 094 094 094 096 096 098 098 098 100 100 102 102 104 104 106 106 108 110 110 115 120 120 130 140 140 150 160 160 180 200 250 300 095 090 Saturated gas Tr 050 055 060 065 070 075 075 080 080 085 085 090 090 092 092 094 094 094 096 096 098 098 098 100 100 102 102 104 104 106 106 108 110 110 115 120 120 130 140 140 150 160 160 180 200 250 300 095 090 Saturated liquid TR PR 080 TR 090 100 110 120 140 160 200 Saturated vapour 080 TR 090 100 110 120 140 160 200 0 005 010 015 020 025 030 0000 0100 0200 0300 0400 0500 Saturated vapor Saturated vapor Saturated vapor Zs FIGURE A30 Generalized entropy departure chart Source of Data Redrawn from Gordon van Wylen and Richard Sontag Fundamentals of Classical Thermodynamics SI version 2d ed Wiley New York 1976 Final PDF to printer 927 APPENDIX 1 cen22672app01881930indd 927 110617 0932 AM FIGURE A31 Psychrometric chart at 1 atm total pressure Reprinted from American Society of Heating Refrigerating and AirConditioning Engineers Inc Atlanta GA Prepared by Center for Applied Thermodynamic Studies University of Idaho 10 095 090 085 080 075 070 065 060 055 050 045 040 036 Enthalpy Humidity ratio Dh D Sensible heat Total heat DHS DHT 60 50 40 30 20 Dry bulb temperature C 10 0 70 80 90 100 110 120 Sensible heat Total heat DHS DHT 10 Σ 50 20 00 10 20 25 30 40 50 100 Σ 01 02 03 04 05 06 08 07 10 15 20 40 40 20 10 05 02 Σ 0 Enthalpy h kilojoules per kilogram dry air 20 30 40 50 60 70 80 90 100 Saturation temperature C 5 10 15 20 25 30 30 094 30 20 15 10 5 0 090 088 086 084 082 080 078 28 90 80 70 60 50 40 30 20 26 24 22 20 18 16 14 12 10 8 10 relative humidity 25 wet bulb temperature C 092 volume cubic meter per kilogram dry air Humidity ratio grams moisture per kilogram dry air 6 4 2 ASHRAE Psychrometric Chart No 1 Normal Temperature Barometric Pressure 101325 kPa 1992 American Society of Heating Refrigerating and AirConditioning Engineers Inc Sea Level Final PDF to printer 928 PROPERTY TABLES AND CHARTS cen22672app01881930indd 928 110617 0932 AM Ma Ma k 1 2 k 1 Ma 2 A A 1 Ma 2 k 1 1 k 1 2 Ma 2 05 k 1 k 1 P P 0 1 k 1 2 Ma 2 k k 1 ρ ρ 0 1 k 1 2 Ma 2 1 k 1 T T 0 1 k 1 2 Ma 2 1 AA Ma TT0 ρρ PP0 25 30 20 15 Compressible flow functions 10 05 0 0 05 10 15 Ma 20 25 30 TABLE A32 Onedimensional isentropic compressibleflow functions for an ideal gas with k 14 Ma Ma AA PP0 ρρ0 TT0 0 0 10000 10000 10000 01 01094 58218 09930 09950 09980 02 02182 29635 09725 09803 09921 03 03257 20351 09395 09564 09823 04 04313 15901 08956 09243 09690 05 05345 13398 08430 08852 09524 06 06348 11882 07840 08405 09328 07 07318 10944 07209 07916 09107 08 08251 10382 06560 07400 08865 09 09146 10089 05913 06870 08606 10 10000 10000 05283 06339 08333 12 11583 10304 04124 05311 07764 14 12999 11149 03142 04374 07184 16 14254 12502 02353 03557 06614 18 15360 14390 01740 02868 06068 20 16330 16875 01278 02300 05556 22 17179 20050 00935 01841 05081 24 17922 24031 00684 01472 04647 26 18571 28960 00501 01179 04252 28 19140 35001 00368 00946 03894 30 19640 42346 00272 00760 03571 50 22361 25000 00019 00113 01667 22495 0 0 0 Final PDF to printer 929 APPENDIX 1 cen22672app01881930indd 929 110617 0932 AM TABLE A33 Onedimensional normalshock functions for an ideal gas with k 14 Ma1 Ma2 P2P1 ρ2ρ1 T2T1 P02P01 P02P1 10 10000 10000 10000 10000 10000 18929 11 09118 12450 11691 10649 09989 21328 12 08422 15133 13416 11280 09928 24075 13 07860 18050 15157 11909 09794 27136 14 07397 21200 16897 12547 09582 30492 15 07011 24583 18621 13202 09298 34133 16 06684 28200 20317 13880 08952 38050 17 06405 32050 21977 14583 08557 42238 18 06165 36133 23592 15316 08127 46695 19 05956 40450 25157 16079 07674 51418 20 05774 45000 26667 16875 07209 56404 21 05613 49783 28119 17705 06742 61654 22 05471 54800 29512 18569 06281 67165 23 05344 60050 30845 19468 05833 72937 24 05231 65533 32119 20403 05401 78969 25 05130 71250 33333 21375 04990 85261 26 05039 77200 34490 22383 04601 91813 27 04956 83383 35590 23429 04236 98624 28 04882 89800 36636 24512 03895 105694 29 04814 96450 37629 25632 03577 113022 30 04752 103333 38571 26790 03283 120610 40 04350 185000 45714 40469 01388 210681 50 04152 29000 50000 58000 00617 326335 03780 60000 0 T 01 T 02 Ma 2 k 1 Ma 1 2 2 2k Ma 1 2 k 1 P 2 P 1 1 k Ma 1 2 1 k Ma 2 2 2k Ma 1 2 k 1 k 1 ρ 2 ρ 1 P 2 P 1 T 2 T 1 k 1 Ma 1 2 2 k 1 Ma 1 2 V 1 V 2 T 2 T 1 2 Ma 1 2 k 1 2 Ma 2 2 k 1 P 02 P 01 Ma 1 Ma 2 1 Ma 2 2 k 1 2 1 Ma 1 2 k 1 2 k 1 2 k 1 P 02 P 01 1 k Ma 1 2 1 Ma 2 2 k 1 2 k k 1 1 k Ma 2 2 Ma2 T2T1 ρ2ρ1 P02P1 P02P01 P2P1 30 40 50 20 Normal shock functions 10 0 10 15 20 25 Ma1 30 Final PDF to printer 930 PROPERTY TABLES AND CHARTS cen22672app01881930indd 930 110617 0932 AM TABLE A34 Rayleigh flow functions for an ideal gas with k 14 Ma T 0 T 0 P 0 P 0 TT PP VV 00 00000 12679 00000 24000 00000 01 00468 12591 00560 23669 00237 02 01736 12346 02066 22727 00909 03 03469 11985 04089 21314 01918 04 05290 11566 06151 19608 03137 05 06914 11141 07901 17778 04444 06 08189 10753 09167 15957 05745 07 09085 10431 09929 14235 06975 08 09639 10193 10255 12658 08101 09 09921 10049 10245 11246 09110 10 10000 10000 10000 10000 10000 12 09787 10194 09118 07958 11459 14 09343 10777 08054 06410 12564 16 08842 11756 07017 05236 13403 18 08363 13159 06089 04335 14046 20 07934 15031 05289 03636 14545 22 07561 17434 04611 03086 14938 24 07242 20451 04038 02648 15252 26 06970 24177 03556 02294 15505 28 06738 28731 03149 02004 15711 30 06540 34245 02803 01765 15882 T 0 T 0 k 1 Ma 2 2 k 1 Ma 2 1 k Ma 2 2 P 0 P 0 k 1 1 k Ma 2 2 k 1 Ma 2 k 1 k k 1 T T Ma 1 k 1 k Ma 2 2 P P 1 k 1 k Ma 2 V V ρ ρ 1 k Ma 2 1 k Ma 2 P0P0 T0T0 TT PP VV 35 30 25 15 20 Rayleigh flow functions 10 05 0 0 05 10 15 Ma 20 25 30 Final PDF to printer 931 cen22672app02931972indd 931 110617 0920 AM 2 APPENDIX P R O P E RT Y TA B L E S AN D C H ARTS EN G L I S H U N I TS Table A1E Molar mass gas constant and criticalpoint properties 932 Table A2E Idealgas specific heats of various common gases 933 Table A3E Properties of common liquids solids and foods 936 Table A4E Saturated waterTemperature table 938 Table A5E Saturated waterPressure table 940 Table A6E Superheated water 942 Table A7E Compressed liquid water 946 Table A8E Saturated icewater vapor 947 Figure A9E Ts diagram for water 948 Figure A10E Mollier diagram for water 949 Table A11E Saturated refrigerant134aTemperature table 950 Table A12E Saturated refrigerant134aPressure table 951 Table A13E Superheated refrigerant134a 952 Figure A14E Ph diagram for refrigerant134a 954 Table A16E Properties of the atmosphere at high altitude 955 Table A17E Idealgas properties of air 956 Table A18E Idealgas properties of nitrogen N2 958 Table A19E Idealgas properties of oxygen O2 960 Table A20E Idealgas properties of carbon dioxide CO2 962 Table A21E Idealgas properties of carbon monoxide CO 964 Table A22E Idealgas properties of hydrogen H2 966 Table A23E Idealgas properties of water vapor H2O 967 Table A26E Enthalpy of formation Gibbs function of formation and absolute entropy at 77F 1 atm 969 Table A27E Properties of some common fuels and hydrocarbons 970 Figure A31E Psychrometric chart at 1 atm total pressure 971 Final PDF to printer 932 PROPERTY TABLES AND CHARTS cen22672app02931972indd 932 110617 0920 AM TABLE A1E Molar mass gas constant and criticalpoint properties Substance Formula Molar mass M lbmlbmol Gas constant R Criticalpoint properties Btu lbmR psiaft3 lbmR Temperature R Pressure psia Volume ft3lbmol Air 2897 006855 03704 2385 547 141 Ammonia NH3 1703 01166 06301 7298 1636 116 Argon Ar 39948 004971 02686 272 705 120 Benzene C6H6 78115 002542 01374 1012 714 417 Bromine Br2 159808 001243 006714 1052 1500 217 nButane C4H10 58124 003417 01846 7652 551 408 Carbon dioxide CO2 4401 004513 02438 5475 1071 151 Carbon monoxide CO 28011 007090 03831 240 507 149 Carbon tetrachloride CCl4 15382 001291 006976 10015 661 442 Chlorine Cl2 70906 002801 01517 751 1120 199 Chloroform CHCl3 11938 001664 008988 9658 794 385 Dichlorodifluoromethane R12 CCl2F2 12091 001643 008874 6924 582 349 Dichlorofluoromethane R21 CHCl2F 10292 001930 01043 8130 749 316 Ethane C2H6 30020 006616 03574 5498 708 237 Ethyl alcohol C2H5OH 4607 004311 02329 9290 926 268 Ethylene C2H4 28054 007079 03825 5083 742 199 Helium He 4003 04961 26809 95 332 0926 nHexane C6H14 86178 002305 01245 9142 439 589 Hydrogen normal H2 2016 09851 53224 599 1881 104 Krypton Kr 8380 002370 01280 3769 798 148 Methane CH4 16043 01238 06688 3439 673 159 Methyl alcohol CH3OH 32042 006198 03349 9237 1154 189 Methyl chloride CH3Cl 50488 003934 02125 7493 968 229 Neon Ne 20183 009840 05316 801 395 0668 Nitrogen N2 28013 007090 03830 2271 492 144 Nitrous oxide N2O 44013 004512 02438 5574 1054 154 Oxygen O2 31999 006206 03353 2786 736 125 Propane C3H8 44097 004504 02433 6659 617 320 Propylene C3H6 42081 004719 02550 6569 670 290 Sulfur dioxide SO2 64063 003100 11675 7752 1143 195 Tetrafluoroethane R134a CF3CH2F 10203 001946 01052 6736 5887 319 Trichlorofluoromethane R11 CCl3F 13737 001446 007811 8481 635 397 Water H2O 18015 01102 05956 11648 3200 090 Xenon Xe 13130 001513 008172 52155 852 190 Calculated from R RuM where Ru 198588 BtulbmolR 107316 psiaft3lbmolR and M is the molar mass Source of Data K A Kobe and R E Lynn Jr Chemical Review 52 1953 pp 117236 and ASHRAE Handbook of Fundamentals Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 pp 164 and 361 Final PDF to printer 933 APPENDIX 2 cen22672app02931972indd 933 110617 0920 AM TABLE A2E Idealgas specific heats of various common gases a At 80F Gas Formula Gas constant R BtulbmR cp BtulbmR c v BtulbmR k Air 006855 0240 0171 1400 Argon Ar 004971 01253 00756 1667 Butane C4H10 003424 0415 0381 109 Carbon dioxide CO2 004513 0203 0158 1285 Carbon monoxide CO 007090 0249 0178 1399 Ethane C2H6 006616 0427 0361 1183 Ethylene C2H4 007079 0411 0340 1208 Helium He 04961 125 0753 1667 Hydrogen H2 09851 343 244 1404 Methane CH4 01238 0532 0403 132 Neon Ne 009840 0246 01477 1667 Nitrogen N2 007090 0248 0177 1400 Octane C8H18 001742 0409 0392 1044 Oxygen O2 006206 0219 0157 1395 Propane C3H8 004504 0407 0362 1124 Steam H2O 01102 0445 0335 1329 Source of Data Gordon J Van Wylen and Richard E Sonntag Fundamentals of Classical Thermodynamics EnglishSI Version 3rd ed New York John Wiley Sons 1986 p 687 Table A8E Final PDF to printer 934 PROPERTY TABLES AND CHARTS cen22672app02931972indd 934 110617 0920 AM TABLE A2E Idealgas specific heats of various common gases Continued b At various temperatures Temp F cp BtulbmR c v BtulbmR k cp BtulbmR c v BtulbmR k cp BtulbmR c v BtulbmR k Air Carbon dioxide CO2 Carbon monoxide CO 40 0240 0171 1401 0195 0150 1300 0248 0177 1400 100 0240 0172 1400 0205 0160 1283 0249 0178 1399 200 0241 0173 1397 0217 0172 1262 0249 0179 1397 300 0243 0174 1394 0229 0184 1246 0251 0180 1394 400 0245 0176 1389 0239 0193 1233 0253 0182 1389 500 0248 0179 1383 0247 0202 1223 0256 0185 1384 600 0250 0182 1377 0255 0210 1215 0259 0188 1377 700 0254 0185 1371 0262 0217 1208 0262 0191 1371 800 0257 0188 1365 0269 0224 1202 0266 0195 1364 900 0259 0191 1358 0275 0230 1197 0269 0198 1357 1000 0263 0195 1353 0280 0235 1192 0273 0202 1351 1500 0276 0208 1330 0298 0253 1178 0287 0216 1328 2000 0286 0217 1312 0312 0267 1169 0297 0226 1314 Hydrogen H2 Nitrogen N2 Oxygen O2 40 3397 2412 1409 0248 0177 1400 0219 0156 1397 100 3426 2441 1404 0248 0178 1399 0220 0158 1394 200 3451 2466 1399 0249 0178 1398 0223 0161 1387 300 3461 2476 1398 0250 0179 1396 0226 0164 1378 400 3466 2480 1397 0251 0180 1393 0230 0168 1368 500 3469 2484 1397 0254 0183 1388 0235 0173 1360 600 3473 2488 1396 0256 0185 1383 0239 0177 1352 700 3477 2492 1395 0260 0189 1377 0242 0181 1344 800 3494 2509 1393 0262 0191 1371 0246 0184 1337 900 3502 2519 1392 0265 0194 1364 0249 0187 1331 1000 3513 2528 1390 0269 0198 1359 0252 0190 1326 1500 3618 2633 1374 0283 0212 1334 0263 0201 1309 2000 3758 2773 1355 0293 0222 1319 0270 0208 1298 Note The unit BtulbmR is equivalent to BtulbmF Source of Data Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 p 830 Table A4 Originally published in Tables of Properties of Gases NBS Circular 564 1955 Final PDF to printer 935 APPENDIX 2 cen22672app02931972indd 935 110617 0920 AM TABLE A2E Idealgas specific heats of various common gases Concluded c As a function of temperature c p a bT c T 2 d T 3 T in R cp in BtulbmolR Substance Formula a b c d Temperature range R error Max Avg Nitrogen N2 6903 002085 102 005957 105 01176 109 4913240 059 034 Oxygen O2 6085 02017 102 005275 105 005372 109 4913240 119 028 Air 6713 002609 102 003540 105 008052 109 4913240 072 033 Hydrogen H2 6952 002542 102 002952 105 003565 109 4913240 102 026 Carbon monoxide CO 6726 002222 102 003960 105 009100 109 4913240 089 037 Carbon dioxide CO2 5316 079361 102 02581 105 03059 109 4913240 067 022 Water vapor H2O 7700 002552 102 007781 105 01472 109 4913240 053 024 Nitric oxide NO 7008 001247 102 007185 105 01715 109 4912700 097 036 Nitrous oxide N2O 5758 07780 102 02596 105 04331 109 4912700 059 026 Nitrogen dioxide NO2 548 07583 102 0260 105 0322 109 4912700 046 018 Ammonia NH3 65846 034028 102 0073034 105 027402 109 4912700 091 036 Sulfur S 6499 02943 102 01200 105 01632 109 4913240 099 038 Sulfur dioxide SO2 6157 07689 102 02810 105 03527 109 4913240 045 024 Sulfur trioxide SO3 3918 1935 102 08256 105 1328 109 4912340 029 013 Acetylene C2H2 521 12227 102 04812 105 07457 109 4912700 146 059 Benzene C6H6 8650 64322 102 2327 105 3179 109 4912700 034 020 Methanol CH4O 455 1214 102 00898 105 0329 109 4911800 018 008 Ethanol C2H6O 475 2781 102 07651 105 0821 109 4912700 040 022 Hydrogen chloride HCl 7244 01011 102 009783 105 01776 109 4912740 022 008 Methane CH4 4750 06666 102 009352 105 04510 109 4912740 133 057 Ethane C2H6 1648 2291 102 04722 105 02984 109 4912740 083 028 Propane C3H8 0966 4044 102 1159 105 1300 109 4912740 040 012 nButane C4H10 0945 4929 102 1352 105 1433 109 4912740 054 024 iButane C4H10 1890 5520 102 1696 105 2044 109 4912740 025 013 nPentane C5H12 1618 6028 102 1656 105 1732 109 4912740 056 021 nHexane C6H14 1657 7328 102 2112 105 2363 109 4912740 072 020 Ethylene C2H4 0944 2075 102 06151 105 07326 109 4912740 054 013 Propylene C3H6 0753 3162 102 08981 105 1008 109 4912740 073 017 Source of Data BG Kyle Chemical and Process Thermodynamics 3rd ed Upper Saddle River NJ Prentice Hall 2000 Final PDF to printer 936 PROPERTY TABLES AND CHARTS cen22672app02931972indd 936 110617 0920 AM TABLE A3E Properties of common liquids solids and foods a Liquids Boiling data at 1 atm Freezing data Liquid properties Substance Normal boiling point F Latent heat of vaporization hfg Btulbm Freezing point F Latent heat of fusion hif Btulbm Temperature F Density ρ lbmft3 Specific heat cp BtulbmR Ammonia 279 2454 1079 1386 279 426 106 0 413 1083 40 395 1103 80 375 1135 Argon 3026 695 3087 120 3026 870 0272 Benzene 1764 1694 419 542 68 549 0411 Brine 20 sodium chloride by mass 2190 07 68 718 0743 nButane 311 1656 2173 345 311 375 0552 Carbon dioxide 1092 996 at 32F 698 32 578 0583 Ethanol 1728 3605 1736 469 77 489 0588 Ethyl alcohol 1735 368 2488 464 68 493 0678 Ethylene glycol 3886 3440 126 779 68 692 0678 Glycerine 3558 419 660 863 68 787 0554 Helium 4521 980 4521 913 545 Hydrogen 4230 1917 4345 256 4230 441 239 Isobutane 109 1578 2555 455 109 371 0545 Kerosene 399559 108 128 68 512 0478 Mercury 6741 1267 380 490 77 847 0033 Methane 2587 2196 2960 251 2587 264 0834 160 200 1074 Methanol 1481 473 1439 427 77 491 0609 Nitrogen 3204 854 3460 109 3204 505 0492 260 382 0643 Octane 2566 1317 715 779 68 439 0502 Oil light 77 568 0430 Oxygen 2973 915 3618 59 2973 712 0408 Petroleum 99165 68 400 0478 Propane 437 1840 3058 344 437 363 0538 32 330 0604 100 294 0673 Refrigerant134a 150 933 1419 40 885 0283 15 860 0294 32 809 0318 90 736 0348 Water 212 9701 32 1435 32 624 101 90 621 100 150 612 100 212 598 101 Sublimation temperature At pressures below the triplepoint pressure of 751 psia carbon dioxide exists as a solid or gas Also the freezingpoint temperature of carbon dioxide is the triplepoint temperature of 698F Final PDF to printer 937 APPENDIX 2 cen22672app02931972indd 937 110617 0920 AM TABLE A3E Properties of common liquids solids and foods Concluded b Solids values are for room temperature unless indicated otherwise Substance Density ρ lbmft3 Specific heat cp BtulbmR Substance Density ρ lbmft3 Specific heat cp BtulbmR Metals Nonmetals Aluminum Asphalt 132 0220 100F 0192 Brick common 120 0189 32F 0212 Brick fireclay 500C 144 0229 100F 170 0218 Concrete 144 0156 200F 0224 Clay 624 0220 300F 0229 Diamond 151 0147 400F 0235 Glass window 169 0191 500F 0240 Glass pyrex 139 0200 Bronze 76 Cu 2 Zn 2 Al 517 00955 Graphite 156 0170 Granite 169 0243 Brass yellow 65 Cu 35 Zn 519 00955 Gypsum or plaster board Ice 50 0260 Copper 50F 0424 60F 00862 0F 0471 0F 00893 20F 0491 100F 555 00925 32F 575 0502 200F 00938 Limestone 103 0217 390F 00963 Marble 162 0210 Iron 490 0107 Plywood Douglas fir 340 Lead 705 0030 Rubber hard 687 Magnesium 108 0239 Rubber soft 718 Nickel 555 0105 Sand 949 Silver 655 0056 Stone 936 Steel mild 489 0119 Woods hard maple oak etc 450 Tungsten 1211 0031 Woods soft fir pine etc 320 c Foods Food Water content mass Freezing point F Specific heat BtulbmR Latent heat of fusion Btulbm Food Water content mass Freezing point F Specific heat BtulbmR Latent heat of fusion Btulbm Above freezing Below freezing Above freezing Below freezing Apples 84 30 0873 0453 121 Lettuce 95 32 0961 0487 136 Bananas 75 31 0801 0426 108 Milk whole 88 31 0905 0465 126 Beef round 67 0737 0402 96 Oranges 87 31 0897 0462 125 Broccoli 90 31 0921 0471 129 Potatoes 78 31 0825 0435 112 Butter 16 0249 23 Salmon fish 64 28 0713 0393 92 Cheese Swiss 39 14 0513 0318 56 Shrimp 83 28 0865 0450 119 Cherries 80 29 0841 0441 115 Spinach 93 31 0945 0481 134 Chicken 74 27 0793 0423 106 Strawberries 90 31 0921 0471 129 Corn sweet 74 31 0793 0423 106 Tomatoes ripe 94 31 0953 0484 135 Eggs whole 74 31 0793 0423 106 Turkey 64 0713 0393 92 Ice cream 63 22 0705 0390 90 Watermelon 93 31 0945 0481 134 Source of Data Values are obtained from various handbooks and other sources or are calculated Water content and freezingpoint data of foods are from ASHRAE Handbook of Fundamentals IP version Atlanta GA American Society of Heating Refrigerating and AirConditioning Engineers Inc 1993 Chap 30 Table 1 Freezing point is the temperature at which freezing starts for fruits and vegetables and the average freezing temperature for other foods Final PDF to printer 938 PROPERTY TABLES AND CHARTS cen22672app02931972indd 938 110617 0920 AM TABLE A4E Saturated waterTemperature table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Temp T F Sat press Psat psia Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 32018 008871 001602 32999 0000 10210 10210 0000 10752 10752 000000 218672 21867 35 009998 001602 29457 3004 10190 10220 3004 10735 10765 000609 217011 21762 40 012173 001602 24436 8032 10156 10237 8032 10707 10787 001620 214271 21589 45 014756 001602 20358 1305 10122 10253 1305 10678 10809 002620 211587 21421 50 017812 001602 17031 1807 10089 10269 1807 10650 10831 003609 208956 21256 55 021413 001603 14304 2307 10055 10286 2307 10622 10853 004586 206377 21096 60 025638 001604 12061 2808 10021 10302 2808 10594 10874 005554 203847 20940 65 030578 001604 10208 3308 99876 10318 3308 10565 10896 006511 201366 20788 70 036334 001605 86718 3808 99539 10335 3808 10537 10918 007459 198931 20639 75 043016 001606 73927 4307 99202 10351 4307 10509 10939 008398 196541 20494 80 050745 001607 63241 4806 98865 10367 4807 10480 10961 009328 194196 20352 85 059659 001609 54280 5306 98528 10383 5306 10452 10983 010248 191892 20214 90 069904 001610 46740 5805 98190 10400 5805 10424 11004 011161 189630 20079 95 081643 001612 40374 6304 97852 10416 6304 10395 11026 012065 187408 19947 100 095052 001613 34983 6803 97514 10432 6803 10367 11047 012961 185225 19819 110 12767 001617 26496 7801 96836 10464 7802 10310 11090 014728 180970 19570 120 16951 001620 20294 8800 96156 10496 8800 10252 11132 016466 176856 19332 130 22260 001625 15709 9799 95473 10527 9799 10194 11174 018174 172877 19105 140 28931 001629 12281 10798 94787 10559 10799 10136 11216 019855 169024 18888 150 37234 001634 96929 11798 94098 10590 11799 10078 11257 021508 165291 18680 160 47474 001639 77185 12798 93405 10620 12800 10018 11298 023136 161670 18481 170 59999 001645 61982 13800 92708 10651 13802 99588 11339 024739 158155 18289 180 75197 001651 50172 14802 92006 10681 14804 98985 11379 026318 154741 18106 190 93497 001657 40920 15805 91299 10710 15808 98376 11418 027874 151421 17930 200 11538 001663 33613 16810 90587 10740 16813 97760 11457 029409 148191 17760 210 14136 001670 27798 17815 89868 10768 17820 97135 11495 030922 145046 17597 212 14709 001671 26782 18016 89724 10774 18021 97009 11503 031222 144427 17565 220 17201 001677 23136 18822 89143 10796 18828 96502 11533 032414 141980 17439 230 20795 001684 19374 19831 88410 10824 19837 95859 11570 033887 138989 17288 240 24985 001692 16316 20841 87670 10851 20849 95206 11605 035342 136069 17141 250 29844 001700 13816 21854 86921 10877 21863 94541 11640 036779 133216 16999 260 35447 001708 11760 22868 86162 10903 22879 93865 11674 038198 130425 16862 270 41877 001717 10059 23885 85394 10928 23898 93176 11707 039601 127694 16730 280 49222 001726 86439 24904 84616 10952 24920 92474 11739 040989 125018 16601 290 57573 001735 74607 25926 83827 10975 25945 91757 11770 042361 122393 16475 300 67028 001745 64663 26951 83025 10998 26973 91024 11800 043720 119818 16354 310 77691 001755 56266 27979 82211 11019 28005 90275 11828 045065 117289 16235 320 89667 001765 49144 29011 81384 11040 29040 89509 11855 046396 114802 16120 330 10307 001776 43076 30046 80543 11059 30080 88725 11881 047716 112355 16007 340 11802 001787 37885 31085 79687 11077 31124 87922 11905 049024 109945 15897 350 13463 001799 33425 32129 78816 11094 32173 87098 11927 050321 107570 15789 360 15303 001811 29580 33176 77928 11110 33228 86253 11948 051607 105227 15683 370 17336 001823 26252 34229 77023 11125 34288 85386 11967 052884 102914 15580 380 19574 001836 23361 35287 76100 11139 35353 84496 11985 054152 100628 15478 390 22033 001850 20842 36350 75158 11151 36425 83581 12001 055411 098366 15378 Final PDF to printer 939 APPENDIX 2 cen22672app02931972indd 939 110617 0920 AM TABLE A4E Saturated waterTemperature table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Temp T F Sat press Psat psia Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 400 24726 001864 18639 37419 74197 11162 37504 82639 12014 056663 096127 15279 410 27669 001878 16706 38494 73214 11171 38590 81671 12026 057907 093908 15182 420 30876 001894 15006 39576 72208 11178 39684 80674 12036 059145 091707 15085 430 34364 001910 13505 40665 71180 11184 40786 79646 12043 060377 089522 14990 440 38149 001926 12178 41761 70126 11189 41897 78587 12048 061603 087349 14895 450 42247 001944 10999 42866 69047 11191 43018 77494 12051 062826 085187 14801 460 46675 001962 099510 43979 67939 11192 44148 76365 12051 064044 083033 14708 470 51452 001981 090158 45101 66802 11190 45290 75198 12049 065260 080885 14615 480 56596 002001 081794 46234 65634 11187 46443 73991 12043 066474 078739 14521 490 62124 002022 074296 47377 64432 11181 47609 72740 12035 067686 076594 14428 500 68056 002044 067558 48532 63194 11173 48789 71444 12023 068899 074445 14334 510 74411 002067 061489 49699 61917 11162 49984 70099 12008 070112 072290 14240 520 81211 002092 056009 50880 60599 11148 51194 68701 11990 071327 070126 14145 530 88474 002118 051051 52076 59235 11131 52423 67247 11967 072546 067947 14049 540 96224 002146 046553 53288 57823 11111 53670 65731 11940 073770 065751 13952 550 10448 002176 042465 54518 56358 11088 54939 64147 11909 075000 063532 13853 560 11327 002207 038740 55768 54833 11060 56231 62491 11872 076238 061284 13752 570 12262 002242 035339 57040 53245 11028 57549 60755 11830 077486 059003 13649 580 13255 002279 032225 58337 51584 10992 58895 58929 11782 078748 056679 13543 590 14308 002319 029367 59661 49843 10950 60275 57004 11728 080026 054306 13433 600 15425 002362 026737 61018 48010 10903 61692 54967 11666 081323 051871 13319 610 16609 002411 024309 62411 46073 10848 63152 52803 11595 082645 049363 13201 620 17862 002464 022061 63847 44014 10786 64662 50492 11515 083998 046765 13076 630 19189 002524 019972 65335 41812 10715 66232 48007 11424 085389 044056 12944 640 20593 002593 018019 66886 39436 10632 67874 45314 11319 086828 041206 12803 650 22078 002673 016184 68516 36844 10536 69608 42365 11197 088332 038177 12651 660 23649 002767 014444 70248 33974 10422 71459 39084 11054 089922 034906 12483 670 25312 002884 012774 72123 30722 10285 73474 35354 10883 091636 031296 12293 680 27073 003035 011134 74211 26900 10111 75732 30957 10669 093541 027163 12070 690 28941 003255 009451 76681 22077 9876 78424 25396 10382 095797 022089 11789 700 30930 003670 007482 80175 14650 9483 82276 16832 9911 099023 014514 11354 70510 32001 004975 004975 86661 0 8666 89607 0 8961 105257 0 10526 Source of Data Tables A4E through A8E are generated using the Engineering Equation Solver EES software developed by S A Klein and F L Alvarado The routine used in calculations is the highly accurate SteamIAPWS which incorporates the 1995 Formulation for the Thermodynamic Properties of Ordinary Water Substance for General and Scientific Use issued by The International Association for the Properties of Water and Steam IAPWS This formulation replaces the 1984 formulation of Haar Gallagher and Kell NBSNRC Steam Tables Hemisphere Publishing Co 1984 which is also available in EES as the routine STEAM The new formulation is based on the correlations of Saul and Wagner J Phys Chem Ref Data 16 893 1987 with modifications to adjust to the International Temperature Scale of 1990 The modifications are described by Wagner and Pruss J Phys Chem Ref Data 22 783 1993 The properties of ice are based on Hyland and Wexler Formulations for the Thermodynamic Properties of the Saturated Phases of H2O from 17315 K to 47315 K ASHRAE Trans Part 2A Paper 2793 1983 Concluded Final PDF to printer 940 PROPERTY TABLES AND CHARTS cen22672app02931972indd 940 110617 0920 AM TABLE A5E Saturated waterPressure table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Press P psia Sat temp Tsat F Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 1 10169 001614 33349 6972 97399 10437 6972 10357 11054 013262 184495 19776 2 12602 001623 17371 9402 95745 10515 9402 10217 11158 017499 174444 19194 3 14141 001630 11870 10939 94690 10563 10940 10128 11222 020090 168489 18858 4 15291 001636 90629 12089 93897 10599 12090 10060 11269 021985 164225 18621 5 16218 001641 73525 13017 93253 10627 13018 10005 11307 023488 160894 18438 6 17000 001645 61982 13800 92708 10651 13802 99588 11339 024739 158155 18289 8 18281 001652 47347 15083 91808 10689 15086 98815 11390 026757 153800 18056 10 19316 001659 38425 16122 91075 10720 16125 98182 11431 028362 150391 17875 14696 21195 001671 26805 18012 89727 10774 18016 97012 11503 031215 144441 17566 15 21299 001672 26297 18116 89652 10777 18121 96947 11507 031370 144441 17549 20 22792 001683 20093 19621 88563 10818 19627 95993 11562 033582 139606 17319 25 24003 001692 16307 20845 87667 10851 20852 95203 11606 035347 136060 17141 30 25030 001700 13749 21884 86898 10878 21893 94521 11641 036821 133132 16995 35 25925 001708 11901 22792 86219 10901 22803 93916 11672 038093 130632 16872 40 26722 001715 10501 23602 85609 10921 23614 93369 11698 039213 128448 16766 45 27441 001721 94028 24334 85052 10939 24349 92868 11722 040216 126506 16672 50 28099 001727 85175 25005 84539 10954 25021 92403 11742 041125 124756 16588 55 28705 001732 77882 25625 84061 10969 25642 91970 11761 041958 123162 16512 60 29269 001738 71766 26201 83613 10981 26220 91561 11778 042728 121697 16442 65 29795 001743 66560 26741 83190 10993 26762 91175 11794 043443 120341 16378 70 30291 001748 62075 27250 82790 11004 27272 90808 11808 044112 119078 16319 75 30759 001752 58167 27731 82409 11014 27755 90458 11821 044741 117895 16264 80 31202 001757 54733 28187 82045 11023 28213 90122 11834 045335 116783 16212 85 31624 001761 51689 28622 81697 11032 28650 89800 11845 045897 115732 16163 90 32026 001765 48972 29038 81362 11040 29067 89489 11856 046431 114737 16117 95 32411 001770 46532 29436 81040 11048 29467 89189 11866 046941 113791 16073 100 32781 001774 44327 29819 80729 11055 29851 88899 11875 047427 112888 16032 110 33477 001781 40410 30541 80137 11068 30578 88344 11892 048341 111201 15954 120 34125 001789 37289 31216 79579 11079 31255 87820 11908 049187 109646 15883 130 34732 001796 34557 31848 79051 11090 31892 87321 11921 049974 108204 15818 140 35303 001802 32202 32445 78549 11099 32492 86845 11934 050711 106858 15757 150 35842 001809 30150 33011 78069 11108 33061 86388 11945 051405 105595 15700 160 36354 001815 28347 33549 77610 11116 33602 85949 11955 052061 104405 15647 170 36841 001821 26749 34062 77168 11123 34119 85525 11964 052682 103279 15596 180 37307 001827 25322 34553 76742 11130 34614 85116 11973 053274 102210 15548 190 37752 001833 24040 35024 76331 11136 35089 84719 11981 053839 101191 15503 200 38180 001839 22882 35478 75932 11141 35546 84333 11988 054379 100219 15460 250 40097 001865 18440 37523 74102 11163 37609 82547 12016 056784 095912 15270 300 41735 001890 15435 39289 72477 11177 39394 80941 12033 058818 092289 15111 350 43174 001912 13263 40855 70998 11185 40979 79465 12044 060590 089143 14973 400 44462 001934 11617 42270 69631 11190 42413 78087 12050 062168 086350 14852 450 45631 001955 10324 43567 68352 11192 43730 76786 12052 063595 083828 14742 500 46704 001975 092819 44768 67142 11191 44951 75548 12050 064900 081521 14642 550 47697 001995 084228 45890 65991 11188 46093 74360 12045 066107 079388 14550 600 48624 002014 077020 46946 64888 11183 47170 73215 12039 067231 077400 14463 Final PDF to printer 941 APPENDIX 2 cen22672app02931972indd 941 110617 0920 AM TABLE A5E Saturated waterPressure table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Press P psia Sat temp Tsat F Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 700 50313 002051 065589 48896 62798 11169 49162 71029 12019 069279 073771 14305 800 51827 002087 056920 50674 60830 11150 50983 68948 11993 071117 070502 14162 900 53202 002124 050107 52319 58954 11127 52673 66946 11962 072793 067505 14030 1000 54465 002159 044604 53858 57149 11101 54257 65003 11926 074341 064722 13906 1200 56726 002232 036241 56689 53687 11038 57185 61239 11842 077143 059632 13677 1400 58714 002307 030161 59279 50350 10963 59876 57566 11744 079658 054991 13465 1600 60493 002386 025516 61699 47069 10877 62406 53918 11632 081972 050645 13262 1800 62107 002470 021831 64003 43786 10779 64826 50235 11506 084144 046482 13063 2000 63585 002563 018815 66233 40446 10668 67182 46460 11364 086224 042409 12863 2500 66817 002860 013076 71767 31353 10312 73090 36079 10917 091311 031988 12330 3000 69541 003433 008460 78339 18641 9698 80245 21432 10168 097321 018554 11587 32001 70510 004975 004975 86661 0 8666 89607 0 8961 105257 0 10526 Concluded Final PDF to printer 942 PROPERTY TABLES AND CHARTS cen22672app02931972indd 942 110617 0920 AM TABLE A6E Superheated water T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 10 psia 10169F P 50 psia 16218F P 10 psia 19316F Sat 33349 10437 11054 19776 73525 10627 11307 18438 38425 10720 11431 17875 200 39253 10775 11501 20509 78153 10762 11485 18716 38849 10745 11464 17926 240 41644 10912 11683 20777 83009 10903 11671 18989 41326 10891 11655 18207 280 44033 11050 11865 21030 87838 11043 11856 19246 43774 11034 11844 18469 320 46420 11189 12048 21271 92650 11184 12041 19490 46205 11176 12031 18716 360 48807 11329 12233 21502 97452 11325 12226 19722 48624 11319 12218 18950 400 51192 11471 12418 21722 10225 11467 12413 19944 51035 11462 12406 19174 440 53577 11613 12604 21934 10703 11609 12600 20156 53441 11605 12594 19388 500 57154 11828 12886 22237 11421 11826 12882 20461 57041 11822 12878 19693 600 63114 12194 13362 22709 12615 12192 13359 20933 63029 12190 13356 20167 700 69073 12568 13846 23146 13809 12567 13844 21371 69007 12565 13842 20605 800 75031 12951 14339 23553 15002 12949 14337 21778 74980 12948 14335 21013 1000 86947 13742 15351 24299 17386 13742 15350 22524 86913 13741 15349 21760 1200 98862 14571 16400 24972 19770 14570 16400 23198 98840 14570 16399 22433 1400 11078 15437 17487 25590 22154 15437 17487 23816 110762 15436 17486 23052 P 15 psia 21299F P 20 psia 22792F P 40 psia 26722F Sat 26297 10777 11507 17549 20093 10818 11562 17319 10501 10921 11698 16766 240 27429 10878 11639 17742 20478 10865 11623 17406 280 29085 11024 11832 18010 21739 11014 11819 17679 10713 10973 11766 16858 320 30722 11169 12022 18260 22980 11161 12012 17933 11363 11129 11971 17128 360 32348 11313 12211 18496 24209 11307 12202 18171 11999 11281 12169 17376 400 33965 11457 12399 18721 25429 11451 12393 18398 12625 11431 12365 17610 440 35576 11601 12588 18936 26644 11597 12583 18614 13244 11579 12560 17831 500 37986 11819 12873 19243 28458 11816 12869 18922 14165 11802 12850 18143 600 41988 12187 13353 19718 31467 12185 13349 19398 15686 12175 13336 18625 700 45981 12563 13839 20156 34467 12561 13837 19837 17197 12553 13826 19067 800 49967 12946 14333 20565 37461 12945 14331 20247 18702 12939 14323 19478 1000 57930 13740 15348 21312 43438 13738 15346 20994 21700 13734 15341 20227 1200 65885 14569 16398 21986 49407 14568 16397 21668 24691 14565 16393 20902 1400 73836 15436 17485 22604 55373 15435 17484 22287 27678 15433 17481 21522 1600 81784 16340 18610 23178 61335 16339 18609 22861 30662 16337 18607 22096 P 60 psia 29269F P 80 psia 31202F P 100 psia 32781F Sat 71766 10981 11778 16442 54733 11023 11834 16212 44327 11055 11875 16032 320 74863 11096 11927 16636 55440 11059 11879 16271 360 79259 11255 12135 16897 58876 11227 12099 16545 46628 11198 12061 16263 400 83548 11409 12337 17138 62187 11387 12308 16794 49359 11364 12278 16521 440 87766 11561 12536 17364 65420 11543 12512 17026 52006 11524 12487 16759 500 94005 11788 12831 17682 70177 11773 12812 17350 55876 11759 12793 17088 600 104256 12165 13322 18168 77951 12154 13308 17841 62167 12144 13294 17586 700 114401 12545 13816 18613 85616 12538 13805 18289 68344 12530 13795 18037 800 124484 12933 14315 19026 93218 12926 14306 18704 74457 12920 14298 18453 1000 144543 13730 15335 19777 108313 13726 15329 19457 86575 13722 15324 19208 1200 164525 14562 16389 20454 123331 14559 16385 20135 98615 14556 16381 19887 1400 184464 15430 17478 21073 138306 15428 17475 20755 110612 15426 17472 20508 1600 20438 16335 18605 21648 153257 16333 18602 21330 122584 16332 18600 21083 1800 22428 17276 19766 22187 168192 17275 19765 21869 134541 17273 19763 21622 2000 24417 18252 20963 22694 183117 18250 20961 22376 146487 18249 20960 22130 The temperature in parentheses is the saturation temperature at the specified pressure Properties of saturated vapor at the specified pressure Final PDF to printer 943 APPENDIX 2 cen22672app02931972indd 943 110617 0920 AM TABLE A6E Superheated water T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 120 psia 34125F P 140 psia 35303F P 160 psia 36354F Sat 37289 11079 11908 15883 32202 11099 11934 15757 28347 11116 11955 15647 360 38446 11167 12021 16023 32584 11134 11978 15811 400 40799 11340 12246 16292 34676 11315 12214 16092 30076 11290 12180 15914 450 43613 11545 12514 16594 37147 11526 12489 16403 32293 11507 12463 16234 500 46340 11744 12773 16872 39525 11729 12753 16686 34412 11714 12732 16522 550 49010 11939 13028 17131 41845 11927 13011 16948 36469 11914 12994 16788 600 51642 12134 13280 17375 44124 12123 13266 17195 38484 12113 13252 17037 700 56829 12522 13784 17829 48604 12514 13773 17652 42434 12506 13763 17498 800 61950 12914 14290 18247 53017 12908 14281 18072 46316 12902 14273 17920 1000 72083 13717 15318 19005 61732 13713 15313 18832 53968 13709 15307 18682 1200 82137 14553 16377 19684 70367 14550 16373 19512 61540 14547 16369 19363 1400 92149 15423 17469 20305 78961 15421 17466 20134 69070 15418 17463 19986 1600 102135 16330 18598 20881 87529 16328 18595 20711 76574 16326 18593 20563 1800 112106 17272 19761 21420 96082 17270 19759 21250 84063 17269 19757 21102 2000 122067 18248 20958 21928 104624 18246 20957 21758 91542 18245 20955 21610 P 180 psia 37307F P 200 psia 38180F P 225 psia 39180F Sat 25322 11130 11973 15548 22882 11141 11988 15460 20423 11153 12003 15360 400 26490 11263 12145 15752 23615 11235 12109 15602 20728 11197 12060 15427 450 28514 11487 12437 16082 25488 11467 12410 15943 22457 11441 12376 15783 500 30433 11698 12712 16376 27247 11682 12690 16243 24059 11662 12663 16091 550 32286 11902 12977 16646 28939 11889 12960 16516 25590 11872 12938 16370 600 34097 12102 13238 16897 30586 12091 13223 16771 27075 12077 13205 16628 700 37635 12498 13752 17361 33796 12490 13741 17238 29956 12480 13727 17099 800 41104 12895 14265 17785 36934 12889 14256 17664 32765 12881 14245 17528 900 44531 13297 14780 18179 40031 13292 14773 18059 35530 13285 14765 17925 1000 47929 13705 15301 18549 43099 13701 15296 18430 38268 13695 15289 18296 1200 54674 14543 16365 19231 49182 14540 16361 19113 43689 14536 16356 18981 1400 61377 15416 17460 19855 55222 15414 17457 19737 49068 15411 17454 19606 1600 68054 16324 18591 20432 61238 16322 18588 20315 54422 16320 18586 20184 1800 74716 17267 19756 20971 67238 17265 19754 20855 59760 17264 19752 20724 2000 81367 18244 20954 21479 73227 18243 20953 21363 65087 18241 20951 21232 P 250 psia 40097F P 275 psia 40945F P 300 psia 41735F Sat 18440 11163 12016 15270 16806 11170 12026 15187 15435 11177 12033 15111 450 20027 11413 12340 15636 18034 11385 12303 15499 16369 11356 12264 15369 500 21506 11641 12636 15953 19415 11620 12608 15825 17670 11598 12579 15706 550 22910 11856 12915 16237 20715 11839 12893 16115 18885 11821 12870 16001 600 24264 12063 13186 16499 21964 12049 13167 16380 20046 12035 13148 16270 650 25586 12268 13451 16743 23179 12256 13435 16627 21172 12244 13419 16520 700 26883 12470 13714 16974 24369 12460 13700 16860 22273 12449 13686 16755 800 29429 12873 14235 17406 26699 12865 14224 17294 24424 12857 14213 17192 900 31930 13279 14756 17804 28984 13273 14748 17694 26529 13266 14739 17593 1000 34403 13690 15282 18177 31241 13685 15274 18068 28605 13679 15267 17968 1200 39295 14533 16350 18863 35700 14529 16345 18755 32704 14525 16340 18657 1400 44144 15408 17450 19488 40116 15405 17446 19381 36759 15402 17442 19284 1600 48969 16317 18583 20066 44507 16315 18580 19960 40789 16313 18577 19863 1800 53777 17262 19749 20607 48882 17260 19747 20501 44803 17258 19745 20404 2000 58575 18239 20949 21116 53247 18238 20947 21010 48807 18236 20946 20913 Continued Final PDF to printer 944 PROPERTY TABLES AND CHARTS cen22672app02931972indd 944 110617 0920 AM TABLE A6E Superheated water T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 350 psia 43174F P 400 psia 44462F P 450 psia 45631F Sat 13263 11185 12044 14973 11617 11190 12050 14852 10324 11192 12052 14742 450 13739 11293 12183 15128 11747 11225 12094 14901 500 14921 11552 12519 15487 12851 11504 12456 15288 11233 11454 12389 15103 550 16004 11786 12822 15795 13840 11749 12773 15610 12152 11711 12723 15441 600 17030 12006 13109 16073 14765 11976 13069 15897 13001 11946 13028 15737 650 18018 12219 13386 16328 15650 12194 13353 16158 13807 12169 13319 16005 700 18979 12428 13658 16567 16507 12407 13629 16401 14584 12385 13600 16253 800 20848 12841 14191 17009 18166 12825 14170 16849 16080 12808 14147 16706 900 22671 13253 14722 17414 19777 13240 14704 17257 17526 13227 14686 17117 1000 24464 13669 15253 17791 21358 13658 15239 17636 18942 13647 15224 17499 1200 27996 14517 16330 18483 24465 14509 16320 18331 21718 14501 16310 18196 1400 31484 15396 17435 19111 27527 15390 17427 18960 24450 15384 17420 18827 1600 34947 16308 18571 19691 30565 16303 18565 19541 27157 16298 18560 19409 1800 38394 17254 19740 20233 33586 17250 19736 20084 29847 17246 19732 19952 2000 41830 18233 20942 20742 36597 18230 20939 20594 32527 18226 20935 20462 P 500 psia 46704F P 600 psia 48624F P 700 psia 50313F Sat 092815 11191 12050 14642 077020 11183 12039 14463 065589 11169 12019 14305 500 099304 11401 12319 14928 079526 11282 12165 14596 550 107974 11671 12670 15284 087542 11587 12559 14996 072799 11495 12438 14730 600 115876 11914 12986 15590 094605 11849 12899 15325 079332 11779 12807 15087 650 123312 12143 13284 15865 101133 12090 13213 15614 085242 12034 13138 15393 700 130440 12364 13570 16117 107316 12319 13510 15877 090769 12272 13448 15666 800 144097 12792 14125 16576 119038 12758 14080 16348 101125 12724 14034 16150 900 157252 13214 14669 16992 130230 13187 14633 16771 110921 13160 14597 16581 1000 170094 13636 15210 17376 141097 13614 15181 17160 120381 13592 15152 16974 1100 182726 14062 15753 17735 151749 14044 15729 17522 129621 14025 15704 17341 1200 195211 14494 16300 18075 162252 14478 16279 17865 138709 14462 16259 17685 1400 21988 15378 17412 18708 182957 15366 17397 18501 156580 15354 17382 18324 1600 24430 16294 18554 19291 20340 16284 18542 19085 174192 16275 18531 18911 1800 26856 17242 19727 19834 22369 17234 19718 19630 191643 17227 19709 19457 2000 29271 18223 20931 20345 24387 18217 20924 20141 208987 18210 20917 19969 P 800 psia 51827F P 1000 psia 54465F P 1250 psia 57245F Sat 056920 11150 11993 14162 044604 11101 11926 13906 034549 11020 11819 13623 550 061586 11394 12305 14476 045375 11152 11992 13972 600 067799 11705 12709 14866 051431 11541 12493 14457 037894 11295 12172 13961 650 073279 11976 13060 15191 056411 11851 12895 14827 042703 11675 12663 14414 700 078330 12224 13384 15476 060844 12124 13250 15140 046735 11987 13068 14771 750 083102 12460 13691 15735 064944 12376 13578 15418 050344 12264 13429 15076 800 087678 12689 13987 15975 068821 12617 13890 15670 053687 12522 13764 15347 900 096434 13133 14560 16413 076136 13077 14486 16126 059876 13005 14390 15826 1000 104841 13570 15122 16812 083078 13525 15062 16535 065656 13467 14986 16249 1100 113024 14007 15680 17181 089783 13969 15631 16911 071184 13922 15568 16635 1200 121051 14446 16238 17528 096327 14414 16197 17263 076545 14374 16145 16993 1400 136797 15342 17367 18170 109101 15318 17337 17911 086944 15287 17298 17649 1600 152283 16265 18519 18759 121610 16246 18496 18504 097072 16222 18467 18246 1800 167606 17219 19700 19306 133956 17203 19682 19053 107036 17184 19660 18799 2000 182823 18204 20910 19819 146194 18191 20896 19568 116892 18175 20879 19315 Continued Final PDF to printer 945 APPENDIX 2 cen22672app02931972indd 945 110617 0920 AM TABLE A6E Superheated water T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 1500 psia 59626F P 1750 psia 61717F P 2000 psia 63585F Sat 027695 10921 11690 13362 022681 10805 11539 13112 018815 10668 11364 12863 600 028189 10972 11754 13423 650 033310 11472 12397 14016 026292 11228 12079 13607 020586 10914 11676 13146 700 037198 11836 12869 14433 030252 11668 12647 14108 024894 11476 12398 13783 750 040535 12144 13269 14771 033455 12015 13098 14489 028074 11874 12913 14218 800 043550 12422 13631 15064 036266 12317 13491 14807 030763 12205 13343 14567 850 046356 12682 13969 15328 038835 12593 13851 15088 033169 12500 13728 14867 900 049015 12931 14292 15569 041238 12854 14190 15341 035390 12775 14085 15134 1000 054031 13409 14908 16007 045719 13349 14829 15796 039479 13287 14749 15606 1100 058781 13873 15505 16402 049917 13824 15441 16201 043266 13775 15376 16021 1200 063355 14333 16092 16767 053932 14292 16039 16572 046864 14251 15985 16400 1400 072172 15257 17260 17432 061621 15226 17221 17245 053708 15195 17183 17081 1600 080714 16198 18438 18033 069031 16174 18409 17852 060269 16150 18380 17693 1800 089090 17164 19637 18589 076273 17145 19615 18410 066660 17125 19592 18255 2000 097358 18159 20861 19108 083406 18142 20843 18931 072942 18126 20826 18778 P 2500 psia 66817F P 3000 psia 69541F P 3500 psia Sat 013076 10312 10917 12330 008460 9698 10168 11587 650 002492 6637 6799 08632 700 016849 10984 11763 13072 009838 10053 10599 11960 003065 7600 7799 09511 750 020327 11549 12490 13686 014840 11141 11965 13118 010460 10576 11254 12434 800 022949 11959 13020 14116 017601 11675 12653 13676 013639 11343 12226 13224 850 025174 12301 13466 14463 019771 12082 13179 14086 015847 11838 12865 13721 900 027165 12607 13864 14761 021640 12428 13629 14423 017659 12234 13378 14106 950 029001 12891 14233 15028 023321 12739 14033 14716 019245 12578 13824 14428 1000 030726 13161 14582 15271 024876 13028 14409 14978 020687 12890 14230 14711 1100 033949 13673 15244 15710 027732 13568 15108 15441 023289 13461 14969 15201 1200 036966 14166 15876 16103 030367 14080 15766 15850 025654 13993 15654 15627 1400 042631 15133 17105 16802 035249 15070 17027 16567 029978 15007 16948 16364 1600 048004 16101 18322 17424 039830 16053 18264 17199 033994 16004 18205 17006 1800 053205 17086 19548 17991 044237 17047 19503 17773 037833 17008 19458 17586 2000 058295 18094 20791 18518 048532 18061 20756 18304 041561 18029 20721 18121 P 4000 psia P 5000 psia P 6000 psia 650 002448 6579 6761 08577 002379 6483 6703 08485 002325 6403 6661 08408 700 002871 7423 7636 09347 002678 7218 7466 09156 002564 7081 7365 09028 750 006370 9621 10092 11410 003373 8218 8530 10054 002981 7887 8218 09747 800 010520 10942 11721 12734 005937 9869 10418 11581 003949 8971 9410 10711 850 012848 11567 12518 13355 008551 10924 11715 12593 005815 10186 10831 11819 900 014647 12025 13109 13799 010390 11559 12521 13198 007584 11035 11877 12603 950 016176 12407 13605 14157 011863 12039 13136 13643 009010 11637 12637 13153 1000 017538 12746 14044 14463 013128 12440 13655 14004 010208 12114 13247 13578 1100 019957 13351 14828 14983 015298 13122 14538 14590 012211 12884 14240 14237 1200 022121 13903 15541 15426 017185 13721 15311 15070 013911 13534 15078 14758 1300 024128 14430 16216 15821 018902 14278 16027 15490 015434 14125 15838 15203 1400 026028 14943 16870 16182 020508 14814 16711 15868 016841 14684 16554 15598 1600 029620 15955 18147 16835 023505 15856 18031 16542 019438 15757 17915 16294 1800 033033 16968 19414 17422 026320 16890 19325 17142 021853 16811 19237 16907 2000 036335 17997 20686 17961 029023 17932 20617 17689 024155 17867 20549 17463 Concluded Final PDF to printer 946 PROPERTY TABLES AND CHARTS cen22672app02931972indd 946 110617 0920 AM TABLE A7E Compressed liquid water T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 500 psia 46704F P 1000 psia 54465F P 1500 psia 59626F Sat 0019750 44768 44951 064900 0021595 53858 54257 074341 0023456 60507 61158 080836 32 0015994 001 149 000001 0015966 003 299 000005 0015939 005 448 000008 50 0015998 1803 1951 003601 0015972 1799 2095 003593 0015946 1795 2238 003584 100 0016107 6786 6935 012930 0016083 6769 7067 012899 0016059 6753 7198 012869 150 0016317 11770 11921 021462 0016292 11742 12043 021416 0016267 11714 12166 021369 200 0016607 16770 16924 029349 0016580 16731 17038 029289 0016553 16692 17152 029229 250 0016972 21804 21961 036708 0016941 21751 22065 036634 0016911 21700 22169 036560 300 0017417 26892 27053 043641 0017380 26824 27146 043551 0017345 26757 27239 043463 350 0017954 32064 32230 050240 0017910 31977 32308 050132 0017866 31891 32387 050025 400 0018609 37361 37533 056595 0018552 37248 37591 056463 0018496 37137 37651 056333 450 0019425 42844 43024 062802 0019347 42693 43051 062635 0019271 42547 43082 062472 500 0020368 48403 48780 068764 0020258 48201 48763 068550 550 0021595 54250 54850 074731 P 2000 psia 63585F P 3000 psia 69541F P 5000 psia Sat 0025634 66233 67182 086224 0034335 78339 80245 097321 32 0015912 007 596 000010 0015859 010 890 000011 0015756 013 1471 000002 50 0015921 1791 2380 003574 0015870 1783 2664 003554 0015773 1765 3225 003505 100 0016035 6736 7330 012838 0015988 6704 7591 012776 0015897 6641 8112 012652 200 0016527 16654 17266 029170 0016475 16579 17494 029053 0016375 16436 17951 028824 300 0017310 26692 27333 043376 0017242 26565 27522 043204 0017112 26324 27907 042874 400 0018442 37030 37712 056205 0018338 36822 37841 055959 0018145 36435 38114 055492 450 0019199 42406 43116 062314 0019062 42136 43194 062010 0018812 41640 43380 061445 500 0020154 48008 48754 068346 0019960 47645 48753 067958 0019620 46994 48810 067254 560 0021739 55221 56026 075692 0021405 54659 55847 075126 0020862 53708 55638 074154 600 0023317 60577 61440 080898 0022759 59742 61006 080086 0021943 58442 60472 078803 640 0024765 65452 66827 085476 0023358 63495 65656 083603 680 0028821 72863 74464 092288 0025366 69067 71414 088745 700 0026777 72178 74656 091564 Final PDF to printer 947 APPENDIX 2 cen22672app02931972indd 947 110617 0920 AM TABLE A8E Saturated icewater vapor Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Temp T F Sat press Psat psia Sat ice vi Sat vapor vg Sat ice ui Subl uig Sat vapor ug Sat ice hi Subl hig Sat vapor hg Sat ice si Subl sig Sat vapor sg 32018 008871 001747 32996 14334 11642 10209 14334 12183 10750 029146 24779 21864 32 008864 001747 33026 14335 11642 10209 14335 12184 10750 029148 24779 21865 30 008086 001747 36058 14435 11646 10202 14435 12185 10742 029353 24883 21948 25 006405 001746 45058 14685 11654 10186 14685 12188 10720 029865 25146 22160 20 005049 001746 56576 14932 11662 10169 14932 12191 10698 030377 25414 22376 15 003960 001745 71389 15176 11670 10152 15176 12193 10676 030889 25687 22598 10 003089 001744 90540 15418 11678 10136 15418 12195 10654 031401 25965 22825 5 002397 001743 11543 15657 11685 10119 15657 12197 10631 031913 26248 23057 0 001850 001743 14797 15894 11692 10103 15894 12199 10609 032426 26537 23295 5 001420 001742 19075 16128 11699 10086 16128 12200 10587 032938 26832 23538 10 001083 001741 24731 16360 11706 10070 16360 12201 10565 033451 27133 23788 15 000821 001740 32257 16590 11712 10053 16590 12202 10543 033964 27440 24044 20 000619 001740 42335 16816 11718 10036 16816 12203 10521 034478 27754 24306 25 000463 001739 55917 17041 11724 10020 17041 12203 10499 034991 28074 24575 30 000344 001738 74345 17263 11730 10003 17263 12203 10477 035505 28401 24850 35 000254 001738 99526 17483 11735 9987 17483 12203 10455 036019 28735 25133 40 000186 001737 134182 17700 11740 9970 17700 12203 10433 036534 29076 25423 Final PDF to printer 948 PROPERTY TABLES AND CHARTS cen22672app02931972indd 948 110617 0920 AM FIGURE A9E Ts diagram for water Source of Data Joseph H Keenan Frederick G Keyes Philip G Hill and Joan G Moore Steam Tables New York John Wiley Sons 1969 100 60 32 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 24 20 15 10 05 00 12 40 1 22 0 1 2 0 0 1160 1140 1120 1100 12 60 12 80 130 0 Constant Enthalpy Btulbm 1320 1340 1360 1380 1400 1420 1440 1460 1480 1500 1520 1540 1560 1580 1600 1620 1640 1660 1680 1700 1720 1740 1760 1780 1220 1200 1180 12 40 1 22 0 1 2 0 0 1160 1140 1120 1100 12 60 12 80 130 0 1320 1340 1360 1380 1400 1420 1440 1460 1480 1500 1520 1540 1560 1580 1600 1620 1640 1660 1680 1700 1720 1740 1760 1780 1800 1800 1220 1200 1180 100 60 30 14696 10 6 4 2 1 05 02 01 200 300 400 600 10 00 15 00 2 5 0 0 100 60 30 14696 10 6 4 2 200 300 400 600 Constant Pressure lbfin 2 1000 15000 20 00 40 0 0 50 0 0 60 0 0 80 00 10000 3 0 0 0 32 04 4000 5000 6000 8000 10000 3000 3204 15000 2500 100 60 30 14696 10 6 4 2 1 05 02 01 200 300 400 600 10 00 15 00 2 5 0 0 100 60 30 14696 10 6 4 2 200 300 400 600 Constant Pressure lbfin 2 1000 15000 20 00 40 0 0 50 0 0 60 0 0 80 00 10000 3 0 0 0 32 04 4000 5000 6000 8000 10000 3000 3204 15000 2500 45 40 35 30 25 20 Constant Quality Percent 15 10 5 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 Constant Quality Percent 15 10 5 95 90 85 80 75 70 65 60 55 50 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 Constant Superh eat F 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000 1050 1100 1150 1200 1250 1300 Constant Superh eat F 1100 1050 1000 950 900 850 750 700 650 600 550 500 450 400 350 300 250 200 150 100 50 1100 1050 1000 950 900 850 800 Consta nt Enth alpy B tulbm 700 650 600 550 500 450 400 350 300 250 200 150 100 50 1150 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1100 1050 1000 950 900 850 750 700 650 600 550 500 450 400 350 300 250 200 150 100 50 1100 1050 1000 950 900 850 800 Consta nt Enth alpy B tulbm 700 650 600 550 500 450 400 350 300 250 200 150 100 50 1150 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 20 50 100 200 500 1000 10 5 Constant Volume ft3lbm 2 1 06 04 02 01 006053 003 20 50 100 200 500 1000 10 5 Constant Volume ft3lbm 2 1 06 04 02 01 005053 003 100 50 20 5 Constant Volume ft3lbm 2 1 06 04 02 01 005053 200 500 1000 2000 100 50 20 10 5 Constant Volume ft3lbm 2 1 06 04 02 01 005053 200 500 1000 2000 2000 2000 Entropy BtulbmR Temperature F Final PDF to printer 949 APPENDIX 2 cen22672app02931972indd 949 110617 0920 AM FIGURE A10E Mollier diagram for water Source of Data Joseph H Keenan Frederick G Keyes Philip G Hill and Joan G Moore Steam Tables New York John Wiley Sons 1969 1600 1550 1500 1450 1400 1350 1300 1250 1200 1150 1100 1050 1000 950 13 14 15 16 17 18 15 16 17 18 19 19 20 20 21 21 22 22 02 05 10 2 psia 3 4 5 6 8 10 20 50 psia 100 200 300 500 1000 2000 3204 5000 10000 psia x 098 saturation line 096 094 092 090 088 086 084 082 080 100 200F 300 400 500 600 700F 800 900 1000 1100F 02 05 10 2 psia 3 4 5 6 8 10 20 50 psia 100 200 300 500 1000 2000 3204 5000 10000 psia x 098 saturation line 096 094 092 090 088 086 084 082 080 100 200F 300 400 500 600 700F 800 900 1000 1100F Entropy BtulbmR Entropy BtulbmR Enthalpy Btulbm Final PDF to printer 950 PROPERTY TABLES AND CHARTS cen22672app02931972indd 950 110617 0920 AM TABLE A11E Saturated refrigerant134aTemperature table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtulbmR Temp T F Sat press Psat psia Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 40 7432 001130 57769 0016 89174 8916 0000 97104 9710 000000 023136 023136 35 8581 001136 50489 1483 88360 8984 1501 96360 9786 000355 022689 023044 30 9869 001143 44286 2987 87542 9053 3008 95608 9862 000707 022250 022957 25 11306 001149 38980 4497 86717 9121 4522 94849 9937 001057 021819 022876 20 12906 001156 34424 6014 85887 9190 6041 94080 10012 001404 021396 022800 15 14680 001163 30495 7536 85050 9259 7568 93303 10087 001748 020981 022729 10 16642 001170 27097 9065 84206 9327 9102 92515 10162 002090 020572 022662 5 18806 001178 24146 10601 83355 9396 10642 91717 10236 002430 020171 022600 0 21185 001185 21575 12143 82496 9464 12190 90907 10310 002767 019775 022542 5 23793 001193 19328 13693 81628 9532 13745 90085 10383 003103 019385 022488 10 26646 001200 17358 15249 80751 9600 15308 89251 10456 003436 019001 022437 15 29759 001208 15625 16813 79865 9668 16879 88403 10528 003767 018623 022390 20 33147 001216 14097 18384 78969 9735 18459 87541 10600 004097 018249 022345 25 36826 001225 12746 19963 78062 9803 20047 86665 10671 004424 017880 022304 30 40813 001233 11548 21550 77144 9869 21643 85772 10742 004750 017515 022265 35 45124 001242 10482 23145 76214 9936 23249 84863 10811 005074 017154 022228 40 49776 001251 095323 24749 75272 10002 24864 83937 10880 005397 016797 022194 45 54787 001261 086837 26361 74317 10068 26489 82993 10948 005718 016443 022162 50 60175 001270 079236 27983 73347 10133 28124 82029 11015 006038 016093 022131 55 65957 001280 072414 29614 72363 10198 29770 81046 11082 006357 015746 022103 60 72152 001290 066277 31254 71364 10262 31426 80041 11147 006674 015401 022075 65 78780 001301 060744 32904 70348 10325 33094 79014 11211 006991 015058 022049 70 85858 001311 055746 34565 69315 10388 34773 77964 11274 007306 014718 022024 75 93408 001323 051222 36237 68264 10450 36465 76889 11335 007621 014379 022000 80 10145 001334 047119 37920 67193 10511 38170 75788 11396 007934 014042 021976 85 11000 001346 043391 39614 66102 10572 39888 74660 11455 008247 013706 021953 90 11908 001359 039997 41321 64989 10631 41620 73503 11512 008560 013371 021931 95 12872 001372 036902 43041 63852 10689 43367 72315 11568 008872 013036 021908 100 13893 001386 034074 44774 62690 10746 45130 71094 11622 009183 012702 021885 105 14973 001400 031486 46521 61501 10802 46909 69838 11675 009495 012367 021862 110 16116 001415 029113 48284 60284 10857 48706 68544 11725 009806 012031 021838 115 17323 001430 026933 50063 59035 10910 50521 67210 11773 010118 011694 021813 120 18596 001446 024928 51858 57753 10961 52356 65833 11819 010430 011356 021786 130 21353 001482 021373 55505 55075 11058 56091 62935 11903 011056 010672 021728 140 24406 001522 018331 59237 52221 11146 59925 59813 11974 011686 009973 021660 150 27779 001567 015707 63070 49151 11222 63875 56419 12029 012324 009253 021577 160 31494 001619 013423 67022 45811 11283 67965 52690 12066 012971 008502 021473 170 35580 001682 011413 71139 42101 11324 72246 48509 12075 013637 007703 021340 180 40066 001759 009619 75464 37893 11336 76768 43721 12049 014327 006834 021161 190 44990 001861 007982 80093 32929 11302 81642 38025 11967 015057 005852 020909 200 50400 002010 006441 85297 26629 11193 87172 30761 11793 015872 004662 020534 210 56376 002309 004722 91993 16498 10849 94402 19015 11342 016924 002839 019763 Source of Data Tables A11E through A13E are generated using the Engineering Equation Solver EES software developed by S A Klein and F L Alvarado The routine used in calculations is the R134a which is based on the fundamental equation of state developed by R TillnerRoth and HD Baehr An International Standard Formulation for the Thermodynamic Properties of 1112Tetrafluoroethane HFC134a for temperatures from 170 K to 455 K and pressures up to 70 MPa J Phys Chem Ref Data Vol 23 No 5 1994 The enthalpy and entropy values of saturated liquid are set to zero at 40C and 40F Final PDF to printer 951 APPENDIX 2 cen22672app02931972indd 951 110617 0920 AM TABLE A12E Saturated refrigerant134aPressure table Specific volume ft3lbm Internal energy Btulbm Enthalpy Btulbm Entropy BtuIbmR Press P psia Sat temp Tsat F Sat liquid vf Sat vapor vg Sat liquid uf Evap ufg Sat vapor ug Sat liquid hf Evap hfg Sat vapor hg Sat liquid sf Evap sfg Sat vapor sg 5 5309 001113 83740 3914 91283 8737 3903 99021 9512 000944 024353 023409 10 2952 001143 43740 3132 87463 9059 3153 95536 9869 000741 022208 022949 15 1415 001164 29882 7796 84907 9270 7828 93170 10100 001806 020911 022717 20 243 001181 22781 11393 82915 9431 11436 91302 10274 002603 019967 022570 25 717 001196 18442 14367 81249 9562 14422 89725 10415 003247 019218 022465 30 1537 001209 15506 16929 79799 9673 16996 88340 10534 003792 018595 022386 35 2257 001221 13382 19195 78504 9770 19274 87093 10637 004265 018058 022324 40 2901 001232 11773 21236 77326 9856 21327 85950 10728 004686 017586 022272 45 3486 001242 10510 23101 76240 9934 23205 84889 10809 005065 017164 022229 50 4023 001252 094909 24824 75228 10005 24939 83894 10883 005412 016780 022192 55 4520 001261 086509 26428 74277 10070 26556 82954 10951 005732 016429 022160 60 4984 001270 079462 27932 73378 10131 28073 82060 11013 006028 016104 022132 65 5420 001278 073462 29351 72523 10187 29505 81205 11071 006306 015801 022107 70 5830 001287 068290 30696 71705 10240 30862 80385 11125 006567 015518 022084 75 6219 001295 063784 31975 70921 10290 32155 79594 11175 006813 015251 022064 80 6589 001303 059822 33198 70167 10336 33391 78830 11222 007047 014998 022045 85 6941 001310 056309 34369 69438 10381 34575 78089 11266 007269 014758 022027 90 7278 001318 053173 35494 68733 10423 35713 77369 11308 007481 014529 022011 95 7602 001325 050356 36577 68048 10463 36810 76668 11348 007684 014311 021995 100 7912 001332 047811 37623 67383 10501 37870 75984 11385 007879 014101 021981 110 8500 001346 043390 39614 66102 10572 39888 74660 11455 008247 013706 021953 120 9049 001360 039681 41489 64878 10637 41791 73388 11518 008590 013338 021928 130 9564 001374 036523 43263 63704 10697 43594 72159 11575 008912 012993 021905 140 10051 001387 033800 44951 62570 10752 45311 70967 11628 009215 012668 021883 150 10512 001400 031426 46563 61473 10804 46952 69807 11676 009502 012359 021861 160 10950 001413 029339 48109 60406 10851 48527 68674 11720 009776 012064 021840 170 11369 001426 027487 49595 59366 10896 50043 67564 11761 010036 011783 021819 180 11769 001439 025833 51027 58349 10938 51507 66475 11798 010286 011513 021799 190 12153 001452 024346 52412 57353 10976 52922 65402 11832 010526 011252 021778 200 12522 001464 023001 53753 56375 11013 54295 64345 11864 010757 011000 021757 220 13221 001490 020662 56321 54462 11078 56927 62267 11919 011195 010519 021714 240 13873 001516 018694 58757 52596 11135 59430 60225 11965 011606 010063 021669 260 14485 001543 017012 61082 50763 11184 61824 58205 12003 011994 009627 021622 280 15062 001570 015555 63313 48951 11226 64126 56197 12032 012364 009207 021571 300 15609 001598 014279 65460 47154 11261 66347 54195 12054 012717 008800 021517 350 16864 001672 011673 70567 42632 11320 71651 49109 12076 013545 007815 021360 400 17986 001758 009643 75401 37957 11336 76702 43794 12050 014317 006847 021164 450 19002 001860 007979 80112 32909 11302 81662 38003 11967 015060 005849 020909 500 19929 001997 006533 84900 27096 11200 86748 31292 11804 015810 004748 020558 Final PDF to printer 952 PROPERTY TABLES AND CHARTS cen22672app02931972indd 952 110617 0920 AM TABLE A13E Superheated refrigerant134a T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 10 psia Tsat 2952F P 15 psia Tsat 1415F P 20 psia Tsat 243F Sat 43740 9059 9869 022949 29882 9270 10100 022717 22781 9431 10274 022570 20 44856 9214 10044 023351 0 47135 9542 10414 024175 31001 9508 10369 023312 22922 9473 10321 022673 20 49380 9877 10791 024978 32551 9849 10752 024129 24130 9819 10712 023506 40 51600 10221 11176 025763 34074 10196 11142 024924 25306 10171 11107 024313 60 53802 10573 11568 026533 35577 10551 11538 025702 26461 10529 11508 025099 80 55989 10933 11969 027290 37064 10914 11942 026465 27600 10894 11915 025868 100 58165 11302 12378 028035 38540 11285 12354 027214 28726 11267 12330 026623 120 60331 11680 12796 028768 40006 11664 12775 027952 29842 11648 12753 027364 140 62490 12066 13223 029492 41464 12052 13203 028678 30950 12038 13183 028094 160 64642 12462 13658 030205 42915 12449 13640 029395 32051 12435 13622 028814 180 66789 12866 14101 030910 44361 12853 14085 030102 33146 12841 14068 029523 200 68930 13278 14554 031606 45802 13267 14538 030800 34237 13256 14523 030223 220 71068 13699 15014 032293 47239 13689 15000 031489 35324 13678 14986 030914 P 30 psia Tsat 1537F P 40 psia Tsat 2901F P 50 psia Tsat 4023F Sat 15506 9673 10534 022386 11773 9856 10728 022272 09491 10005 10883 022192 20 15691 9756 10627 022583 40 16528 10118 11035 023416 12126 10061 10959 022740 60 17338 10483 11445 024220 12768 10435 11380 023567 10019 10385 11312 023033 80 18130 10854 11860 025003 13389 10812 11803 024365 10540 10769 11744 023849 100 18908 11231 12281 025769 13995 11194 12230 025142 11043 11156 12178 024639 120 19675 11616 12708 026519 14588 11583 12663 025902 11534 11549 12616 025408 140 20434 12008 13143 027256 15173 11979 13102 026646 12015 11948 13060 026160 160 21185 12409 13585 027981 15750 12382 13547 027377 12488 12354 13509 026898 180 21931 12817 14034 028695 16321 12792 14000 028096 12955 12767 13966 027622 200 22671 13233 14492 029399 16887 13210 14460 028805 13416 13187 14429 028335 220 23408 13658 14957 030094 17449 13637 14928 029503 13873 13615 14899 029037 240 24141 14090 15430 030780 18007 14070 15403 030192 14326 14051 15376 029730 260 24871 14530 15911 031458 18562 14512 15886 030873 14776 14494 15861 030413 280 25598 14979 16400 032128 19114 14962 16377 031545 15223 14945 16353 031087 P 60 psia Tsat 4984F P 70 psia Tsat 5830F P 80 psia Tsat 6589F Sat 07946 10131 11013 022132 06829 10240 11125 022084 05982 10336 11222 022045 60 08179 10331 11239 022572 06857 10274 11162 022157 80 08636 10724 11682 023408 07271 10677 11618 023018 06243 10627 11551 022663 100 09072 11117 12124 024212 07662 11077 12069 023838 06601 11035 12012 023501 120 09495 11514 12569 024992 08037 11479 12520 024630 06941 11443 12470 024305 140 09908 11917 13017 025753 08401 11886 12974 025399 07270 11853 12929 025084 160 10312 12326 13471 026497 08756 12298 13432 026151 07589 12269 13392 025843 180 10709 12742 13931 027227 09105 12716 13895 026886 07900 12689 13859 026585 200 11101 13164 14397 027945 09447 13140 14364 027608 08206 13117 14331 027312 220 11489 13594 14869 028651 09785 13572 14840 028318 08507 13550 14809 028026 240 11872 14031 15349 029346 10118 14011 15322 029017 08803 13991 15294 028728 260 12252 14476 15836 030032 10449 14457 15810 029706 09096 14438 15785 029420 280 12629 14928 16330 030709 10776 14910 16306 030386 09386 14893 16282 030102 300 13004 15388 16831 031378 11101 15371 16809 031057 09674 15355 16787 030775 320 13377 15855 17340 032039 11424 15840 17320 031720 09959 15825 17299 031440 Final PDF to printer 953 APPENDIX 2 cen22672app02931972indd 953 110617 0920 AM TABLE A13E Superheated refrigerant134a T F v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR v ft3lbm u Btulbm h Btulbm s BtulbmR P 90 psia Tsat 7278F P 100 psia Tsat 7912F P 120 psia Tsat 9049F Sat 053173 10423 11308 022011 047811 10501 11385 021981 039681 10637 11518 021928 80 054388 10575 11481 022332 047906 10519 11406 022018 100 057729 10991 11953 023191 051076 10946 11891 022902 041013 10849 11759 022364 120 060874 11405 12419 024009 054022 11366 12366 023735 043692 11285 12255 023234 140 063885 11820 12884 024799 056821 11786 12838 024535 046190 11716 12742 024059 160 066796 12239 13351 025565 059513 12209 13310 025310 048563 12147 13225 024853 180 069629 12663 13822 026313 062122 12636 13785 026065 050844 12580 13709 025621 200 072399 13092 14298 027045 064667 13068 14264 026802 053054 13018 14196 026370 220 075119 13528 14779 027763 067158 13505 14748 027525 055206 13460 14686 027102 240 077796 13970 15266 028469 069605 13950 15238 028234 057312 13908 15180 027819 260 080437 14419 15759 029164 072016 14400 15733 028932 059379 14362 15680 028523 280 083048 14875 16258 029849 074396 14858 16234 029620 061413 14822 16186 029216 300 085633 15339 16765 030524 076749 15322 16742 030297 063420 15289 16697 029898 320 088195 15809 17278 031191 079079 15794 17257 030966 065402 15762 17215 030571 P 140 psia Tsat 10051F P 160 psia Tsat 10950F P 180 psia Tsat 11769F Sat 033800 10752 11628 021883 029339 10851 11720 021840 025833 10938 11798 021799 120 036243 11197 12136 022775 030578 11101 12007 022339 026083 10995 11864 021912 140 038551 11642 12640 023630 032774 11563 12533 023232 028231 11478 12418 022852 160 040711 12082 13137 024444 034790 12014 13044 024070 030154 11943 12947 023720 180 042766 12523 13631 025229 036686 12463 13549 024872 031936 12401 13465 024542 200 044743 12966 14125 025990 038494 12913 14052 025647 033619 12858 13977 025332 220 046657 13413 14622 026731 040234 13365 14556 026399 035228 13316 14489 026095 240 048522 13865 15122 027457 041921 13821 15062 027133 036779 13776 15001 026838 260 050345 14322 15626 028168 043564 14282 15572 027851 038284 14241 15516 027564 280 052134 14785 16136 028866 045171 14748 16086 028555 039751 14711 16035 028275 300 053895 15255 16651 029553 046748 15221 16605 029248 041186 15186 16558 028972 320 055630 15731 17172 030230 048299 15699 17129 029929 042594 15667 17085 029658 340 057345 16214 17699 030898 049828 16184 17659 030600 043980 16153 17618 030333 360 059041 16703 18233 031557 051338 16675 18195 031262 045347 16647 18157 030998 P 200 psia Tsat 12522F P 300 psia Tsat 15609F P 400 psia Tsat 17986F Sat 023001 11013 11864 021757 014279 11261 12054 021517 009643 11336 12050 021164 140 024541 11386 12294 022483 160 026412 11867 12844 023386 014656 11382 12196 021747 180 028115 12336 13377 024231 016355 11953 12861 022803 009658 11342 12056 021174 200 029704 12801 13900 025037 017776 12479 13466 023734 011440 12053 12899 022473 220 031212 13265 14420 025813 019044 12986 14043 024596 012746 12645 13588 023502 240 032658 13731 14939 026566 020211 13483 14605 025412 013853 13196 14221 024420 260 034054 14199 15460 027300 021306 13977 15160 026193 014844 13727 14826 025272 280 035410 14673 15983 028017 022347 14471 15711 026949 015756 14248 15415 026079 300 036733 15150 16510 028720 023346 14966 16262 027683 016611 14765 15995 026853 320 038029 15634 17041 029410 024310 15463 16813 028399 017423 15281 16571 027601 340 039300 16123 17577 030089 025246 15965 17366 029100 018201 15797 17145 028328 360 040552 16618 18119 030758 026159 16471 17923 029788 018951 16316 17719 029037 Concluded Final PDF to printer 954 PROPERTY TABLES AND CHARTS cen22672app02931972indd 954 110617 0920 AM FIGURE A14E Ph diagram for refrigerant134a Reprinted by permission of American Society of Heating Refrigerating and AirConditioning Engineers Inc Atlanta GA 620 580 540 500 460 420 380 340 T 300 F 260 220 180 140 100 60 20 20 60 2 2 0 18 0 140 T 100 F 60 20 20 60 620 580 540 500 460 420 380 340 T 300 F 260 220 180 140 100 60 20 20 60 2 2 0 18 0 140 T 100F 60 20 20 60 T 100F T 100F 95 24 30 35 40 90 85 80 75 lbmft 3 70 6 5 60 55 50 45 Density 18 lb mft3 0012 0018 0024 0072 0096 012 018 024 036 048 072 096 12 18 24 36 48 72 96 12 95 24 30 35 40 90 85 80 75 lbmft 3 70 6 5 60 55 50 45 Density 18 lb mft3 0012 0018 0024 0036 0048 0036 0048 0072 0096 012 018 024 036 048 072 096 12 18 24 36 48 72 96 12 032 s 030 BtulbmR 028 026 024 022 020 018 016 014 012 010 008 006 004 002 000 002 004 034 036 038 040 042 044 046 s 042 032 s 030 BtulbmR 028 026 024 022 020 018 016 014 012 010 008 006 004 002 000 002 004 034 036 038 040 042 044 046 s 042 saturated liquid 01 02 03 04 05 06 07 08 09 saturated vapour saturated liquid 01 02 03 04 05 06 07 08 09 saturated vapour 20 20 140 60 100 180 220 260 20 20 140 60 100 180 220 260 Enthalpy Btulbm 1 2 4 10 20 40 100 200 400 1000 2000 3000 1 2 4 10 20 40 100 200 400 1000 2000 3000 Pressure psia R134a R134a Final PDF to printer 955 APPENDIX 2 cen22672app02931972indd 955 110617 0920 AM TABLE A16E Properties of the atmosphere at high altitude Altitude ft Temperature F Pressure psia Gravity g fts2 Speed of sound fts Density lbmft3 Viscosity μ lbmfts Thermal conductivity BtuhftR 0 5900 147 32174 1116 007647 1202 105 00146 500 5722 144 32173 1115 007536 1199 105 00146 1000 5543 142 32171 1113 007426 1196 105 00146 1500 5365 139 32169 1111 007317 1193 105 00145 2000 5187 137 32168 1109 007210 1190 105 00145 2500 5009 134 32166 1107 007104 1186 105 00144 3000 4830 132 32165 1105 006998 1183 105 00144 3500 4652 129 32163 1103 006985 1180 105 00143 4000 4474 127 32162 1101 006792 1177 105 00143 4500 4296 125 32160 1099 006690 1173 105 00142 5000 4117 122 32159 1097 006590 1170 105 00142 5500 3939 120 32157 1095 006491 1167 105 00141 6000 3761 118 32156 1093 006393 1164 105 00141 6500 3583 116 32154 1091 006296 1160 105 00141 7000 3405 113 32152 1089 006200 1157 105 00140 7500 3226 111 32151 1087 006105 1154 105 00140 8000 3048 109 32149 1085 006012 1150 105 00139 8500 2870 107 32148 1083 005919 1147 105 00139 9000 2692 105 32146 1081 005828 1144 105 00138 9500 2514 103 32145 1079 005738 1140 105 00138 10000 2336 101 32145 1077 005648 1137 105 00137 11000 1979 972 32140 1073 005473 1130 105 00136 12000 1623 934 32137 1069 005302 1124 105 00136 13000 1267 899 32134 1065 005135 1117 105 00135 14000 912 863 32131 1061 004973 1110 105 00134 15000 555 829 32128 1057 004814 1104 105 00133 16000 199 797 32125 1053 004659 1097 105 00132 17000 158 765 32122 1049 004508 1090 105 00132 18000 514 734 32119 1045 004361 1083 105 00130 19000 870 705 32115 1041 004217 1076 105 00129 20000 122 676 32112 1037 004077 1070 105 00128 22000 194 621 32106 1029 003808 1056 105 00126 24000 265 570 32100 1020 003553 1042 105 00124 26000 336 522 32094 1012 003311 1028 105 00122 28000 407 478 32088 1003 003082 1014 105 00121 30000 478 437 32082 995 002866 1000 105 00119 32000 549 399 3208 987 002661 0986 105 00117 34000 620 363 3207 978 002468 0971 105 00115 36000 692 330 3206 969 002285 0956 105 00113 38000 697 305 3206 968 002079 0955 105 00113 40000 697 273 3205 968 001890 0955 105 00113 45000 697 2148 3204 968 001487 0955 105 00113 50000 697 1691 3202 968 001171 0955 105 00113 55000 697 1332 3200 968 000922 0955 105 00113 60000 697 1048 3199 968 000726 0955 105 00113 Source of Data US Standard Atmosphere Supplements US Government Printing Office 1966 Based on yearround mean conditions at 45 latitude and varies with the time of the year and the weather patterns The conditions at sea level z 0 are taken to be P 14696 psia T 59F ρ 0076474 lbmft3 g 321741 ft2s Final PDF to printer 956 PROPERTY TABLES AND CHARTS cen22672app02931972indd 956 110617 0920 AM TABLE A17E Idealgas properties of air T R h Btulbm Pr u Btulbm vr s BtulbmR T R h Btulbm Pr u Btulbm vr s BtulbmR 360 8597 03363 6129 3966 050369 1600 39574 7113 28606 8263 087130 380 9075 04061 6470 3466 051663 1650 40913 8089 29603 7556 087954 400 9553 04858 6811 3050 052890 1700 42259 9095 30606 6924 088758 420 10032 05760 7152 2701 054058 1750 43612 10198 31616 6357 089542 440 10511 06776 7493 2406 055172 1800 44971 1140 32632 5847 090308 460 10990 07913 7836 21533 056235 1850 46337 1272 33655 5388 091056 480 11469 09182 8177 19365 057255 1900 47709 1415 34685 4974 091788 500 11948 10590 8520 17490 058233 1950 49088 1571 35720 4598 092504 520 12427 12147 8862 15858 059173 2000 50471 1740 36761 4258 093205 537 12810 13593 9153 14634 059945 2050 51871 1923 37808 3949 093891 540 12906 13860 9204 14432 060078 2100 53255 2121 38860 3667 094564 560 13386 15742 9547 13178 060950 2150 54654 2235 39917 3410 095222 580 13866 17800 9890 12070 061793 2200 56059 2566 40978 3176 095919 600 14347 2005 10234 11088 062607 2250 57469 2814 42046 2961 096501 620 14828 2249 10578 10212 063395 2300 58882 3081 43116 2765 097123 640 15309 2514 10921 9430 064159 2350 60300 3368 44191 2585 097732 660 15792 2801 11267 8727 064902 2400 61722 3676 45270 2419 098331 680 16273 3111 11612 8096 065621 2450 63148 4005 46354 2266 098919 700 16756 3446 11958 7525 066321 2500 64578 4357 47440 2125 099497 720 17239 3806 12304 7007 067002 2550 66012 4733 48531 1996 100064 740 17723 4193 12651 6538 067665 2600 67449 5135 49626 1876 100623 760 18208 4607 12999 6110 068312 2650 68890 5563 50725 1765 101172 780 18694 5051 13347 5720 068942 2700 70335 6019 51826 1662 101712 800 19181 5526 13697 5363 069558 2750 71783 6504 52931 1566 102244 820 19669 6033 14047 5035 070160 2800 73233 7020 54040 1478 102767 840 20156 6573 14398 4734 070747 2850 74688 7567 55152 1395 103282 860 20646 7149 14750 4457 071323 2900 76145 8148 56266 1318 103788 880 21135 7761 15102 4201 071886 2950 77605 8764 57384 1247 104288 900 21626 8411 15457 3964 072438 3000 79068 9414 58504 1180 104779 920 22118 9102 15812 3744 072979 3050 80534 1011 59628 1118 105264 940 22611 9834 16168 3541 073509 3100 82003 1083 60753 1060 105741 960 23106 1061 16526 3352 074030 3150 83475 1161 61882 1006 106212 980 23602 1143 16883 3176 074540 3200 84948 1242 63012 0955 106676 1000 24098 1230 17243 3012 075042 3250 86424 1328 64146 0907 107134 1040 25095 1418 17966 2717 076019 3300 87902 1418 65281 08621 107585 1080 26097 1628 18693 2458 076964 3350 89383 1513 66420 08202 108031 1120 27103 1860 19425 2230 077880 3400 90866 1613 67560 07807 108470 1160 28114 2118 20163 2029 078767 3450 92352 1719 68704 07436 108904 1200 29130 2401 20905 1851 079628 3500 93840 1829 69848 07087 109332 1240 30152 2713 21653 1693 080466 3550 95330 1946 70995 06759 109755 1280 31179 3055 22405 1552 081280 3600 96821 2068 72144 06449 110172 1320 32211 3431 23163 1425 082075 3650 98315 2196 73295 06157 110584 1360 33248 3841 23925 1312 082848 3700 99811 2330 74448 05882 110991 1400 34290 4288 24693 1210 083604 3750 10131 2471 75604 05621 111393 1440 35337 4775 25466 1117 084341 3800 10281 2618 76760 05376 111791 1480 36389 5304 26244 1034 085062 3850 10431 2773 77919 05143 112183 1520 37447 5878 27026 9578 085767 3900 10581 2934 79080 04923 112571 1560 38508 6500 27813 8890 086456 3950 10732 3103 80243 04715 112955 Final PDF to printer 957 APPENDIX 2 cen22672app02931972indd 957 110617 0920 AM TABLE A17E Idealgas properties of air T R h Btulbm Pr u Btulbm vr s BtulbmR T R h Btulbm Pr u Btulbm vr s BtulbmR 4000 10883 3280 81406 04518 113334 4600 12704 6089 95504 02799 117575 4050 11034 3464 82572 04331 113709 4700 13009 6701 97873 02598 118232 4100 11185 3656 83740 04154 114079 4800 13315 7362 10025 02415 118876 4150 11336 3858 84909 03985 114446 4900 13622 8073 10263 02248 119508 4200 11487 4067 86081 03826 114809 5000 13929 8837 10501 02096 120129 4300 11790 4513 88428 03529 115522 5100 14236 9658 10740 01956 120738 4400 12094 4997 90781 03262 116221 5200 14544 10539 10980 01828 121336 4500 12399 5521 93139 03019 116905 5300 14853 11481 11220 01710 12192 Note The properties Pr relative pressure and vr relative specific volume are dimensionless quantities used in the analysis of isentropic processes and should not be confused with the properties pressure and specific volume Source of Data Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 pp 83233 Table A5 Originally published in J H Keenan and J Kaye Gas Tables New York John Wiley Sons 1948 Concluded Final PDF to printer 958 PROPERTY TABLES AND CHARTS cen22672app02931972indd 958 110617 0920 AM TABLE A18E Idealgas properties of nitrogen N2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 20820 14862 41695 1080 75510 54062 50651 320 22210 15855 42143 1100 76950 55105 50783 340 23600 16844 42564 1120 78393 56152 50912 360 24989 17840 42962 1140 79840 57201 51040 380 26380 18834 43337 1160 81290 58254 51167 400 27770 19826 43694 1180 82744 59310 51291 420 29161 20820 44034 1200 84200 60370 51143 440 30551 21813 44357 1220 85661 61434 51534 460 31941 22806 44665 1240 87126 62501 51653 480 33331 23799 44962 1260 88593 63572 51771 500 34722 24793 45246 1280 90064 64645 51887 520 36113 25786 45519 1300 91539 65723 51001 537 37295 26631 45743 1320 93018 66804 52114 540 37503 26780 45781 1340 94500 67889 52225 560 38895 27774 46034 1360 95986 68978 52335 580 40287 28769 46278 1380 97475 70070 52444 600 41679 29764 46514 1400 98969 71167 52551 620 43071 30759 46742 1420 100466 72267 52658 640 44464 31755 46964 1440 101966 73370 52763 660 45858 32752 47178 1460 103470 74476 52867 680 47253 33749 47386 1480 104978 75587 52969 700 48649 34748 47588 1500 106480 76701 53071 720 50045 35747 47785 1520 108004 77819 53171 740 51443 36747 47977 1540 109522 78939 53271 760 52841 37749 48164 1560 111043 80064 53369 780 54242 38752 48345 1580 112569 81192 53465 800 55644 39757 48522 1600 114097 82323 53561 820 57047 40763 48696 1620 115628 83457 53656 840 58453 41771 48865 1640 117164 84596 53751 860 59859 42781 49031 1660 118702 85736 53844 880 61269 43794 49193 1680 120243 86881 53936 900 62681 44808 49352 1700 121789 88029 54028 920 64096 45826 49507 1720 123337 89180 54118 940 65512 46845 49659 1740 124888 90334 54208 960 66931 47867 49808 1760 126443 91492 54297 980 68354 48893 49955 1780 128002 92653 54385 1000 69779 49920 50099 1800 129563 93817 54472 1020 71207 50951 50241 1820 131127 94984 54559 1040 72638 51985 50380 1840 132695 96155 54645 1060 74072 53022 50516 1860 134265 97328 54729 Final PDF to printer 959 APPENDIX 2 cen22672app02931972indd 959 110617 0920 AM TABLE A18E Idealgas properties of nitrogen N2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 1900 13742 9968 54896 3500 27016 20065 59944 1940 14058 10205 55061 3540 27359 20329 60041 1980 14375 10443 55223 3580 27703 20593 60138 2020 14694 10682 55383 3620 28046 20858 60234 2060 15013 10923 55540 3660 28391 21122 60328 2100 15334 11164 55694 3700 28735 21387 60422 2140 15656 11406 55846 3740 29080 21653 60515 2180 15978 11649 55995 3780 29425 21919 60607 2220 16302 11893 56141 3820 29771 22185 60698 2260 16626 12138 56286 3860 30117 22451 60788 2300 16951 12384 56429 3900 30463 22718 60877 2340 17277 12630 56570 3940 30809 22985 60966 2380 17604 12878 56708 3980 31156 23252 61053 2420 17392 13126 56845 4020 31503 23520 61139 2460 18260 13375 56980 4060 31850 23788 61225 2500 18590 13625 57112 4100 32198 24056 61310 2540 18919 13875 57243 4140 32546 24324 61395 2580 19250 14127 57372 4180 32894 24593 61479 2620 19582 14379 57499 4220 33242 24862 61562 2660 19914 14631 57625 4260 33591 25131 61644 2700 20246 14885 57750 4300 33940 25401 61726 2740 20580 15139 57872 4340 34289 25670 61806 2780 20914 15393 57993 4380 34638 25940 61887 2820 21248 15648 58113 4420 34988 26210 61966 2860 21584 15905 58231 4460 35338 26481 62045 2900 21920 16161 58348 4500 35688 26751 62123 2940 22256 16417 58463 4540 36038 27022 62201 2980 22593 16675 58576 4580 36389 27293 62278 3020 22930 16933 58688 4620 36739 27565 62354 3060 23268 17192 58800 4660 37090 27836 62429 3100 23607 17451 58910 4700 37441 28108 62504 3140 23946 17710 59019 4740 37792 28379 62578 3180 24285 17970 59126 4780 38144 28651 62652 3220 24625 18231 59232 4820 38495 28924 62725 3260 24965 18491 59338 4860 38847 29196 62798 3300 25306 18753 59442 4900 39199 29468 62870 3340 25647 19014 59544 5000 40080 30151 63049 3380 25989 19277 59646 5100 40962 30834 63223 3420 26331 19539 59747 5200 41844 31518 63395 3460 26673 19802 59846 5300 42728 32203 63563 Source of Data Tables A18E through A23E are adapted from Kenneth Wark Thermodynamics 4th ed New York McGrawHill 1983 pp 83444 Originally published in J H Keenan and J Kaye Gas Tables New York John Wiley Sons 1945 Concluded Final PDF to printer 960 PROPERTY TABLES AND CHARTS cen22672app02931972indd 960 110617 0920 AM TABLE A19E Idealgas properties of oxygen O2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 20735 14778 44927 1080 76968 55521 54064 320 22126 15771 45375 1100 78504 56659 54204 340 23517 16765 45797 1120 80045 57803 54343 360 24908 17759 46195 1140 81591 58952 54480 380 26300 18753 46571 1160 83142 60106 54614 400 27691 19748 46927 1180 84698 61265 54748 420 29083 20743 47267 1200 86258 62428 54879 440 30475 21738 47591 1220 87824 63596 55008 460 31869 22734 47900 1240 89394 64769 55136 480 33265 23733 48198 1260 90967 65945 55262 500 34662 24732 48483 1280 92546 67127 55386 520 36061 25734 48757 1300 94129 68313 55508 537 37251 26587 48982 1320 95719 69502 55630 540 37462 26738 49021 1340 97307 70696 55750 560 38866 27745 49276 1360 98902 71894 55867 580 40273 28755 49522 1380 100501 73096 55984 600 41683 29768 49762 1400 102104 74301 56099 620 43097 30784 49993 1420 103710 75511 56213 640 44514 31804 50218 1440 105320 76724 56326 660 45935 32829 50437 1460 106933 77939 56437 680 47362 33858 50650 1480 108551 79160 56547 700 48793 34892 50858 1500 110171 80383 56656 720 50229 35931 51059 1520 111796 81611 56763 740 51670 36974 51257 1540 113424 82842 56869 760 53114 38024 51450 1560 115054 84074 56975 780 54564 39075 51638 1580 116688 85311 57079 800 56020 40133 51821 1600 118325 86551 57182 820 57481 41197 52002 1620 119966 87795 57284 840 58948 42266 52179 1640 121609 89041 57385 860 60419 43341 52352 1660 123255 90290 57484 880 61896 44420 52522 1680 124904 91541 57582 900 63379 45506 52688 1700 126556 92796 57680 920 64867 46597 52852 1720 128211 94054 57777 940 66361 47694 53012 1740 129869 95315 57873 960 67860 48795 53170 1760 131530 96579 57968 980 69364 49903 53326 1780 133192 97844 58062 1000 70875 51016 53477 1800 134858 99112 58155 1020 72389 52133 53628 1820 136525 100382 58247 1040 73910 53257 53775 1840 138196 101656 58339 1060 75436 54386 53921 1860 139868 102931 58428 Final PDF to printer 961 APPENDIX 2 cen22672app02931972indd 961 110617 0920 AM TABLE A19E Idealgas properties of oxygen O2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 1900 14322 10549 58607 3500 28273 21323 63914 1940 14658 10806 58782 3540 28633 21603 64016 1980 14995 11063 58954 3580 28994 21884 64114 2020 15333 11321 59123 3620 29354 22165 64217 2060 15672 11581 59289 3660 29716 22447 64316 2100 16011 11841 59451 3700 30078 22730 64415 2140 16351 12101 59612 3740 30440 23013 64512 2180 16692 12363 59770 3780 30803 23296 64609 2220 17036 12625 59926 3820 31166 23580 64704 2260 17376 12888 60077 3860 31529 23864 64800 2300 17719 13151 60228 3900 31894 24149 64893 2340 18062 13416 60376 3940 32258 24434 64986 2380 18407 13680 60522 3980 32623 24720 65078 2420 18572 13946 60666 4020 32989 25006 65169 2460 19097 14212 60808 4060 33355 25292 65260 2500 19443 14479 60946 4100 33722 25580 65350 2540 19790 14746 61084 4140 34089 25867 64439 2580 20138 15014 61220 4180 34456 26155 65527 2620 20485 15282 61354 4220 34824 26144 65615 2660 20834 15551 61486 4260 35192 26733 65702 2700 21183 15821 61616 4300 35561 27022 65788 2740 21533 16091 61744 4340 35930 27312 65873 2780 21883 16362 61871 4380 36300 27602 65958 2820 22232 16633 61996 4420 36670 27823 66042 2860 22584 16905 62120 4460 37041 28184 66125 2900 22936 17177 62242 4500 37412 28475 66208 2940 23288 17450 62363 4540 37783 28768 66290 2980 23641 17723 62483 4580 38155 29060 66372 3020 23994 17997 62599 4620 38528 29353 66453 3060 24348 18271 62716 4660 38900 29646 66533 3100 24703 18546 62831 4700 39274 29940 66613 3140 25057 18822 62945 4740 39647 30234 66691 3180 25413 19098 63057 4780 40021 30529 66770 3220 25769 19374 63169 4820 40396 30824 66848 3260 26175 19651 63279 4860 40771 31120 66925 3300 26412 19928 63386 4900 41146 31415 67003 3340 26839 20206 63494 5000 42086 32157 67193 3380 27197 20485 63601 5100 43021 32901 67380 3420 27555 20763 63706 5200 43974 33648 67562 3460 27914 21043 63811 5300 44922 34397 67743 Concluded Final PDF to printer 962 PROPERTY TABLES AND CHARTS cen22672app02931972indd 962 110617 0920 AM TABLE A20E Idealgas properties of carbon dioxide CO2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 21082 15124 46353 1080 95758 74311 58072 320 22566 16211 46832 1100 98026 76181 58281 340 24073 17321 47289 1120 100306 78064 58485 360 25605 18456 47728 1140 102601 79962 58689 380 27164 19618 48148 1160 104906 81870 58889 400 28747 20804 48555 1180 107223 83790 59088 420 30357 22017 48947 1200 109553 85723 59283 440 31994 23256 49329 1220 111894 87666 59477 460 33657 24522 49698 1240 114246 89621 59668 480 35347 25815 50058 1260 116610 91588 59858 500 37062 27133 50408 1280 118984 93565 60044 520 38803 28477 50750 1300 121369 95553 60229 537 40275 29638 51032 1320 123764 97550 60412 540 40568 29844 51082 1340 126170 99559 60593 560 42358 31237 51408 1360 128585 101577 60772 580 44172 32654 51726 1380 131010 103605 60949 600 46009 34094 52038 1400 133447 105645 61124 620 47866 35556 52343 1420 135891 107692 61298 640 49749 37040 52641 1440 138345 109748 61469 660 51652 38546 52934 1460 140808 111814 61639 680 53576 40072 53225 1480 143280 113889 61800 700 55520 41619 53503 1500 145760 115972 61974 720 57484 43186 53780 1520 148249 118064 62138 740 59468 44773 54051 1540 150747 120165 62302 760 61470 46379 54319 1560 153253 122273 62464 780 63491 48001 54582 1580 155767 124390 62624 800 65529 49642 54839 1600 158290 126516 62783 820 67583 51299 55093 1620 160819 128648 62939 840 69657 52976 55343 1640 163357 130789 63095 860 71747 54669 55589 1660 165902 132937 63250 880 73853 56377 55831 1680 168455 135092 63403 900 75976 58103 56070 1700 171014 137254 63555 920 78114 59844 56305 1720 173581 139424 63704 940 80268 61601 56536 1740 176155 141601 63853 960 82438 63374 56765 1760 178735 143784 64001 980 84622 65161 56990 1780 181322 145974 64147 1000 86821 66962 57212 1800 183915 148169 64292 1020 89034 68778 57432 1820 186515 150372 64435 1040 91262 70609 57647 1840 189122 152582 64578 1060 93503 72453 57861 1860 191734 154797 64719 Final PDF to printer 963 APPENDIX 2 cen22672app02931972indd 963 110617 0920 AM TABLE A20E Idealgas properties of carbon dioxide CO2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 1900 19698 15925 64999 3500 41965 35015 73462 1940 20224 16372 65272 3540 42543 35513 73627 1980 20753 16821 65543 3580 43121 36012 73789 2020 21284 17273 65809 3620 43701 36512 73951 2060 21818 17727 66069 3660 44280 37012 74110 2100 22353 18182 66327 3700 44861 37513 74267 2140 22890 18640 66581 3740 45442 38014 74423 2180 23429 19101 66830 3780 46023 38517 74578 2220 23970 19561 67076 3820 46605 39019 74732 2260 24512 20024 67319 3860 47188 39522 74884 2300 25056 20489 67557 3900 47771 40026 75033 2340 25602 20955 67792 3940 48355 40531 75182 2380 26150 21423 68025 3980 48939 41035 75330 2420 26699 21893 68253 4020 49524 41541 75477 2460 27249 22364 68479 4060 50109 42047 75622 2500 27801 22837 68702 4100 50695 42553 75765 2540 28355 23310 68921 4140 51282 43060 75907 2580 28910 23786 69138 4180 51868 43568 76048 2620 29465 24262 69352 4220 52456 44075 76188 2660 30023 24740 69563 4260 53044 44584 76327 2700 30581 25220 69771 4300 53632 45093 76464 2740 31141 25701 69977 4340 54221 45602 76601 2780 31702 26181 70181 4380 54810 46112 76736 2820 32264 26664 70382 4420 55400 46622 76870 2860 32827 27148 70580 4460 55990 47133 77003 2900 33392 27633 70776 4500 56581 47645 77135 2940 33957 28118 70970 4540 57172 48156 77266 2980 34523 28605 71160 4580 57764 48668 77395 3020 35090 29093 71350 4620 58356 49181 77581 3060 35659 29582 71537 4660 58948 49694 77652 3100 36228 30072 71722 4700 59541 50208 77779 3140 36798 30562 71904 4740 60134 50721 77905 3180 37369 31054 72085 4780 60728 51236 78029 3220 37941 31546 72264 4820 61322 51750 78153 3260 38513 32039 72441 4860 61916 52265 78276 3300 39087 32533 72616 4900 62511 52781 78398 3340 39661 33028 72788 5000 64000 54071 78698 3380 40236 33524 72960 5100 65491 55363 78994 3420 40812 34020 73129 5200 66984 56658 79284 3460 41388 34517 73297 5300 68471 57954 79569 Concluded Final PDF to printer 964 PROPERTY TABLES AND CHARTS cen22672app02931972indd 964 110617 0920 AM TABLE A21E Idealgas properties of carbon monoxide CO T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 20819 14861 43223 1080 75711 54264 52203 320 22209 15854 43672 1100 77168 55323 52337 340 23599 16847 44093 1120 78629 56387 52468 360 24988 17839 44490 1140 80092 57454 52598 380 26379 18833 44866 1160 81561 58515 52726 400 27769 19826 45223 1180 83033 59600 52852 420 29160 20819 45563 1200 84508 60678 52976 440 30550 21812 45886 1220 85988 61760 53098 460 31940 22805 46194 1240 87472 62847 53218 480 33330 23798 46491 1260 88960 63938 53337 500 34721 24792 46775 1280 90450 65031 53455 520 36112 25786 47048 1300 91946 66130 53571 537 37251 26631 47272 1320 93446 67232 53685 540 37503 26779 47310 1340 94948 68337 53799 560 38895 27774 47563 1360 96455 69447 53910 580 40287 28769 47807 1380 97966 70561 54021 600 41680 29765 48044 1400 99481 71679 54129 620 43074 30762 48272 1420 101000 72801 54237 640 44469 31759 48494 1440 102522 73926 54344 660 45866 32758 48709 1460 104048 75054 54448 680 47262 33758 48917 1480 105578 76187 54522 700 48860 34759 49120 1500 107111 77323 54665 720 50061 35763 49317 1520 108649 78464 54757 740 51464 36769 49509 1540 110190 79608 54858 760 52868 37775 49697 1560 111734 80754 54958 780 54274 38784 49880 1580 113282 81905 55056 800 55682 39795 50058 1600 114834 83060 55154 820 57094 40810 50232 1620 116389 84218 55251 840 58507 41826 50402 1640 117947 85379 55347 860 59923 42845 50569 1660 119509 86544 55411 880 61342 43866 50732 1680 121075 87712 55535 900 62764 44891 50892 1700 122643 88883 55628 920 64190 45920 51048 1720 124214 90057 55720 940 65617 46950 51202 1740 125790 91236 55811 960 67049 47985 51353 1760 127367 92416 55900 980 68484 49023 51501 1780 128949 93600 55990 1000 69922 50063 51646 1800 130532 94786 56078 1020 71364 51108 51788 1820 132120 95977 56166 1040 72810 52157 51929 1840 133710 97170 56253 1060 74259 53209 52067 1860 135302 98365 56339 Final PDF to printer 965 APPENDIX 2 cen22672app02931972indd 965 110617 0920 AM TABLE A21E Idealgas properties of carbon monoxide CO T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 1900 13850 10077 56509 3500 27262 20311 61612 1940 14170 10318 56677 3540 27608 20576 61710 1980 14492 10560 56841 3580 27954 20844 61807 2020 14815 10803 57007 3620 28300 21111 61903 2060 15139 11048 57161 3660 28647 21378 61998 2100 15463 11293 57317 3700 28994 21646 62093 2140 15789 11539 57470 3740 29341 21914 62186 2180 16116 11787 57621 3780 29688 22182 62279 2220 16443 12035 57770 3820 30036 22450 62370 2260 16722 12284 57917 3860 30384 22719 62461 2300 17101 12534 58062 3900 30733 22988 62511 2340 17431 12784 58204 3940 31082 23257 62640 2380 17762 13035 58344 3980 31431 23527 62728 2420 18093 13287 58482 4020 31780 23797 62816 2460 18426 13541 58619 4060 32129 24067 62902 2500 18759 13794 58754 4100 32479 24337 62988 2540 19093 14048 58885 4140 32829 24608 63072 2580 19427 14303 59016 4180 33179 24878 63156 2620 19762 14559 59145 4220 33530 25149 63240 2660 20098 14815 59272 4260 33880 25421 63323 2700 20434 15072 59398 4300 34231 25692 63405 2740 20771 15330 59521 4340 34582 25934 63486 2780 21108 15588 59644 4380 34934 26235 63567 2820 21446 15846 59765 4420 35285 26508 63647 2860 21785 16105 59884 4460 35637 26780 63726 2900 22124 16365 60002 4500 35989 27052 63805 2940 22463 16225 60118 4540 36341 27325 63883 2980 22803 16885 60232 4580 36693 27598 63960 3020 23144 17146 60346 4620 37046 27871 64036 3060 23485 17408 60458 4660 37398 28144 64113 3100 23826 17670 60569 4700 37751 28417 64188 3140 24168 17932 60679 4740 38104 28691 64263 3180 24510 18195 60787 4780 38457 28965 64337 3220 24853 18458 60894 4820 38811 29239 64411 3260 25196 18722 61000 4860 39164 29513 64484 3300 25539 18986 61105 4900 39518 29787 64556 3340 25883 19250 61209 5000 40403 30473 64735 3380 26227 19515 61311 5100 41289 31161 64910 3420 26572 19780 61412 5200 42176 31849 65082 3460 26917 20045 61513 5300 43063 32538 65252 Concluded Final PDF to printer 966 PROPERTY TABLES AND CHARTS cen22672app02931972indd 966 110617 0920 AM TABLE A22E Idealgas properties of hydrogen H2 T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 20635 14677 27337 1400 96738 68936 37883 320 21894 15539 27742 1500 103815 74027 38372 340 23172 16420 28130 1600 110925 79151 38830 360 24468 17319 28501 1700 118074 84314 39264 380 25778 18232 28856 1800 125268 89522 39675 400 27102 19158 29195 1900 132509 94778 40067 420 28437 20096 29520 2000 139801 100084 40441 440 29781 21043 29833 2100 147145 105442 40799 460 31135 22000 30133 2200 154544 110855 41143 480 32494 22962 20424 2300 161998 116323 41475 500 33861 23932 30703 2400 169506 121845 41794 520 35232 24906 30972 2500 177073 127426 42104 537 36403 25739 31194 2600 184697 133064 42403 540 36609 25885 31232 2700 192378 138760 42692 560 37988 26867 31482 2800 200118 144514 42973 580 39371 27853 31724 2900 207915 150325 43247 600 40756 28841 31959 3000 215769 156193 43514 620 42143 29831 32187 3100 223677 162115 43773 640 43531 30821 32407 3200 231641 168093 44026 660 44921 31814 32621 3300 239655 174121 44273 680 46311 32807 32829 3400 247719 180199 44513 700 47702 33801 33031 3500 255829 186324 44748 720 49095 34796 33226 3600 263985 192494 44978 740 50488 35792 33417 3700 272185 198708 45203 760 51881 36788 33603 3800 280428 204965 45423 780 53276 37786 33784 3900 288711 211262 45638 800 54671 38784 33961 4000 297035 217600 45849 820 56067 39783 34134 4100 305398 223977 46056 840 57463 40782 34302 4200 313798 230392 46257 860 58859 41780 34466 4300 322235 236843 46456 880 60256 42780 34627 4400 330709 243331 46651 900 61653 43780 34784 4500 339216 249852 46842 920 63051 44781 34938 4600 347757 256407 47030 940 64449 45781 35087 4700 356330 262994 47215 960 65847 46783 35235 4800 364934 269612 47396 980 67246 47784 35379 4900 353569 276261 47574 1000 68645 48786 35520 5000 382233 282940 47749 1100 75646 53801 36188 5100 390928 289649 47921 1200 82658 58828 36798 5200 399651 296386 48090 1300 89687 63871 37360 5300 408402 303151 48257 Final PDF to printer 967 APPENDIX 2 cen22672app02931972indd 967 110617 0920 AM TABLE A23E Idealgas properties of water vapor H2O T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 300 23676 17718 40439 1080 87682 66235 50854 320 25268 18913 40952 1100 89420 67575 51013 340 26860 20108 41435 1120 91164 68922 51171 360 28451 21302 41889 1140 92914 70275 51325 380 30044 22498 42320 1160 94671 71635 51478 400 31638 23694 42728 1180 96434 73001 51360 420 33232 24891 43117 1200 98204 74374 51777 440 34827 26089 43487 1220 99980 75752 51925 460 36423 27288 43841 1240 101761 77136 52070 480 38020 28488 44182 1260 103549 78527 52212 500 39620 29691 44508 1280 105344 79925 52354 520 41220 30894 44821 1300 107145 81329 52494 537 42580 31919 45079 1320 108953 82740 52631 540 42824 32100 45124 1340 110766 84155 52768 560 44428 33307 45415 1360 112587 85579 52903 580 46037 34519 45696 1380 114414 87009 53037 600 47647 35732 45970 1400 116248 88446 53168 620 49261 36949 46235 1420 118088 89889 53299 640 50878 38168 46492 1440 119934 91338 53428 660 52500 39393 46741 1460 121788 92794 53556 680 54125 40621 46984 1480 123648 94257 53682 700 55754 41853 47219 1500 125514 95727 53808 720 57388 43090 47450 1520 127388 97203 53932 740 59026 44331 47673 1540 129268 98686 54055 760 60669 45576 47893 1560 131156 100176 54117 780 62317 46827 48106 1580 133050 101673 54298 800 63969 48082 48316 1600 134944 103176 54418 820 65626 49342 48520 1620 136857 104686 54535 840 67289 50608 48721 1640 138770 106202 54653 860 68956 51878 48916 1660 140692 107727 54770 880 70629 53153 49109 1680 142619 109256 54886 900 72309 54436 49298 1700 144554 110794 54999 920 73994 55724 49483 1720 146495 112338 55113 940 75684 57017 49665 1740 148443 113889 55226 960 77380 58316 49843 1760 150398 115447 55339 980 79082 59620 50019 1780 152361 117012 55449 1000 80782 60930 50191 1800 154330 118584 55559 1020 82504 62248 50360 1820 156306 120163 55668 1040 84224 63571 50528 1840 158287 121747 55777 1060 85950 64900 50693 1860 160276 123339 55884 Final PDF to printer 968 PROPERTY TABLES AND CHARTS cen22672app02931972indd 968 110617 0920 AM TABLE A23E Idealgas properties of water vapor H2O T R h Btulbmol u Btulbmol s BtulbmolR T R h Btulbmol u Btulbmol s BtulbmolR 1900 16428 12654 56097 3500 34324 27373 62876 1940 16830 12977 56307 3540 34809 27779 63015 1980 17235 13303 56514 3580 35296 28187 63153 2020 17643 13632 56719 3620 35785 28596 63288 2060 18054 13963 56920 3660 36274 29006 63423 2100 18467 14297 57119 3700 36765 29418 63557 2140 18883 14633 57315 3740 37258 29831 63690 2180 19301 14972 57509 3780 37752 30245 63821 2220 19722 15313 57701 3820 38247 30661 63952 2260 20145 15657 57889 3860 38743 31077 64082 2300 20571 16003 58077 3900 39240 31495 64210 2340 20999 16352 58261 3940 39739 31915 64338 2380 21429 16703 58445 3980 40239 32335 64465 2420 21862 17057 58625 4020 40740 32757 64591 2460 22298 17413 58803 4060 41242 33179 64715 2500 22735 17771 58980 4100 41745 33603 64839 2540 23175 18131 59155 4140 42250 34028 64962 2580 23618 18494 59328 4180 42755 34454 65084 2620 24062 18859 59500 4220 43267 34881 65204 2660 24508 19226 59669 4260 43769 35310 65325 2700 24957 19595 59837 4300 44278 35739 65444 2740 25408 19967 60003 4340 44788 36169 65563 2780 25861 20340 60167 4380 45298 36600 65680 2820 26316 20715 60330 4420 45810 37032 65797 2860 26773 21093 60490 4460 46322 37465 65913 2900 27231 21472 60650 4500 46836 37900 66028 2940 27692 21853 60809 4540 47350 38334 66142 2980 28154 22237 60965 4580 47866 38770 66255 3020 28619 22621 61120 4620 48382 39207 66368 3060 29085 23085 61274 4660 48899 39645 66480 3100 29553 23397 61426 4700 49417 40083 66591 3140 30023 23787 61577 4740 49936 40523 66701 3180 30494 24179 61727 4780 50455 40963 66811 3220 30967 24572 61874 4820 50976 41404 66920 3260 31442 24968 62022 4860 51497 41856 67028 3300 31918 25365 62167 4900 52019 42288 67135 3340 32396 25763 62312 5000 53327 43398 67401 3380 32876 26164 62454 5100 54640 44512 67662 3420 33357 26565 62597 5200 55957 45631 67918 3460 33839 26968 62738 5300 57279 46754 68172 Concluded Final PDF to printer 969 APPENDIX 2 cen22672app02931972indd 969 110617 0920 AM TABLE A26E Enthalpy of formation Gibbs function of formation and absolute entropy at 77F 1 atm Substance Formula h f Btulbmol g f Btulbmol s BtulbmolR Carbon Cs 0 0 136 Hydrogen H2g 0 0 3121 Nitrogen N2g 0 0 4577 Oxygen O2g 0 0 4900 Carbon monoxide COg 47540 59010 4721 Carbon dioxide CO2g 169300 169680 5107 Water vapor H2Og 104040 98350 4511 Water H2Ol 122970 102040 1671 Hydrogen peroxide H2O2g 58640 45430 5560 Ammonia NH3g 19750 7140 4597 Methane CH4g 32210 21860 4449 Acetylene C2H2g 97540 87990 4800 Ethylene C2H4g 22490 29306 5254 Ethane C2H6g 36420 14150 5485 Propylene C3H6g 8790 26980 6380 Propane C3H8g 44680 10105 6451 nButane C4H10g 54270 6760 7411 nOctane C8H18g 89680 7110 11155 nOctane C8H18l 107530 2840 8623 nDodecane C12H26g 125190 21570 14886 Benzene C6H6g 35680 55780 6434 Methyl alcohol CH3OHg 86540 69700 5729 Methyl alcohol CH3OHl 102670 71570 3030 Ethyl alcohol C2H5OHg 101230 72520 6754 Ethyl alcohol C2H5OHl 119470 75240 3840 Oxygen Og 107210 99710 3847 Hydrogen Hg 93780 87460 2739 Nitrogen Ng 203340 195970 3661 Hydroxyl OHg 16790 14750 4392 Source of Data From JANAF Thermochemical Tables Midland MI Dow Chemical Co 1971 Selected Values of Chemical Thermodynamic Properties NBS Technical Note 2703 1968 and API Research Project 44 Carnegie Press 1953 Final PDF to printer 970 PROPERTY TABLES AND CHARTS cen22672app02931972indd 970 110617 0920 AM TABLE A27E Properties of some common fuels and hydrocarbons Fuel phase Formula Molar mass lbmlbmol Density1 lbmft3 Enthalpy of vaporization2 Btulbm Specific heat1 cp BtulbmF Higher heating value3 Btulbm Lower heating value3 Btulbm Carbon s C 12011 125 0169 14100 14100 Hydrogen g H2 2016 344 60970 51600 Carbon monoxide g CO 28013 0251 4340 4340 Methane g CH4 16043 219 0525 23880 21520 Methanol l CH4O 32042 493 502 0604 9740 8570 Acetylene g C2H2 26038 0404 21490 20760 Ethane g C2H6 30070 74 0418 22320 20430 Ethanol l C2H6O 46069 493 395 0583 12760 11530 Propane l C3H8 44097 312 144 0662 21640 19930 Butane l C4H10 58123 361 156 0578 21130 19510 1Pentene l C5H10 70134 400 156 0525 20540 19190 Isopentane l C5H12 72150 391 0554 20890 19310 Benzene l C6H6 78114 547 186 0411 17970 17240 Hexene l C6H12 84161 420 169 0439 20430 19090 Hexane l C6H14 86177 412 157 0542 20770 19240 Toluene l C7H8 92141 541 177 0408 18230 17420 Heptane l C7H16 100204 427 157 0535 20680 19180 Octane l C8H18 114231 439 156 0533 20590 19100 Decane l C10H22 142285 456 155 0528 20490 19020 Gasoline l CnH187n 100110 4549 151 057 20300 18900 Light diesel l CnH18n 170 4952 116 053 19800 18600 Heavy diesel l CnH17n 200 5155 99 045 19600 18400 Natural gas g CnH38nN01n 18 048 21500 19400 1At 1 atm and 68F 2At 77F for liquid fuels and 1 atm and normal boiling temperature for gaseous fuels 3At 77F Multiply by molar mass to obtain heating values in Btulbmol Final PDF to printer 971 APPENDIX 2 cen22672app02931972indd 971 110617 0920 AM FIGURE A31E Psychrometric chart at 1 atm total pressure From the American Society of Heating Refrigerating and AirConditioning Engineers Atlanta GA ASHRAE Psychrometric Chart No 1 Normal Temperature Barometric Pressure 29921 inches of mercury 1992 American Society of Heating Refrigerating and AirConditioning Engineers Inc Sea Level Prepared by Center for Applied Thermodynamic Studies University of Idaho 10 15 20 25 Dry bulb temperature F 30 35 40 45 50 55 60 Enthalpy Btu per pound of dry air 15 20 25 30 35 40 45 50 60 55 028 026 024 150 145 140 135 130 125 Volume cu ft per lb dry air 022 020 018 016 014 Humidity ratio pounds moisture per pound dry air 012 010 008 20 30 40 50 60 70 80 90 10 Relative humidity 006 004 002 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Dry bulb temperature F 105 110 115 120 35 30 25 35 40 45 50 55 60 65 70 75 80 85F Wet bulb temperature 90 40 45 50 55 60 65 70 75 80 85 Saturation temperature F Enthalpy Humidity ratio Δh Δ Sensible heat Total heat QS QT 10 1000 0 500 1000 1500 2000 3000 5000 20 40 40 20 10 05 04 03 02 01 10 0 08 06 05 04 03 02 01 Final PDF to printer cen22672app02931972indd 972 110617 0920 AM Final PDF to printer 973 cen22672idx973980indd 973 111017 0528 PM I N D E X Absolute or specific humidity 713 Absolute pressure 2223 Absolute temperature 95 296 Absorption refrigeration system 622626 Absorptivity 94 Acid rain 8788 Adiabatic combustion temperature 767 Adiabatic efficiency 361 Adiabatic flame temperature 767769 Adiabatic mixing of airstreams 730 Adiabatic process 6061 Adiabatic saturation 717720 Adiabatic saturation temperature 717 Afterburner 515 Air conditioners 282283 Air conditioning 711745 adiabatic mixing of airstreams and 730 mixing conditioned air with outdoor air in 731732 Airconditioning process 720734 components of 723734 steadyflow process for 724 use of psychrometric chart and 720 Airfuel ratio 750 Airstandard assumptions 480 Airstandard cycle 480481 Amagats law of additive volumes 678679 Ampere 5 Annual fuel utilization efficiency AFUE 79 Atmospheric air 712 714 Autoignition 486 Average velocity 213 Back pressure 836837 842 Back work ratio 499 Barometer 2629 Barometric pressure 26 BeattieBridgeman equation of state 139142 679680 BenedictWebbRubin equation of state 139142 679680 Bernoulli equation 355 832 Binary vapor cycle 574579 Biological systems 185 Blackbody 94 radiation 94 Body mass index BMI 191 Boiler pressure 552 Boiling 146147 Boltzmann relation 338 Bore 482 Bottom dead center BDC 482483 Boundary 1011 162167 212 219 328 Boundary work 162165 Bourdon tube 32 Bow wave 852 854 Brayton cycle 496516 gasturbine engine and 497499 with intercooling 506 with regeneration 504506 with reheating 506 British thermal unit Btu 7 Caloric theory 6162 Calorie 7 Candela 5 Carnot cycle 291293 478 493495 516 567 599 refrigerator or heat pump 599 reversed 293 599 totally reversible ideal 567 totally reversible processes and the 478 495 516 Carnot efficiency 297 431 Carnot heat engine 291 297298 431 Carnot principles 293294 Carnot refrigeration cycle 293 301 Carnot refrigerator and heat pump 300303 599 Carnot vapor cycle 544 Cascade refrigerator system 613614 Celsius scale 18 20 Chemical energy 5355 Chemical equilibrium 14 791795 802 criterion for 792 794 definition 14 for simultaneous reactions 802 of an ideal gas mixture 795 of an isolated system 791 Chemical potential 691 694 793 Chemical reactions 273 434 747789 irreversibility of that destroys exergy 434 second law of thermodynamics and 273 Choked flow 836 ClapeyronClausius equation 652 Clapeyron equation 650653 Clausius inequality 324325 Clausius statement 284 Clearance ratio 482 Closed systems 10 371 427 763764 Coefficient of performance COP 280281 598 Cogeneration 569572 Coldairstandard assumptions 480 Combined cycle or combined gasvapor cycle 574 Combustion processes 752753 complete 752 incomplete 752 Combustion equipment efficiency 78 Combustion 749750 Component pressure Pi of gas in a mixture 678 Component volume Vi of gas in a mixture 678 Compressed liquid 112 Compressed liquid region 116 Compressibility factor Z 136137 Compressionignition CI engine 482 Conduction 62 91 Final PDF to printer 974 INDEX cen22672idx973980indd 974 111017 0528 PM Conservation of energy principle 2 52 7071 172173 244 definition 2 244 equation for 172173 example of 52 Conservation of mass principle 212217 724 750 chemical equations and 212 equation 724 for a bathtub 214 for a closed system 212 for a control volume 214 for a steadyflow system 216 principle 214 750 Constantpressure process 182 Constanttemperature process 182 Constantvolume gas thermometer 18 Continuum 1213 Control mass 10 Control surface 11 Control volume 11 51 54 211270 372 change in mass 214 conservation of mass 214 definition 11 energy transferred by mass 51 fluid flow 54 212 mass and energy analysis 211270 moving or deformed 216 rate of mass flow 214 Convection 62 9293 definition of heat transfer by 62 forced 93 freenatural 93 heat transfer by 92 Convection heat transfer coefficient 93 Convergingdiverging nozzles 840842 Conversion efficiency of a power plant 546 Cooking appliances efficiency 80 Cooling See also Evaporative cooling Vacuum cooling air conditioning process and simple 723725 conservation of mass equation for 724 evaporative 728729 of air 725 with dehumidification 727728 Cooling pond 733 Coulomb repulsion 57 Critical point 116 Critical properties 834835 Critical ratios 835 Cutoff ratio rc 490 Cycle 16 74 168 See also Name of specific cycle Refrigeration cycles Cyclic relation 648 Daily calorie needs 188 Daltons law of additive pressures 678679 Dead state 414415 425 Deadweight tester 33 Decrease of exergy principle 434 Deficiency of air 752 percent deficiency of air 752 Degree kelvin K 5 18 Density 13 Desalination process 698 Dewpoint temperature 715716 750 Diesel cycle 489 516 Diffuser 226227 335 510 Dimensions 38 Displacement volume 482 Dodecane 748 Dry air 712713 Drybulb temperature 714 Dual cycle 491 Duct 824833 Duct flow with heat transfer 858861 Dynamic temperature 825 Effectiveness 504 Efficiency 7885 275277 361365 511 551555 See also Thermal efficiency compressor power input and effect on 364365 cooking appliances 81 definition and equation for overall 79 definition of adiabatic 361 definition of isentropic 361 definition of thermal 275277 effect of boiler pressure and temperature on 553555 equations 78 83 511 generator 79 isothermal 364 lighting system 80 mechanical and electrical devices 8285 pump 82 simple ideal Rankine cycle and increased 551555 turbine 82 water heater 78 Elastic solid bars 6768 Electrical polarization work 70 Electrical power 65 Electrical work 70 Emissivity 94 Energy 2 5160 7374 8596 219 change in stationary systems internal 54 convected 219 definition 2 dynamic 56 environment and 8596 forms 5154 heat transfer 60 internal 53 transfer 7374 transfer via mass flow for control volume 51 work and transfer of 62 Energy analysis 222240 545546 of ideal Rankine cycle 545546 of steadyflow systems 222225 of unsteadyflow process 239240 Energy balance 7172 Energy efficiency ratio 282 Energy transfer 63 370 Engineering Equation Solver 3637 Engine knock 486 English system 37 Enthalpy 122123 183185 225 651 663 682 692 712714 758762 changes 183185 225 Final PDF to printer 975 INDEX cen22672idx973980indd 975 111017 0528 PM definition 122 departure 663 departure factor 663 equation for 122 of a chemical component 762 of a gas mixture 682 of atmospheric air 714 of combustion 758 of formation 759 of mixing 692 of reaction 759 of vaporization 123 of vaporization and Clapeyron equation 651 of water vapor 712 on a unitmole basis 762 reasons for negative and positive 760 Entropy 324412 664665 682 692 771 absolute 339 770 Boltzmann and 338 change during an isothermal process 327 change of a thermal energy reservoir 327 definition 326 departure factor 665 ideal gas and absolute 771 increase during heat transfer process 340 of a gas mixture 682 of ideal gas variations 348 of mixing 692 of pure crystalline substance and absolute zero 339 principle increase of 328329 property diagram 336337 system and change 368 transfer with heat 328 transferred only by heat 369 Entropy balance 370372 equation for 370 equation for control volume 372 equation for steadyflow process 372 in a rate form 370 on a unitmass basis 370 Entropy change of reacting system 769771 Entropy departure 665 Entropy generation 328329 Environment 81 8596 414 Equation of state 133 139148 Equilibrium 14 chemical 14 mechanical 14 phase 14 thermal 14 Equilibrium constant 796 Equivalence ratio 752 Ericsson cycle 493496 Evaporation 146148 Evaporative cooler 146 Evaporative cooling 728730 Exact differentials 63 Excess air percent of 752 Exergy 413474 balance 435 closed system nonflow 427 definition 415416 destroyed 434 771 flow 428 property of systemenvironment combination 415 thermomechanical 425 transfer by heat 431 transfer by mass 432433 Exergy change 425430 of a system 425430 of steadyflow devices 427 Exhaust blowdown 483 Exhaust valve 482 Expansion fan 854 858 Extensive properties 12 682 Externally reversible 291 Fahrenheit scale 18 Feedwater heater FWH 559562 closed 561562 open or directcontact 559560 Firstlaw analysis of reacting systems 762767 First law of thermodynamics 7071 244 Fixed mass 428 Flow 27 213 823879 See also Isentropic flow compressible 213 823879 compressing 854 expanding 854 gravitydriven 27 incompressible 213 Rayleigh 858861 subsonic versus supersonic 832 Flow energy or flow work 5859 219222 Force F 66 Formal sign convention 63 Fouriers law 92 Fourstroke engines 482 Friction 289 Fuel 748 Fuelair ratio 751 Fuel cells 776778 Gage pressure 22 Gas 111 133 Gas constant 133 of a gas mixture 676677 of different substances 133 Gas mixtures 675709 apparent or average molar mass of a 676 enthalpy and entropy of 682 gas constant for 676677 ideal and real gases and properties of 682690 intensive properties of 683 mass and mole fractions of 676677 total internal energy of 682 Gas phase 111 Gas power cycles 475542 Gas refrigeration cycles 619620 Gas turbine engine 498500 Gasvapor mixtures 711745 General Conference of Weights and Measures 5 Final PDF to printer 976 INDEX cen22672idx973980indd 976 111017 0528 PM Generalized compressibility chart 137 139 Generalized enthalphy departure chart 663 Generalized entropy departure chart 663 Generator efficiency 79 83 GibbsDalton law 684 Gibbs equations 649 Gibbs formulation 338 Gibbs function 649 691 771 793795 Gibbs phase rule 807808 Global climate change and warming 8890 Gravimetric analysis of a gas mixture 676 Gravity force 7 Greenhouse effect 8890 Heat 60 Heat engines 274279 298 422 example that violates KelvinPlanck statement 279 thermal efficiency 277 298 422 working fluid 274 Heat exchangers 235 Heat pumps 281282 300302 598 611612 coefficient of performance 281 300302 612 common energy sources 611 energy efficiency ratio EER 282 geothermal 282 groundsource 282 611 heating and cooling modes 612 Heat rate of power plants 546 Heat reservoirs 273 Heat transfer 56 61 9196 225 290 330331 368369 as a mechanism for entropy transfer 368369 energy interaction of 56 entropy generation and process of 330331 from a person 9596 per unit mass 61 rate 225 types of mechanisms of 9195 Heating value of a fuel 78 760 Helmholtz function 649 Henrys constant 809810 Henrys law 809810 Higher heating value HHV 79 760 Human comfort and air conditioning 721723 Humidity ratio 713 Hydrocarbon fuels 748 Hypersonic flow 829 Ice point 1820 Ideal cycle 476 Ideal refrigeration cycle 600 Ideal gas 133 174 See also Gas Idealgas behavior 136138 Idealgas equation of state 133 677 Idealgas mixtures 678679 683684 693694 Idealgas relation 133 Idealgas temperature scale 18 175 656 Ideal mixture 693 Ideal solution 693 Ideal vaporcompression refrigeration cycle 600 Idealizations used in analysis of a power cycle 477 Ignition temperature 750 Incompressible substance 181 Independent property 15 Inexact differentials 63 Intake valve 482 Intensive properties 12 683 Internal energy 5355 182 423 658659 682 763 binding forces between molecules 55 changes in 182 definition 55 equals sum of kinetic and potential energies 55 exergy and 423 gas mixture and total 682 latent energy and 55 of a chemical component 763 one of the components of total energy 54 specific heat of ideal gas and 659 temperature and 658659 Internal energy changes 654 658 664 Internally reversible 290 International Temperature Scale ITS90 1 20 Inversion line 661 Inversion temperature 661 Irreversibilities 289 418 Irreversible processes 288 Isentropic efficiencies 361367 compressors 363364 nozzles 365366 pumps 363364 turbines 361363 steadyflow devices 361367 Isentropic flow 832844 differential form of energy equation for steady 832 ideal gases and property relations for 833835 onedimensional 829836 through a nozzle 836844 Isentropic process 334336 Isobaric process 16 Isochoricisometric process 16 Isolated system 10 Isothermal compressibility 657 Isothermal efficiency 364 Isothermal process 16 291 327 358 Jetpropulsion cycles 510 Joule 7 JouleThomson coefficient 660661 Kays rule 679680 688 KelvinPlanck statement 279 325 Kelvin scale 18 296 Kilogram kg 5 Kinetic energy KE 5355 5860 Kirchhoffs law 94 Latent energy 55 425 758 Latent heat 114 Latent heat of fusion 114 Latent heat of vaporization 114 123 Lighting efficacy 80 Liquefaction of gases 618619 Liquid 111112 343346 compressed or subcooled 112 entropy change of 343346 Liquid phase 110111 Liquidvapor saturation curve 114 Lower heating value LHV 79 760 Final PDF to printer 977 INDEX cen22672idx973980indd 977 111017 0528 PM Mach number Ma 829 Mach wave 854 Macroscopic energy 53 Magnetic work 70 Manometer 2930 Mass balance 214 216 223 724 750 Mass flow rate 54 59 212217 220221 382 definition and equation 216 221 differential 212 equation for differential 215 equation for pipe or duct 213 fluid stream with uniform properties 220221 in a steadyflow process 217 into a control volume 214 mass flowing per unit time 216 net 215 of air 382 of fluid 59 out of a control volume 214 Mass fraction gas mixtures 676 Mass transfer 212 Maximum inversion temperature 661 Maxwell relations 649650 Mayer relation 657658 Mean effective pressure MEP 482 Mechanical efficiency 82 Mechanical energy 53 58 Mechanical equilibrium 14 Metabolism 185186 Meter m 5 Methane 344 Methyl alcohol 748 Microscopic energy 53 Mixing chambers 232234 Molar analysis of a gas mixture 676 Molar mass M 133 676 Mole mol 5 Mole fraction 676 683 Molecules 2 175 Mollier chart 123 Motor efficiency 83 384 Moving boundary work 162167 Multistage compression refrigeration systems 615 Multistage compression with intercooling 358 Naturaldraft cooling tower 732 Newton N 56 Newtons law of cooling 93 Normal shock 846847 Nozzles 222 226 365366 823 836844 867868 converging 823 convergingdiverging 823 840842 definition 226 isentropic efficiency of 365366 isentropic flow through 836844 steadyflow device 222 steam 867868 Nuclear energy 5558 Oblique shock 842 850854 Oblique shock angles 857 Octane 748 Open cycle 480 Open system or control volume 1012 Orsat gas analyzer 753 Osmotic pressure 698699 Osmotic rise 698699 Otto cycle 482489 advantages and disadvantages of ideal 484 four strokes in the ideal 485 ideal cycle for a sparkignition engine 482489 internally reversible processes in the 484 modified 485 monatomic gas and the ideal 487489 specific heat ratio and the ideal 487 thermal efficiency of the ideal 485486 Overall or combined efficiency 79 83 Ozone 8687 Package icing 116 Partial derivative 645 Partial differentials 645 648 Partial molar properties 691 Partial pressure 678 712 Partial volume 678 Pascal Pa 21 Pascals law 25 Path 15 Path functions 63 212 Peltier effect 627 Perpetualmotion machine 286287 of the first kind PMM1 286 of the second kind PMM2 286 Per unit mass 74 168 Ph diagram of ideal vaporcompression refrigeration cycle 601 Phase 110 Phasechange processes 111122 Phase equilibrium 14 146 806807 definition and equation for 14 146 for a multicomponent system 808 for a saturated mixture 807 for a singlecomponent system 806 Phases of a pure substance 110111 Piezoelectric transducers 33 Pipe transport of liquid or gas in a 237 Point function 63 Polytropic process 166 Potential energy PE 54 Pound lb 6 Poundforce lbf 6 Power 7 62 Power cycle 476 574575 Brayton and Rankine combined 574 gasturbine and steamturbine combined 574575 thermodynamic cycle and the 476 PrandtlMeyer expansion wave 854858 expansion fan 855 calculations for 858 Pressure 2122 26 116 136 144 350351 698699 712 atmospheric 26 144 barometric 26 critical 116 definition and equation for 21 Final PDF to printer 978 INDEX cen22672idx973980indd 978 111017 0528 PM definition of absolute 22 definition of gage 22 definition of vacuum 22 desalination and osmotic 698699 reduced 136 relative 350351 standard atmospheric 26 vapor 712 Pressure change 3032 Pressure fraction 678 Pressure measurement devices 2633 Pressure ratio 498 Pressure transducers 3233 Primary or fundamental dimension 3 Principle of corresponding states 136 Problemsolving techniques 3339 Process 15 Products 750 Properties definition 12 extensive 12 intensive 12 of saturated liquid 123 A4 of saturated liquidvapor mixture 127128 of saturated vapor 123 A5 point functions and 63 problemsolving and determining unknown 34 specific 12 using steam tables to determine 131132 Property diagrams entropy 336337 phasechange process 116122 Property tables 122123 SI and English units 122123 thermodynamic 122133 Propjet engine 515 Propulsive efficiency 511 Propulsive power W p 511 Pseudoreduced specific volume 138 Psychrometric charts 720721 726 729730 adiabatic mixing of airstreams 730 for evaporative cooling 729 for heating and humidification 726 use of 720721 PT phase diagram 120121 Pump efficiency 82 Pure substances 16 110122 127 165 485 487 490491 497498 definition and examples 110 phasechange process 111116 property diagrams 116122 compression process of a gas on 16 ideal Brayton cycle and 497498 ideal Diesel cycle and 490 ideal dual cycle and 491 ideal Otto cycle 485 487 isothermal compression of ideal gas on 165 saturated liquidvapor mixture on 127 PvT behavior 676682 PvT surface 121122 Qout heat loss 63 Quality x definition and equation for 125 Quasiequilibrium expansion process 163 Quasiequilibrium process 15 162 164 Quasistatic process 15 162 Radiation 62 93 Ramjet engine 515516 Rankine cycle 478 545549 559564 cavitation 549 component equipment 545 energy analysis of the ideal 545546 equation for isentropic efficiencies 549 fluid friction and heat loss 548 ideal cycle for steam power plant 478 ideal for vapor power cycle 545548 ideal regenerative 559564 ideal reheat 555559 thermal efficiency 546 Rankine scale 18 Raoults law 810 Rarefied gas flow theory 13 Rate form 74 167 370 435 446 Rate of heat transfer 225 Rayleigh flow 858861 choked 865866 property relations 864865 Rayleigh line 860 Reactants 750 Realgas mixtures 679680 687688 analysis of properties 687688 BeattieBridgeman equation 679 BenedictWebbRubin equation 679 T ds relation for 687 van der Waals equation 679 Reciprocating engine 481482 Reduced pressure 136 Reduced temperature 136 Reference state 130132 Reference values 130132 Refrigerants 280 598 609610 Refrigeration 598 Refrigeration cycles 597641 actual vaporcompression 603604 ammonia absorption 623624 cascade 613614 compression 616617 diagram of temperatureentropy 607 ideal gas 621 ideal vaporcompression 600602 regeneration with gas 621 reversed Brayton cycle and gas 619 reversed Carnot cycle not suitable for 599600 simple ideal gas 621622 621 Refrigerators 279284 300 305 423 598600 610 626627 Clausius statement 284 coefficient of performance COPR 280281 definition 279280 EER value 283 nofrost 305 objective and schematic 598599 reversed Carnot cycle 300 secondlaw efficiency for a cyclic 423 thermoelectric 626627 Final PDF to printer 979 INDEX cen22672idx973980indd 979 111017 0528 PM use of R134a in a household 610 vaporcompression refrigeration cycle 600 Regeneration 493 504506 Regenerator 504 559 Relative humidity 145 713714 723724 Relative pressure P 351 Relative specific volume 351 Reversed Carnot cycle 293 599 Reversible adiabatic compression 292 Reversible adiabatic expansion 292 Reversible adiabatic process 335 Reversible isothermal compression 292 Reversible isothermal expansion 292 Reversible mixing process 696697 Reversible process 288291 Reversible steadyflow work 354 Reversible work 418 428 771773 exergy of a closed system 428 Rocket 516 Saturated air 145 713 Saturated liquid 112 Saturated liquid line 116 Saturated liquidvapor mixture 112 125 properties 127128 Saturated liquidvapor mixture region 116 Saturated vapor 112113 123 Saturated vapor line 116 Saturation pressure Psat 113 Saturation table 332 332 A5 Saturation temperature Tsat 113114 Scramjet engine 516 Seasonal energy efficiency ratio SEER 282 Secondary or derived dimensions 3 Secondlaw analysis 567569 605609 771776 of adiabatic combustion 773775 of ideal Rankine cycle 568569 of isothermal combustion 775776 of reacting system 771776 of vaporcompression refrigeration cycle 605609 of vapor power cycle 567569 Secondlaw efficiency 289 422423 697 definition and equation 422423 equation for mixing process 697 equation for separation process 697 for workconsuming noncyclic and cyclic devices 423 of reversible devices 423 Seebeck effect 626627 Sensible energy 55 Separation twocomponent mixture and work 697698 Set point 190 Shaft work 66 Shock waves 844854 SI system 57 18 Significant digits 3738 Simple compressible system 15 Sink 273275 Solids 110 343346 Solubility 809 Sonic flow 829 Sonic speed 827 Source 273 Sparkignition SI engine 482 Specific gravity or relative density 1314 Specific heat 172176 181 657661 average 175 common units 173 constantpressure 172 181 657 661 constantvolume 172 181 definition 172174 difference for an ideal gas 659660 idealgas 175 Specific heat at constant pressure 172177 Specific heat at constant volume 172176 181 Specific heat ratio 177 Specific humidity 713714 Specific properties 12 Specific volume 12 Specific weight 14 Speed of sound 823 827829 Spray pond 733 Spring work 67 Stagnation enthalpy 824 Stagnation pressure 825 Stagnation properties 824826 Stagnation state 823 Stagnation temperature 825 837 Standard atmosphere atm 22 Standard reference state 758 762 State 15 119 130133 idealgas equation 133 independent and dependent properties 15 reference 130132 triplephase 119 State postulate 1415 Static enthalpy 824 Stationary systems 54 73 Steadyflow devices 222 418 Steadyflow process 16 222 447 Steadyflow systems 222225 478479 762763 boundary work 223 Carnot cycle 478479 energy analysis 222225 energy balance 223224 equation for mass balance 223 firstlaw analysis 762763 Steam nozzles 867868 Steam point 20 StefanBoltzmann law 94 Stirling cycle 493496 516 Stirling engine 496 Stoichiometric air 752 Stoichiometric combustion 752 Straingage pressure transducer 3233 Stroke 482 Subcooled liquid 112 Sublimation 120 Subsonic flow 832 Superheated vapor 112113 Superheated vapor region 116 128 Superheated water vapor 128129 Supersaturation 868 Supersonic flow 832 Final PDF to printer 980 INDEX cen22672idx973980indd 980 111017 0528 PM Surface tension 68 Surroundings 10 414 immediate 414 work 418 Systems 1013 52 54 73 223 239 371372 622626 absorption refrigeration 622626 adiabatic closed 52 boundary work and steadyflow 223 closed or open 1012 definition of a closed 10 371372 definition of a stationary 54 definition of an isolated 10 different between steady and unsteadyflow 239 equation for mass balance and a steadyflow 223 properties 1213 stationary 73 T ds relations 341342 Temperature 21 95 119120 296 325 829 absolute 95 idealgas tables and reference 175 Kelvin scale 296 Mach number 829 thermodynamic 325 triplepoint 119120 units of measurement 21 Theoretical air percent of 752 Theoretical combustion 752 Thermal conductivity 9192 Thermal efficiency 275277 475479 Thermal energy 56 Thermal energy reservoirs 273 Thermal equilibrium 17 Thermal process 16 Thermodynamic property relations 643673 Thermodynamic property tables 122133 Thermodynamic temperature scale 18 295297 328 Thermodynamics 23 17 52 7071 185 244 284322 326 339 application areas 3 classical 2 definition and examples 23 first law of 2 52 7071 244 of a biological system 185 second law of 2 284322 statistical 2 third law of 326 339 770 zeroth law of 17 Thermoelectric circuit or device 626 Thermoelectric generator 627 Thermoelectric power generation 626 Thermoelectric refrigeration 303 Thermoelectric refrigerator 626 Throat 831 Throttling valve 232 603 Tons of refrigeration 598 Top dead center TDC 482483 490 Torr 26 Total energy 5354 220 of a flowing fluid 220 Total differential 646 Totally reversible 291 Transonic flow 829 Trap of feedwater heater 562 Triple line 119 Triple point 1920 119120 Ts diagrams 552 561 Turbine efficiency 82 Turbofan engine 514515 Tv diagram 116118 Twostroke engines 483 Uniformflow process 240 Unitmass basis 370 Units 39 Btu thermal 7 definition and examples 3 English system 3 metric SI 3 newtonmeter 7 poundforce 6 secondary 9 weight 6 Unity conversion ratios 9 Universal gas constant 133 176 Unrestrained expansion of a gas 290 Unsteadyflow process 239240 Useful pumping power 82 Useful work Wu 418 Vacuum cooling 115116 Vacuum freezing 116 Vacuum pressures 22 van der Waals equation of state 139140 vant Hoff equation 805 Vapor cycle category of a thermodynamic cycle 475 Vapor power cycles 544569 actual versus ideal 548549 secondlaw analysis of 567569 steam and 544 Vapor pressure 712 phase equilibrium and 144147 Virial equation of state 141142 Volume expansivity 657 Volume flow rate 54 213 Volume fraction 678 Water heater efficiency 78 Water vapor 712715 Watt W 7 Wave angle 850 Weight 6 Wetbulb temperature 717720 Wet cooling towers 732 Wet region 116 Wilson line 868 Work 56 63 6670 definition 70 examples of nonmechanical forms 70 path function 63 types of mechanical 6670 Work transfer 212 Working fluid 274 Zeroth law of thermodynamics 17 Final PDF to printer a Acceleration ms2 a Specific Helmholtz function u Ts kJkg A Area m2 A Helmholtz function U TS kJ AF Airfuel ratio c Speed of sound ms c Specific heat kJkgK cp Constantpressure specific heat kJkgK c v Constantvolume specific heat kJkgK COP Coefficient of performance COPHP Coefficient of performance of a heat pump COPR Coefficient of performance of a refrigerator d D Diameter m e Specific total energy kJkg E Total energy kJ EER Energy efficiency rating F Force N FA Fuelair ratio g Gravitational acceleration ms2 g Specific Gibbs function h Ts kJkg G Total Gibbs function H TS kJ h Convection heat transfer coefficient Wm2K h Specific enthalpy u Pv kJkg H Total enthalpy U PV kJ h C Enthalpy of combustion kJkmol fuel h f Enthalpy of formation kJkmol h R Enthalpy of reaction kJkmol HHV Higher heating value kJkg fuel i Specific irreversibility kJkg I Electric current A I Total irreversibility kJ k Specific heat ratio cp c v k Spring constant Nm k Thermal conductivity WmK KP Equilibrium constant ke Specific kinetic energy V22 kJkg KE Total kinetic energy mV22 kJ LHV Lower heating value kJkg fuel m Mass kg m Mass flow rate kgs M Molar mass kgkmol Ma Mach number MEP Mean effective pressure kPa mf Mass fraction n Polytropic exponent N Number of moles kmol P Pressure kPa Pcr Critical pressure kPa Pi Partial pressure kPa Pm Mixture pressure kPa Pr Relative pressure PR Reduced pressure Pv Vapor pressure kPa P0 Surroundings pressure kPa pe Specific potential energy gz kJkg PE Total potential energy mgz kJ q Heat transfer per unit mass kJkg Q Total heat transfer kJ Q Heat transfer rate kW QH Heat transfer with hightemperature body kJ QL Heat transfer with lowtemperature body kJ r Compression ratio R Gas constant kJkgK rc Cutoff ratio rp Pressure ratio Ru Universal gas constant kJkmolK s Specific entropy kJkgK S Total entropy kJK sgen Specific entropy generation kJkgK Sgen Total entropy generation kJK SG Specific gravity or relative density t Time s T Temperature C or K T Torque Nm Tcr Critical temperature K Tdb Drybulb temperature C Tdp Dewpoint temperature C Tf Bulk fluid temperature C TH Temperature of hightemperature body K TL Temperature of lowtemperature body K TR Reduced temperature Twb Wetbulb temperature C T0 Surroundings temperature C or K u Specific internal energy kJkg U Total internal energy kJ v Specific volume m3kg vcr Critical specific volume m3kg vr Relative specific volume vR Pseudoreduced specific volume V Total volume m3 V Volume flow rate m3s V Voltage V V Velocity ms Vavg Average velocity w Work per unit mass kJkg W Total work kJ N O M E N C L A T U R E 981 cen22672nomenclature981982 981 110817 1145 AM Final PDF to printer 982 NOMENCLATURE cen22672nomenclature981982 982 110817 1145 AM W Power kW Wrev Reversible work kJ x Quality x Specific exergy kJkg X Total exergy kJ xdest Specific exergy destruction kJkg Xdest Total exergy destruction kJ X dest Rate of total exergy destruction kW y Mole fraction z Elevation m Z Compressibility factor Zh Enthalpy departure factor Zs Entropy departure factor Greek Letters α Absorptivity α Isothermal compressibility 1kPa β Volume expansivity 1K Δ Finite change in quantity ε Emissivity ϵ Effectiveness ηth Thermal efficiency ηII Secondlaw efficiency θ Total energy of a flowing fluid kJkg μJT JouleThomson coefficient KkPa μ Chemical potential kJkg ν Stoichiometric coefficient ρ Density kgm3 σ StefanBoltzmann constant σn Normal stress Nm2 σs Surface tension Nm ϕ Relative humidity ϕ Specific closed system exergy kJkg Φ Total closed system exergy kJ ψ Stream exergy kJkg γs Specific weight Nm3 ω Specific or absolute humidity kg H2Okg dry air Subscripts a Air abs Absolute act Actual atm Atmospheric avg Average c Combustion crosssection cr Critical point CV Control volume e Exit conditions f Saturated liquid fg Difference in property between saturated liquid and saturated vapor g Saturated vapor gen Generation H High temperature as in TH and QH i Inlet conditions i ith component in input as in Qin and Win L Low temperature as in TL and QL m Mixture out output as in Qout and Wout r Relative R Reduced rev Reversible s Isentropic sat Saturated surr Surroundings sys System v Water vapor 0 Dead state 1 Initial or inlet state 2 Final or exit state Superscripts overdot Quantity per unit time overbar Quantity per unit mole circle Standard reference state asterisk Quantity at 1 atm pressure Final PDF to printer cen22672conversionfactor983984 983 110117 1043 AM Conversion Factors DIMENSION METRIC METRICENGLISH Acceleration 1 ms2 100 cms2 1 ms2 32808 fts2 1 fts2 03048 ms2 Area 1 m2 104 cm2 106 mm2 106 km2 1 m2 1550 in2 10764 ft2 1 ft2 144 in2 009290304 m2 Density 1 gcm3 1 kgL 1000 kgm3 1 gcm3 62428 lbmft3 0036127 lbmin3 1 lbmin3 1728 lbmft3 1 kgm3 0062428 lbmft3 Energy heat work 1 kJ 1000 J 1000 Nm 1 kPam3 1 kJ 094782 Btu internal energy 1 kJkg 1000 m2s2 1 Btu 1055056 kJ enthalpy 1 kWh 3600 kJ 540395 psiaft3 778169 lbfft 1 cal 4184 J 1 Btulbm 25037 ft2s2 2326 kJkg 1 IT cal 41868 J 1 kJkg 0430 Btulbm 1 Cal 41868 kJ 1 kWh 341214 Btu 1 therm 105 Btu 1055 105 kJ natural gas Force 1 N 1 kgms2 105 dyne 1 N 022481 lbf 1 kgf 980665 N 1 lbf 32174 lbmfts2 444822 N Heat flux 1 Wcm2 104 Wm2 1 Wm2 03171 Btuhft2 Heat transfer 1 Wm2C 1 Wm2K 1 Wm2C 017612 Btuhft2F coefficient Length 1 m 100 cm 1000 mm 106 μm 1 m 39370 in 32808 ft 10926 yd 1 km 1000 m 1 ft 12 in 03048 m 1 mile 5280 ft 16093 km 1 in 254 cm Mass 1 kg 1000 g 1 kg 22046226 lbm 1 metric ton 1000 kg 1 lbm 045359237 kg 1 ounce 283495 g 1 slug 32174 lbm 145939 kg 1 short ton 2000 lbm 9071847 kg Power 1 W 1 Js 1 kW 341214 Btuh heat transfer rate 1 kW 1000 W 1341 hp 73756 lbffts 1 hp 7457 W 1 hp 550 lbffts 07068 Btus 4241 Btumin 25445 Btuh 074570 kW 1 boiler hp 33475 Btu 1 Btuh 1055056 kJh 1 ton of refrigeration 200 Btumin Pressure 1 Pa 1 Nm2 1 Pa 14504 104 psia 1 kPa 103 Pa 103 MPa 0020886 lbfft2 1 atm 101325 kPa 101325 bars 1 psi 144 lbfft2 6894757 kPa 760 mm Hg at 0C 1 atm 14696 psia 2992 in Hg at 30F 103323 kgfcm2 1 in Hg 3387 kPa 1 mm Hg 01333 kPa Specific heat 1 kJkgC 1 kJkgK 1 JgC 1 BtulbmF 41868 kJkgC 1 BtulbmolR 41868 kJkmolK 1 kJkgC 023885 BtulbmF 023885 BtulbmR Exact conversion factor between metric and English units Calorie is originally defined as the amount of heat needed to raise the temperature of 1 g of water by 1C but it varies with temperature The international steam table IT calorie generally preferred by engineers is exactly 41868 J by definition and corresponds to the specific heat of water at 15C The thermochemical calorie generally pre ferred by physicists is exactly 4184 J by definition and corresponds to the specific heat of water at room temperature The difference between the two is about 006 percent which is negligible The capitalized Calorie used by nutritionists is actually a kilocalorie 1000 IT calories 983 Final PDF to printer 984 CONVERSION FACTORS cen22672conversionfactor983984 984 110117 1043 AM DIMENSION METRIC METRICENGLISH Specific volume 1 m3kg 1000 Lkg 1000 cm3g 1 m3kg 1602 ft3lbm 1 ft3lbm 0062428 m3kg Temperature TK TC 27315 TR TF 45967 18TK ΔTK ΔTC TF 18TC 32 ΔTF ΔTR 18ΔTK Thermal 1 WmC 1 WmK 1 WmC 057782 BtuhftF conductivity Velocity 1 ms 360 kmh 1 ms 32808 fts 2237 mih 1 mih 146667 fts 1 mih 16093 kmh Volume 1 m3 1000 L 106 cm3 cc 1 m3 61024 104 in3 35315 ft3 26417 gal US 1 US gallon 231 in3 37854 L 1 fl ounce 295735 cm3 00295735 L 1 US gallon 128 fl ounces Volume flow rate 1 m3s 60000 Lmin 106 cm3s 1 m3s 15850 galmin gpm 35315 ft3s 21189 ft3min cfm Mechanical horsepower The electrical horsepower is taken to be exactly 746 W Some Physical Constants Universal gas constant Ru 831447 kJkmolK 831447 kPam3kmolK 00831447 barm3kmolK 8205 LatmkmolK 19858 BtulbmolR 154537 ftlbflbmolR 1073 psiaft3lbmolR Standard acceleration of gravity g 980665 ms2 32174 fts2 Standard atmospheric pressure 1 atm 101325 kPa 101325 bar 14696 psia 760 mm Hg 0C 299213 in Hg 32F 103323 m H2O 4C StefanBoltzmann constant 𝛼 56704 108 Wm2K4 01714 108 Btuhft2R4 Boltzmanns constant k 1380650 1023 JK Speed of light in vacuum co 29979 108 ms 9836 108 fts Speed of sound in dry air at 0C and 1 atm c 33136 ms 1089 fts Heat of fusion of water at 1 atm hif 3337 kJkg 1435 Btulbm Enthalpy of vaporization of water at 1 atm hfg 22565 kJkg 97012 Btulbm Final PDF to printer