Thermodynamics 8th Edition

Fundamentals of Thermodynamics

BORGNAKKE t SONNTAG

8e

P1: PCX/OVY P2: PCX/OVY QC: PCX/OVY T1: PCX

WB00776-FM JWCL672-Borgnakke-v1 October 25, 2012 17:24

xviii

This page is intentionally left blank


WB00776-IFC JWCL672-Borgnakke-v1 October 1, 2012 19:20

Fundamental Physical Constants

Avogadro N0 = 6.022 1415 1023 mol1Boltzmann k = 1.380 6505 1023 J K1Planck h = 6.626 0693 1034 JsGas Constant R = N0 k = 8.314 472 J mol1 K1Atomic Mass Unit m0 = 1.660 538 86 1027 kgVelocity of light c = 2.997 924 58 108 ms1Electron Charge e = 1.602 176 53 1019 CElectron Mass me = 9.109 3826 1031 kgProton Mass mp = 1.672 621 71 1027 kgGravitation (Std.) g = 9.806 65 ms2Stefan Boltzmann = 5.670 400 108 W m2 K4

Mol here is gram mol.

Prefixes

101 deci d102 centi c103 milli m106 micro 109 nano n1012 pico p1015 femto f101 deka da102 hecto h103 kilo k106 mega M109 giga G1012 tera T1015 peta P

Concentration

106 parts per million ppm

i



xviii




8/eFundamentals ofThermodynamics

Claus Borgnakke

Richard E. SonntagUniversity of Michigan

i


WB00776-FM JWCL672-Borgnakke-v1 November 19, 2012 21:43

FINAL PAGES

PUBLISHER Don FowleyACQUISITIONS EDITOR Linda RattsMARKETING MANAGER Christopher RuelCREATIVE DIRECTOR Harry NolanSENIOR DESIGNER Jim OSheaPRODUCTION MANAGEMENT SERVICES Aptara, Inc.SENIOR PRODUCTION EDITOR Sujin HongPHOTO EDITOR Sheena GoldsteinCOVER IMAGE Dr. Hong Im, University of Michigan

This book was set in Times New Roman by Aptara, Inc. and printed and bound by Quad/Graphics. The coverwas printed by Quad/Graphics.

This book is printed on acid free paper.

Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for morethan 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company isbuilt on a foundation of principles that include responsibility to the communities we serve and where we live andwork. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental,social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbonimpact, paper specifications and procurement, ethical conduct within our business and among our vendors, andcommunity and charitable support. For more information, please visit ourwebsite: www.wiley.com/go/citizenship.

Copyright c 2013, 2009, 2002, 1998 John Wiley & Sons, Inc. All rights reserved.

No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted underSection 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of thePublisher or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center,Inc., 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher forpermission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street,Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008, website www.wiley.com/go/permissions.

Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in theircourses during the next academic year. These copies are licensed and may not be sold or transferred to a thirdparty. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions anda free of charge return mailing label are available at www.wiley.com/go/returnlabel. If you have chosen to adoptthis textbook for use in your course, please accept this book as your complimentary desk copy. Outside of theUnited States, please contact your local sales representative.

ISBN 978-1-118-13199-2

Printed in the United States of America.

10 9 8 7 6 5 4 3 2 1

ii



Preface

In this eighth edition the basic objective of the earlier editions have been retained:

to present a comprehensive and rigorous treatment of classical thermodynamics whileretaining an engineering perspective, and in doing so

to lay the groundwork for subsequent studies in such fields as fluid mechanics, heattransfer, and statistical thermodynamics, and also

to prepare the student to effectively use thermodynamics in the practice of engineering.

The presentation is deliberately directed to students. New concepts and definitions arepresented in the context where they are first relevant in a natural progression. The introduc-tion has been reorganized with a very short introduction followed by the first thermodynamicproperties to be defined (Chapter 1), which are those that can be readily measured: pressure,specific volume, and temperature. In Chapter 2, tables of thermodynamic properties are in-troduced, but only in regard to these measurable properties. Internal energy and enthalpy areintroduced in connection with the energy equation and the first law, entropy with the secondlaw, and the Helmholtz and Gibbs functions in the chapter on thermodynamic relations.Many real-world realistic examples have been included in the book to assist the student ingaining an understanding of thermodynamics, and the problems at the end of each chapterhave been carefully sequenced to correlate with the subject matter, and are grouped andidentified as such. The early chapters in particular contain a large number of examples,illustrations, and problems, and throughout the book, chapter-end summaries are included,followed by a set of concept/study problems that should be of benefit to the students.

This is the first edition I have prepared without the thoughtful comments from mycolleague and coauthor, the late Professor Richard E. Sonntag, who substantially contributedto earlier versions of this textbook. I am grateful for the collaboration and fruitful discussionswith my friend and trusted colleague, whom I have enjoyed the privilege of working with overthe last three decades. Professor Sonntag consistently shared generously his vast knowledgeand experience in conjunction with our mutual work on previous editions of this book andon various research projects, advising PhD students and performing general professionaltasks at our department. In honor of my colleagues many contributions, Professor Sonntagstill appears as a coauthor of this edition.

NEW FEATURES IN THIS EDITION

Chapter Reorganization and RevisionsThe introduction and the first five chapters in the seventh edition have been completelyreorganized. A much shorter introduction leads into the description of some backgroundmaterial from physics, thermodynamic properties, and units all of which is in the newChapter 1. To have the tools for the analysis, the order of the presentation has been kept

iii



iv PREFACE ...............................................................................................................................................................................

from the previous editions, so the behavior of pure substances is presented in Chapter 2,with a slight expansion and separation of the different domains for solid, liquid, and gasphase behavior. Some new figures and explanations have been added to show the ideal gasregion as a limit behavior for a vapor at low density.

Discussion about work and heat is now included in Chapter 3 with the energy equationto emphasize that they are transfer terms of energy explaining how energy for mass at onelocation can change because of energy exchange with a mass at another location. The energyequation is presented first for a control mass as a basic principle accounting for energy ina control volume as

Change of storage = transfer in transfer out

The chapter then discusses the form of energy storage as various internal energies associatedwith the mass and its structure to better understand how the energy is actually stored. Thisalso helps in understanding why internal energy and enthalpy can vary nonlinearly with tem-perature, leading to nonconstant specific heats. Macroscopic potential and kinetic energythen naturally add to the internal energy for the total energy. The first law of thermodynam-ics, which often is taken as synonymous with the energy equation, is shown as a naturalconsequence of the energy equation applied to a cyclic process. In this respect, the currentpresentation follows modern physics rather than the historical development presented in theprevious editions.

After discussion about the storage of energy, the left-hand side of the energy equation,the transfer terms as work and heat transfer are discussed, so the whole presentation is shorterthan that in the previous editions. This allows less time to be spent on the material used forpreparation before the energy equation is applied to real systems.

All the balance equations for mass, momentum, energy, and entropy follow the sameformat to show the uniformity in the basic principles and make the concept something tobe understood and not merely memorized. This is also the reason to use the names energyequation and entropy equation for the first and second laws of thermodynamics to stressthat they are universally valid, not just used in the field of thermodynamics but apply to allsituations and fields of study with no exceptions. Clearly, special cases require extensionsnot covered in this text, like effects of surface tension in drops or for liquid in small pores,relativity, and nuclear processes, to mention a few.

The energy equation applied to a general control volume is retained from the previousedition with the addition of a section on multiflow devices. Again, this is done to reinforceto students that the analysis is done by applying the basic principles to systems underinvestigation. This means that the actual mathematical form of the general laws follows thesketches and figures of the system, and the analysis is not a question about finding a suitableformula in the text.

To show the generality of the entropy equation, a small example is presented applyingthe energy and entropy equations to heat engines and heat pumps shown in Chapter 6.This demonstrates that the historical presentation of the second law in Chapter 5 can becompletely substituted by the postulation of the entropy equation and the existence of theabsolute temperature scale. Carnot cycle efficiencies and the fact that real devices havelower efficiency follow from the basic general laws. Also, the direction of heat transferfrom a higher temperature domain toward a lower temperature domain is predicted by theentropy equation due to the requirement of a positive entropy generation. These are examplesthat show the application of the general laws for specific cases and improve the studentsunderstanding of the material.



PREFACE v.............................................................................................................................................................. ................

The rest of the chapters have been updated to improve the students understandingof the material. The word availability has been substituted by exergy as a general concept,though it is not strictly in accordance with the original definition. The chapters concerningcycles have been expanded, with a few details for specific cycles and some extensions shownto tie the theory to industrial applications with real systems. The same is done for Chapter 13with combustion to emphasize an understanding of the basic physics of what happens, whichmay not be evident in the more abstract definition of terms like enthalpy of combustion.

Web-Based MaterialSeveral new documents will be available from Wileys website for the book. The followingmaterial will be accessible for students, with additional material reserved for instructors ofthe course.

Notes for classical thermodynamics. A very short set of notes covers the basic ther-modynamic analysis with the general laws (continuity, energy, and entropy equations) andsome of the specific laws like device equations, process equations, and so on. This is usefulfor students doing review of the course or for exam preparation, as it gives a comprehensivepresentation in a condensed form.

Extended set of study examples. This document includes a collection of additionalexamples for students to study. These examples have slightly longer and more detailedsolutions than the examples printed in the book and thus are excellent for self-study. Thereare about 8 SI unit problems with 34 English unit problems for each chapter covering mostof the material in the chapters.

How-to notes. Frequently asked questions are listed for each of the set of subject areasin the book with detailed answers. These are questions that are difficult to accommodate inthe book. Examples:

How do I find a certain state for R-410a in the B-section tables?

How do I make a linear interpolation?

Should I use internal energy (u) or enthalpy (h) in the energy equation?

When can I use the ideal gas law?

Instructor material. The material for instructors covers typical syllabus and homeworkassignments for a first and a second course in thermodynamics. Additionally, examples oftwo standard 1-hour midterm exams and a 2-hour final exam are given for typical Thermo-dynamics I and Thermodynamics II classes.

FEATURES CONTINUED FROM THE SEVENTH EDITION

In-Text-Concept QuestionsThe in-text concept questions appear in the text after major sections of material to allowstudent to reflect on the material just presented. These questions are intended to be quickself-tests for students or used by teachers as wrap-up checks for each of the subjects covered,and most of them emphasize the understanding of the material without being memory facts.



vi PREFACE ...............................................................................................................................................................................

End-of-Chapter Engineering ApplicationsThe last section in each chapter, called Engineering Applications, has been revised withupdated illustrations and a few more examples. These sections are intended to be motivatingmaterial, consisting mostly of informative examples of how this particular chapter material isbeing used in actual engineering. The vast majority of these sections do not have any materialwith equations or developments of theory, but they do contain figures and explanations ofa few real physical systems where the chapter material is relevant for the engineeringanalysis and design. These sections are deliberately kept short and not all the details in thedevices shown are explained, but the reader can get an idea about the applications relativelyquickly.

End-of-Chapter Summaries with Main Concepts and FormulasThe end-of-chapter summaries provide a review of the main concepts covered in the chapter,with highlighted key words. To further enhance the summary, a list of skills that the studentshould have mastered after studying the chapter is presented. These skills are among theoutcomes that can be tested with the accompanying set of study-guide problems in additionto the main set of homework problems. Main concepts and formulas are included after thesummary for reference, and a collection of these will be available on Wileys website.

Concept-Study Guide ProblemsAdditional concept questions are placed as problems in the first section of the end-of-chapter homework problems. These problems are similar to the in-text concept questionsand serve as study guide problems for each chapter. They are a little like homework problemswith numbers to provide a quick check of the chapter material. These questions are shortand directed toward very specific concepts. Students can answer all of these questions toassess their level of understanding and determine if any of the subjects need to be studiedfurther. These problems are also suitable for use with the rest of the homework problems inassignments and are included in the solution manual.

Homework ProblemsThe number of homework problems now exceeds 2800, with more than 700 new andmodified problems. A large number of introductory problems cover all aspects of the chaptermaterial and are listed according to the subject sections for easy selection according to theparticular coverage given. They are generally ordered to be progressively more complexand involved. The later problems in many sections are related to real industrial processesand devices, and the more comprehensive problems are retained and grouped at the end asreview problems.

TablesThe tables of the substances have been carried over from the seventh edition with alternativerefrigerant R-410a, which is the replacement for R-22, and carbon dioxide, which is anatural refrigerant. Several more substances have been included in the software. The idealgas tables have been printed on a mass basis as well as a mole basis, to reflect their use on amass basis early in the text and a mole basis for the combustion and chemical equilibriumchapters.



PREFACE vii.............................................................................................................................................................. ................

Software IncludedThe software CATT3 includes a number of additional substances besides those included inthe printed tables in Appendix B. The current set of substances for which the software canprovide the complete tables are:

Water Refrigerants: R-11, 12, 13, 14, 21, 22, 23, 113, 114, 123, 134a, 152a, 404a,407c, 410a, 500, 502, 507a, and C318

Cryogenics: Ammonia, argon, ethane, ethylene, isobutane, methane, neon,nitrogen, oxygen, and propane

Ideal Gases: air, CO2, CO, N, N2, NO, NO2, H, H2, H2O, O, O2, and OH

Some of these are printed in the booklet Thermodynamic and Transport Properties,by Claus Borgnakke and Richard E. Sonntag, John Wiley and Sons, 1997. Besides theproperties of the substances just mentioned, the software can provide the psychrometricchart and the compressibility and generalized charts using the Lee-Keslers equation-of-state, including an extension for increased accuracy with the acentric factor. The softwarecan also plot a limited number of processes in the Ts and log Plog v diagrams, giving thereal process curves instead of the sketches presented in the text material.

FLEXIBILITY IN COVERAGE AND SCOPEThe book attempts to cover fairly comprehensively the basic subject matter of classical ther-modynamics, and I believe that it provides adequate preparation for study of the applicationof thermodynamics to the various professional fields as well as for study of more advancedtopics in thermodynamics, such as those related to materials, surface phenomena, plasmas,and cryogenics. I also recognize that a number of colleges offer a single introductory coursein thermodynamics for all departments, and I have tried to cover those topics that the var-ious departments might wish to have included in such a course. However, since specificcourses vary considerably in prerequisites, specific objectives, duration, and backgroundof the students, the material is arranged in sections, particularly in the later chapters, soconsiderable flexibility exist in the amount of material that may be covered.

The book covers more material than required for a two-semester course sequence,which provides flexibility for specific choices of topic coverage. Instructors may want tovisit the publishers website at www.wiley.com/college/borgnakke for information and sug-gestions on possible course structure and schedules, and the additional material mentionedas Web material that will be updated to include current errata for the book.

ACKNOWLEDGMENTSI acknowledge with appreciation the suggestions, counsel, and encouragement of manycolleagues, both at the University of Michigan and elsewhere. This assistance has beenvery helpful to me during the writing of this edition, as it was with the earlier editions ofthe book. Both undergraduate and graduate students have been of particular assistance, fortheir perceptive questions have often caused me to rewrite or rethink a given portion of thetext, or to try to develop a better way of presenting the material in order to anticipate such



viii PREFACE ...............................................................................................................................................................................

questions or difficulties. Finally, the encouragement and patience of my wife and familyhave been indispensable, and have made this time of writing pleasant and enjoyable, inspite of the pressures of the project. A special thanks to a number of colleagues at otherinstitutions who have reviewed the earlier editions of the book and provided input to therevisions. Some of the reviewers are

Ruhul Amin, Montana State UniversityEdward E. Anderson, Texas Tech UniversityCory Berkland, University of KansasEugene Brown, Virginia Polytechnic Institute and State UniversitySung Kwon Cho, University of PittsburghSarah Codd, Montana State UniversityRam Devireddy, Louisiana State UniversityFokion Egolfopoulos, University of Southern CaliforniaHarry Hardee, New Mexico State UniversityHong Huang, Wright State UniversitySatish Ketkar, Wayne State UniversityBoris Khusid, New Jersey Institute of TechnologyJoseph F. Kmec, Purdue UniversityRoy W. Knight, Auburn UniversityDaniela Mainardi, Louisiana Tech UniversityRandall Manteufel, University of Texas, San AntonioHarry J. Sauer, Jr., Missouri University of Science and TechnologyJ. A. Sekhar, University of CincinnatiAhned Soliman, University of North Carolina, CharlotteReza Toossi, California State University, Long BeachThomas Twardowski, Widener UniversityEtim U. Ubong, Kettering UniversityYanhua Wu, Wright State UniversityWalter Yuen, University of California at Santa Barbara

I also wish to welcome the new editor, Linda Ratts, and thank her for encouragement andhelp during the production of this edition.

I hope that this book will contribute to the effective teaching of thermodynamics tostudents who face very significant challenges and opportunities during their professionalcareers. Your comments, criticism, and suggestions will also be appreciated, and you maycommunicate those to me at [email protected].

CLAUS BORGNAKKEAnn Arbor, Michigan

July 2012



Contents

1 Introduction and Preliminaries 11.1 A Thermodynamic System and the Control Volume, 21.2 Macroscopic versus Microscopic Points of View, 51.3 Properties and State of a Substance, 61.4 Processes and Cycles, 61.5 Units for Mass, Length, Time, and Force, 81.6 Specific Volume and Density, 101.7 Pressure, 131.8 Energy, 191.9 Equality of Temperature, 221.10 The Zeroth Law of Thermodynamics, 221.11 Temperature Scales, 231.12 Engineering Applications, 24Summary, 28Problems, 29

2 Properties of a Pure Substance 392.1 The Pure Substance, 402.2 The Phase Boundaries, 402.3 The PvT Surface, 442.4 Tables of Thermodynamic Properties, 472.5 The Two-Phase States, 492.6 The Liquid and Solid States, 512.7 The Superheated Vapor States, 522.8 The Ideal Gas States, 552.9 The Compressibility Factor, 592.10 Equations of State, 632.11 Computerized Tables, 642.12 Engineering Applications, 65Summary, 68Problems, 69

3 First Law of Thermodynamics andEnergy Equation 813.1 The Energy Equation, 813.2 The First Law of Thermodynamics, 84

ix



FINAL PAGES

x CONTENTS .............................................................................................................................................................................

3.3 The Definition of Work, 853.4 Work Done at the Moving Boundary of a Simple

Compressible System, 903.5 Definition of Heat, 983.6 Heat Transfer Modes, 993.7 Internal Energya Thermodynamic Property, 1013.8 Problem Analysis and Solution Technique, 1033.9 The Thermodynamic Property Enthalpy, 1093.10 The Constant-Volume and Constant-Pressure

Specific Heats, 1123.11 The Internal Energy, Enthalpy, and Specific Heat of

Ideal Gases, 1143.12 General Systems That Involve Work, 1213.13 Conservation of Mass, 1233.14 Engineering Applications, 125Summary, 132Problems, 135

4 Energy Analysis for a Control Volume 1604.1 Conservation of Mass and the Control Volume, 1604.2 The Energy Equation for a Control Volume, 1634.3 The Steady-State Process, 1654.4 Examples of Steady-State Processes, 1674.5 Multiple Flow Devices, 1804.6 The Transient Process, 1824.7 Engineering Applications, 189Summary, 194Problems, 196

5 The Second Law of Thermodynamics 2165.1 Heat Engines and Refrigerators, 2165.2 The Second Law of Thermodynamics, 2225.3 The Reversible Process, 2255.4 Factors That Render Processes Irreversible, 2265.5 The Carnot Cycle, 2295.6 Two Propositions Regarding the Efficiency of a

Carnot Cycle, 2315.7 The Thermodynamic Temperature Scale, 2325.8 The Ideal-Gas Temperature Scale, 2335.9 Ideal versus Real Machines, 2375.10 Engineering Applications, 240Summary, 243Problems, 245



CONTENTS xi............................................................................................................................................................ ................

6 Entropy 2586.1 The Inequality of Clausius, 2586.2 Entropya Property of a System, 2626.3 The Entropy of a Pure Substance, 2646.4 Entropy Change in Reversible Processes, 2666.5 The Thermodynamic Property Relation, 2716.6 Entropy Change of a Solid or Liquid, 2726.7 Entropy Change of an Ideal Gas, 2736.8 The Reversible Polytropic Process for an Ideal Gas, 2776.9 Entropy Change of a Control Mass During an

Irreversible Process, 2816.10 Entropy Generation and the Entropy Equation, 2826.11 Principle of the Increase of Entropy, 2856.12 Entropy as a Rate Equation, 2886.13 Some General Comments about Entropy and Chaos, 292Summary, 294Problems, 296

7 Second-Law Analysis for a Control Volume 3157.1 The Second Law of Thermodynamics for a

Control Volume, 3157.2 The Steady-State Process and the Transient Process, 3177.3 The Steady-State Single-Flow Process, 3267.4 Principle of the Increase of Entropy, 3307.5 Engineering Applications; Efficiency, 3337.6 Summary of General Control Volume Analysis, 339Summary, 340Problems, 342

8 Exergy 3628.1 Exergy, Reversible Work, and Irreversibility, 3628.2 Exergy and Second-Law Efficiency, 3748.3 Exergy Balance Equation, 3828.4 Engineering Applications, 387Summary, 388Problems, 389

9 Power and Refrigeration SystemswithPhase Change 4039.1 Introduction to Power Systems, 4049.2 The Rankine Cycle, 406



xii CONTENTS .............................................................................................................................................................................

9.3 Effect of Pressure and Temperature on theRankine Cycle, 409

9.4 The Reheat Cycle, 4149.5 The Regenerative Cycle and Feedwater Heaters, 4179.6 Deviation of Actual Cycles from Ideal Cycles, 4249.7 Combined Heat and Power: Other Configurations, 4309.8 Introduction to Refrigeration Systems, 4329.9 The Vapor-Compression Refrigeration Cycle, 4339.10 Working Fluids for Vapor-Compression Refrigeration

Systems, 4369.11 Deviation of the Actual Vapor-Compression Refrigeration

Cycle from the Ideal Cycle, 4379.12 Refrigeration Cycle Configurations, 4399.13 The Absorption Refrigeration Cycle, 442Summary, 443Problems, 444

10 Power and Refrigeration SystemsGaseousWorking Fluids 46210.1 Air-Standard Power Cycles, 46210.2 The Brayton Cycle, 46310.3 The Simple Gas-Turbine Cycle with a Regenerator, 47010.4 Gas-Turbine Power Cycle Configurations, 47310.5 The Air-Standard Cycle for Jet Propulsion, 47710.6 The Air-Standard Refrigeration Cycle, 48010.7 Reciprocating Engine Power Cycles, 48310.8 The Otto Cycle, 48410.9 The Diesel Cycle, 48910.10 The Stirling Cycle, 49210.11 The Atkinson and Miller Cycles, 49210.12 Combined-Cycle Power and Refrigeration Systems, 495Summary, 497Problems, 499

11 Gas Mixtures 51311.1 General Considerations and Mixtures of

Ideal Gases, 51311.2 A Simplified Model of a Mixture Involving Gases

and a Vapor, 52111.3 The Energy Equation Applied to GasVapor

Mixtures, 52611.4 The Adiabatic Saturation Process, 53011.5 Engineering ApplicationsWet-Bulb and Dry-Bulb

Temperatures and the Psychrometric Chart, 532Summary, 539Problems, 540



CONTENTS xiii............................................................................................................................................................ ................

12 Thermodynamic Relations 55712.1 The Clapeyron Equation, 55712.2 Mathematical Relations for a Homogeneous Phase, 56112.3 The Maxwell Relations, 56312.4 Thermodynamic Relations Involving Enthalpy,

Internal Energy, and Entropy, 56512.5 Volume Expansivity and Isothermal and

Adiabatic Compressibility, 57112.6 Real-Gas Behavior and Equations of State, 57312.7 The Generalized Chart for Changes of Enthalpy at

Constant Temperature, 57812.8 The Generalized Chart for Changes of Entropy at

Constant Temperature, 58112.9 The Property Relation for Mixtures, 58512.10 Pseudopure Substance Models for Real Gas Mixtures, 58812.11 Engineering ApplicationsThermodynamic Tables, 593Summary, 596Problems, 598

13 Chemical Reactions 60913.1 Fuels, 60913.2 The Combustion Process, 61313.3 Enthalpy of Formation, 62113.4 Energy Analysis of Reacting Systems, 62313.5 Enthalpy and Internal Energy of Combustion;

Heat of Reaction, 63013.6 Adiabatic Flame Temperature, 63513.7 The Third Law of Thermodynamics and Absolute Entropy, 63713.8 Second-Law Analysis of Reacting Systems, 63813.9 Fuel Cells, 64313.10 Engineering Applications, 647Summary, 652Problems, 653

14 Introduction to Phase and Chemical Equilibrium 67014.1 Requirements for Equilibrium, 67014.2 Equilibrium Between Two Phases of a Pure Substance, 67214.3 Metastable Equilibrium, 67614.4 Chemical Equilibrium, 67714.5 Simultaneous Reactions, 68714.6 Coal Gasification, 69114.7 Ionization, 69214.8 Engineering Applications, 694Summary, 697Problems, 698



xiv CONTENTS .............................................................................................................................................................................

15 Compressible Flow 70815.1 Stagnation Properties, 70815.2 The Momentum Equation for a Control Volume, 71015.3 Forces Acting on a Control Surface, 71315.4 Adiabatic, One-Dimensional, Steady-State Flow of an

Incompressible Fluid through a Nozzle, 71515.5 Velocity of Sound in an Ideal Gas, 71715.6 Reversible, Adiabatic, One-Dimensional Flow of an

Ideal Gas through a Nozzle, 72015.7 Mass Flow Rate of an Ideal Gas through an

Isentropic Nozzle, 72315.8 Normal Shock in an Ideal Gas Flowing through a Nozzle, 72815.9 Nozzle and Diffuser Coefficients, 73315.10 Nozzles and Orifices as Flow-Measuring Devices, 736Summary, 740Problems, 745

Contents of Appendix 753Appendix A SI Units: Single-State Properties 755Appendix B SI Units: Thermodynamic Tables 775Appendix C Ideal Gas Specific Heat 825Appendix D Equations of State 827Appendix E Figures 832Appendix F English Unit Tables 837

Answers to Selected Problems 878

Index 889



Symbols

a accelerationA areaa, A specific Helmholtz function and total Helmholtz functionAF air-fuel ratioBS adiabatic bulk modulusBT isothermal bulk modulusc velocity of soundc mass fractionCD coefficient of dischargeC p constant-pressure specific heatCv constant-volume specific heatC po zero-pressure constant-pressure specific heatCvo zero-pressure constant-volume specific heatCOP coefficient of performanceCR compression ratioe, E specific energy and total energyEMF electromotive forceF forceFA fuel-air ratiog acceleration due to gravityg, G specific Gibbs function and total Gibbs functionh, H specific enthalpy and total enthalpyHV heating valuei electrical currentI irreversibilityJ proportionality factor to relate units of work to units of heatk specific heat ratio: C p/CvK equilibrium constantKE kinetic energyL lengthm massm mass flow rateM molecular massM Mach numbern number of molesn polytropic exponentP pressurePi partial pressure of component i in a mixturePE potential energy

xv



xvi SYMBOLS ...............................................................................................................................................................................

Pr reduced pressure P/PcPr relative pressure as used in gas tablesq, Q heat transfer per unit mass and total heat transferQ rate of heat transferQ H , QL heat transfer with high-temperature body and heat transfer with

low-temperature body; sign determined from contextR gas constantR universal gas constants, S specific entropy and total entropySgen entropy generationSgen rate of entropy generationt timeT temperatureTr reduced temperature T /T cu, U specific internal energy and total internal energyv, V specific volume and total volumevr relative specific volume as used in gas tablesV velocityw, W work per unit mass and total workW rate of work, or powerw rev reversible work between two statesx qualityy gas-phase mole fractiony extraction fractionZ elevationZ compressibility factorZ electrical charge

Script Letters e electrical potentials surface tensiont tension

Greek Letters residual volume dimensionless Helmholtz function a/RTp volume expansivity coefficient of performance for a refrigerator coefficient of performance for a heat pumpS adiabatic compressibilityT isothermal compressibility dimensionless density /c efficiency chemical potential stoichiometric coefficient density dimensionless temperature variable Tc/T0 dimensionless temperature variable 1 Tr equivalence ratio relative humidity



SYMBOLS xvii.............................................................................................................................................................. ................

, exergy or availability for a control mass exergy, flow availability humidity ratio or specific humidity acentric factor

Subscripts c property at the critical pointc.v. control volumee state of a substance leaving a control volumef formationf property of saturated liquidfg difference in property for saturated vapor and saturated liquidg property of saturated vapori state of a substance entering a control volumei property of saturated solidif difference in property for saturated liquid and saturated solidig difference in property for saturated vapor and saturated solidr reduced propertys isentropic process0 property of the surroundings0 stagnation property

Superscripts bar over symbol denotes property on a molal basis (over V , H , S, U , A, G,the bar denotes partial molal property)

property at standard-state condition ideal gas property at the throat of a nozzleirr irreversibler real gas partrev reversible



xviii


P1: OSO/OVY P2: OSO/OVY QC: SCF/OVY T1: SCF

WB0077601 JWCL672-Borgnakke-v1 September 15, 2012 19:4

1Introduction andPreliminariesThe field of thermodynamics is concerned with the science of energy focusing on energystorage and energy conversion processes. We will study the effects on different substances,as we may expose a mass to heating/cooling or to volumetric compression/expansion. Duringsuch processes we are transferring energy into or out of the mass, so it changes its conditionsexpressed by properties like temperature, pressure, and volume. We use several processessimilar to this in our daily lives; we heat water to make coffee or tea or cool it in a refrigeratorto make cold water or ice cubes in a freezer. In nature, water evaporates from oceans andlakes and mixes with air where the wind can transport it, and later the water may drop outof the air as either rain (liquid water) or snow (solid water). As we study these processesin detail, we will focus on situations that are physically simple and yet typical of real-lifesituations in industry or nature.

By a combination of processes, we are able to illustrate more complex devices orcomplete systemsfor instance, a simple steam power plant that is the basic system thatgenerates the majority of our electric power. A power plant that produces electric powerand hot water for district heating burns coal, as shown in Fig. 1.1. The coal is suppliedby ship, and the district heating pipes are located in underground tunnels and thus are notvisible. A more technical description and a better understanding are obtained from thesimple schematic of the power plant, as shown in Fig. 1.2. This includes various outputsfrom the plant as electric power to the net, warm water for district heating, slag from burningcoal, and other materials like ash and gypsum; the last output is a flow of exhaust gases outof the chimney.

Another set of processes forms a good description of a refrigerator that we use tocool food or apply it at very low temperatures to produce a flow of cold fluid for cryogenicsurgery by freezing tissue for minimal bleeding. A simple schematic for such a system isshown in Fig. 1.3. The same system can also function as an air conditioner with the dualpurpose of cooling a building in summer and heating it in winter; in this last mode of use, itis also called a heat pump. For mobile applications, we can make simple models for gasolineand diesel engines typically used for ground transportation and gas turbines in jet enginesused in aircraft, where low weight and volume are of prime concern. These are just a fewexamples of familiar systems that the theory of thermodynamics allows us to analyze. Oncewe learn and understand the theory, we will be able to extend the analysis to other cases wemay not be familiar with.

Beyond the description of basic processes and systems, thermodynamics is extendedto cover special situations like moist atmospheric air, which is a mixture of gases, andthe combustion of fuels for use in the burning of coal, oil, or natural gas, which is achemical and energy conversion process used in nearly all power-generating devices. Manyother extensions are known; these can be studied in specialty texts. Since all the processes

1



2 CHAPTER ONE INTRODUCTION AND PRELIMINARIES ..........................................................................................................................

(Courtesy of Dong Energy A/S, Denmark.)

FIGURE 1.1 A powerstation in Esbjerg,Denmark.

engineers deal with have an impact on the environment, we must be acutely aware of theways in which we can optimize the use of our natural resources and produce the minimalamount of negative consequences for our environment. For this reason, the treatment ofefficiencies for processes and devices is important in a modern analysis and is requiredknowledge for a complete engineering consideration of system performance and operation.

Before considering the application of the theory, we will cover a few basic conceptsand definitions for our analysis and review some material from physics and chemistry thatwe will need.

1.1 A THERMODYNAMIC SYSTEM ANDTHE CONTROL VOLUME

A thermodynamic system is a device or combination of devices containing a quantity ofmatter that is being studied. To define this more precisely, a control volume is chosen sothat it contains the matter and devices inside a control surface. Everything external to the



A THERMODYNAMIC SYSTEM AND THE CONTROL VOLUME 3................................................................................................... ................

Powergrid

purifier

Chimney

GypsumFlyash

Coalgrinder

Oil

Air Slag

Coalsilo

Turbine Generator

Districtheating

Heatexchanger

Gas Ashseparator

Steamdrum

Flue gas

Pump

FIGURE 1.2 Schematic diagram of a steam power plant.

control volume is the surroundings, with the separation provided by the control surface.The surface may be open or closed to mass flows, and it may have flows of energy in termsof heat transfer and work across it. The boundaries may be movable or stationary. In thecase of a control surface that is closed to mass flow, so that no mass can escape or enter

Heat to room

3

1

2

4

Compressor

Condenser

Work

Evaporator

Cold vapor

Warm vapor

Warm liquid

Cold liquid + vapor

Expansion valveor

capillary tube

Heat from coldrefrigerated space

FIGURE 1.3Schematic diagramof a refrigerator.




Weights

Piston

Systemboundary

g

P0

Gas

FIGURE 1.4 Exampleof a control mass.

the control volume, it is called a control mass containing the same amount of matter at alltimes.

Selecting the gas in the cylinder of Fig. 1.4 as a control volume by placing a controlsurface around it, we recognize this as a control mass. If a Bunsen burner is placed underthe cylinder, the temperature of the gas will increase and the piston will move out. As thepiston moves, the boundary of the control mass also changes. As we will see later, heatand work cross the boundary of the control mass during this process, but the matter thatcomposes the control mass can always be identified and remains the same.

An isolated system is one that is not influenced in any way by the surroundings sothat no mass, heat, or work is transferred across the boundary of the system. In a moretypical case, a thermodynamic analysis must be made of a device like an air compressorwhich has a flow of mass into and out of it, as shown schematically in Fig. 1.5. The realsystem includes possibly a storage tank, as shown later in Fig. 1.20. In such an analysis,we specify a control volume that surrounds the compressor with a surface that is called thecontrol surface, across which there may be a transfer of mass, and momentum, as well asheat and work.

Thus, the more general control surface defines a control volume, where mass mayflow in or out, with a control mass as the special case of no mass flow in or out. Hence,the control mass contains a fixed mass at all times, which explains its name. The generalformulation of the analysis is considered in detail in Chapter 4. The terms closed system(fixed mass) and open system (involving a flow of mass) are sometimes used to make thisdistinction. Here, we use the term system as a more general and loose description for amass, device, or combination of devices that then is more precisely defined when a controlvolume is selected. The procedure that will be followed in presenting the first and second

Controlsurface

Heat

High-pressureair out

to storage tank

Work

Aircompressor

Low-pressureair in

MotorFIGURE 1.5 Exampleof a control volume.



MACROSCOPIC VERSUS MICROSCOPIC POINTS OF VIEW 5...................................................................................................... ................

laws of thermodynamics is first to present these laws for a control mass and then to extendthe analysis to the more general control volume.

1.2 MACROSCOPIC VERSUS MICROSCOPICPOINTS OF VIEW

The behavior of a system may be investigated from either a microscopic or macroscopicpoint of view. Let us briefly describe a system from a microscopic point of view. Consider asystem consisting of a cube 25 mm on a side and containing a monatomic gas at atmosphericpressure and temperature. This volume contains approximately 1020 atoms. To describe theposition of each atom, we need to specify three coordinates; to describe the velocity of eachatom, we specify three velocity components.

Thus, to describe completely the behavior of this system from a microscopic pointof view, we must deal with at least 6 1020 equations. Even with a modern computer,this is a hopeless computational task. However, there are two approaches to this problemthat reduce the number of equations and variables to a few that can be computed relativelyeasily. One is the statistical approach, in which, on the basis of statistical considerationsand probability theory, we deal with average values for all particles under consideration.This is usually done in connection with a model of the atom under consideration. This isthe approach used in the disciplines of kinetic theory and statistical mechanics.

The other approach to reducing the number of variables to a few that can be handledrelatively easily involves the macroscopic point of view of classical thermodynamics. Asthe word macroscopic implies, we are concerned with the gross or average effects of manymolecules. These effects can be perceived by our senses and measured by instruments.However, what we really perceive and measure is the time-averaged influence of manymolecules. For example, consider the pressure a gas exerts on the walls of its container.This pressure results from the change in momentum of the molecules as they collide withthe wall. From a macroscopic point of view, however, we are concerned not with the actionof the individual molecules but with the time-averaged force on a given area, which canbe measured by a pressure gauge. In fact, these macroscopic observations are completelyindependent of our assumptions regarding the nature of matter.

Although the theory and development in this book are presented from a macroscopicpoint of view, a few supplementary remarks regarding the significance of the microscopicperspective are included as an aid to understanding the physical processes involved. Anotherbook in this series, Introduction to Thermodynamics: Classical and Statistical, by R. E.Sonntag and G. J. Van Wylen, includes thermodynamics from the microscopic and statisticalpoint of view.

A few remarks should be made regarding the continuum approach. We are normallyconcerned with volumes that are very large compared to molecular dimensions and withtime scales that are very large compared to intermolecular collision frequencies. For thisreason, we deal with very large numbers of molecules that interact extremely often duringour observation period, so we sense the system as a simple uniformly distributed mass in thevolume called a continuum. This concept, of course, is only a convenient assumption thatloses validity when the mean free path of the molecules approaches the order of magnitudeof the dimensions of the vessel, as, for example, in high-vacuum technology. In muchengineering work the assumption of a continuum is valid and convenient, consistent withthe macroscopic point of view.




1.3 PROPERTIES AND STATE OF A SUBSTANCEIf we consider a given mass of water, we recognize that this water can exist in various forms.If it is a liquid initially, it may become a vapor when it is heated or a solid when it is cooled.Thus, we speak of the different phases of a substance. A phase is defined as a quantity ofmatter that is homogeneous throughout. When more than one phase is present, the phases areseparated from each other by the phase boundaries. In each phase the substance may exist atvarious pressures and temperatures or, to use the thermodynamic term, in various states. Thestate may be identified or described by certain observable, macroscopic properties; somefamiliar ones are temperature, pressure, and density. In later chapters, other properties will

THERMONET

be introduced. Each of the properties of a substance in a given state has only one definitevalue, and these properties always have the same value for a given state, regardless of howthe substance arrived at the state. In fact, a property can be defined as any quantity thatdepends on the state of the system and is independent of the path (that is, the prior history)by which the system arrived at the given state. Conversely, the state is specified or describedby the properties. Later we will consider the number of independent properties a substancecan have, that is, the minimum number of properties that must be specified to fix the stateof the substance.

Thermodynamic properties can be divided into two general classes: intensive andextensive. An intensive property is independent of the mass; the value of an extensive prop-erty varies directly with the mass. Thus, if a quantity of matter in a given state is divided intotwo equal parts, each part will have the same value of intensive properties as the originaland half the value of the extensive properties. Pressure, temperature, and density are exam-ples of intensive properties. Mass and total volume are examples of extensive properties.Extensive properties per unit mass, such as specific volume, are intensive properties.

Frequently we will refer not only to the properties of a substance but also to theproperties of a system. When we do so, we necessarily imply that the value of the prop-erty has significance for the entire system, and this implies equilibrium. For example, ifthe gas that composes the system (control mass) in Fig. 1.4 is in thermal equilibrium, thetemperature will be the same throughout the entire system, and we may speak of the tem-perature as a property of the system. We may also consider mechanical equilibrium, whichis related to pressure. If a system is in mechanical equilibrium, there is no tendency forthe pressure at any point to change with time as long as the system is isolated from thesurroundings. There will be variation in pressure with elevation because of the influence ofgravitational forces, although under equilibrium conditions there will be no tendency forthe pressure at any location to change. However, in many thermodynamic problems, thisvariation in pressure with elevation is so small that it can be neglected. Chemical equilib-rium is also important and will be considered in Chapter 14. When a system is in equilib-rium regarding all possible changes of state, we say that the system is in thermodynamicequilibrium.

1.4 PROCESSES AND CYCLESWhenever one or more of the properties of a system change, we say that a change in statehas occurred. For example, when one of the weights on the piston in Fig. 1.6 is removed,the piston rises and a change in state occurs, for the pressure decreases and the specific



PROCESSES AND CYCLES 7........................................................................................................................................... ................

Weights

Piston

Systemboundary

g

P0

GasFIGURE 1.6 Exampleof a system that mayundergo a quasi-equilibrium process.

volume increases. The path of the succession of states through which the system passes iscalled the process.

Let us consider the equilibrium of a system as it undergoes a change in state. Themoment the weight is removed from the piston in Fig. 1.6, mechanical equilibrium doesnot exist; as a result, the piston is moved upward until mechanical equilibrium is restored.The question is this: Since the properties describe the state of a system only when it isin equilibrium, how can we describe the states of a system during a process if the actualprocess occurs only when equilibrium does not exist? One step in finding the answer tothis question concerns the definition of an ideal process, which we call a quasi-equilibriumprocess. A quasi-equilibrium process is one in which the deviation from thermodynamicequilibrium is infinitesimal, and all the states the system passes through during a quasi-equilibrium process may be considered equilibrium states. Many actual processes closelyapproach a quasi-equilibrium process and may be so treated with essentially no error. Ifthe weights on the piston in Fig. 1.6 are small and are taken off one by one, the processcould be considered quasi-equilibrium. However, if all the weights are removed at once, thepiston will rise rapidly until it hits the stops. This would be a nonequilibrium process, andthe system would not be in equilibrium at any time during this change of state.

For nonequilibrium processes, we are limited to a description of the system beforethe process occurs and after the process is completed and equilibrium is restored. We areunable to specify each state through which the system passes or the rate at which the processoccurs. However, as we will see later, we are able to describe certain overall effects thatoccur during the process.

Several processes are described by the fact that one property remains constant. Theprefix iso- is used to describe such a process. An isothermal process is a constant-temperatureprocess, an isobaric process is a constant-pressure process, and an isochoric process is aconstant-volume process.

When a system in a given initial state goes through a number of different changes ofstate or processes and finally returns to its initial state, the system has undergone a cycle.Therefore, at the conclusion of a cycle, all the properties have the same value they had atthe beginning. Steam (water) that circulates through a steam power plant undergoes a cycle.

A distinction should be made between a thermodynamic cycle, which has just beendescribed, and a mechanical cycle. A four-stroke-cycle internal-combustion engine goesthrough a mechanical cycle once every two revolutions. However, the working fluid doesnot go through a thermodynamic cycle in the engine, since air and fuel are burned andchanged to products of combustion that are exhausted to the atmosphere. In this book, theterm cycle will refer to a thermodynamic cycle unless otherwise designated.




1.5 UNITS FOR MASS, LENGTH, TIME, AND FORCESince we are considering thermodynamic properties from a macroscopic perspective, weare dealing with quantities that can, either directly or indirectly, be measured and counted.Therefore, the matter of units becomes an important consideration. In the remaining sec-tions of this chapter we will define certain thermodynamic properties and the basic units.Because the relation between force and mass is often difficult for students to understand, itis considered in this section in some detail.

Force, mass, length, and time are related by Newtons second law of motion, which

THERMONET

states that the force acting on a body is proportional to the product of the mass and theacceleration in the direction of the force:

F maThe concept of time is well established. The basic unit of time is the second (s), which

in the past was defined in terms of the solar day, the time interval for one complete revolutionof the earth relative to the sun. Since this period varies with the season of the year, an averagevalue over a 1-year period is called the mean solar day, and the mean solar second is 1/86 400of the mean solar day. In 1967, the General Conference of Weights and Measures (CGPM)adopted a definition of the second as the time required for a beam of cesium-133 atoms toresonate 9 192 631 770 cycles in a cesium resonator.

For periods of time less than 1 s, the prefixes milli, micro, nano, pico, or femto, as listedin Table 1.1, are commonly used. For longer periods of time, the units minute (min), hour (h),or day (day) are frequently used. It should be pointed out that the prefixes in Table 1.1 areused with many other units as well.

The concept of length is also well established. The basic unit of length is the meter (m),which used to be marked on a platinumiridium bar. Currently, the CGPM has adopted amore precise definition of the meter in terms of the speed of light (which is now a fixedconstant): The meter is the length of the path traveled by light in a vacuum during a timeinterval of 1/299 792 458 of a second.

The fundamental unit of mass is the kilogram (kg). As adopted by the first CGPM in1889 and restated in 1901, it is the mass of a certain platinumiridium cylinder maintainedunder prescribed conditions at the International Bureau of Weights and Measures. A relatedunit that is used frequently in thermodynamics is the mole (mol), defined as an amount of sub-stance containing as many elementary entities as there are atoms in 0.012 kg of carbon-12.These elementary entities must be specified; they may be atoms, molecules, electrons, ions,or other particles or specific groups. For example, 1 mol of diatomic oxygen, having a

TABLE 1.1Unit Prefixes

Factor Prefix Symbol Factor Prefix Symbol

1015 peta P 103 milli m1012 tera T 106 micro 109 giga G 109 nano n106 mega M 1012 pico p103 kilo k 1015 femto f



UNITS FOR MASS, LENGTH, TIME, AND FORCE 9................................................................................................................... ................

molecular mass of 32 (compared to 12 for carbon), has a mass of 0.032 kg. The mole isoften termed a gram mole, since it is an amount of substance in grams numerically equal tothe molecular mass. In this book, when using the metric SI system, we will find it preferableto use the kilomole (kmol), the amount of substance in kilograms numerically equal to themolecular mass, rather than the mole.

The system of units in use presently throughout most of the world is the metricInternational System, commonly referred to as SI units (from Le Syste`me InternationaldUnites). In this system, the second, meter, and kilogram are the basic units for time,length, and mass, respectively, as just defined, and the unit of force is defined directly fromNewtons second law.

Therefore, a proportionality constant is unnecessary, and we may write that law as anequality:

F = ma (1.1)The unit of force is the newton (N), which by definition is the force required to acceleratea mass of 1 kg at the rate of 1 m/s2:

1 N = 1 kg m/s2

It is worth noting that SI units derived from proper nouns use capital letters for symbols;others use lowercase letters. The liter, with the symbol L, is an exception.

The traditional system of units used in the United States is the English EngineeringSystem. In this system the unit of time is the second, which was discussed earlier. The basicunit of length is the foot (ft), which at present is defined in terms of the meter as

1 ft = 0.3048 mThe inch (in.) is defined in terms of the foot:

12 in. = 1 ftThe unit of mass in this system is the pound mass (lbm). It was originally defined as themass of a certain platinum cylinder kept in the Tower of London, but now it is defined interms of the kilogram as

1 lbm = 0.453 592 37 kgA related unit is the pound mole (lb mol), which is an amount of substance in pounds massnumerically equal to the molecular mass of that substance. It is important to distinguishbetween a pound mole and a mole (gram mole).

In the English Engineering System of Units, the unit of force is the pound force(lbf), defined as the force with which the standard pound mass is attracted to the earthunder conditions of standard acceleration of gravity, which is that at 45 latitude and sealevel elevation, 9.806 65 m/s2 or 32.1740 ft/s2. Thus, it follows from Newtons second lawthat

1 lbf = 32.174 lbm ft/s2

which is a necessary factor for the purpose of units conversion and consistency. Note thatwe must be careful to distinguish between an lbm and an lbf, and we do not use the termpound alone.

The term weight is often used with respect to a body and is sometimes confused withmass. Weight is really correctly used only as a force. When we say that a body weighs so




much, we mean that this is the force with which it is attracted to the earth (or some otherbody), that is, the product of its mass and the local gravitational acceleration. The mass ofa substance remains constant with elevation, but its weight varies with elevation.

Example 1.1What is the weight of a 1-kg mass at an altitude where the local acceleration of gravity is9.75 m/s2?

SolutionWeight is the force acting on the mass, which from Newtons second law is

F = mg = 1 kg 9.75 m/s2 [1 N s2/kg m] = 9.75 N

Example 1.1EWhat is the weight of a 1-lbm mass at an altitude where the local acceleration of gravityis 32.0 ft/s2?

SolutionWeight is the force acting on the mass, which from Newtons second law is

F = mg = 1 lbm 32.0 ft/s2 [lbf s2/32.174 lbm ft] = 0.9946 lbf

In-Text Concept Questions

a. Make a control volume around the turbine in the steam power plant in Fig. 1.2 andlist the flows of mass and energy located there.

b. Take a control volume around your kitchen refrigerator, indicate where the compo-nents shown in Fig. 1.3 are located, and show all energy transfers.

1.6 SPECIFIC VOLUME AND DENSITYThe specific volume of a substance is defined as the volume per unit mass and is giventhe symbol v. The density of a substance is defined as the mass per unit volume, and itis therefore the reciprocal of the specific volume. Density is designated by the symbol .Specific volume and density are intensive properties.

The specific volume of a system in a gravitational field may vary from point to point.For example, if the atmosphere is considered a system, the specific volume increases as



SPECIFIC VOLUME AND DENSITY 11.................................................................................................................................. ................

v

V

V m

V

FIGURE 1.7 Thecontinuum limit for thespecific volume.

the elevation increases. Therefore, the definition of specific volume involves the specificvolume of a substance at a point in a system.

Consider a small volume V of a system, and let the mass be designated m. Thespecific volume is defined by the relation

v = limV V

V

m

where V is the smallest volume for which the mass can be considered a continuum.Volumes smaller than this will lead to the recognition that mass is not evenly distributedin space but is concentrated in particles as molecules, atoms, electrons, and so on. This istentatively indicated in Fig. 1.7, where in the limit of a zero volume the specific volume maybe infinite (the volume does not contain any mass) or very small (the volume is part of anucleus).

Thus, in a given system, we should speak of the specific volume or density at a pointin the system and recognize that this may vary with elevation. However, most of the systemsthat we consider are relatively small, and the change in specific volume with elevation isnot significant. Therefore, we can speak of one value of specific volume or density for theentire system.

In this book, the specific volume and density will be given either on a mass or a molebasis. A bar over the symbol (lowercase) will be used to designate the property on a molebasis. Thus, v will designate molal specific volume and will designate molal density.In SI units, those for specific volume are m3/kg and m3/mol (or m3/kmol); for densitythe corresponding units are kg/m3 and mol/m3 (or kmol/m3). In English units, those forspecific volume are ft3/lbm and ft3/lb mol; the corresponding units for density are lbm/ft3 andlb mol/ft3.

Although the SI unit for volume is the cubic meter, a commonly used volume unitis the liter (L), which is a special name given to a volume of 0.001 m3, that is, 1 L =103 m3. The general ranges of density for some common solids, liquids, and gases areshown in Fig. 1.8. Specific values for various solids, liquids, and gases in SI units arelisted in Tables A.3, A.4, and A.5, respectively, and in English units in Tables F.2, F.3,and F.4.




FiberAtm.air

Gas invacuum

Wood Al LeadCottonWool

Propane Water Hg

IceRock Ag Au

102 101 100 101Density [kg/m3]

102 103 104

Solids

Liquids

Gases

FIGURE 1.8 Densityof common substances.

Example 1.2A 1-m3 container, shown in Fig. 1.9, is filled with 0.12 m3 of granite, 0.15 m3 of sand,and 0.2 m3 of liquid 25C water; the rest of the volume, 0.53 m3, is air with a density of1.15 kg/m3. Find the overall (average) specific volume and density.

SolutionFrom the definition of specific volume and density we have

v = V/m and = m/V = 1/vWe need to find the total mass, taking density from Tables A.3 and A.4:

mgranite = Vgranite = 2750 kg/m3 0.12 m3 = 330 kgmsand = sand Vsand = 1500 kg/m3 0.15 m3 = 225 kg

mwater = water Vwater = 997 kg/m3 0.2 m3 = 199.4 kgmair = air Vair = 1.15 kg/m3 0.53 m3 = 0.61 kg

Air

FIGURE 1.9 Sketch for Example 1.2.

Now the total mass becomes

m tot = mgranite + msand + mwater + mair = 755 kg



PRESSURE 13............................................................................................................................................................. ................

and the specific volume and density can be calculated:

v = Vtot/m tot = 1 m3/755 kg = 0.001325 m3/kg = m tot/Vtot = 755 kg/1 m3 = 755 kg/m3

Remark: It is misleading to include air in the numbers for and V , as the air is separatefrom the rest of the mass.


c. Why do people float high in the water when swimming in the Dead Sea as comparedwith swimming in a freshwater lake?

d. The density of liquid water is = 1008 T /2 [kg/m3] with T in C. If the temperatureincreases, what happens to the density and specific volume?

1.7 PRESSUREWhen dealing with liquids and gases, we ordinarily speak of pressure; for solids we speakof stresses. The pressure in a fluid at rest at a given point is the same in all directions, andwe define pressure as the normal component of force per unit area. More specifically, if Ais a small area, A is the smallest area over which we can consider the fluid a continuum,and Fn is the component of force normal to A, we define pressure, P, as

P = limAA

FnA

where the lower limit corresponds to sizes as mentioned for the specific volume, shown inFig. 1.7. The pressure P at a point in a fluid in equilibrium is the same in all directions. Ina viscous fluid in motion, the variation in the state of stress with orientation becomes animportant consideration. These considerations are beyond the scope of this book, and wewill consider pressure only in terms of a fluid in equilibrium.

The unit for pressure in the International System is the force of one newton acting ona square meter area, which is called the pascal (Pa). That is,

1 Pa = 1 N/m2

Two other units, not part of the International System, continue to be widely used.These are the bar, where

1 bar = 105 Pa = 0.1 MPaand the standard atmosphere, where

1 atm = 101 325 Pa = 14.696 lbf/in.2




GasP Fext

FIGURE 1.10 Thebalance of forces on amovable boundaryrelates to inside gaspressure.

which is slightly larger than the bar. In this book, we will normally use the SI unit, the pascal,and especially the multiples of kilopascal and megapascal. The bar will be utilized oftenin the examples and problems, but the atmosphere will not be used, except in specifyingcertain reference points.

Consider a gas contained in a cylinder fitted with a movable piston, as shown inFig. 1.10. The pressure exerted by the gas on all of its boundaries is the same, assumingthat the gas is in an equilibrium state. This pressure is fixed by the external force actingon the piston, since there must be a balance of forces for the piston to remain stationary.Thus, the product of the pressure and the movable piston area must be equal to the externalforce. If the external force is now changed in either direction, the gas pressure inside mustaccordingly adjust, with appropriate movement of the piston, to establish a force balanceat a new equilibrium state. As another example, if the gas in the cylinder is heated by anoutside body, which tends to increase the gas pressure, the piston will move instead, suchthat the pressure remains equal to whatever value is required by the external force.

Example 1.3The hydraulic piston/cylinder system shown in Fig. 1.11 has a cylinder diameter of D =0.1 m with a piston and rod mass of 25 kg. The rod has a diameter of 0.01 m with anoutside atmospheric pressure of 101 kPa. The inside hydraulic fluid pressure is 250 kPa.How large a force can the rod push with in the upward direction?

SolutionWe will assume a static balance of forces on the piston (positive upward), so

Fnet = ma = 0= Pcyl Acyl P0(Acyl Arod) F mpg

P0

Arod

Pcyl

F




PRESSURE 15............................................................................................................................................................. ................

Solve for F:

F = Pcyl Acyl P0(Acyl Arod) m pgThe areas are

Acyl = r2 = D2/4 = 4

0.12 m2 = 0.007 854 m2

Arod = r2 = D2/4 = 4

0.012 m2 = 0.000 078 54 m2

So the force becomes

F = [250 0.007 854 101(0.007 854 0.000 078 54)]1000 25 9.81= 1963.5 785.32 245.25= 932.9 N

Note that we must convert kPa to Pa to get units of N.

In most thermodynamic investigations we are concerned with absolute pressure. Mostpressure and vacuum gauges, however, read the difference between the absolute pressureand the atmospheric pressure existing at the gauge. This is referred to as gauge pressure.It is shown graphically in Fig. 1.12, and the following examples illustrate the principles.Pressures below atmospheric and slightly above atmospheric, and pressure differences (forexample, across an orifice in a pipe), are frequently measured with a manometer, whichcontains water, mercury, alcohol, oil, or other fluids.

Pabs,1

Patm

Pabs,2

Ordinary pressure gaugeP = Pabs,1 Patm

Ordinary vacuum gaugeP = Patm Pabs,2

Barometer readsatmospheric pressure

O

P

FIGURE 1.12Illustration of termsused in pressuremeasurement.




FluidP

A

Patm = P0

gH

BFIGURE 1.13Example of pressuremeasurement using acolumn of fluid.

Consider the column of fluid of height H standing above point B in the manometershown in Fig. 1.13. The force acting downward at the bottom of the column is

P0 A + mg = P0 A + AgHwhere m is the mass of the fluid column, A is its cross-sectional area, and is its density.This force must be balanced by the upward force at the bottom of the column, which is PBA.Therefore,

PB P0 = gHSince points A and B are at the same elevation in columns of the same fluid, their pressures

P 0

Patm g

H0

FIGURE 1.14Barometer.

must be equal (the fluid being measured in the vessel has a much lower density, such thatits pressure P is equal to PA). Overall,

P = P P0 = gH (1.2)For distinguishing between absolute and gauge pressure in this book, the term pascal

will always refer to absolute pressure. Any gauge pressure will be indicated as such.Consider the barometer used to measure atmospheric pressure, as shown in Fig. 1.14.

Since there is a near vacuum in the closed tube above the vertical column of fluid, usu-ally mercury, the height of the fluid column gives the atmospheric pressure directly fromEq. 1.2:

Patm = gH0 (1.3)

Example 1.4A mercury barometer located in a room at 25C has a height of 750 mm. What is theatmospheric pressure in kPa?

SolutionThe density of mercury at 25C is found from Table A.4 to be 13 534 kg/m3. UsingEq. 1.3,

Patm = gH0 = 13 534 kg/m3 9.807 m/s2 0.750 m/1000= 99.54 kPa



PRESSURE 17............................................................................................................................................................. ................

Example 1.5A mercury (Hg) manometer is used to measure the pressure in a vessel as shown inFig. 1.13. The mercury has a density of 13 590 kg/m3, and the height difference between thetwo columns is measured to be 24 cm. We want to determine the pressure inside the vessel.

SolutionThe manometer measures the gauge pressure as a pressure difference. From Eq. 1.2,

P = Pgauge = gH = 13 590 kg/m3 9.807 m/s2 0.24 m

= 31 985 Pa = 31.985 kPa= 0.316 atm

To get the absolute pressure inside the vessel, we have

PA = Pvessel = PB = P + PatmWe need to know the atmospheric pressure measured by a barometer (absolute pressure). As-sume that this pressure is known to be 750 mm Hg. The absolute pressure in the vessel becomes

Pvessel = P + Patm = 31 985 Pa + 13 590 kg/m3 0.750 m 9.807 m/s2

= 31 985 + 99 954 = 131 940 Pa = 1.302 atm

Example 1.5EA mercury (Hg) manometer is used to measure the pressure in a vessel as shown inFig. 1.13. The mercury has a density of 848 lbm/ft3, and the height difference between thetwo columns is measured to be 9.5 in. We want to determine the pressure inside the vessel.

SolutionThe manometer measures the gauge pressure as a pressure difference. From Eq. 1.2,

P = Pgauge = gH

= 848 lbmft3

32.174 fts2

9.5 in. 11728

ft3

in.3

[1 lbf s2

32.174 lbm ft

]

= 4.66 lbf/in.2To get the absolute pressure inside the vessel, we have

PA = Pvessel = P0 = P + PatmWe need to know the atmospheric pressure measured by a barometer (absolute pressure). As-sume that this pressure is known to be 29.5 in. Hg. The absolute pressure in the vessel becomes

Pvessel = P + Patm= 848 32.174 29.5 1

1728

(1

32.174

)+ 4.66

= 19.14 lbf/in.2




Example 1.6What is the pressure at the bottom of the 7.5-m-tall storage tank of fluid at 25C shownin Fig. 1.15? Assume that the fluid is gasoline with atmospheric pressure 101 kPa on thetop surface. Repeat the question for the liquid refrigerant R-134a when the top surfacepressure is 1 MPa.

H

FIGURE 1.15 Sketchfor Example 1.6.

SolutionThe densities of the liquids are listed in Table A.4:

gasoline = 750 kg/m3; R-134a = 1206 kg/m3

The pressure difference due to gravity is, from Eq. 1.2,

P = gHThe total pressure is

P = Ptop + PFor the gasoline we get

P = gH = 750 kg/m3 9.807 m/s2 7.5 m = 55 164 PaNow convert all pressures to kPa:

P = 101 + 55.164 = 156.2 kPaFor the R-134a we get

P = gH = 1206 kg/m3 9.807 m/s2 7.5 m = 88 704 PaNow convert all pressures to kPa:

P = 1000 + 88.704 = 1089 kPa

Example 1.7A piston/cylinder with a cross-sectional area of 0.01 m2 is connected with a hydraulicline to another piston/cylinder with a cross-sectional area of 0.05 m2. Assume that bothchambers and the line are filled with hydraulic fluid of density 900 kg/m3 and the largersecond piston/cylinder is 6 m higher up in elevation. The telescope arm and the buckets havehydraulic piston/cylinders moving them, as seen in Fig. 1.16. With an outside atmosphericpressure of 100 kPa and a net force of 25 kN on the smallest piston, what is the balancingforce on the second larger piston?

H

F2

F1P1

P2




ENERGY 19............................................................................................................................................................... ................

SolutionWhen the fluid is stagnant and at the same elevation, we have the same pressure throughoutthe fluid. The force balance on the smaller piston is then related to the pressure (we neglectthe rod area) as

F1 + P0 A1 = P1 A1from which the fluid pressure is

P1 = P0 + F1/A1 = 100 kPa + 25 kN/0.01 m2 = 2600 kPaThe pressure at the higher elevation in piston/cylinder 2 is, from Eq. 1.2,

P2 = P1 gH = 2600 kPa 900 kg/m3 9.81 m/s2 6 m/(1000 Pa/kPa)= 2547 kPa

where the second term is divided by 1000 to convert from Pa to kPa. Then the force balanceon the second piston gives

F2 + P0 A2 = P2 A2F2 = (P2 P0)A2 = (2547 100) kPa 0.05 m2 = 122.4 kN

If the density is variable, we should consider Eq. 1.2 in differential form as

dP = g dhincluding the sign, so pressure drops with increasing height. Now the finite differencebecomes

P = P0 H

0g dh (1.4)

with the pressure P0 at zero height.


e. A car tire gauge indicates 195 kPa; what is the air pressure inside?

f. Can I always neglect P in the fluid above location A in Fig. 1.13? What circumstancesdoes that depend on?

g. A U tube manometer has the left branch connected to a box with a pressure of110 kPa and the right branch open. Which side has a higher column of fluid?

1.8 ENERGYA macroscopic amount of mass can possess energy in the form of internal energy inherentin its internal structure, kinetic energy in its motion, and potential energy associated withexternal forces acting on the mass. We write the total energy as

E = Internal + Kinetic + Potential = U + KE + PE




and the specific total energy becomes

e = E/m = u + ke + pe = u + 1/2V 2 + gz (1.5)where the kinetic energy is taken as the translational energy and the potential energy iswritten for the external force being the gravitational force assumed constant. If the mass isrotating, we should add a rotational kinetic energy (1/2 I 2) to the translational term. Whatis called internal energy on the macroscale has a similar set of energies associated with themicroscale motion of the individual molecules. This enables us to write

u = uext molecule + utranslation + uint molecule (1.6)as a sum of the potential energy from intermolecular forces between molecules, the moleculetranslational kinetic energy, and the energy associated with the molecular internal and atomicstructure.

Without going into detail, we realize that there is a difference between the intermolec-ular forces. Thus, the first term of the energy for a configuration where the molecules areclose together, as in a solid or liquid (high density), contrasts with the situation for a gaslike air, where the distance between the molecules is large (low density). In the limit of avery thin gas, the molecules are so far apart that they do not sense each other, unless theycollide and the first term becomes near zero. This is the limit we have when we consider asubstance to be an ideal gas, as will be covered in Chapter 2.

The translational energy depends only on the mass and center of mass velocity of themolecules, whereas the last energy term depends on the detailed structure. In general, wecan write the energy as

uint molecule = upotential + urotation + uvibration + uatoms (1.7)To illustrate the potential energy associated with the intermolecular forces, consider anoxygen molecule of two atoms, as shown in Fig. 1.17. If we want to separate the two atoms,we pull them apart with a force and thereby we do some work on the system, as explainedin Chapter 3. That amount of work equals the binding (potential) energy associated withthe two atoms as they are held together in the oxygen molecule.

Consider a simple monatomic gas such as helium. Each molecule consists of ahelium atom. Such an atom possesses electronic energy as a result of both orbital an-gular momentum of the electrons about the nucleus and angular momentum of the electronsspinning on their axes. The electronic energy is commonly very small compared with thetranslational energies. (Atoms also possess nuclear energy, which, except in the case ofnuclear reactions, is constant. We are not concerned with nuclear energy at this time.)When we consider more complex molecules, such as those composed of two or three

x

y

z

FIGURE 1.17 Thecoordinate system for adiatomic molecule.



ENERGY 21............................................................................................................................................................... ................

O

H H

O

H H

O

HH

FIGURE 1.18 The three principal vibrational modes for the H2O molecule.

atoms, additional factors must be considered. In addition to having electronic energy, amolecule can rotate about its center of gravity and thus have rotational energy. Further-more, the atoms may vibrate with respect to each other and have vibrational energy. Insome situations there may be an interaction between the rotational and vibrational modesof energy.

In evaluating the energy of a molecule, we often refer to the degree of freedom, f , ofthese energy modes. For a monatomic molecule such as helium, f = 3, which represents thethree directions x, y, and z in which the molecule can move. For a diatomic molecule, such asoxygen, f = 6. Three of these are the translation of the molecule as a whole in the x, y, andz directions, and two are for rotation. The reason that there are only two modes of rotationalenergy is evident from Fig. 1.17, where we take the origin of the coordinate system at thecenter of gravity of the molecule and the y-axis along the molecules internuclear axis. Themolecule will then have an appreciable moment of inertia about the x-axis and the z-axisbut not about the y-axis. The sixth degree of freedom of the molecule is vibration, whichrelates to stretching of the bond joining the atoms.

For a more complex molecule such as H2O, there are additional vibrational degreesof freedom. Fig. 1.18 shows a model of the H2O molecule. From this diagram, it is evidentthat there are three vibrational degrees of freedom. It is also possible to have rotationalenergy about all three axes. Thus, for the H2O molecule, there are nine degrees of freedom( f = 9): three translational, three rotational, and three vibrational.

Most complex molecules, such as typical polyatomic molecules, are usually three-dimensional in structure and have multiple vibrational modes, each of which contributes tothe energy storage of the molecule. The more complicated the molecule is, the larger thenumber of degrees of freedom that exist for energy storage. The modes of energy storageand their evaluation are discussed in some detail in Appendix C for those interested infurther development of the quantitative effects from a molecular viewpoint.

This general discussion can be summarized by referring to Fig. 1.19. Let heat beVapor H2O(steam)

Liquid H2O

Heat

FIGURE 1.19 Heattransfer to H2O.

transferred to H2O. During this process the temperature of the liquid and vapor (steam)will increase, and eventually all the liquid will become vapor. From the macroscopic pointof view, we are concerned only with the energy that is transferred as heat, the change inproperties such as temperature and pressure, and the total amount of energy (relative tosome base) that the H2O contains at any instant. Thus, questions about how energy is storedin the H2O do not concern us. From a microscopic viewpoint, we are concerned about theway in which

Thermodynamics 8th Edition

Documents

Transcript of Thermodynamics 8th Edition