Chapter 1 Introduction and Basic Concepts 11 Thermodynamics and Energy Application Areas of Thermodynamics 12 Importance of Dimensions and Units Some SI and English Units Dimensional Homogeneity Unity Conversion Ratios 13 Systems and Control Volumes 14 Properties of a System Continuum 15 Density and Specific Gravity 16 State and Equilibrium The State Postulate 17 Processes and Cycles The SteadyFlow Process 18 Temperature and the Zeroth Law of Thermodynamics Temperature Scales The International Temperature Scale of 1990 (ITS90) 19 Pressure Variation of Pressure with Depth 110 The Manometer Other Pressure Measurement Devices 111 The Barometer and Atmospheric Pressure 112 ProblemSolving Technique Step 1: Problem Statement Step 2: Schematic Step 3: Assumptions and Approximations Step 4: Physical Laws Step 5: Properties Step 6: Calculations Step 7: Reasoning, Verification, and Discussion Engineering Software Packages A Remark on Significant Digits Summary References and Suggested Reading Problems
Chapter 2 Energy Conversion and General Energy Analysis 21 Introduction 22 Forms of Energy Some Physical Insight to Internal Energy Mechanical Energy More on Nuclear Energy 23 Energy Transfer by Heat Historical Background on Heat 24 Energy Transfer by Work Electrical Work 25 Mechanical Forms of Work Shaft Work Spring Work Work Done on Elastic Solid Bars Work Associated with the Stretching of a Liquid Film Work Done to Raise or to Accelerate a Body Nonmechanical Forms of Work 26 The First Law of Thermodynamics Energy Balance Energy Change of a System, ∆Esystem Mechanisms of Energy Transfer, Ein and Eout 27 Energy Conversion Efficiencies 28 Energy and Environment Ozone and Smog Acid Rain The Greenhouse Effect: Global Warming and Climate Change Topic of Special Interest: Mechanisms of Heat Transfer Summary References and Suggested Reading Problems Chapter 3 Properties of Pure Substances 31 Pure Substance
32 Phases of a Pure Substance 33 PhaseChange Processes of Pure Substances Compressed Liquid and Saturated Liquid Saturated Vapor and Superheated Vapor Saturation Temperature and Saturation Pressure Some Consequences of Tsat and Psat Dependence 34 Property Diagrams for PhaseChange Processes 1 The Tv Diagram 2 The Pv Diagram Extending the Diagrams to Include the Solid Phase 3 The PT Diagram The PvT Surface 35 Property Tables Enthalpy—A Combination Property 1a Saturated Liquid and Saturated Vapor States 1b Saturated Liquid–Vapor Mixture 2 Superheated Vapor 3 Compressed Liquid Reference State and Reference Values 36 The IdealGas Equation of State Is Water Vapor an Ideal Gas? 37 Compressibility Factor—A Measure of Deviation from IdealGas Behavior 38 Other Equations of State Van der Waals Equation of State BeattieBridgeman Equation of State BenedictWebbRubin Equation of State Virial Equation of State Topic of Special Interest Vapor Pressure and Phase Equilibrium Summary References and Suggested Reading Problems Chapter 4 Energy Analysis of Closed Systems 41 Moving Boundary Work Polytropic Process
42 Energy Balance for Closed Systems 43 Specific Heats 44 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases Specific Heat Relations of Ideal Gases 45 Internal Energy, Enthalpy, and Specific Heat of Solids and Liquids Internal Energy Changes Enthalpy Changes Topic of Special Interest: Thermodynamic Aspects of Biological Systems Summary References and Suggested Reading Problems Chapter 5 Mass and Energy Analysis of Control Volumes 51 Conservation of Mass Mass and Volume Flow Rates Conservation of Mass Principle Mass Balance for SteadyFlow Processes Special Case: Incompressible Flow 52 Flow Work and the Energy of a Flowing Fluid Total Energy of a Flowing Fluid Energy Transport by Mass 53 Energy Analysis of SteadyFlow Systems Energy Balance 54 Some SteadyFlow Engineering Devices 1 Nozzles and Diffusers 2 Turbines and Compressors 3 Throttling Valves 4a Mixing Chambers 4b Heat Exchangers 5 Pipe and Duct Flow 55 Energy Analysis of UnsteadyFlow Processes Mass Balance Energy Balance Topic of Special Interest: General Energy Equation Summary References and Suggested Reading Problems
Chapter 6 The Second Law of Thermodynamics 61 Introduction to the Second Law 62 Thermal Energy Reservoirs 63 Heat Engines Thermal Efficiency Can We Save Qout ? The Second Law of Thermodynamics: Kelvin–Planck Statement 65 Refrigerators and Heat Pumps Coefficient of Performance Heat Pumps The Second Law of Thermodynamics: Clausius Statement Equivalence of the Two Statements 66 PerpetualMotion Machines 67 Reversible and Irreversible Processes Irreversibilities Internally and Externally Reversible Processes 68 The Carnot Cycle The Reversed Carnot Cycle 69 The Carnot Principles 610 The Thermodynamic Temperature Scale 611 The Carnot Heat Engine The Quality of Energy Quantity versus Quality in Daily Life 612 The Carnot Refrigerator and Heat Pump Topics of Special Interest: Household Refrigerators Summary References and Suggested Reading Problems Chapter 7 Entropy 71 Entropy A Special Case: Internally Reversible Isothermal Heat Transfer Processes
72 The Increase of Entropy Principle Some Remarks about Entropy 73 Entropy Change of Pure Substances 74 Isentropic Processes 75 Property Diagrams Involving Entropy 76 What Is Entropy? Entropy and Entropy Generation in Daily Life 77 The T ds Relations 78 Entropy Change of Liquids and Solids 79 The Entropy Change of Ideal Gases Constant Specific Heats (Approximate Analysis) Variable Specific Heats (Exact Analysis) Isentropic Processes of Ideal Gases Constant Specific Heats (Approximate Analysis) Variable Specific Heats (Exact Analysis) Relative Pressure and Relative Specific Volume 710 Reversible SteadyFlow Work Proof that SteadyFlow Devices Deliver the Most and Consume the Least Work when the Process Is Reversible 711 Minimizing the Compressor Work Multistage Compression with Intercooling 712 Isentropic Efficiencies of SteadyFlow Devices Isentropic Efficiency of Turbines Isentropic Efficiencies of Compressors and Pumps Isentropic Efficiency of Nozzles 713 Entropy Balance Entropy Change of a System, ∆S system Mechanisms of Entropy Transfer, Sin and Sout 1 Heat Transfer 2 Mass Flow Entropy Generation, Sgen Closed Systems Control Volumes Entropy Generation Associated with a Heat Transfer Process Topics of Special Interest: Reducing the Cost of Compressed Air Summary References and Suggested Reading Problems
Chapter 8 Exergy: A Measure of Work Potential 81 Exergy: Work Potential of Energy Exergy (Work Potential) Associated with Kinetic and Potential Energy 82 Reversible Work and Irreversibility 83 SecondLaw Efficiency, ηII 84 Exergy Change of a System Exergy of a Fixed Mass: Nonflow (or Closed System) Exergy Exergy of a Flow Stream: Flow (or Stream) Exergy 85 Exergy Transfer by Heat, Work, and Mass Exergy Transfer by Heat Transfer, Q Exergy Transfer by Work, W Exergy Transfer by Mass, m 86 The Decrease of Exergy Principle and Exergy Destruction Exergy Destruction 87 Exergy Balance: Closed Systems 88 Exergy Balance: Control Volumes Exergy Balance for SteadyFlow Systems Reversible Work, W rev SecondLaw Efficiency of SteadyFlow Devices, ηII Topics of Special Interest: SecondLaw Aspects of Daily Life Summary References and Suggested Reading Problems Chapter 9 Gas Power Cycles 91 Basic Considerations in the Analysis of Power Cycles 92 The Carnot Cycle and Its Value in Engineering 93 AirStandard Assumptions 94 An Overview of Reciprocating Engines 95 Otto Cycle: The Ideal Cycle for SparkIgnition Engines 96 Diesel Cycle: The Ideal Cycle for CompressionIgnition Engines 97 Stirling and Ericsson Cycles 98 Brayton Cycle: The Ideal Cycle for GasTurbine Engines
Development of Gas Turbines Deviation of Actual GasTurbine Cycles from Idealized Ones 99 The Brayton Cycle with Regeneration 910 The Brayton Cycle with Intercooling, Reheating, and Regeneration 911 Ideal JetPropulsion Cycles Modifications to Turbojet Engines 912 SecondLaw Analysis of Gas Power Cycles Topics of Special Interest: Saving Fuel and Money by Driving Sensibly Summary References and Suggested Reading Problems Chapter 10 Vapor and Combined Power Cycles 101 The Carnot Vapor Cycle 102 Rankine Cycle: The Ideal Cycle for Vapor Power Cycles Energy Analysis of the Ideal Rankine Cycle 103 Deviation of Actual Vapor Power Cycles from Idealized Ones 104 How Can We Increase the Efficiency of the Rankine Cycle? Lowering the Condenser Pressure (Lowers T low,av) Superheating the Steam to High Temperatures (Increases Thigh,av) Increasing the Boiler Pressure (Increases Thigh,av) 105 The Ideal Reheat Rankine Cycle 106 The Ideal Regenerative Rankine Cycle Open Feedwater Heaters Closed Feedwater Heaters 107 SecondLaw Analysis of Vapor Power Cycles 108 Cogeneration 109 Combined Gas–Vapor Power Cycles Topics of Special Interest: Binary Vapor Cycles Summary References and Suggested Reading Problems Chapter 11 Refrigeration Cycles 111 Refrigerators and Heat Pumps
112 The Reversed Carnot Cycle 113 The Ideal VaporCompression Refrigeration Cycle 114 Actual VaporCompression Refrigeration Cycle 115 Selecting the Right Refrigerant 116 Heat Pump Systems 117 Innovative VaporCompression Refrigeration Systems Cascade Refrigeration Systems Multistage Compression Refrigeration Systems Multipurpose Refrigeration Systems with a Single Compressor Liquefaction of Gases 118 Gas Refrigeration Cycles 119 Absorption Refrigeration Systems Topics of Special Interest: Thermoelectric Power Generation and Refrigeration Systems Summary References and Suggested Reading Problems Chapter 12 Thermodynamic Property Relations 121 A Little Math—Partial Derivatives and Associated Relations Partial Differentials Partial Differential Relations 122 The Maxwell Relations 123 The Clapeyron Equation 124 General Relations for du, dh, ds, Cv, and Cp Internal Energy Changes Enthalpy Changes Entropy Changes Specific Heats Cv and Cp 125 The JouleThomson Coefficient 126 The ∆h, ∆u, and ∆s of Real Gases Enthalpy Changes of Real Gases Internal Energy Changes of Real Gases Entropy Changes of Real Gases Summary References and Suggested Reading
Problems Chapter 13 Gas Mixtures 131 Composition of a Gas Mixture: Mass and Mole Fractions 132 PvT Behavior of Gas Mixtures: Ideal and Real Gases IdealGas Mixtures RealGas Mixtures 133 Properties of Gas Mixtures: Ideal and Real Gases IdealGas Mixtures RealGas Mixtures Topics of Special Interest: Chemical Potential and the Separation Work of Mixtures Ideal Gas Mixtures and Ideal Solutions Minimum Work of Separation of Mixtures Reversible Mixing Processes SecondLaw Efficiency SpecialCase: Separation of a TwoComponent Mixture An Application: Desalination Processes Chapter 14 Gas–Vapor Mixtures and AirConditioning 141 Dry and Atmospheric Air 142 Specific and Relative Humidity of Air 143 DewPoint Temperature 144 Adiabatic Saturation and WetBulb Temperatures 145 The Psychrometric Chart 146 Human Comfort and AirConditioning 147 AirConditioning Processes Simple Heating and Cooling (w = constant) Heating with Humidification Cooling with Dehumidification Evaporative Cooling Adiabatic Mixing of Airstreams Wet Cooling Towers Summary
References and Suggested Reading Problems Chapter 15 Chemical Reactions 151 Fuels and Combustion 152 Theoretical and Actual Combustion Processes 153 Enthalpy of Formation and Enthalpy of Combustion 154 FirstLaw Analysis of Reacting Systems SteadyFlow Systems Closed Systems 155 Adiabatic Flame Temperature 156 Entropy Change of Reacting Systems 157 SecondLaw Analysis of Reacting systems Topics of Special Interest: Fuel Cells Summary References and Suggested Reading Problems Chapter 16 Chemical and Phase Equilibrium 161 Criterion for Chemical Equilibrium 162 The Equilibrium Constant for IdealGas Mixtures 163 Some Remarks about the KP of IdealGas Mixtures 164 Chemical Equilibrium for Simultaneous Reactions 165 Variation of KP with Temperature 166 Phase Equilibrium Phase Equilibrium for a SingleComponent System The Phase Rule Phase Equilibrium for a Multicomponent System Summary References and Suggested Reading Problems Chapter 17 Compressible Flow
171 Stagnation Properties 172 Speed of Sound and Mach Number 173 OneDimensional Isentropic Flow Variation of Fluid Velocity with Flow Area Property Relations for Isentropic Flow of Ideal Gases 174 Isentropic Flow through Nozzles Converging Nozzles Converging–Diverging Nozzles 175 Shock Waves and Expansion Normal Shocks Oblique Shocks Prandtl–Meyer Expansion Waves 176 Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) Property Relations for Rayleigh Flow Choked Rayleigh Flow 177 Steam Nozzles Summary References and Suggested Reading Problems Appendix 1 Property Tables and Charts (SI Units) Table A1
Molar mass, gas constant, and criticalpoint properties
Table A2
Idealgas specific heats of various common gases
Table A3
Properties of common liquids, solids, and foods
Table A4
Saturated water—Temperature table
Table A5
Saturated water—Pressure table
Table A6
Superheated water
Table A7
Compressed liquid water
Table A8
Saturated ice—water vapor
Figure A9 Ts diagram for water Figure A10 Mollier diagram for water Table A11
Saturated refrigerant134a—Temperature table
Table A12 Saturated refrigerant134a—Pressure table Table A13 Superheated refrigerant134a Figure A14 Ph diagram for refrigerant134a Figure A15 Nelson–Obert generalized compressibility chart Table A16 Properties of the atmosphere at high altitude Table A17 Idealgas properties of air Table A18 Idealgas properties of nitrogen, N2 Table A19 Idealgas properties of oxygen, O2 Table A20 Idealgas properties of carbon dioxide, CO2 Table A21 Idealgas properties of carbon monoxide, CO Table A22 Idealgas properties of hydrogen, H2 Table A23 Idealgas properties of water vapor, H2O Table A24 Idealgas properties of monatomic oxygen, O Table A25 Idealgas properties of hydroxyl, OH Table A26 Enthalpy of formation, Gibbs function of formation, and absolute entropy at 25°C, 1 atm
Table A27 Properties of some common fuels and hydrocarbons Table A28 Natural Logarithms of the equilibrium constant Kp Figure A29 Generalized enthalpy departure chart Figure A30 Generalized entropy departure chart Figure A31 Psychrometric chart at 1 atm total pressure Table A32 Onedimensional isentropic compressibleflow functions for an ideal gas with k = 1.4 Table A33 Onedimensional normalshock functions for an ideal gas with k =1.4 Table A34 Rayleigh flow functions for an ideal gas with k = 1.4
Appendix 2 Property Tables and Charts (English Units) Table A1E Molar mass, gas constant, and criticalpoint properties Table A2E Idealgas specific heats of various common gases Table A3E Properties of common liquids, solids, and foods Table A4E Saturated water—Temperature table Table A5E Saturated water—Pressure table Table A6E Superheated water Table A7E Compressed liquid water Table A8E Saturated ice—water vapor Figure A9E Ts diagram for water Figure A10E Mollier diagram for water Table A11E Saturated refrigerant134a—Temperature table Table A12E Saturated refrigerant134a—Pressure table Table A13E Superheated refrigerant134a Figure A14E Ph diagram for refrigerant134a Table A15E Table A16E Properties of the atomosphere at high altitude Table A17E Idealgas properties of air
Table A18E Idealgas properties of nitrogen, N2 Table A19E Idealgas properties of oxygen, O2 Table A20E Idealgas properties of carbon dioxide, CO2 Table A21E Idealgas properties of carbon monoxide, CO Table A22E Idealgas properties of hydrogen, H2 Table A23E Idealgas properties of water vapor, H2O Table A24E Table A25E Table A26E Enthalpy of formation, Gibbs function of formation, and absolute entropy at 77°C, 1 atm Table A27E
Properties of some common fuels and hydrocarbons
Figure A31E Psycrometric chart at 1 atm total pressure
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PREFACE BACKGROUND Thermodynamics is an exciting and fascinating subject that deals with energy, which is essential for sustenance of life, 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.
OBJECTIVES 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 emphasizing the physics and physical arguments. 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.
PHILOSOPHY AND GOAL 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 tomorrow’s 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.

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Preface 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 editions—from small colleges to large universities all over the world— indicate that our objectives have largely been achieved. It is our philosophy that the best way to learn is by practice. Therefore, special effort is made throughout the book to reinforce material that was presented earlier. Yesterday’s 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. Tomorrow’s 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 EDITION All the popular features of the previous editions are retained while new ones are added. With the exception of reorganizing the first law coverage and updating the steam and refrigerant properties, the main body of the text remains largely unchanged. The most significant changes in this fifth edition are highlighted below.
EARLY INTRODUCTION OF THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics is now introduced early in the new 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, thermoeconomics, 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.
SEPARATE COVERAGE OF CLOSED SYSTEMS AND CONTROL VOLUME ENERGY ANALYSES The energy analysis of closed systems is now presented in a separate chapter, Chapter 4, together with the boundary work and the discussion of specific heats for both ideal gases and incompressible substances. The conservation of mass is now covered together with conservation of energy in new Chapter 5. A formal derivation of the general energy equation is also given in this chapter as the Topic of Special Interest.
REVISED COVERAGE OF COMPRESSIBLE FLOW The chapter on compressible flow that deals with compressibility effects (now Chapter 17) is greatly revised and expanded. This chapter now includes
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coverage of oblique shocks and flow with heat transfer (Rayleigh flow) with some exciting photographs and extended discussions of shock waves.
UPDATED STEAM AND REFRIGERANT134A TABLES The steam and refrigerant134a tables are updated using the most current property data from EES. Tables A4 through A8 and A11 through A13, as well as their counterparts in English units, have all been revised. All the examples and homework problems in the text that involve steam or refrigerant134a are also revised to reflect the small changes in steam and refrigerant properties. An added advantage of this update is that students will get the same result when solving problems whether they use steam or refrigerant properties from EES or property tables in the appendices.
OVER 300 NEW COMPREHENSIVE PROBLEMS This edition includes over 300 new comprehensive problems that come mostly from industrial applications. Problems whose solutions require parametric investigations, and thus the use of a computer, are identified by a computerEES icon, as before.
CONTENT CHANGES AND REORGANIZATION The noteworthy changes in various chapters are summarized below for those who are familiar with the previous edition. • Chapter 1 is greatly revised, and its title is changed to “Introduction and Basic Concepts.” A new section Density and Specific Gravity and a new subsection The International Temperature Scale of 1990 are added. The sections Forms of Energy and Energy and the Environment are moved to new Chapter 2, and the Topic of Special Interest Thermodynamic Aspects of Biological Systems is moved to new Chapter 4. • The new Chapter 2 “Energy, Energy Transfer, and General Energy Analysis” mostly consists of the sections Forms of Energy and Energy and the Environment moved from Chapter 1, Energy Transfer by Heat and Energy Transfer by Work, and Mechanical Forms of Energy from Chapter 3, The First Law of Thermodynamics from Chapter 4, and Energy Conversion Efficiencies from Chapter 5. The Topic of Special Interest in this chapter is Mechanisms of Heat Transfer moved from Chapter 3. • Chapter 3 “Properties of Pure Substance” is essentially the previous edition Chapter 2, except that the last three sections on specific heats are moved to new Chapter 4. • Chapter 4 “Energy Analysis of Closed Systems” consists of Moving Boundary Work from Chapter 3, sections on Specific Heats from Chapter 2, and Energy Balance for Closed Systems from Chapter 4. Also, the Topic of Special Interest Thermodynamic Aspects of Biological Systems is moved here from Chapter 1. • Chapter 5 “Mass and Energy Analysis of Control Volumes” consists of Mass Balance for Control Volumes and Flow Work and the Energy of a Flowing Fluid from Chapter 3 and the sections on Energy Balance for Steady and UnsteadyFlow Systems from Chapter 4. The
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•
• •
•
•
Topic of Special Interest Refrigeration and Freezing of Foods is deleted and is replaced by a formal derivation of the General Energy Equation. Chapter 6 “The Second Law of Thermodynamics” is identical to the previous edition Chapter 5, except the section Energy Conversion Efficiencies is moved to Chapter 2. Chapters 7 through 15 are essentially identical to the previous edition Chapters 6 through 14, respectively. Chapter 17 “Compressible Flow” is an updated version of the previous edition Chapter 16. The entire chapter is greatly revised, the section Flow Through Actual Nozzles and Diffusers is deleted, and a new section Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) is added. In Appendices 1 and 2, the steam and refrigerant134a tables (Tables 4 through 8 and 11 through 13) are entirely revised, but the table numbers are kept the same. The tables for isentropic compressible flow functions and the normal shock functions (Tables A32 and A33) are updated and plots of functions are now included. Also, Rayleigh flow functions are added as Table A34. Appendix 3 Introduction to EES is moved to the Student Resources DVD that comes packaged free with the text. The conversion factors on the inner cover pages and the physical constants are updated, and some nomenclature symbols are revised.
LEARNING TOOLS 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.
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 physical, 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
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Preface 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. The fifth edition of Thermodynamics: An Engineering Approach contains more figures and illustrations than any other book in this category. This edition incorporates an expanded photo program and updated art style. 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. The popular cartoon feature “Blondie” is used to make some important points in a humorous way and also to break the ice and ease the nerves. Who says studying thermodynamics can’t be fun?
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 properties 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 discussed following the solutions. This approach is also used consistently in the solutions presented in the instructor’s solutions manual.
A WEALTH OF REALWORLD ENDOFCHAPTER PROBLEMS The endofchapter problems are grouped under specific topics to make problem 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 section of a chapter—in 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

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Preface their findings in a professional manner. Problems designated by an “E” are in English units, and SI users can ignore them. Problems with the are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the are comprehensive in nature and are intended to be solved with a computer, preferably using the EES software that accompanies this text. Several economics and safetyrelated problems are incorporated throughout to enhance cost and safety awareness among engineering students. Answers to selected problems are listed immediately following the problem for convenience 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 multiplechoice 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 directions of interactions.
PHYSICALLY MEANINGFUL FORMULAS The physically meaningful forms of the balance equations rather than formulas 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 min mout ¢msystem
Mass balance:
E in E out
Energy balance:
Entropy balance:
Change in internal, kinetic, potential, etc., energies
S in S out S gen ¢S system
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎬ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
Exergy balance:
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭ Net energy transfer by heat, work, and mass
Net entropy transfer by heat and mass
Entropy generation
Change in entropy
X in X out X destroyed ¢X system Exergy destruction
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭ Net exergy transfer by heat, work, and mass
Change in exergy
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.
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A CHOICE OF SI ALONE OR SI/ENGLISH 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 SI/English 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 introduced and defined, it appears in boldface type. Fundamental thermodynamic terms and concepts also appear in a glossary located on our accompanying website (www.mhhe.com/cengel). 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. In addition, students can test their knowledge of these fundamental terms by using the flash cards and other interactive resources.
CONVERSION FACTORS Frequently used conversion factors and physical constants are listed on the inner cover pages of the text for easy reference.
SUPPLEMENTS The following supplements are available to the adopters of the book.
STUDENT RESOURCES DVD Packaged free with every new copy of the text, this DVD provides a wealth of resources for students including Physical Experiments in Thermodynamics, an Interactive Thermodynamics Tutorial, and EES Software. Physical Experiments in Thermodynamics: A new feature of this book is the addition of Physical Experiments in Thermodynamics created by Ronald Mullisen of the Mechanical Engineering Department at California Polytechnic State University (Cal Poly), San Luis Obispo. At appropriate places in the margins of Chapters 1, 3, and 4, photos with captions show physical experiments that directly relate to material covered on that page. The captions

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Preface refer the reader to endofchapter problems that give a brief description of the experiments. These experiments cover thermodynamic properties, thermodynamic processes, and thermodynamic laws. The Student Resources DVD contains complete coverage of the nine experiments. Each experiment contains a video clip, a complete writeup including historical background, and actual data (usually in an Excel file). The results are also provided on the website that accompanies the text, and they are password protected for instructor use. After viewing the video and reading the writeup, the student will be ready to reduce the data and obtain results that directly connect with material presented in the chapters. For all of the experiments the final results are compared against published information. Most of the experiments give final results that come within 10 percent or closer to these published values. Interactive Thermodynamics Tutorial: Also included on the Student Resources DVD is the Interactive Thermodynamics Tutorial developed by Ed Anderson of Texas Tech University. The revised tutorial is now tied directly to the text with an icon to indicate when students should refer to the tutorial to further explore specific topics such as energy balance and isentropic processes. Engineering Equation Solver (EES): Developed by Sanford Klein and William Beckman from the University of Wisconsin–Madison, this software combines equationsolving capability and engineering property data. EES can do optimization, parametric analysis, and linear and nonlinear regression, and provides publicationquality plotting capabilities. Thermodynamics and transport properties for air, water, and many other fluids are built in, and EES allows the user to enter property data or functional relationships.
ONLINE LEARNING CENTER (OLC) Web support is provided for the book on our Online Learning Center at www.mhhe.com/cengel. Visit this robust site for book and supplement information, errata, author information, and further resources for instructors and students.
INSTRUCTOR’S RESOURCE CDROM (AVAILABLE TO INSTRUCTORS ONLY) This CD, available to instructors only, offers a wide range of classroom preparation and presentation resources including the solutions manual in PDF files by chapter, all text chapters and appendices as downloadable PDF files, and all text figures in JPEG format.
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Errata Sheet Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 5th ed, McGrawHill, 2006. (Last update: Dec. 29, 2005) Page Location Error Correction 42 45 50 177 327 461 462 462 462 462 619 916
Prob. 150E, 2nd line Prob. 176, Answer Prob. 1128 Example 46, Assumptions Prob. 6138, 2nd line Last line 1st calculation 1st line after 1st calculation 2nd calculation 1st line after 2nd calculation Fig. 119 Table A20, the line T=760 K
29.1 mm Hg 5.0 It is a repeat of Prob. 1127 Assumption #4 Bobbler (spelling error) 4665 4665, 0.923, 92.3% 7.7 46654306=359 359 Lightgray and lightcolor lines 20,135
29.1 in Hg 1.34 Delete it. Delete it (Temp doesn’t have to be const) bubbler 5072 5072, 0.849, 84.9% 15.1 50724306=766 766 They should be switched in both figures. 30,135
Note: Changes that will be made in the next edition are designated by “6th ed”.
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Chapter 1 INTRODUCTION AND BASIC CONCEPTS
E
very science has a unique vocabulary associated with it, and thermodynamics 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 thermodynamics and the unit systems, and continue with a discussion of some basic concepts such as system, state, state postulate, equilibrium, and process. 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 discuss absolute and gage pressures, the variation of pressure with depth, and pressure measurement devices, such as manometers 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. • Review concepts of temperature, temperature scales, pressure, and absolute and gage pressure. • Introduce an intuitive systematic problemsolving technique.

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Thermodynamics INTERACTIVE TUTORIAL
SEE TUTORIAL CH. 1, SEC. 1 ON THE DVD.
PE = 10 units KE = 0
Potential energy
PE = 7 units KE = 3 units
Kinetic energy
FIGURE 1–1 Energy cannot be created or destroyed; it can only change forms (the first law). Energy storage (1 unit) Energy in (5 units)
Energy out (4 units)
FIGURE 1–2 Conservation of energy principle for the human body.
1–1
■
THERMODYNAMICS AND ENERGY
Thermodynamics can be defined as the science of energy. Although everybody 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, refrigeration, 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 constant. 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. 1–1). 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. 1–2). 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 output, and the energy balance is expressed as Ein Eout E. The first law of thermodynamics is simply an expression of the conservation 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 eventually cools, but a cup of cool coffee in the same room never gets hot by itself (Fig. 1–3). The hightemperature energy of the coffee is degraded (transformed 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 development 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 thermodynamic textbook was written in 1859 by William Rankine, a professor at the University of Glasgow. It is wellknown 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
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Chapter 1 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 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 the solution of 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 the supporting role.
1–2
■
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.
3
Cool environment 20°C Hot coffee 70°C
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. The 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 ordinary house is, in some respects, an exhibition hall filled with wonders of thermodynamics (Fig. 1–4). Many ordinary household utensils and appliances are designed, in whole or in part, by using the principles of thermodynamics. Some examples include the electric or gas range, the heating and airconditioning 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 analysis 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. 1–5). 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.

Heat
FIGURE 1–3 Heat flows in the direction of decreasing temperature.
Solar collectors
Shower Hot water Hot water tank
Cold water Heat exchanger
Pump
FIGURE 1–4 The design of many engineering systems, such as this solar hot water system, involves thermodynamics.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 2 ON THE DVD.
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Thermodynamics
The human body
Air conditioning systems
Airplanes
Car radiators
Power plants
Refrigeration systems
FIGURE 1–5 Some application areas of thermodynamics. A/C unit, fridge, radiator: © The McGrawHill Companies, Inc./Jill Braaten, photographer; Plane: © Vol. 14/PhotoDisc; Humans: © Vol. 121/PhotoDisc; Power plant: © Corbis Royalty Free
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 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 gal, etc.), which makes it confusing and difficult to learn. The United States is the only industrialized country that has not yet fully converted to the metric system. 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
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Chapter 1 did not find universal 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 candela (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 1–1). 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 (1647–1723), 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 abbreviations 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 promote a voluntary switch to the metric system by passing the Metric Conversion 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. However, 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 in converting 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. Presently 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 today’s 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 1–2. They are standard for all units, and the student is encouraged to memorize them because of their widespread use (Fig. 1–6).

TABLE 1–1 The seven fundamental (or primary) dimensions and their units in SI Dimension
Unit
Length Mass Time Temperature Electric current Amount of light Amount of matter
meter (m) kilogram (kg) second (s) kelvin (K) ampere (A) candela (cd) mole (mol)
TABLE 1–2 Standard prefixes in SI units Multiple
Prefix
1012
tera, T giga, G mega, M kilo, k hecto, h deka, da deci, d centi, c milli, m micro, m nano, n pico, p
109 106 103 102 101 101 102 103 106 109 1012
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Thermodynamics
FIGURE 1–6 The SI unit prefixes are used in all branches of engineering.
200 mL (0.2 L)
1 kg (103 g)
1 M (106 )
Some SI and English Units
m = 1 kg
a = 1 m/s2
m = 32.174 lbm
F=1N
a = 1 ft/s2 F = 1 lbf
FIGURE 1–7 The definition of the force units.
1 kgf
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 0.45359 kg 1 ft 0.3048 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 formulas. To avoid this nuisance, we consider force to be a secondary dimension whose unit is derived from Newton’s second law, that is, or
F ma
10 apples m = 1 kg 1 apple m = 102 g
1N
Force 1Mass2 1Acceleration2
4 apples m = 1 lbm
1 lbf
(1–1)
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 m/s2. In the English system, the force unit is the poundforce (lbf) and is defined as the force required to accelerate a mass of 32.174 lbm (1 slug) at a rate of 1 ft/s2 (Fig. 1–7). That is, 1 N 1 kg # m>s2 1 lbf 32.174 lbm # ft>s2
FIGURE 1–8 The relative magnitudes of the force units newton (N), kilogramforce (kgf), and poundforce (lbf).
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. 1–8. 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 9.807 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 Newton’s second law, W mg¬1N2
(1–2)
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Chapter 1 where m is the mass of the body, and g is the local gravitational acceleration (g is 9.807 m/s2 or 32.174 ft/s2 at sea level and 45° latitude). An ordinary bathroom scale measures the gravitational force acting on a body. The weight of a unit volume of a substance is called the specific weight g and is determined from g rg, where r is density. The mass of a body remains the same regardless of its location in the universe. Its weight, however, changes with a change in gravitational acceleration. A body weighs less on top of a mountain since g decreases with altitude. On the surface of the moon, an astronaut weighs about onesixth of what she or he normally weighs on earth (Fig. 1–9). At sea level a mass of 1 kg weighs 9.807 N, as illustrated in Fig. 1–10. A mass of 1 lbm, however, weighs 1 lbf, which misleads people to believe 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 attraction 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 the local density of the earth’s crust, 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 9.8295 m/s2 at 4500 m below sea level to 7.3218 m/s2 at 100,000 m above sea level. However, at altitudes up to 30,000 m, the variation of g from the sealevel value of 9.807 m/s2 is less than 1 percent. Therefore, for most practical purposes, the gravitational acceleration can be assumed to be constant at 9.81 m/s2. 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 red shift, 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 measuring precious metals. Work, which is a form of energy, can simply be defined as force times distance; therefore, it has the unit “newtonmeter (N · m),” which is called a joule (J). That is, 1J1N#m
(1–3)
A more common unit for energy in SI is the kilojoule (1 kJ 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 68°F by 1°F. In the metric system, the amount of energy needed to raise the temperature of 1 g of water at 14.5°C by 1°C is defined as 1 calorie (cal), and 1 cal 4.1868 J. The magnitudes of the kilojoule and Btu are almost identical (1 Btu 1.0551 kJ). 103

7
FIGURE 1–9 A body weighing 150 lbf on earth will weigh only 25 lbf on the moon.
kg g = 9.807 m/s2 W = 9.807 kg · m/s2 = 9.807 N = 1 kgf
lbm
g = 32.174 ft/s2 W = 32.174 lbm · ft/s2 = 1 lbf
FIGURE 1–10 The weight of a unit mass at sea level.
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Thermodynamics
Dimensional Homogeneity We all know from grade school that apples and oranges do not add. But we somehow manage 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 (Fig. 1–11). 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. EXAMPLE 1–1
Spotting Errors from Unit Inconsistencies
While solving a problem, a person ended up with the following equation at some stage:
FIGURE 1–11 To be dimensionally homogeneous, all the terms in an equation must have the same unit. © Reprinted with special permission of King Features Syndicate.
E 25 kJ 7 kJ>kg where E is the total energy and has the unit of kilojoules. Determine how to correct the error and discuss what may have caused it.
Solution During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable cause of the error is to be determined. Analysis The two terms on the righthand side do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit. Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage.
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; they can even be used to derive formulas, as explained in the following example. EXAMPLE 1–2
Obtaining Formulas from Unit Considerations
A tank is filled with oil whose density is r 850 kg/m3. 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 deter
OIL
V = 2 m3 ρ = 850 kg/m3 m=?
FIGURE 1–12 Schematic for Example 1–2.
mined. Assumptions Oil is an incompressible substance and thus its density is constant. Analysis A sketch of the system just described is given in Fig. 1–12. Suppose we forgot 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
r 850 kg>m3¬and¬V 2 m3
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Chapter 1

9
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 rV Thus,
m 1850 kg>m3 2 12 m3 2 1700 kg Discussion formulas.
Note that this approach may not work for more complicated
You should keep in mind that a formula that is not dimensionally homogeneous is definitely wrong, 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 m ft N kg¬ 2 ¬and¬lbf 32.174 lbm¬ 2 s s
They can also be expressed more conveniently as unity conversion ratios as N
lbf
1¬and¬ 1 kg # m>s2 32.174 lbm # ft>s2
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. Students are encouraged to always use unity conversion ratios such as those given here when converting units. Some textbooks insert the archaic gravitational constant gc defined as gc 32.174 lbm · ft/lbf · s2 kg · m/N · s2 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 students instead use unity conversion ratios.
EXAMPLE 1–3
The Weight of One PoundMass
lbm
Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth (Fig. 1–13).
Solution A mass of 1.00 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 32.174 ft/s2.
FIGURE 1–13 A mass of 1 lbm weighs 1 lbf on earth.
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Thermodynamics Analysis We apply Newton’s 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, Net weight: One pound (454 grams)
W mg 11.00 lbm2 132.174 ft>s2 2 a
1 lbf b 1.00 lbf 32.174 lbm # ft>s2
Discussion 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.
When you buy a box of breakfast cereal, the printing may say “Net weight: One pound (454 grams).” (See Fig. 1–14.) Technically, this means that the cereal inside the box weighs 1.00 lbf on earth and has a mass of 453.6 g (0.4536 kg). Using Newton’s second law, the actual weight of the cereal in the metric system is W mg 1453.6 g2 19.81 m>s2 2 a
FIGURE 1–14 A quirk in the metric system of units. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 3 ON THE DVD. SURROUNDINGS
SYSTEM
BOUNDARY
FIGURE 1–15 System, surroundings, and boundary.
CLOSED SYSTEM
Mass
NO
m = constant Energy YES
FIGURE 1–16 Mass cannot cross the boundaries of a closed system, but energy can.
1–3
■
1 kg 1N ba b 4.45 N 2 1000 g 1 kg # m>s
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 surroundings is called the boundary. These terms are illustrated in Fig. 1–15. 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 boundary 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) consists of a fixed amount of mass, and no mass can cross its boundary. That is, no mass can enter or leave a closed system, as shown in Fig. 1–16. 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. Consider the pistoncylinder device shown in Fig. 1–17. 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
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Chapter 1 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 analyzed as control volumes (open systems) instead of as control masses (closed systems). In general, any arbitrary region in space can be selected as a control 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. 1–18a). 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. 1–18b. 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. As an example of an open system, consider the water heater shown in Fig. 1–19. 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.

11
Moving boundary GAS 2 kg 3 m3
GAS 2 kg 1 m3
Fixed boundary
FIGURE 1–17 A closed system with a moving boundary.
Hot water out
Control surface WATER HEATER
Imaginary boundary
(control volume)
Real boundary
CV (a nozzle)
Cold water in
Moving boundary CV Fixed boundary
(a) A control volume with real and imaginary boundaries
(b) A control volume with fixed and moving boundaries
FIGURE 1–18 A control volume can involve fixed, moving, real, and imaginary boundaries.
FIGURE 1–19 An open system (a control volume) with one inlet and one exit.
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Thermodynamics
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 4 ON THE DVD.
m V T P ρ
–12 m –12 V T P ρ
–12 m –12 V T P ρ
Extensive properties Intensive properties
FIGURE 1–20 Criterion to differentiate intensive and extensive properties. O2
1 atm, 20°C
3 × 1016 molecules/mm3
VOID
FIGURE 1–21 Despite the large gaps between molecules, a substance can be treated as a continuum because of the very large number of molecules even in an extremely small volume.
In an engineering analysis, the system under study must be defined carefully. 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.
1–4
■
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, modulus 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 temperature, pressure, and density. Extensive properties are those whose values depend on the size—or extent—of 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. 1–20. 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 V/m) and specific total energy (e E/m).
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.” 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 5.3 1026 kg. Also, the mean free path of oxygen at 1 atm pressure and 20°C is 6.3 108 m. That is, an oxygen molecule travels, on average, a distance of 6.3 108 m (about 200 times of 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 20°C (Fig. 1–21). The continuum model is applicable as long as the characteristic length of the system (such as its
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13
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 0.1 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.
1–5
■
INTERACTIVE TUTORIAL
DENSITY AND SPECIFIC GRAVITY
SEE TUTORIAL CH. 1, SEC. 5 ON THE DVD.
Density is defined as mass per unit volume (Fig. 1–22). Density:
r
m ¬¬1kg>m3 2 V
(1–4)
EXPERIMENT
The reciprocal of density is the specific volume v, which is defined as volume per unit mass. That is, v
1 V r m
(1–5)
For a differential volume element of mass dm and volume dV, density can be expressed as r dm/dV. The density of a substance, in general, depends on temperature and pressure. The density of most gases is proportional to pressure and inversely proportional 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 20°C, for example, the density of water changes from 998 kg/m3 at 1 atm to 1003 kg/m3 at 100 atm, a change of just 0.5 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 kg/m3 at 20°C to 975 kg/m3 at 75°C, a change of 2.3 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 4°C, for which rH2O 1000 kg/m3). That is, Specific gravity:
SG
r r H2O
Use actual data from the experiment shown here to obtain the density of water in the neighborhood of 4°C. See endofchapter problem 1–129. © Ronald Mullisen
V = 12 m 3 m = 3 kg
(1–6)
Note that the specific gravity of a substance is a dimensionless quantity. However, in SI units, the numerical value of the specific gravity of a substance is exactly equal to its density in g/cm3 or kg/L (or 0.001 times the density in kg/m3) since the density of water at 4°C is 1 g/cm3 1 kg/L 1000 kg/m3. The specific gravity of mercury at 0°C, for example, is 13.6. Therefore, its density at 0°C is 13.6 g/cm3 13.6 kg/L 13,600 kg/m3. The specific gravities of some substances at 0°C are given in Table 1–3. Note that substances with specific gravities less than 1 are lighter than water, and thus they would float on water.
ρ = 0.25 kg/m 3 1 3 v =– ρ = 4 m /kg
FIGURE 1–22 Density is mass per unit volume; specific volume is volume per unit mass.
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Thermodynamics
TABLE 1–3
The weight of a unit volume of a substance is called specific weight and is expressed as
Specific gravities of some substances at 0°C
Specific weight:
Substance
SG
Water Blood Seawater Gasoline Ethyl alcohol Mercury Wood Gold Bones Ice Air (at 1 atm)
1.0 1.05 1.025 0.7 0.79 13.6 0.3–0.9 19.2 1.7–2.0 0.92 0.0013
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 6 ON THE DVD.
m = 2 kg T1 = 20°C V1 = 1.5 m3 (a) State 1
m = 2 kg T 2 = 20°C V2 = 2.5 m3
(b) State 2
FIGURE 1–23 A system at two different states.
20° C
23°C ° 30 C
35° C
40° C 42° C
(a) Before
32° C
32° C ° 32 C
32° C 32° C 32° C (b) After
FIGURE 1–24 A closed system reaching thermal equilibrium.
gs rg¬¬1N>m3 2
(1–7)
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.
1–6
■
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. 1–23 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. 1–24. 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 variation of pressure as a result of gravity in most thermodynamic systems is relatively 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 composition 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, 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.
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Chapter 1 A system is called a simple compressible system in the absence of electrical, 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 independent 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. 1–25). Temperature and pressure, however, are independent properties for singlephase systems, but are dependent properties for multiphase systems. At sea level (P 1 atm), water boils at 100°C, 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 system. Phasechange processes are discussed in detail in Chap. 3.

15
Nitrogen T = 25 °C v = 0.9 m 3/kg
FIGURE 1–25 The state of nitrogen is fixed by two independent, intensive properties.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 7 ON THE DVD. Property A
1–7
■
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. 1–26). To describe a process completely, one should specify the initial and final states of the process, 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 infinitesimally 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. 1–27. 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 pressure 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. 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,
State 2
Process path State 1 Property B
FIGURE 1–26 A process between states 1 and 2 and the process path.
(a) Slow compression (quasiequilibrium)
(b) Very fast compression (nonquasiequilibrium)
FIGURE 1–27 Quasiequilibrium and nonquasiequilibrium compression processes.
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Thermodynamics
P Final state 2 Process path Initial state 1
V2
V1
V
System (2)
(1)
FIGURE 1–28 The PV diagram of a compression process. Mass in
300°C
250°C
Control volume
workproducing devices deliver the most work when they operate on quasiequilibrium processes. Therefore, quasiequilibrium processes serve as standards to which actual processes can be compared. Process diagrams plotted by employing thermodynamic properties as coordinates 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 1–28 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 quasiequilibrium processes only. For nonquasiequilibrium processes, we are not able to 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.
225°C 200°C
150°C
Mass out
Time: 1 PM Mass in
300°C
250°C
Control volume 225°C 200°C
150°C
Mass out
Time: 3 PM
FIGURE 1–29 During a steadyflow process, fluid properties within the control volume may change with position but not with time.
The SteadyFlow Process The terms steady and uniform are used frequently 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 girlfriend, 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. 1–29). 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. 1–30). Steadyflow conditions can be closely approximated by devices that are intended for continuous operation such as turbines, pumps, boilers, 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
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Chapter 1 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.
Mass in

17
Control volume mcV = const.
1–8
■
EcV = const.
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 physiological 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. Furthermore, 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 temperature 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 eventually 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. 1–31). 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 temperature 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 recognized 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 measurements, and several have been introduced throughout history. All temperature scales are based on some easily reproducible states such as the freezing and boiling points of water, which are also called the ice point and
Mass out
FIGURE 1–30 Under steadyflow conditions, the mass and energy contents of a control volume remain constant.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 8 ON THE DVD.
IRON
IRON
150°C
60°C
COPPER
COPPER
20°C
60°C
FIGURE 1–31 Two bodies reaching thermal equilibrium after being brought into contact in an isolated enclosure.
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Thermodynamics 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, 1702–1744, who devised it) and the Fahrenheit scale (named after the German instrument maker G. Fahrenheit, 1686–1736), respectively. On the Celsius scale, the ice and steam points were originally assigned the values of 0 and 100°C, respectively. The corresponding values on the Fahrenheit scale are 32 and 212°F. 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 temperature scale is called a thermodynamic temperature scale, which is developed later in conjunction with the second law of thermodynamics. The thermodynamic temperature scale in the SI is the Kelvin scale, named after Lord Kelvin (1824–1907). The temperature unit on this scale is the kelvin, which is designated by K (not °K; the degree symbol was officially dropped from kelvin 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 0.000000002 K in 1989). The thermodynamic temperature scale in the English system is the Rankine scale, named after William Rankine (1820–1872). 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 measured 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 temperature 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
(1–8)
where the values of the constants a and b for a gas thermometer are determined 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 thermal 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 pressures 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 points. Considering that only one straight line passes through two fixed
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Chapter 1 points on a plane, these two measurements are sufficient to determine the constants a and b in Eq. 1–8. Then the unknown temperature T of a medium corresponding 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 thermometer, 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 0°C and 100°C, respectively, then the gas temperature 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 273.15°C 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 passing through the data points in this case will intersect the temperature axis at 273.15°C when extrapolated, as shown in Fig. 1–32. This is the lowest temperature 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. 1–8. In that case Eq. 1–8 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 thermodynamic 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 thermodynamic temperature in the temperature range in which the gas thermometer can be used, and 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 273.15°C on the Celsius scale (Fig. 1–33). The Kelvin scale is related to the Celsius scale by T 1K2 T 1°C2 273.15
(1–9)
The Rankine scale is related to the Fahrenheit scale by T 1R 2 T 1°F2 459.67
(1–10)
It is common practice to round the constant in Eq. 1–9 to 273 and that in Eq. 1–10 to 460. The temperature scales in the two unit systems are related by T 1R2 1.8T 1K2
T 1°F2 1.8T 1°C2 32
(1–11) (1–12)
A comparison of various temperature scales is given in Fig. 1–34. The reference temperature chosen in the original Kelvin scale was 273.15 K (or 0°C), which is the temperature at which water freezes (or ice melts) and water exists as a solid–liquid mixture in equilibrium under standard atmospheric pressure (the ice point). At the Tenth General Conference on Weights and Measures in 1954, the reference point was changed to a much more precisely reproducible point, the triple point of water (the state at which all three phases of water coexist in equilibrium), which is

Measured data points
P
19
Gas A
Gas B Extrapolation
Gas C Gas D
–273.15
T(°C)
FIGURE 1–32 P versus T plots of the experimental data obtained from a constantvolume gas thermometer using four different gases at different (but low) pressures. T (°C) 200 225 250 275
– 273.15
T (K) 75 50 25 0
P (kPa) 120 80 40 0
Absolute vacuum V = constant
FIGURE 1–33 A constantvolume gas thermometer would read 273.15°C at absolute zero pressure.
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Thermodynamics
°C
0.01
–273.15
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K
273.16
°F
32.02
– 459.67
R
Triple 491.69 point of water
0 Absolute zero
FIGURE 1–34 Comparison of temperature scales.
assigned the value 273.16 K. The Celsius scale was also redefined at this conference in terms of the idealgas temperature scale and a single fixed point, which is again the triple point of water with an assigned value of 0.01°C. The boiling temperature of water (the steam point) was experimentally determined to be again 100.00°C, 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 International 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 Conference 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 temperature scale. On this scale, the unit of thermodynamic temperature T is again the kelvin (K), defined as the fraction 1/273.16 of the thermodynamic temperature of the triple point of water, which is 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 definition 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 0°C (273.15°C) in both ITS90 and ITPS68, but the steam point is 99.975°C in ITS90 (with an uncertainly of 0.005°C) whereas it was 100.000°C in IPTS68. 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 0.65 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 0.65 to 5 K, the temperature scale is defined in terms of the vapor pressure—temperature relations for 3He and 4He. Between 3 and 24.5561 K (the triple point of neon), it is defined by means of a properly calibrated helium gas thermometer. From 13.8033 K (the triple point of hydrogen) to 1234.93 K (the freezing point of silver), it is defined by means of platinum resistance thermometers calibrated at specified sets of defining fixed points. Above 1234.93 K, it is defined in terms of the Planck radiation law and a suitable defining fixed point such as the freezing point of gold (1337.33 K).
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Chapter 1 We emphasize that the magnitudes of each division of 1 K and 1°C are identical (Fig. 1–35). Therefore, when we are dealing with temperature differences T, the temperature interval on both scales is the same. Raising the temperature of a substance by 10°C is the same as raising it by 10 K. That is, ¢T 1K2 ¢T 1°C 2 ¢T 1R 2 ¢T 1°F2
1°C
1.8 R
21
1.8°F
(1–13) (1–14)
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 bT), 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 1–4
1K

FIGURE 1–35 Comparison of magnitudes of various temperature units.
Expressing Temperature Rise in Different Units
During a heating process, the temperature of a system rises by 10°C. Express this rise in temperature in K, °F, and R.
Solution The temperature rise of a system is to be expressed in different units. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Then,
¢T 1K2 ¢T 1°C2 10 K
The temperature changes in Fahrenheit and Rankine scales are also identical and are related to the changes in Celsius and Kelvin scales through Eqs. 1–11 and 1–14:
¢T 1R2 1.8 ¢T 1K2 11.8 2 110 2 18 R
and
¢T 1°F2 ¢T 1R2 18°F
Discussion Note that the units °C and K are interchangeable when dealing with temperature differences.
1–9
■
PRESSURE
Pressure is defined as a normal force exerted by a fluid per unit area. We speak of pressure only when we deal with a gas or a liquid. The counterpart of pressure in solids is normal stress. Since pressure is defined as force per unit area, it has the unit of newtons per square meter (N/m2), which is called a pascal (Pa). That is, 1 Pa 1 N>m2
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 9 ON THE DVD.
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Thermodynamics 300 pounds
150 pounds
The pressure unit pascal is too small for 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 105 Pa 0.1 MPa 100 kPa 1 atm 101,325 Pa 101.325 kPa 1.01325 bars
Afeet = 50 in2
1 kgf>cm2 9.807 N>cm2 9.807 104 N>m2 9.807 104 Pa 0.9807 bar 0.9679 atm
P = 3 psi
P = 6 psi
W = –––––– 150 lbf = 3 psi P = sn = –––– Afeet 50 in2
FIGURE 1–36 The normal stress (or “pressure”) on the feet of a chubby person is much greater than on the feet of a slim person.
FIGURE 1–37 Some basic pressure gages. Dresser Instruments, Dresser, Inc. Used by permission.
Note that the pressure units bar, atm, and kgf/cm2 are almost equivalent to each other. In the English system, the pressure unit is poundforce per square inch (lbf/in2, or psi), and 1 atm 14.696 psi. The pressure units kgf/cm2 and lbf/in2 are also denoted by kg/cm2 and lb/in2, respectively, and they are commonly used in tire gages. It can be shown that 1 kgf/cm2 14.223 psi. Pressure is also used for solids as synonymous to normal stress, which is force acting perpendicular to the surface per unit area. For example, a 150pound person with a total foot imprint area of 50 in2 exerts a pressure of 150 lbf/50 in2 3.0 psi on the floor (Fig. 1–36). 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 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 (i.e., absolute zero pressure). Most pressuremeasuring devices, however, are calibrated to read zero in the atmosphere (Fig. 1–37), and so they indicate the difference between the absolute pressure and the local atmospheric pressure. This difference is called the gage pressure. Pressures below atmospheric pressure are called vacuum pressures and are measured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pressure. Absolute, gage, and vacuum pressures are all positive quantities and are related to each other by Pgage Pabs Patm
(1–15)
Pvac Patm Pabs
(1–16)
This is illustrated in Fig. 1–38. 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 32 psi (2.25 kgf/cm2) indicates a pressure of 32 psi above the atmospheric pressure. At a location where the atmospheric pressure is 14.3 psi, for example, the absolute pressure in the tire is 32 14.3 46.3 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.
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Chapter 1

P gage Patm Pvac
P abs P atm
Patm Pabs Absolute vacuum
P abs = 0
Absolute vacuum
FIGURE 1–38 Absolute, gage, and vacuum pressures.
EXAMPLE 1–5
Absolute Pressure of a Vacuum Chamber
A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 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. 1–16 to be
Pabs Patm Pvac 14.5 5.8 8.7 psi Discussion Note that the local value of the atmospheric pressure is used when determining the absolute pressure.
Pressure is the compressive force per unit area, and it gives the impression of being a vector. However, pressure at any point in a fluid is the same in all directions. That is, it has magnitude but not a specific direction, and thus it is a scalar quantity.
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 balanced by an increase in pressure (Fig. 1–39).
Pgage
FIGURE 1–39 The pressure of a fluid at rest increases with depth (as a result of added weight).
23
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Thermodynamics To obtain a relation for the variation of pressure with depth, consider a rectangular fluid element of height z, length x, and unit depth (into the page) in equilibrium, as shown in Fig. 1–40. Assuming the density of the fluid r to be constant, a force balance in the vertical zdirection gives
z
P1
a Fz maz 0:¬¬P2 ¢x P1 ¢x rg ¢x ¢z 0
x
(1–17)
where W mg rg x z is the weight of the fluid element. Dividing by x and rearranging gives
z
¢P P2 P1 rg ¢z gs ¢z
W
(1–18)
Ptop = 1 atm
where gs rg is the specific weight of the fluid. Thus, we conclude that the pressure difference between two points in a constant density fluid is proportional to the vertical distance z between the points and the density r of the fluid. In other words, pressure in a fluid increases linearly with depth. This is what a diver experiences when diving deeper in a lake. For a given fluid, the vertical distance z is sometimes used as a measure of pressure, and it is called the pressure head. We also conclude from Eq. 1–18 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 considered 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 assumed to be constant (Fig. 1–41). If we take point 1 to be at the free surface of a liquid open to the atmosphere (Fig. 1–42), where the pressure is the atmospheric pressure Patm, then the pressure at a depth h from the free surface becomes
AIR
P Patm rgh¬or¬Pgage rgh
P2 x
FIGURE 1–40 Freebody diagram of a rectangular fluid element in equilibrium.
(A 5mhigh room)
P bottom = 1.006 atm
FIGURE 1–41 In a room filled with a gas, the variation of pressure with height is negligible.
(1–19)
Liquids are essentially incompressible substances, and thus the variation of density with depth is negligible. This is also the case for gases when the elevation 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 9.807 m/s2 at sea level to 9.764 m/s2 at an elevation of 14,000 m where large passenger planes cruise. This is a change of just 0.4 percent in this extreme case. Therefore, g can be assumed to be 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. 1–17 by x z, and taking the limit as z → 0. It gives dP rg dz
(1–20)
The negative sign is due to our taking the positive z direction to be upward so that dP is negative when dz is positive since pressure decreases in an upward direction. When the variation of density with elevation is known,
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Chapter 1

25
the pressure difference between points 1 and 2 can be determined by integration to be 2
¢P P2 P1
rg dz
1
(1–21)
P1 = Patm
1
For constant density and constant gravitational acceleration, this relation reduces to Eq. 1–18, as expected. 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 (1548–1620) published in 1586 the principle illustrated in Fig. 1–43. Note that the pressures 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 (i.e., 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, the pressure force exerted by the fluid is always normal to the surface at the specified points. A consequence of the pressure in a fluid remaining constant in the horizontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is called Pascal’s law, after Blaise Pascal (1623–1662). 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
h
2 P2 = Patm + rgh
FIGURE 1–42 Pressure in a liquid at rest increases linearly with distance from the free surface.
Patm
Water
h
A
B
C
D
E
PA = PB = PC = PD = PE = PF = PG = Patm + rgh
Mercury
PH ≠ PI
H
F
G
I
FIGURE 1–43 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.
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Thermodynamics to exert a proportionally greater force than that applied to the smaller. “Pascal’s 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 by one arm, as shown in Fig. 1–44. 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
F2 = P2 A2 F 1 = P 1A 1
1
A1 P1
A2 P2
F1 F2 F2 A2 P1 P2¬¬ S ¬¬ ¬¬ S ¬¬ A1 A2 F1 A1
2
FIGURE 1–44 Lifting of a large weight by a small force by the application of Pascal’s law.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 10 ON THE DVD.
Gas
h
1
2
The area ratio A2/A1 is called the ideal mechanical advantage of the hydraulic lift. Using a hydraulic car jack with a piston area ratio of A2/A1 10, for example, a person can lift a 1000kg car by applying a force of just 100 kgf ( 981 N).
1–10
■
THE MANOMETER
We notice from Eq. 1–18 that an elevation change of z in a fluid at rest corresponds to P/rg, 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 pressure differences. A manometer mainly consists of a glass or plastic Utube containing one or more fluids such as mercury, water, alcohol, or oil. To keep the size of the manometer to a manageable level, heavy fluids such as mercury are used if large pressure differences are anticipated. Consider the manometer shown in Fig. 1–45 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. 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. 1–19 to be P2 Patm rgh
FIGURE 1–45 The basic manometer.
(1–22)
(1–23)
where r is the density of the fluid in the tube. Note that the crosssectional area of the tube has no effect on the differential height h, and thus the pressure exerted by the fluid. However, the diameter of the tube should be large enough (more than a few millimeters) to ensure that the surface tension effect and thus the capillary rise is negligible.
EXAMPLE 1–6
Measuring Pressure with a Manometer
A manometer is used to measure the pressure in a tank. The fluid used has a specific gravity of 0.85, and the manometer column height is 55 cm, as shown in Fig. 1–46. If the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank.
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Chapter 1 spheric pressure are given. The absolute pressure in the tank is to be determined. Assumptions The fluid in the tank is a gas whose density is much lower than the density of manometer fluid. Properties The specific gravity of the manometer fluid is given to be 0.85. We take the standard density of water to be 1000 kg/m3. Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, which is taken to be 1000 kg/m3:
P=?
h = 55 cm
SG = 0.85
r SG 1r H2O 2 10.852 11000 kg>m3 2 850 kg>m3
FIGURE 1–46 Schematic for Example 1–6.
P Patm rgh 96 kPa 1850 kg>m3 2 19.81 m>s2 2 10.55 m2 a
27
Patm = 96 kPa
Solution The reading of a manometer attached to a tank and the atmo
Then from Eq. 1–23,

1N 1 kPa ba b 2 # 1 kg m>s 1000 N>m2
Patm Fluid 1
100.6 kPa Discussion
h1
Note that the gage pressure in the tank is 4.6 kPa. Fluid 2 h2
Many engineering problems and some manometers involve multiple immiscible 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 P rgh, (2) pressure increases downward in a given fluid and decreases upward (i.e., 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 Pascal’s law, allows us to “jump” from one fluid column to the next in manometers without worrying about pressure change as long as we don’t jump over a different fluid, and the fluid is at rest. Then the pressure at any point can be determined by starting with a point of known pressure and adding or subtracting rgh terms as we advance toward the point of interest. For example, the pressure at the bottom of the tank in Fig. 1–47 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
Fluid 3 h3
1
FIGURE 1–47 In stackedup fluid layers, the pressure change across a fluid layer of density r and height h is rgh. A flow section or flow device Fluid
1
2
Patm r1gh1 r2gh2 r3gh3 P1
In the special case of all fluids having the same density, this relation reduces to Eq. 1–23, as expected. 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. 1–48. The working fluid can be either a gas or a liquid whose density is r1. The density of the manometer fluid is r2, and the differential fluid height is h.
a h
r1 A
B
r2
FIGURE 1–48 Measuring the pressure drop across a flow section or a flow device by a differential manometer.
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Thermodynamics 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 rgh terms until we reach point 2, and setting the result equal to P2: P1 r1g 1a h 2 r2gh r1ga P2
(1–24)
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, P1 P2 1r2 r1 2gh
(1–25)
Note that the distance a has no effect on the result, but must be included in the analysis. Also, when the fluid flowing in the pipe is a gas, then r1
r2 and the relation in Eq. 1–25 simplifies to P1 P2 r2gh.
EXAMPLE 1–7
Oil AIR 1 WATER
h1 2 h2
h3
Mercury
FIGURE 1–49 Schematic for Example 1–7. (Drawing not to scale.)
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. 1–49. The tank is located on a mountain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if h1 0.1 m, h2 0.2 m, and h3 0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, 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. Assumption The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air–water interface. Properties The densities of water, oil, and mercury are given to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. Analysis Starting with the pressure at point 1 at the air–water interface, moving along the tube by adding or subtracting the rgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives
P1 rwatergh1 roilgh2 rmercurygh3 Patm Solving for P1 and substituting,
P1 Patm rwatergh1 roilgh2 rmercurygh3 Patm g 1rmercuryh3 rwaterh1 roilh2 2
85.6 kPa 19.81 m>s2 2 3 113,600 kg>m3 2 10.35m2 1000 kg>m3 2 10.1 m2 ¬ 1850 kg>m3 2 10.2 m2 4 a
1N 1 kPa ba b 1 kg # m>s2 1000 N>m2
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.
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29
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 (1808–1884), which consists of a hollow metal tube bent like a hook whose end is closed and connected to a dial indicator needle (Fig. 1–50). 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 pressure). When the fluid inside the tube is pressurized, the tube stretches and moves the needle in proportion to the pressure applied. 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, absolute, and differential pressures in a wide range of applications. Gage pressure 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. The absolute 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 stretches in response to a change in pressure difference across it, the strain gage stretches and a Wheatstone bridge circuit amplifies the output. A capacitance 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 the diaphragm units and are very suitable for highpressure applications, but they are generally not as sensitive as the diaphragmtype transducers.
1–11
■
THE BAROMETER AND ATMOSPHERIC PRESSURE
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 (1608–1647) was the first to conclusively prove that the atmospheric pressure can be measured by inverting a mercuryfilled tube into a mercury container that is open to the atmosphere,
Ctype
Spiral
Twisted tube Helical
Tube cross section
FIGURE 1–50 Various types of Bourdon tubes used to measure pressure.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 11 ON THE DVD.
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Thermodynamics as shown in Fig. 1–51. The pressure at point B is equal to the atmospheric pressure, and the pressure at 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 vertical direction gives
C
A h
h W = rgh A
B Mercury Patm
FIGURE 1–51 The basic barometer.
A1
A2
A3
FIGURE 1–52 The length or the crosssectional area of the tube has 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.
Patm rgh
(1–26)
where r 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. 1–52). 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 0°C (rHg 13,595 kg/m3) under standard gravitational acceleration (g 9.807 m/s2). If water instead of mercury were used to measure the standard atmospheric pressure, a water column of about 10.3 m would be needed. Pressure 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 (29.92 inHg) at 0°C. The unit mmHg is also called the torr in honor of Torricelli. Therefore, 1 atm 760 torr and 1 torr 133.3 Pa. The standard atmospheric pressure Patm changes from 101.325 kPa at sea level to 89.88, 79.50, 54.05, 26.5, and 5.53 kPa at altitudes of 1000, 2000, 5000, 10,000, and 20,000 meters, respectively. The standard atmospheric pressure in Denver (elevation 1610 m), for example, is 83.4 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 ramifications 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 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 2.0L car engine will act like a 1.7L 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. 1–53). A fan or compressor will displace 15 percent less air at that altitude for the same volume 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 for reduced drag and thus better fuel efficiency.
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Chapter 1 EXAMPLE 1–8
Measuring Atmospheric Pressure with a Barometer
Engine

31
Lungs
Determine the atmospheric pressure at a location where the barometric reading is 740 mm Hg and the gravitational acceleration is g 9.81 m/s2. Assume the temperature of mercury to be 10C, at which its density is 13,570 kg/m3.
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 10C. Properties The density of mercury is given to be 13,570 kg/m3. Analysis From Eq. 1–26, the atmospheric pressure is determined to be
Patm rgh 113,570 kg>m3 2 19.81 m>s2 2 10.74 m2 a
FIGURE 1–53 At high altitudes, a car engine generates less power and a person gets less oxygen because of the lower density of air.
1N 1 kPa ba b 2 # 1 kg m>s 1000 N>m2
98.5 kPa Discussion Note that density changes with temperature, and thus this effect should be considered in calculations.
EXAMPLE 1–9
Effect of Piston Weight on Pressure in a Cylinder
The piston of a vertical piston–cylinder device containing a gas has a mass of 60 kg and a crosssectional area of 0.04 m2, as shown in Fig. 1–54. The local atmospheric pressure is 0.97 bar, and the gravitational acceleration is 9.81 m/s2. (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?
Solution A gas is contained in a vertical cylinder with a heavy piston. The pressure inside the cylinder and the effect of volume change on pressure are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis (a) The gas pressure in the piston–cylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the freebody diagram of the piston as shown in Fig. 1–54 and balancing the vertical forces yield
PA Patm A W Solving for P and substituting,
P Patm
mg A
0.97 bar 1.12 bar
Patm = 0.97 bar m = 60 kg
Patm
A = 0.04 m 2
160 kg2 19.81 m>s2 2 10.04 m2 2
a
1N 1 bar ba 5 b 2 # 1 kg m>s 10 N>m2
(b) The volume change will have no effect on the freebody diagram drawn in part (a), and therefore the pressure inside the cylinder will remain the same. Discussion If the gas behaves as an ideal gas, the absolute temperature doubles when the volume is doubled at constant pressure.
P=?
P W = mg
FIGURE 1–54 Schematic for Example 1–9, and the freebody diagram of the piston.
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Thermodynamics EXAMPLE 1–10
Hydrostatic Pressure in a Solar Pond with Variable Density
Solar ponds are small artificial lakes of 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. 1–55, and the density can be expressed as
r r0
B
1 tan2 a
p z b 4 H
where r0 is the density on the water surface, z is the vertical distance measured downward from the top of the gradient zone, and H is the thickness of the gradient zone. For H 4 m, r0 1040 kg/m3, and a thickness of 0.8 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 kg/m3. 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
P1 rgh1 11040 kg>m3 2 19.81 m>s2 2 10.8 m2 a
1 kN b 8.16 kPa 1000 kg # m>s2
since 1 kN/m2 1 kPa. The differential change in hydrostatic pressure across a vertical distance of dz is given by
dP rg dz Integrating from the top of the gradient zone (point 1 where z 0) to any location z in the gradient zone (no subscript) gives
P P1
z
z
rg dz¬¬S ¬¬P P1
r B 1 tan a 4 H b g dz 2
Sun Increasing salinity and density r0 = 1040 kg/m3 Surface zone
1
z H=4m
Gradient zone Storage zone
FIGURE 1–55 Schematic for Example 1–10.
2
p z
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Chapter 1
4 14 m2 p
Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig. 1–56. The dashed line indicates the hydrostatic pressure for the case of constant density at 1040 kg/m3 and is given for reference. Note that the variation of pressure with depth is not linear when density varies with depth.
■
2 1.5 1 0.5 0 0
p4 1 kN ba b 4 4 1000 kg # m>s2
54.0 kPa 1gage2
1–12
z, m
2.5
Then the pressure at the bottom of the gradient zone (z H 4 m) becomes
sinh1 a tan
3.5 3
4H p z sinh1 a tan b p 4 H
P2 8.16 kPa 11040 kg>m3 2 19.81 m>s2 2
33
4
Performing the integration gives the variation of gage pressure in the gradient zone to be
P P1 r0g

10
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 problem and the objectives before you attempt to solve the problem.
30 40 P, kPa
50
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 12 ON THE DVD.
SOLUTION
AY
SY
W
EA
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 elaborate, but it should resemble the actual system and show the key features. 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 temperature during an isothermal process), and indicate them on the sketch.
60
FIGURE 1–56 The variation of gage pressure with depth in the gradient zone of the solar pond.
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 testing this knowledge. This is done by solving significant realworld problems. Solving such problems, especially complicated ones, require a systematic approach. By using a stepbystep approach, an engineer can reduce the solution of a complicated problem into the solution of a series of simple problems (Fig. 1–57). 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.
20
HARD WAY
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FIGURE 1–57 A stepbystep approach can greatly simplify problem solving.
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Step 3: Assumptions and Approximations 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 = 0.83 atm (Appropriate. Ignores only minor effects such as weather.)
FIGURE 1–58 The assumptions made while solving an engineering problem must be reasonable and justifiable.
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 necessary. 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 0.83 atm in Denver (elevation 1610 m) (Fig. 1–58).
Step 4: Physical Laws Apply all the relevant basic physical laws and principles (such as the conservation of mass), and reduce them to their simplest form by utilizing the assumptions 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, don’t give a false implication of high precision by copying all the digits from the screen of the calculator—round the results to an appropriate number of significant digits (see p. 38).
Step 7: Reasoning, Verification, and Discussion
Energy use:
$80/yr
Energy saved by insulation:
$200/yr
IMPOSSIBLE!
FIGURE 1–59 The results obtained from an engineering analysis must be checked for reasonableness.
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. 1–59). Also, point out the significance of the results, and discuss their implications. State the conclusions that can be drawn from the results, and any recommendations 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 $20 insulation jacket will reduce the energy cost by $30 a year, indicate that 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.
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Keep in mind that the solutions you present to your instructors, and any engineering analysis presented to others, is a form of communication. Therefore neatness, organization, completeness, and visual appearance are of utmost importance for maximum effectiveness. 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 without 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 misinformation. 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 without 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 wordprocessing program can also learn how to use those software packages. However, 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 engineering 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. 1–60). Hand calculators did not eliminate the need to teach our children how to add or subtract, and the sophisticated medical software packages did not take the
FIGURE 1–60 An excellent wordprocessing program does not make a person a good writer; it simply makes a good writer a more efficient writer. © Vol. 80/PhotoDisc
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Thermodynamics 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 physics. 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 today’s 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.
Engineering Equation Solver (EES) EES is a program that solves systems of linear or nonlinear algebraic or differential equations numerically. It has a large library of builtin thermodynamic property functions as well as mathematical functions, and allows the user to supply additional property data. Unlike some software packages, EES does not solve engineering problems; it only solves the equations supplied by the user. Therefore, the user must understand the problem and formulate it by applying any relevant physical laws and relations. EES saves 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. EES is a very powerful yet intuitive program that is very easy to use, as shown in Examples 1–11 and 1–12. The use and capabilities of EES are explained in Appendix 3 on the enclosed DVD.
EXAMPLE 1–11
Solving a System of Equations with EES
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. They are to be determined. Analysis We start the EES program by doubleclicking on its icon, open a new file, and type the following on the blank screen that appears:
xy 4 x^2y^2 xy20
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which is an exact mathematical expression of the problem statement with x and y denoting the unknown numbers. The solution to this system of two nonlinear equations with two unknowns is obtained by a single click on the “calculator” icon on the taskbar. It gives
x5¬and¬y1 Discussion Note that all we did is formulate the problem as we would on paper; EES took care of all the mathematical details of 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.
EXAMPLE 1–12
Oil
Analyzing a Multifluid Manometer with EES
Reconsider the multifluid manometer discussed in Example 1–7 and replotted in Fig. 1–61. Determine the air pressure in the tank using EES. Also determine what the differential fluid height h3 would be for the same air pressure if the mercury in the last column were replaced by seawater with a density of 1030 kg/m3.
Solution The pressure in a water tank is measured by a multifluid manometer. The air pressure in the tank and the differential fluid height h3 if mercury is replaced by seawater are to be determined using EES. Analysis We start the EES program by doubleclicking on its icon, open a new file, and type the following on the blank screen that appears (we express the atmospheric pressure in Pa for unit consistency):
g=9.81 Patm=85600 h1=0.1; h2=0.2; h3=0.35 rw=1000; roil=850; rm=13600 P1+rw*g*h1+roil*g*h2rm*g*h3=Patm Here P1 is the only unknown, and it is determined by EES to be
P1 129647 Pa 130 kPa which is identical to the result obtained before. The height of the fluid column h3 when mercury is replaced by seawater is determined easily by replacing “h30.35” by “P1129647” and “r m13600” by “r m1030,” and clicking on the calculator symbol. It gives
h3 4.62 m Discussion Note that we used the screen like a paper pad and wrote down the relevant information together with the applicable relations in an organized manner. EES did the rest. Equations can be written on separate lines or on the same line by separating them by semicolons, and blank or comment lines can be inserted for readability. EES makes it very easy to ask “what if” questions, and to perform parametric studies, as explained in Appendix 3 on the DVD. EES also has the capability to check the equations for unit consistency if units are supplied together with numerical values. Units can be specified
AIR 1 WATER
h1 2 h2
Mercury
FIGURE 1–61 Schematic for Example 1–12.
h3
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Given: Volume: Density:
within brackets [ ] after the specified value. When this feature is utilized, the previous equations would take the following form: V = 3.75 L r = 0.845 kg/L kg
(3 significant digits) Also,
3.75 0.845 = 3.16875
g=9.81 [m/s^2] Patm=85600 [Pa] h1=0.1 [m]; h2=0.2 [m]; h3=0.35 [m] rw=1000 [kg/m^3]; roil=850 [kg/m^3]; rm=13600 [kg/m^3] P1+rw*g*h1+roil*g*h2rm*g*h3=Patm
Find: Mass: m = rV = 3.16875 kg Rounding to 3 significant digits: m = 3.17 kg
FIGURE 1–62 A result with more significant digits than that of given data falsely implies more accuracy.
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 implies greater accuracy than exists, and it should be avoided. For example, consider a 3.75L container filled with gasoline whose density is 0.845 kg/L, and try to determine its mass. Probably the first thought that comes to your mind is to multiply the volume and density to obtain 3.16875 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 3.17 kg instead of what appears in the screen of the calculator. The result 3.16875 kg would be correct only if the volume and density were given to be 3.75000 L and 0.845000 kg/L, respectively. The value 3.75 L implies that we are fairly confident that the volume is accurate within 0.01 L, and it cannot be 3.74 or 3.76 L. However, the volume can be 3.746, 3.750, 3.753, etc., since they all round to 3.75 L (Fig. 1–62). 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 accurate 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 40.0 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 kg/m3 for density, which is the density value of pure water at 0°C. Using this value at 75°C will result in an error of 2.5 percent since the density at this temperature is 975 kg/m3. The minerals and impurities in the water will introduce additional 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.
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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 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. 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 equilibrium. 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 independent, 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 system today are the Celsius scale and the Fahrenheit scale, respectively. They are related to absolute temperature scales by T 1K2 T 1°C2 273.15 T 1R2 T 1°F2 459.67
The magnitudes of each division of 1 K and 1°C are identical, and so are the magnitudes of each division of 1 R and 1°F. Therefore, ¢T 1K 2 ¢T 1°C2 ¬and¬¢T 1R 2 ¢T 1°F2
The normal force exerted by a fluid per unit area is called pressure, and its unit is the pascal, 1 Pa 1 N/m2. 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 below atmospheric pressure are called vacuum pressures. The absolute, gage, and vacuum pressures are related by Pgage Pabs Patm¬¬1for pressures above Patm 2
Pvac Patm Pabs¬¬1for pressures below Patm 2
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 rg dz 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 z is ¢P P2 P1 rg ¢z The absolute and gage pressures in a liquid open to the atmosphere at a depth h from the free surface are P Patm rgh¬or¬Pgage rgh Small to moderate pressure differences are measured by a manometer. The pressure in a stationary fluid remains constant in the horizontal direction. Pascal’s 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 Patm rgh 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. 2nd ed. New York: Wiley, 1997.
3. J. A. Schooley. Thermometry. Boca Raton, FL: CRC Press, 1986.
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PROBLEMS* Thermodynamics 1–1C What is the difference between the classical and the statistical approaches to thermodynamics? 1–2C Why does a bicyclist pick up speed on a downhill road even when he is not pedaling? Does this violate the conservation of energy principle? 1–3C An office worker claims that a cup of cold coffee on his table warmed up to 80°C by picking up energy from the surrounding air, which is at 25°C. Is there any truth to his claim? Does this process violate any thermodynamic laws?
Mass, Force, and Units 1–4C What is the difference between poundmass and poundforce? 1–5C What is the difference between kgmass and kgforce? 1–6C What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
1–12
A 5kg rock is thrown upward with a force of 150 N at a location where the local gravitational acceleration is 9.79 m/s2. Determine the acceleration of the rock, in m/s2. 1–13
Solve Prob. 1–12 using EES (or other) software. Print out the entire solution, including the numerical results with proper units.
1–14 The value of the gravitational acceleration g decreases with elevation from 9.807 m/s2 at sea level to 9.767 m/s2 at an altitude of 13,000 m, where large passenger planes cruise. Determine the percent reduction in the weight of an airplane cruising at 13,000 m relative to its weight at sea level.
Systems, Properties, State, and Processes 1–15C 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.
1–7 A 3kg plastic tank that has a volume of 0.2 m3 is filled with liquid water. Assuming the density of water is 1000 kg/m3, determine the weight of the combined system. 1–8 Determine the mass and the weight of the air contained in a room whose dimensions are 6 m 6 m 8 m. Assume the density of the air is 1.16 kg/m3. Answers: 334.1 kg, 3277 N 1–9 At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g a bz, where a 9.807 m/s2 and b 3.32 106 s2. Determine the height above sea level where the weight of an object will decrease by 1 percent. Answer: 29,539 m 1–10E 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 5.48 ft/s2. Determine how much he will weigh (a) on the spring scale and (b) on the beam scale. Answers: (a) 25.5 lbf; (b) 150 lbf 1–11 The acceleration of highspeed aircraft is sometimes expressed in g’s (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 g’s. *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 solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the icon are comprehensive in nature and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
FIGURE P1–15C © The McGrawHill Companies, Inc./Jill Braaten, photographer
1–16C 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. 1–17C What is the difference between intensive and extensive properties? 1–18C For a system to be in thermodynamic equilibrium, do the temperature and the pressure have to be the same everywhere? 1–19C What is a quasiequilibrium process? What is its importance in engineering? 1–20C Define the isothermal, isobaric, and isochoric processes. 1–21C What is the state postulate? 1–22C Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain.
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1–23C What is a steadyflow process?
Pressure, Manometer, and Barometer
1–24C What is specific gravity? How is it related to density?
1–34C What is the difference between gage pressure and absolute pressure?
1–25
The density of atmospheric air varies with elevation, 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 atmosphere 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. kg/m3
z, km
r,
6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402
1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008
Temperature 1–26C What is the zeroth law of thermodynamics? 1–27C What are the ordinary and absolute temperature scales in the SI and the English system? 1–28C Consider an alcohol and a mercury thermometer that read exactly 0°C at the ice point and 100°C 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, 60°C? Explain. 1–29 The deep body temperature of a healthy person is 37°C. What is it in kelvins? 1–30E Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F.
1–35C Explain why some people experience nose bleeding and some others experience shortness of breath at high elevations. 1–36C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled. Do you agree? Explain. 1–37C A tiny steel cube is suspended in water by a string. If the lengths of the sides of the cube are very small, how would you compare the magnitudes of the pressures on the top, bottom, and side surfaces of the cube? 1–38C Express Pascal’s law, and give a realworld example of it. 1–39C 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? 1–40 A vacuum gage connected to a chamber reads 35 kPa at a location where the atmospheric pressure is 92 kPa. Determine the absolute pressure in the chamber. 1–41E A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the manometer arm with the (a) higher and (b) lower fluid level being attached to the tank. 1–42 The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig. P1–42. Determine the gage pressure of air in the tank if Oil AIR 1 WATER
h1
2
1–31 The temperature of a system rises by 15°C during a heating process. Express this rise in temperature in kelvins. 1–32E The temperature of a system drops by 45°F during a cooling process. Express this drop in temperature in K, R, and °C. 1–33 Consider two closed systems A and B. System A contains 3000 kJ of thermal energy at 20°C, whereas system B contains 200 kJ of thermal energy at 50°C. Now the systems are brought into contact with each other. Determine the direction of any heat transfer between the two systems.
h2
Mercury
FIGURE P1–42
h3
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h1 0.2 m, h2 0.3 m, and h3 0.46 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. 1–43 Determine the atmospheric pressure at a location where the barometric reading is 750 mm Hg. Take the density of mercury to be 13,600 kg/m3. 1–44 The gage pressure in a liquid at a depth of 3 m is read to be 28 kPa. Determine the gage pressure in the same liquid at a depth of 9 m. 1–45 The absolute pressure in water at a depth of 5 m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0.85 at the same location. 1–46E Show that 1 kgf/cm2 14.223 psi. 1–47E 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. 1–48 Consider a 70kg woman who has a total foot imprint area of 400 cm2. She wishes to walk on the snow, but the snow cannot withstand pressures greater than 0.5 kPa. Determine the minimum size of the snowshoes needed (imprint area per shoe) to enable her to walk on the snow without sinking.
FIGURE P1–53
1–49 A vacuum gage connected to a tank reads 15 kPa at a location where the barometric reading is 750 mm Hg. Determine the absolute pressure in the tank. Take rHg 13,590 kg/m3. Answer: 85.0 kPa 1–50E A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 mm Hg. Determine the absolute pressure in the tank. Take rHg 848.4 lbm/ft3. Answer: 64.3 psia
© Vol. 74/Corbis
1–54
Solve Prob. 1–53 using EES (or other) software. Print out the entire solution, including the numerical results with proper units.
1–51 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.
1–55 Determine the pressure exerted on a diver at 30 m below the free surface of the sea. Assume a barometric pressure of 101 kPa and a specific gravity of 1.03 for seawater.
1–52 The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20 kg/m3. Answer: 1274 m
Answer: 404.0 kPa
1–56E Determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. Assume that the barometric pressure is 14.7 psia and the specific gravity of seawater is 1.03.
1–53 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 730 and 755 mm Hg, respectively, determine the height of the building. Take the densities of air and mercury to be 1.18 kg/m3 and 13,600 kg/m3, respectively.
1–57 A gas is contained in a vertical, frictionless piston–cylinder device. The piston has a mass of 4 kg and a crosssectional area of 35 cm2. A compressed spring above the piston exerts a force of 60 N on the piston. If the atmospheric pressure is 95 kPa, determine the pressure inside the cylinder. Answer: 123.4 kPa
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Patm = 95 kPa

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ence in the manometer levels is 15 mm, and the atmospheric pressure is 100 kPa. (a) Judging from Fig. P1–62, determine if the pressure in the duct is above or below the atmospheric pressure. (b) Determine the absolute pressure in the duct.
mP = 4 kg AIR A = 35 cm2 P=?
h = 15 mm
FIGURE P1–57 1–58
Reconsider Prob. 1–57. Using EES (or other) software, 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. 1–59
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 (r 13,600 kg/m3) or (b) water (r 1000 kg/m3). Pg = 80 kPa
Gas
h=?
FIGURE P1–62 1–63 Repeat Prob. 1–62 for a differential mercury height of 45 mm. 1–64 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 mm Hg. The systolic and diastolic pressures of a healthy person are about 120 mm Hg and 80 mm Hg, respectively, and are indicated as 120/80. Express both of these gage pressures in kPa, psi, and meter water column. 1–65 The maximum blood pressure in the upper arm of a healthy person is about 120 mm Hg. If a vertical tube open to the atmosphere is connected to the vein in the arm of the person, determine how high the blood will rise in the tube. Take the density of the blood to be 1050 kg/m3.
FIGURE P1–59 1–60
Reconsider Prob. 1–59. Using EES (or other) software, investigate the effect of the manometer fluid density in the range of 800 to 13,000 kg/m3 on the differential fluid height of the manometer. Plot the differential fluid height against the density, and discuss the results.
h
1–61 A manometer containing oil (r 850 kg/m3) is attached to a tank filled with air. If the oillevel difference between the two columns is 60 cm and the atmospheric pressure is 98 kPa, determine the absolute pressure of the air in the tank. Answer: 103 kPa 1–62 A mercury manometer (r 13,600 kg/m3) is connected to an air duct to measure the pressure inside. The differ
FIGURE P1–65
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Thermodynamics
1–66 Consider a 1.8mtall 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 this man, in kPa.
Air
Natural Gas
1–67 Consider a Utube whose arms are open to the atmosphere. Now water is poured into the Utube from one arm, and light oil (r 790 kg/m3) 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.
10 in 25 in 6 in
Mercury SG = 13.6
1–72E Repeat Prob. 1–71E by replacing air by oil with a specific gravity of 0.69.
Water
1–73 The gage pressure of the air in the tank shown in Fig. P1–73 is measured to be 80 kPa. Determine the differential height h of the mercury column.
FIGURE P1–67 1–68 The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2000 kg. Determine the fluid gage pressure that must be maintained in the reservoir.
75 cm Air
h
Mercury SG = 13.6
FIGURE P1–73 1–74
40 cm 70 cm
Water 30 cm
Air
60 cm
Oil SG = 0.72
80 kPa
1–69 Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double Utube manometer, as shown in Fig. P1–69. Determine the pressure difference between the two pipelines. Take the density of seawater at that location to be r 1035 kg/m3. Can the air column be ignored in the analysis?
Fresh water
Water
FIGURE P1–71E
Oil 70 cm
2 in
Sea water
Repeat Prob. 1–73 for a gage pressure of 40 kPa.
1–75 The top part of a water tank is divided into two compartments, as shown in Fig. P1–75. Now a fluid with an
Unknown liquid
10 cm Mercury
80 cm
FIGURE P1–69 1–70 Repeat Prob. 1–69 by replacing the air with oil whose specific gravity is 0.72. 1–71E The pressure in a natural gas pipeline is measured by the manometer shown in Fig. P1–71E with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure in the pipeline.
WATER 50 cm
FIGURE P1–75
95 cm
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Chapter 1 unknown density is poured into one side, and the water level rises a certain amount on the other side to compensate for this effect. Based on the final fluid heights shown on the figure, determine the density of the fluid added. Assume the liquid does not mix with water.

45
1–78 A multifluid container is connected to a Utube, as shown in Fig. P1–78. For the given specific gravities and fluid column heights, determine the gage pressure at A. Also determine the height of a mercury column that would create the same pressure at A. Answers: 0.471 kPa, 0.353 cm
1–76 Consider a doublefluid manometer attached to an air pipe shown in Fig. P1–76. If the specific gravity of one fluid is 13.55, determine the specific gravity of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kPa. Answer: 5.0
A 70 cm
Air
30 cm
Oil SG = 0.90
Water 90 cm
P = 76 kPa 40 cm 20 cm
Glycerin SG = 1.26
SG2 15 cm
22 cm
FIGURE P1–78 SG1 = 13.55
Solving Engineering Problems and EES 1–79C What is the value of the engineering software packages in (a) engineering education and (b) engineering practice?
FIGURE P1–76 1–77 Consider the system shown in Fig. P1–77. If a change of 0.7 kPa in the pressure of air causes the brine–mercury 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.
1–80
Determine a positive real root of this equation using EES: 2x3 10x0.5 3x 3
1–81
Solve this system of two equations with two unknowns using EES: x3 y2 7.75 3xy y 3.5
Air
Water
Brine pipe SG = 1.1
1–82
Solve this system of three equations with three unknowns using EES: 2x y z 5 3x2 2y z 2
Area, A1
xy 2z 8 Mercury SG = 13.56
Area, A2
1–83
Solve this system of three equations with three unknowns using EES: x2y z 1 x 3y0.5 xz 2
FIGURE P1–77
xyz2
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1–84E
Specific heat is defined as the amount of energy needed to increase the temperature of a unit mass of a substance by one degree. The specific heat of water at room temperature is 4.18 kJ/kg °C in SI unit system. Using the unit conversion function capability of EES, express the specific heat of water in (a) kJ/kg K, (b) Btu/lbm F, (c) Btu/lbm R, and (d) kCal/kg °C units.
Answers: (a) 4.18, (b) (c) (d) 0.9984
1–87 The pilot of an airplane reads the altitude 3000 m and the absolute pressure 58 kPa when flying over a city. Calculate the local atmospheric pressure in that city in kPa and in mm Hg. Take the densities of air and mercury to be 1.15 kg/m3 and 13,600 kg/m3, respectively. Altitude: 3 km P = 58 kPa
Review Problems 1–85 A hydraulic lift is to be used to lift a 2500 kg 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.
Weight 2500 kg
F1 25 kg
F2
D2
10 cm
FIGURE P1–87 1–88 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 in Prob. 1–9, 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). 1–89 A man goes to a traditional market to buy a steak for dinner. He finds a 12oz steak (1 lbm 16 oz) for $3.15. He then goes to the adjacent international market and finds a 320g steak of identical quality for $2.80. Which steak is the better buy? 1–90 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 85,000 lbf. Express this thrust in N and kgf.
FIGURE P1–85
1–86 A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 5 kg and a diameter of 12 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: 95.7 kPa, 115.3 kg
WEIGHTS
1–91E The efficiency of a refrigerator increases by 3 percent for each °C rise in the minimum temperature in the device. What is the increase in the efficiency for each (a) K, (b) °F, and (c) R rise in temperature? 1–92E The boiling temperature of water decreases by about 3°C 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? 1–93E The average body temperature of a person rises by about 2°C during strenuous exercise. What is the rise in the body temperature in (a) K, (b) °F, and (c) R during strenuous exercise? 1–94E Hyperthermia of 5°C (i.e., 5°C rise above the normal body temperature) is considered fatal. Express this fatal level of hyperthermia in (a) K, (b) °F, and (c) R.
GAS
FIGURE P1–86
1–95E A house is losing heat at a rate of 4500 kJ/h 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.
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Chapter 1 1–96 The average temperature of the atmosphere in the world is approximated as a function of altitude by the relation Tatm 288.15 6.5z 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 12,000 m. 1–97 Joe Smith, an oldfashioned engineering student, believes that the boiling point of water is best suited for use as the reference point on temperature scales. Unhappy that the boiling point corresponds to some odd number in the current absolute temperature scales, he has proposed a new absolute temperature scale that he calls the Smith scale. The temperature unit on this scale is smith, denoted by S, and the boiling point of water on this scale is assigned to be 1000 S. From a thermodynamic point of view, discuss if it is an acceptable temperature scale. Also, determine the ice point of water on the Smith scale and obtain a relation between the Smith and Celsius scales.

47
Determine the upward force the water will exert on the duct. Take the densities of air and water to be 1.3 kg/m3 and 1000 kg/m3, respectively. 1–101 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 rairgVballoon, will push the balloon upward. If the balloon has a diameter of 10 m and carries two people, 70 kg each, determine the acceleration of the balloon when it is first released. Assume the density of air is r 1.16 kg/m3, and neglect the weight of the ropes and the cage. Answer: 16.5 m/s2
HELIUM D = 10 m rHe = 17 rair
1–98E It is wellknown 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 temperature in °F is given by [ASHRAE, Handbook of Fundamentals (Atlanta, GA, 1993), p. 8.15] Tequiv 91.4 191.4 Tambient 2
¬ 10.475 0.0203V 0.3042V 2
where V is the wind velocity in mi/h 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 mi/h. The constant 91.4°F 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 km/h and Tambient is the ambient air temperature in °C. Answer:
Tequiv 33.0(33.0Tambient )
– (0.475 0.0126V 0.240V )
1–99E
Reconsider Problem 1–98E. Using EES (or other) software, plot the equivalent wind chill temperatures in °F as a function of wind velocity in the range of 4 to 100 mph for the ambient temperatures of 20, 40, and 60°F. Discuss the results.
1–100 An airconditioning system requires a 20mlong section of 15cm diameter duct work to be laid underwater.
m = 140 kg
FIGURE P1–101
1–102
Reconsider Prob. 1–101. Using EES (or other) software, investigate the effect of the number of people carried in the balloon on acceleration. Plot the acceleration against the number of people, and discuss the results.
1–103 Determine the maximum amount of load, in kg, the balloon described in Prob. 1–101 can carry. Answer: 520.5 kg
1–104E The pressure in a steam boiler is given to be 92 kgf/cm2. Express this pressure in psi, kPa, atm, and bars. 1–105 The basic barometer can be used as an altitudemeasuring device in airplanes. The ground control reports a barometric reading of 753 mm Hg while the pilot’s reading is 690 mm Hg. Estimate the altitude of the plane from ground level if the average air density is 1.20 kg/m3. Answer: 714 m
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1–106 The lower half of a 10mhigh cylindrical container is filled with water (r 1000 kg/m3) and the upper half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and bottom of the cylinder. Answer: 90.7 kPa
1–109 A glass tube is attached to a water pipe, as shown in Fig. P1–109. If the water pressure at the bottom of the tube is 115 kPa and the local atmospheric pressure is 92 kPa, determine how high the water will rise in the tube, in m. Take the density of water to be 1000 kg/m3.
OIL SG = 0.85 h = 10 m
Patm = 92 kPa
WATER ρ = 1000 kg/m3
h=?
FIGURE P1–106
Water
1–107 A vertical, frictionless piston–cylinder device contains a gas at 250 kPa absolute pressure. The atmospheric pressure outside is 100 kPa, and the piston area is 30 cm2. Determine the mass of the piston. 1–108 A pressure cooker cooks a lot faster than an ordinary 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 opening 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 opening crosssectional area of 4 mm2. Assume an atmospheric pressure of 101 kPa, and draw the freebody diagram of the petcock. Answer: 40.8 g Patm = 101 kPa Petcock A = 4 mm2
FIGURE P1–109 1–110 The average atmospheric pressure on earth is approximated as a function of altitude by the relation Patm 101.325 (1 0.02256z)5.256, 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). 1–111 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 35° from the horizontal, as shown in Fig. P1–111. The density of the liquid in the manometer is 0.81 kg/L, and the vertical distance between the fluid levels in the two arms of the manometer is 8 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.
Duct
Air L
8 cm PRESSURE COOKER
35°
FIGURE P1–111 FIGURE P1–108
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Chapter 1 1–112E Consider a Utube whose arms are open to the atmosphere. Now equal volumes of water and light oil (r 49.3 lbm/ft3) 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, determine the gage pressure the person exerts on the oil by blowing.
49
Oil SG = 0.79 Pgage = 370 kPa Gasoline SG = 0.70
Air
45 cm 50 cm
Pipe 22 cm 10 cm
Water
Mercury SG = 13.6
FIGURE P1–114
Air
Oil

Water
30 in
FIGURE P1–112E 1–113 Intravenous infusions are usually driven by gravity by hanging the fluid bottle at sufficient height to counteract the blood pressure in the vein and to force the fluid into the body. 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 1.2 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 kg/m3. Patm IV bottle 1.2 m
FIGURE P1–113 1–114 A gasoline line is connected to a pressure gage through a doubleU manometer, as shown in Fig. P1–114. If the reading of the pressure gage is 370 kPa, determine the gage pressure of the gasoline line.
1–115 Repeat Prob. 1–114 for a pressure gage reading of 180 kPa. 1–116E A water pipe is connected to a doubleU manometer as shown in Fig. P1–116E at a location where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure at the center of the pipe. Oil SG = 0.80 Oil SG = 0.80 35 in Water pipe
60 in
40 in
15 in Mercury SG = 13.6
FIGURE P1–116E 1–117 It is wellknown that the temperature of the atmosphere varies with altitude. In the troposphere, which extends to an altitude of 11 km, for example, the variation of temperature can be approximated by T T0 bz, where T0 is the temperature at sea level, which can be taken to be 288.15 K, and b 0.0065 K/m. The gravitational acceleration also changes with altitude as g(z) g0 /(1 z/6,370,320)2 where g0 9.807 m/s2 and z is the elevation from sea level in m. Obtain a relation for the variation of pressure in the troposphere (a) by ignoring and (b) by considering the variation of g with altitude. 1–118 The variation of pressure with density in a thick gas layer is given by P Crn, where C and n are constants. Noting that the pressure change across a differential fluid layer of thickness dz in the vertical zdirection is given as dP rg dz, obtain a relation for pressure as a function of
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elevation z. Take the pressure and density at z 0 to be P0 and r0, respectively.
kPa. If the density of air is 1.0 kg/m3, the height of the building is
1–119 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. P1–119 can be used to calibrate pressure transducers. 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 pressure 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 P aI b, where a and b are constants, and calculate the pressure that corresponds to a signal of 10 mA.
(a) 17 m (d) 204 m
h, mm I, mA
28.0 4.21
181.5 5.78
297.8 6.97
413.1 8.15
765.9 11.76
h, mm I, mA
1027 14.43
1149 15.68
1362 17.86
1458 18.84
1536 19.64
Multimeter
(a) 1.25 kJ (d) 8.1 kJ
(b) 2.50 kJ (e) 4.1 kJ
(c) 5.0 kJ
1–123 Consider a 2m deep swimming pool. The pressure difference between the top and bottom of the pool is (a) 12.0 kPa (d) 50.8 kPa
(b) 19.6 kPa (e) 200 kPa
(c) 38.1 kPa
1–124 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is (a) 1 lbf (d) 0.1 lbf
(b) 9.81 lbf (e) 0.031 lbf
(c) 32.2 lbf
1–125 During a heating process, the temperature of an object rises by 20°C. This temperature rise is equivalent to a temperature rise of (b) 52°F (e) 293 K
(c) 36 K
Design, Essay, and Experiment Problems 1–126 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 applicability. Which device would you recommend for use in the following cases: taking the temperatures of patients in a doctor’s office, monitoring the variations of temperature of a car engine block at several locations, and monitoring the temperatures in the furnace of a power plant?
Valve
Pressurized air, P
∆h
Rigid container
Manometer
1–127 Write an essay on the various mass and volumemeasurement devices used throughout history. Also, explain the development of the modern units for mass and volume.
Mercury SG = 13.56
1–128 Write an essay on the various mass and volumemeasurement devices used throughout history. Also, explain the development of the modern units for mass and volume.
FIGURE P1–119 Fundamentals of Engineering (FE) Exam Problems 1–120 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 45 m below the free surface is (a) 392 Pa (d ) 392,000 Pa
(c) 170 m
1–122 An apple loses 4.5 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) 20°F (d) 36 R Pressure transducer
(b) 20 m (e) 252 m
(b) 9800 Pa (e) 441,000 Pa
(c) 50,000 Pa
1–121 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0
1–129 Density of Water as a Function of Temperature Experiment The density of water as a function of temperature is obtained with a sensitive cylindrical float constructed from brass tubing. The float is placed in a Thermos bottle filled with water at different temperatures. From 0 to 4°C (water density is a maximum at 4°C) the float rose about 8 mm and from 4 to 25°C the float sank about 40 mm. The analysis includes differential and integral calculus to account for thermal expansion of the float. The final results closely follow the published density curve including the characteristic hump at 4°C. Obtain this density curve using the video clip, the complete writeup, and the data provided on the DVD accompanying this book.
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Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSIS
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.
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. • Discuss the three mechanisms of heat transfer: conduction, convection, and radiation. • 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 and/or work. • Define energy conversion efficiencies. • Discuss the implications of energy conversion on the environment.

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2–1 Room
FIGURE 2–1 A refrigerator operating with its door open in a wellsealed and wellinsulated room.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 1 ON THE DVD.
Wellsealed and wellinsulated room Fan
FIGURE 2–2 A fan running in a wellsealed and wellinsulated room will raise the temperature of air in the room.
■
INTRODUCTION
We are familiar with the conservation of energy principle, which is an expression 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 process; it can only change from one form to another. This seems simple enough, but let’s 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 negligible. Now let’s place a refrigerator in the middle of the room with its door open, and plug it into a wall outlet (Fig. 2–1). 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 temperature 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 temperature 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 room—including the air and the refrigerator—as 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 system 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. 2–2)? Note that energy is conserved during the process of operating the refrigerator placed in a room—the electrical energy is converted into an equivalent 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. Electricity, 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.
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53
Now if asked to name the energy transformations associated with the operation 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.
2–2
■
FORMS OF ENERGY
Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical, and nuclear, 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 ¬¬1kJ>kg2 m
V2 ¬¬1kJ2 2
SEE TUTORIAL CH. 2, SEC. 2 ON THE DVD.
(2–1)
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: macroscopic 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. 2–3). 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 thermodynamics 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 nineteenth 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
INTERACTIVE TUTORIAL
(2–2)
FIGURE 2–3 The macroscopic energy of an object changes with velocity and elevation.
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Thermodynamics or, on a unit mass basis, ke
V2 ¬¬1kJ>kg2 2
(2–3)
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 12 Iv2 where I is the moment of inertia of the body and v is the angular velocity. The energy that a system possesses as a result of its elevation in a gravitational field is called potential energy (PE) and is expressed as PE mgz¬¬1kJ2
(2–4)
pe gz¬¬1kJ>kg2
(2–5)
or, on a unit mass basis, 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
V2 mgz¬¬1kJ 2 2
(2–6)
or, on a unit mass basis, e u ke pe u
Vavg Steam
(2–7)
Most closed systems remain stationary during a process and thus experience no change in their kinetic and potential energies. Closed systems whose velocity and elevation of the center of gravity remain constant during a process are frequently referred to as stationary systems. The change in the total energy E of a stationary system is identical to the change in its internal energy U. 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
Ac = pD 2/4 D
V2 gz¬¬1kJ>kg2 2
m• = rAcVavg
Mass flow rate:
# # m rV rAcVavg¬¬1kg>s2
(2–8)
•
• E = me
FIGURE 2–4 Mass and energy flow rates associated with the flow of steam in a pipe of inner diameter D with an average velocity of Vavg.
which is analogous to m rV. Here r is the fluid density, Ac is the crosssectional 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. 2–4) Energy flow rate:
# # E me¬¬1kJ>s or kW2
which is analogous to E me.
(2–9)
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55
Some Physical Insight to Internal Energy Internal energy is 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 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 significant 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. 2–5). 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, turning 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. 2–6). 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 in the core or nucleus. Therefore, an
Molecular translation
Molecular rotation
– +
Electron translation
Molecular vibration
–
Electron spin
+
Nuclear spin
FIGURE 2–5 The various forms of microscopic energies that make up sensible energy.
Sensible and latent energy
Chemical energy
Nuclear energy
FIGURE 2–6 The internal energy of a system is the sum of all forms of the microscopic energies.
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Thermodynamics
Microscopic kinetic energy of molecules (does not turn the wheel) Water
Dam
Macroscopic kinetic energy (turns the wheel)
FIGURE 2–7 The macroscopic kinetic energy is an organized form of energy and is much more useful than the disorganized microscopic kinetic energies of the molecules.
atom preserves its identity during a chemical reaction but loses it during a nuclear reaction. Atoms may also possess electric and magnetic dipolemoment energies when subjected to external 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 frequently refer to the sensible and latent forms of internal energy as heat, and we talk about heat content of bodies. In thermodynamics, however, we usually refer to those forms of energy as thermal energy to prevent any confusion with heat transfer. 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. 2–7). 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 completely 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 disorganized 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 potential energy of an object as a whole and the microscopic potential energies of the molecules.
More on Nuclear Energy The best known fission reaction involves the split of the uranium atom (the U235 isotope) into other elements and is commonly used to generate electricity in nuclear power plants (440 of them in 2004, generating 363,000 MW worldwide), to power nuclear submarines and aircraft carriers, and even to power spacecraft as well as building nuclear bombs. The percentage of electricity produced by nuclear power is 78 percent in France, 25 percent in Japan, 28 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
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Chapter 2 uranium235 atom absorbs a neutron and splits during a fission process, it produces a cesium140 atom, a rubidium93 atom, 3 neutrons, and 3.2 1011 J of energy. In practical terms, the complete fission of 1 kg of uranium235 releases 6.73 1010 kJ of heat, which is more than the heat released when 3000 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 that involves the combination of two hydrogen atoms into a helium atom. When two heavy hydrogen (deuterium) nuclei combine during a fusion process, they produce a helium3 atom, a free neutron, and 5.1 1013 J of energy (Fig. 2–8). Fusion reactions are much more difficult to achieve in practice because of the strong repulsion between the positively charged nuclei, called the Coulomb 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 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 massive lasers, powerful magnetic fields, and electric currents to generate power have failed. EXAMPLE 2–1
Uranium U235
 3.2 × 10 –11 J Ce140
n n 3 neutrons n n neutron
Rb93
(a) Fission of uranium
H2
He3 n
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 0.75 kg/L. 2 Nuclear fuel is completely converted to thermal energy. Analysis The mass of gasoline used per day by the car is
mgasoline 1rV 2 gasoline 10.75 kg>L2 15 L>day 2 3.75 kg>day
Noting that the heating value of gasoline is 44,000 kJ/kg, the energy supplied to the car per day is
E 1mgasoline 2 1Heating value2
13.75 kg>day 2 144,000 kJ>kg2 165,000 kJ>day
neutron
H2 5.1 × 10 –13 J (b) Fusion of hydrogen
FIGURE 2–8 The fission of uranium and the fusion of hydrogen during nuclear reactions, and the release of nuclear energy.
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 0.68 to 0.78 kg/L, and its lower heating value is about 44,000 kJ/kg (that is, 44,000 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 0.1kg of the nuclear fuel U235, determine if this car will ever need refueling under average driving conditions (Fig. 2–9).
57
Nuclear fuel
FIGURE 2–9 Schematic for Example 2–1.
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Thermodynamics The complete fission of 0.1 kg of uranium235 releases
16.73 1010 kJ>kg 2 10.1 kg2 6.73 109 kJ of heat, which is sufficient to meet the energy needs of the car for
No. of days
Energy content of fuel 6.73 109 kJ 40,790 days Daily energy use 165,000 kJ>day
which is equivalent to about 112 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 partial conversion.
Mechanical Energy Many engineering systems are designed to transport a fluid from one location 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. 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 significant 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 (i.e., 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, however, 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 pressure, 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 N/m2 N · m/m3 J/m3, which is energy per unit volume, and the product Pv or its equivalent P/r has the unit J/kg, 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 P/r 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 emech
V2 P gz r 2
(2–10)
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Chapter 2

where P/r is the flow energy, V 2/2 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 # V2 # # P Emech memech m a gz b r 2
(2–11)
. where m is the mass flow rate of the fluid. Then the mechanical energy change of a fluid during incompressible (r constant) flow becomes ¢emech
V 22 V 12 P2 P1 g 1z2 z1 2¬¬1kJ>kg2 r 2
(2–12)
and # V 22 V 12 # # P2 P1 g 1z 2 z 1 2 b ¬¬1kW2 ¢E mech m ¢emech m a r 2
(2–13)
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 losses, the mechanical energy change represents the mechanical work supplied to the fluid (if emech 0) or extracted from the fluid (if emech 0).
EXAMPLE 2–2
Wind Energy
8.5 m/s
—
A site evaluated for a wind farm is observed to have steady winds at a speed of 8.5 m/s (Fig. 2–10). Determine the wind energy (a) per unit mass, (b) for a mass of 10 kg, and (c) for a flow rate of 1154 kg/s 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
18.5 m>s2 1 J>kg V a b 36.1 J>kg 2 2 1 m2>s2 2
e ke
2
(b) Wind energy for an air mass of 10 kg is
E me 110 kg2 136.1 J>kg2 361 J
(c) Wind energy for a mass flow rate of 1154 kg/s is
# 1 kW # E me 11154 kg>s2 136.1 J>kg2 a b 41.7 kW 1000 J>s Discussion It can be shown that the specified mass flow rate corresponds to a 12m diameter flow section when the air density is 1.2 kg/m3. Therefore, a wind turbine with a wind span diameter of 12 m has a power generation potential of 41.7 kW. Real wind turbines convert about onethird of this potential to electric power.
FIGURE 2–10 Potential site for a wind farm as discussed in Example 2–2. © Vol. 36/PhotoDisc
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Thermodynamics
2–3
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 3 ON THE DVD.
System boundary Heat CLOSED SYSTEM
Work
(m = constant)
FIGURE 2–11 Energy can cross the boundaries of a closed system in the form of heat and work.
Room air 25°C No heat transfer
8 J/s
25°C
15°C
Heat
16 J/s
Heat
5°C
FIGURE 2–12 Temperature difference is the driving force for heat transfer. The larger the temperature difference, the higher is the rate of heat transfer.
■
ENERGY TRANSFER BY HEAT
Energy can cross the boundary of a closed system in two distinct forms: heat and work (Fig. 2–11). 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 transfer 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 temperature 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 systems (or a system and its surroundings) by virtue of a temperature difference (Fig. 2–12). 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 today—such 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, sensible heat, latent heat, waste heat, body heat, process heat, heat sink, and heat source—are 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 substance 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 frequently referred to as heat addition and the transfer of heat out of a system as heat rejection. Perhaps there are thermodynamic reasons for being so reluctant to replace heat by 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 contains 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. 2–13. 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.
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Chapter 2 A process during which there is no heat transfer is called an adiabatic process (Fig. 2–14). The word adiabatic comes from the Greek word adiabatos, 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 difference) 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 common 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 ¬¬1kJ>kg2 m

61
2 kJ thermal energy HEAT
SURROUNDING AIR
2 kJ heat
BAKED POTATO System boundary
2 kJ thermal energy
FIGURE 2–13 Energy is recognized as heat transfer only as it crosses the system boundary.
(2–14) Insulation
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. 2–15). The heat transfer rate is denoted Q, where the overdot . stands for the time derivative, or “per unit time.” The .heat transfer rate Q has the unit kJ/s, 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: t2
Q
Q dt¬¬1kJ2
Q=0 ADIABATIC SYSTEM
#
(2–15)
t1
. When Q remains constant during a process, this relation reduces to # Q Q¬¢t¬¬1kJ2
(2–16)
FIGURE 2–14 During an adiabatic process, a system exchanges no heat with its surroundings.
where t 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 sensation 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 nineteenth 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 eighteenth and early nineteenth 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 nineteenth century was based on the caloric theory proposed by the French chemist Antoine Lavoisier (1744–1794) 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. 2–16). When caloric was added to a body, its
Q = 30 kJ k m=2k kg ∆t = 5 s
k 30 kJ heat
Q = 6 kW q = 15 kJ/ kJ/kg
FIGURE 2–15 . The relationships among q, Q, and Q.
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Thermodynamics Contact surface Hot body
Cold body
Caloric
FIGURE 2–16 In the early nineteenth century, heat was thought to be an invisible fluid called the caloric that flowed from warmer bodies to the cooler ones.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 4 ON THE DVD.
W = 30 kJ k m = 2 kg ∆t = 5 s
30 kJ work
W = 6 kW w = 15 kJ/k k J/kg
temperature increased; and when caloric was removed from a body, its temperature decreased. When a body could not contain any more caloric, much the same way 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 maintained that heat is a substance that could not be created or destroyed. Yet it was known that heat can be generated indefinitely by rubbing one’s hands together or rubbing two pieces of wood together. In 1798, the American Benjamin Thompson (Count Rumford) (1754–1814) 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 (1818–1889) 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 nineteenth century, it contributed greatly to the development of thermodynamics and heat transfer. Heat is transferred by three mechanisms: conduction, convection, and radiation. 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 surface 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.
2–4
■
ENERGY TRANSFER BY WORK
Work, like heat, is an energy interaction between a system and its surroundings. 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 surroundings 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
FIGURE 2–17 # The relationships among w, W, and W.
W ¬¬1kJ>kg2 m
(2–17)
. The work done per unit time is called power and is denoted W (Fig. 2–17). The unit of power is kJ/s, or kW.
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Chapter 2 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 generally 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. 2–18). 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 interaction 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 mechanisms between a system and its surroundings, and there are many similarities between them:

63
Surroundings
Qin Qout System Win Wout
FIGURE 2–18 Specifying the directions of heat and work.
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 (i.e., their magnitudes depend on the path followed during a process as well as the end states).
1
A
2 m3
2
dW W
12¬¬1not
¢W2
B
ss
ce
dV V2 V1 ¢V
1
∆VB = 3 m3; WB = 12 kJ
2
2
That is, the volume change during process 1–2 is always the volume at state 2 minus the volume at state 1, regardless of the path followed (Fig. 2–19). The total work done during process 1–2, however, is
1
∆VA = 3 m3; WA = 8 kJ
s es oc Pr
P
o Pr
Path functions have inexact differentials designated by the symbol d. Therefore, a differential amount of heat or work is represented by dQ or dW, respectively, instead of dQ or dW. Properties, however, are point functions (i.e., 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 during a process between states 1 and 2 is
5 m3
V
FIGURE 2–19 Properties are point functions; but heat and work are path functions (their magnitudes depend on the path followed).
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Thermodynamics That is, the total work is obtained by following the process path and adding the differential amounts of work (dW) done along the way. The integral of dW is not W2 W1 (i.e., 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 2–3
Burning of a Candle in an Insulated Room
A candle is burning in a wellinsulated room. Taking the room (the air plus the candle) as the system, determine (a) if there is any heat transfer during this burning process and (b) if there is any change in the internal energy of the system.
(Insulation) Room
Solution A candle burning in a wellinsulated room is considered. It is to
FIGURE 2–20 Schematic for Example 2–3.
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 indicated by the dashed lines in Fig. 2–20. 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 internal energy of the system, U 0 for this process.
(Insulation)
EXAMPLE 2–4
OVEN Heat POTATO 25°C
Heating of a Potato in an Oven
A potato initially at room temperature (25°C) is being baked in an oven that is maintained at 200°C, as shown in Fig. 2–21. Is there any heat transfer during this baking process?
Solution A potato is being baked in an oven. It is to be determined 200°C
FIGURE 2–21 Schematic for Example 2–4.
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 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 2–5
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.
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Chapter 2 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 boundary, as shown in Fig. 2–22. The energy content of the oven obviously increases during 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.
EXAMPLE 2–6

65
System boundary
Electric oven
Heating element
FIGURE 2–22 Schematic for Example 2–5.
Heating of an Oven by Heat Transfer
Answer the question in Example 2–5 if the system is taken as only the air in the oven without the heating element. System boundary
Solution The question in Example 2–5 is to be reconsidered by taking the system to be only the air in the oven. Analysis This time, the system boundary will include the outer surface of the heating element and will not cut through it, as shown in Fig. 2–23. 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.
Electric oven
Heating element
FIGURE 2–23 Schematic for Example 2–6.
Electrical Work It was pointed out in Example 2–5 that electrons crossing the system boundary 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 We VN
which can also be expressed in the rate form as # We VI¬¬1W2
2
VI dt¬¬1kJ2
(2–19)
1
When both V and I remain constant during the time interval t, it reduces to We VI¬¢t¬¬1kJ2
We = VI = I 2R = V2/R
R
V
(2–18)
. where We is the electrical power and I is the number of electrical charges flowing per unit time, that is, the current (Fig. 2–24). In general, both V and I vary with time, and the electrical work done during a time interval t is expressed as We
I
(2–20)
FIGURE 2–24 Electrical power in terms of resistance R, current I, and potential difference V.
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Thermodynamics
2–5
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 5 ON THE DVD.
F
F
■
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. 2–25). In elementary mechanics, the work done by a constant force F on a body displaced a distance s in the direction of the force is given by W Fs¬¬1kJ2
(2–21)
If the force F is not constant, the work done is obtained by adding (i.e., integrating) the differential amounts of work,
s
FIGURE 2–25 The work done is proportional to the force applied (F ) and the distance traveled (s).
FIGURE 2–26 If there is no movement, no work is done. © Reprinted with special permission of King Features Syndicate.
2
W
F ds¬¬1kJ2 ¬
Obviously one needs to know how the force varies with displacement to perform this integration. Equations 2–21 and 2–22 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 boundary 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 system or with the movement of the entire system as a whole (Fig. 2–26). Some common forms of mechanical work are discussed next.
Shaft Work Energy transmission with a rotating shaft is very common in engineering practice (Fig. 2–27). 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 follows: A force F acting through a moment arm r generates a torque T of (Fig. 2–28) T Fr¬S ¬F
Boat
(2–22)
1
T r
(2–23)
This force acts through a distance s, which is related to the radius r by s 12pr2 n
(2–24)
Then the shaft work is determined from Engine
FIGURE 2–27 Energy transmission through rotating shafts is commonly encountered in practice.
Wsh Fs a
T b 12prn 2 2pnT¬¬1kJ2 r
(2–25)
The power transmitted through the shaft is the shaft work done per unit time, which can be expressed as # # Wsh 2pnT¬¬1kW2
. where n is the number of revolutions per unit time.
(2–26)
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Chapter 2

·
EXAMPLE 2–7
· Wsh = 2πnT
Power Transmission by the Shaft of a Car
Determine the power transmitted through the shaft of a car when the torque applied is 200 N · m and the shaft rotates at a rate of 4000 revolutions per minute (rpm).
r n· F Torque = Fr
Solution The torque and the rpm for a car engine are given. The power transmitted is to be determined. Analysis A sketch of the car is given in Fig. 2–29. The shaft power is determined directly from
# 1 1 min 1 kJ # b 1200 N # m2 a ba b Wsh 2pnT 12p2 a 4000 min 60 s 1000 N # m
FIGURE 2–28 Shaft work is proportional to the torque applied and the number of revolutions of the shaft.
83.8 kW¬¬1or 112 hp2 Discussion Note that power transmitted by a shaft is proportional to torque and the rotational speed. n· = 4000 rpm T = 200 N • m
Spring Work It is common knowledge that when a force is applied on a spring, the length of the spring changes (Fig. 2–30). When the length of the spring changes by a differential amount dx under the influence of a force F, the work done is dWspring F dx
FIGURE 2–29 Schematic for Example 2–7.
(2–27)
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. 2–31). That is, F kx¬¬1kN2
(2–28)
where k is the spring constant and has the unit kN/m. The displacement x is measured from the undisturbed position of the spring (that is, x 0 when F 0). Substituting Eq. 2–28 into Eq. 2–27 and integrating yield Wspring 12k 1x22 x21 2¬¬1kJ2
(2–29)
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. 2–32, and when the force is lifted, they return to their original lengths, like a spring. This is true as long as the force is in the elastic range, that is, not large enough to cause permanent (plastic) deformations. Therefore, the equations given for a linear spring can also be used for elastic solid bars. Alternately, we can determine
Rest position
dx
x
F
FIGURE 2–30 Elongation of a spring under the influence of a force.
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Thermodynamics the work associated with the expansion or contraction of an elastic solid bar by replacing pressure P by its counterpart in solids, normal stress sn F/A, in the work expression: Rest position
x1 = 1 mm
x2 = 2 mm
Welastic
2
2
F dx
1
s A dx¬¬1kJ2 n
(2–30)
1
where A is the crosssectional area of the bar. Note that the normal stress has pressure units.
F1 = 300 N
Work Associated with the Stretching of a Liquid Film F2 = 600 N
FIGURE 2–31 The displacement of a linear spring doubles when the force is doubled.
Consider a liquid film such as soap film suspended on a wire frame (Fig. 2–33). 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 overcome the microscopic forces between molecules at the liquid–air 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 ss, whose unit is N/m. Therefore, the work associated with the stretching of a film is also called surface tension work. It is determined from 2
Wsurface
s dA¬¬1kJ2 s
(2–31)
1
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 2bss where ss is the surface tension force per unit length.
x
Work Done to Raise or to Accelerate a Body
F
FIGURE 2–32 Solid bars behave as springs under the influence of a force.
Rigid wire frame Surface of film
Movable wire F
b dx x
FIGURE 2–33 Stretching a liquid film with a movable wire.
When a body is raised in a gravitational field, its potential energy increases. Likewise, when a body is accelerated, its kinetic energy increases. The conservation 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 temperature 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. 2–34). Similarly, 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 from a given water reservoir, which is simply the potential energy of the water relative to the location of the hydraulic turbine.
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Chapter 2 EXAMPLE 2–8
Power Needs of a Car to Climb a Hill

69
Motor
Consider a 1200kg car cruising steadily on a level road at 90 km/h. Now the car starts climbing a hill that is sloped 30° from the horizontal (Fig. 2–35). If the velocity of the car is to remain constant during climbing, determine the additional power that must be delivered by the engine.
Solution A car is to climb a hill while maintaining a constant velocity. The
Elevator car
additional power needed is to be determined. Analysis The additional power required is simply the work that needs to be done per unit time to raise the elevation of the car, which is equal to the change in the potential energy of the car per unit time:
# Wg mg¬¢z> ¢t mgVvertical 11200 kg2 19.81 m>s2 2 190 km>h2 1sin 30°2 a
1 m>s 3.6 km>h
ba
147 kJ>s 147 kW¬¬1or 197 hp2
1 kJ>kg
1000 m2>s2
b
Discussion Note that the car engine will have to produce almost 200 hp of additional power while climbing the hill if the car is to maintain its velocity.
EXAMPLE 2–9
FIGURE 2–34 The energy transferred to a body while being raised is equal to the change in its potential energy.
Power Needs of a Car to Accelerate
Determine the power required to accelerate a 900kg car shown in Fig. 2–36 from rest to a velocity of 80 km/h in 20 s on a level road.
Solution The power required to accelerate a car to a specified velocity is to
m = 1200 kg 90 km/h
be determined. Analysis The work needed to accelerate a body is simply the change in the kinetic energy of the body,
Wa 12m 1V 22 V 21 2 12 1900 kg 2 c a 222 kJ
1 kJ>kg 80,000 m 2 b 02 d a b 3600 s 1000 m2>s2
The average power is determined from
# Wa 222 kJ Wa 11.1 kW¬¬1or 14.9 hp2 ¢t 20 s
30°
FIGURE 2–35 Schematic for Example 2–8.
Discussion This is in addition to the power required to overcome friction, rolling resistance, and other imperfections.
Nonmechanical Forms of Work The treatment in Section 2–5 represents a fairly comprehensive coverage of mechanical forms of work except the moving boundary work that is covered in Chap. 4. But 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
80 km/h m = 900 kg
FIGURE 2–36 Schematic for Example 2–9.
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Thermodynamics of a generalized displacement x. Then the work associated with the differential displacement under the influence of this force is determined from dW F dx. 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.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 6 ON THE DVD.
m
PE1 = 10 kJ KE1 = 0
∆z
m
PE 2 = 7 kJ KE2 = 3 kJ
FIGURE 2–37 Energy cannot be created or destroyed; it can only change forms.
2–6
■
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. 2–37). Experimental data show that the decrease in potential energy (mg z) exactly equals the increase in kinetic energy 3 m 1V 22 V 21 2>2 4 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 specified 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 adiabatic conditions, this statement appears to be very powerful, with a potential for farreaching implications. This statement, which is largely based on the experiments of Joule in the first half of the nineteenth century, cannot be drawn 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 definition 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 prop
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Chapter 2 erty 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 principle. 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. 2–38). 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. 2–39). If 15 kJ of heat is transferred to the water from the heating 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 (i.e., adiabatic) room heated by an electric heater as our system (Fig. 2–40). 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. 2–41). 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. 2–42). 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. 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. 2–43). 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

71
Qin = 5 kJ
POTATO ∆E = 5 kJ
FIGURE 2–38 The increase in the energy of a potato in an oven is equal to the amount of heat transferred to it. Qout = 3 kJ
∆E = Q net = 12 kJ
Qin = 15 kJ
FIGURE 2–39 In the absence of any work interactions, the energy change of a system is equal to the net heat transfer. (Adiabatic) Win = 5 kJ
∆E = 5 kJ
–
+ Battery
FIGURE 2–40 The work (electrical) done on an adiabatic system is equal to the increase in the energy of the system.
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Thermodynamics between the total energy entering and the total energy leaving the system during that process. That is,
(Adiabatic)
a
∆E = 8 kJ Wsh, in = 8 kJ
Total energy Total energy Change in the total b a b a b entering the system leaving the system energy of the system
or Ein Eout ¢Esystem
FIGURE 2–41 The work (shaft) done on an adiabatic system is equal to the increase in the energy of the system. Wb,in = 10 kJ
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, Esystem 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 ¢Esystem Efinal Einitial E2 E1 ∆E = 10 kJ
(Adiabatic)
FIGURE 2–42 The work (boundary) done on an adiabatic system is equal to the increase in the energy of the system.
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 (i.e., 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 ¢E ¢U ¢KE ¢PE
Qout = 3 kJ
where ∆E = (15 – 3) + 6 = 18 kJ
(2–32)
(2–33)
¢U m 1u2 u1 2
¢KE 12 m 1V 22 V 12 2 Wsh, in = 6 kJ
Qin = 15 kJ
FIGURE 2–43 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings.
¢PE mg 1z2 z1 2
When the initial and final states are specified, the values of the specific internal 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. 2–44). Thus, for stationary systems, the changes in kinetic and potential energies are zero (that is, KE PE 0), and the total energy change relation in Eq. 2–33 reduces to E U for such systems. Also, the energy
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Chapter 2 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 process. 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 transfer 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 (i.e., work done on a system) increases the energy of the system, and work transfer from a system (i.e., work done by the system) decreases it since the energy transferred out as work comes from the energy contained 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 additional 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 leaving 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. 2–45). 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 1Q in Q out 2 1Win Wout 2 1E mass,in E mass,out 2 ¢E system (2–34)
where the subscripts “in” and “out” denote quantities that enter and leave the system, 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.” 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 boundaries (i.e., closed systems).

73
Stationary Systems z 1 = z 2← ∆PE = 0 V1 = V2 ← ∆KE = 0 ∆E = ∆U
FIGURE 2–44 For stationary systems, KE PE 0; thus E U.
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Thermodynamics
Mass in
W
Energy balance for any system undergoing any kind of process can be expressed more compactly as
Mass out
⎫ ⎪ ⎬ ⎪ ⎭
Q
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬ ¢E system¬¬1kJ2
Control volume
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
or, in the rate form, as
Rate of net energy transfer by heat, work, and mass
(2–36)
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
. . E in E out¬ ¬dE system>dt¬¬1kW2
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
FIGURE 2–45 The energy content of a control volume can be changed by mass flow as well as heat and work interactions.
(2–35)
Rate of change in internal, kinetic, potential, etc., energies
For constant rates, the total quantities during a time interval t are related to the quantities per unit time as # # Q Q ¢t,¬W W¬¢t,¬and¬¢E 1dE>dt2 ¢t¬¬1kJ2
(2–37)
The energy balance can be expressed on a per unit mass basis as ein eout ¢esystem¬¬1kJ>kg2
(2–38)
which is obtained by dividing all the quantities in Eq. 2–35 by the mass m of the system. Energy balance can also be expressed in the differential form as
P
dEin dEout dEsystem¬or¬dein deout desystem
(2–39)
For a closed system undergoing a cycle, the initial and final states are identical, and thus Esystem 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
Qnet = Wnet
V
FIGURE 2–46 For a cycle E 0, thus Q W.
# # Wnet,out Q net,in¬or¬Wnet,out Q net,in¬¬1for a cycle 2
That is, the net work output during a cycle is equal to net heat input (Fig. 2–46). EXAMPLE 2–10
Qout = 500 kJ
(2–40)
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 looses heat while being stirred. The final U1 = 800 kJ U2 = ? Wsh, in = 100 kJ Fluid
FIGURE 2–47 Schematic for Example 2–10.
internal energy of the fluid is to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero, KE PE 0. Therefore, E U and internal energy is the only form of the system’s 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. 2–47). 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 moving 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
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Chapter 2
75
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬

Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Wsh,in Q out ¢U U2 U1 100 kJ 500 kJ U2 800 kJ U2 400 kJ Therefore, the final internal energy of the system is 400 kJ.
EXAMPLE 2–11
Acceleration of Air by a Fan 8 m/s
A fan that consumes 20 W of electric power when operating is claimed to discharge air from a ventilated room at a rate of 0.25 kg/s at a discharge velocity of 8 m/s (Fig. 2–48). 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, let’s 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
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
S
# # E in E out
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
dE system > dt →0 1steady2 0
# # E in E out
2 # # # Vout Welect, in mair keout mair 2
Solving for Vout and substituting gives the maximum air outlet velocity to be
Vout
# 20 J>s 1 m2>s2 Welect,in a b 6.3 m>s # B 2mair B 2 10.25 kg>s 2 1 J>kg
which is less than 8 m/s. Therefore, the claim is false. Discussion The conservation of energy principle requires the energy to be preserved 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 6.3 m/s—but 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 6.3 m/s 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.
Air
FIGURE 2–48 Schematic for Example 2–11. © Vol. 0557/PhotoDisc
Fan
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Thermodynamics •
Qout
Room •
Welect. in Fan
EXAMPLE 2–12
Heating Effect of a Fan
A room is initially at the outdoor temperature of 25°C. Now a large fan that consumes 200 W of electricity when running is turned on (Fig. 2–49). The heat transfer rate between the room and the outdoor air is given as · Q UA(Ti To) where U 6 W/m2 · °C 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. Determine the indoor air temperature when steady operating conditions are established.
Solution A large fan is turned on and kept on in a room that looses heat to
FIGURE 2–49 Schematic for Example 2–12.
the outdoors. 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
dE system > dt →0 1steady2 0
# # E in E out
S
# # E in E out
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Welect,in Q out UA 1Ti To 2 Substituting,
200 W 16 W>m2 # °C2 130 m2 2 1Ti 25°C2
It gives
Ti 26.1°C Therefore, the room air temperature will remain constant after it reaches 26.1°C. 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 mechanical 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 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 2–13 FIGURE 2–50 Fluorescent lamps lighting a classroom as discussed in Example 2–13. © Vol. 24/PhotoDisc
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. 2–50). The lights in the classroom are kept on for 12 hours a day and 250 days a year. For a unit electricity cost of
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77
7 cents per kWh, determine 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 lighting’s effect on the heating and airconditioning requirements is to be discussed. Assumptions The effect of voltage fluctuations is negligible so that each fluorescent lamp consumes its rated power. Analysis The electric power consumed by the lamps when all are on and the number of hours they are kept on per year are
Lighting power 1Power consumed per lamp2 1No. of lamps2 180 W>lamp 2 130 lamps 2 2400 W 2.4 kW
Operating hours 112 h>day 2 1250 days>year2 3000 h>year Then the amount and cost of electricity used per year become
Lighting energy 1Lighting power2 1Operating hours2
12.4 kW2 13000 h>year2 7200 kWh>year
Lighting cost 1Lighting energy 2 1Unit cost2
17200 kWh>year2 1$0.07>kWh2 $504>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 2.4 kW of electric power consumed by the lamps eventually becomes part of thermal energy of the classroom. Therefore, the lighting system in this room reduces the heating requirements by 2.4 kW, but increases the airconditioning load by 2.4 kW. Discussion Note that the annual lighting cost of this classroom alone is over $500. This shows the importance of energy conservation measures. If incandescent light bulbs 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.
EXAMPLE 2–14
Conservation of Energy for an Oscillating Steel Ball
The motion of a steel ball in a hemispherical bowl of radius h shown in Fig. 2–51 is to be analyzed. The ball is initially held at the highest location at point A, and then it is released. Obtain relations for the conservation of energy of the ball for the cases of frictionless and actual motions.
z
A
h
Steel ball
C
1
Solution A steel ball is released in a bowl. Relations for the energy balance are to be obtained. Assumptions The motion is frictionless, and thus friction between the ball, the bowl, and the air is negligible. Analysis When the ball is released, it accelerates under the influence of gravity, reaches a maximum velocity (and minimum elevation) at point B at
B
2
FIGURE 2–51 Schematic for Example 2–14.
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Thermodynamics the bottom of the bowl, and moves up toward point C on the opposite side. In the ideal case of frictionless motion, the ball will oscillate between points A and C. The actual motion involves the conversion of the kinetic and potential energies of the ball to each other, together with overcoming resistance to motion due to friction (doing frictional work). The general energy balance for any system undergoing any process is
E in E out¬ ¬
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Then the energy balance for the ball for a process from point 1 to point 2 becomes
wfriction 1ke2 pe2 2 1ke1 pe1 2
or
V 22 V 12 gz1 gz2 wfriction 2 2 since there is no energy transfer by heat or mass and no change in the internal energy of the ball (the heat generated by frictional heating is dissipated to the surrounding air). The frictional work term wfriction is often expressed as eloss to represent the loss (conversion) of mechanical energy into thermal energy. For the idealized case of frictionless motion, the last relation reduces to
V 21 V 22 V2 gz1 gz2¬or¬ gz C constant 2 2 2 where the value of the constant is C gh. That is, when the frictional effects are negligible, the sum of the kinetic and potential energies of the ball remains constant. Discussion This is certainly a more intuitive and convenient form of the conservation of energy equation for this and other similar processes such as the swinging motion of the pendulum of a wall clock.
2–7
FIGURE 2–52 The definition of performance is not limited to thermodynamics only. © Reprinted with special permission of King Features Syndicate.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 7 ON THE DVD.
■
ENERGY CONVERSION EFFICIENCIES
Efficiency is one of the most frequently used terms in thermodynamics, and it indicates how well an energy conversion or transfer process is accomplished. Efficiency is also one of the most frequently 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. Performance or efficiency, in general, can be expressed in terms of the desired output and the required input as (Fig. 2–52) Performance
Desired output Required output
(2–41)
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. 2–53). You may find this confusing, since the heating elements of electric water heaters are resistance heaters, and the efficiency of
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Chapter 2 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 surrounding 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 delivered 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. 2–54). Then the performance of combustion equipment can be characterized by combustion efficiency, defined as hcombustion
Amount of heat released during combustion Q HV Heating value of the fuel burned
79
Water heater
Type
(2–42)
A combustion efficiency of 100 percent indicates that the fuel is burned completely and the stack gases leave the combustion chamber at room temperature, and thus the amount of heat released during a combustion process is equal to the heating value of the fuel. 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 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 temperature. For example, the lower and higher heating values of gasoline are 44,000 kJ/kg and 47,300 kJ/kg, 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 recuperate the heat of vaporization. Efficiencies of furnaces, on the other hand, are based on higher heating values. The efficiency of space heating systems of residential and commercial buildings is usually expressed in terms of the annual fuel utilization efficiency, or AFUE, which accounts for the combustion 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

Efficiency
Gas, conventional Gas, highefficiency Electric, conventional Electric, highefficiency
55% 62% 90% 94%
FIGURE 2–53 Typical efficiencies of conventional and highefficiency electric and natural gas water heaters. © The McGrawHill Companies, Inc./Jill Braaten, photographer
Combustion gases 25°C CO2, H2O, etc.
Air 25°C
Combustion chamber
LHV = 44,000 kJ/kg 1 kg Gasoline 25°C
FIGURE 2–54 The definition of the heating value of gasoline.
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Thermodynamics
TABLE 2–1 The efficacy of different lighting systems Type of lighting Combustion Candle
Efficacy, lumens/W 0.2
Incandescent Ordinary Halogen
6–20 16–25
Fluorescent Ordinary High output Compact
40–60 70–90 50–80
Highintensity discharge Mercury vapor 50–60 Metal halide 56–125 Highpressure sodium 100–150 Lowpressure sodium up to 200
15 W
60 W
FIGURE 2–55 A 15W compact fluorescent lamp provides as much light as a 60W incandescent lamp.
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 temperatures as low as 38°C (or 100°F) instead of about 200°C (or 400°F) 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 mechanical 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, # Wnet,electric h overall h combustion h thermal h generator # HHV mnet
(2–43)
The overall efficiencies are about 26–30 percent for gasoline automotive engines, 34–40 percent for diesel engines, and 40–60 percent for large power plants. We are all familiar with the conversion of electrical energy to light by incandescent 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 example, common incandescent lightbulbs convert about 10 percent of the electrical energy they consume to light; the rest of the energy consumed is dissipated as heat, which adds to the cooling load of the air conditioner in summer. However, it is more common to express the effectiveness of this conversion process 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 2–1. 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. 2–55). Also, a compact fluorescent bulb lasts about 10,000 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 considerably 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
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Chapter 2 the food to the energy consumed by the appliance (Fig. 2–56). 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 2–2). 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 conventional 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. Using energyefficient appliances and practicing energy conservation measures help our pocketbooks by reducing our utility bills. It 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 6.4 kg of carbon dioxide, which causes global climate change; 4.7 g of nitrogen oxides and 0.54 g of hydrocarbons, which cause smog; 2.0 g of carbon monoxide, which is toxic; and 0.030 g of sulfur dioxide, which causes acid rain. Each therm of natural gas saved eliminates the emission of these pollutants while saving $0.60 for the average consumer in the United States. Each kWh of electricity conserved saves 0.4 kg of coal and 1.0 kg of CO2 and 15 g of SO2 from a coal power plant.
TABLE 2–2 Energy costs of cooking a casserole with different appliances* [From A. Wilson and J. Morril, Consumer Guide to Home Energy Savings, Washington, DC: American Council for an EnergyEfficient Economy, 1996, p. 192.]
Cooking appliance Electric oven Convection oven (elect.) Gas oven Frying pan Toaster oven Electric slow cooker Microwave oven
Cooking temperature 350F (177C) 325F (163C) 350F (177C) 420F (216C) 425F (218C) 200F (93C) “High”
Cooking time 1 45 1 1 50 7 15
h min h h min h min
Energy used
Cost of energy
2.0 kWh 1.39 kWh 0.112 therm 0.9 kWh 0.95 kWh 0.7 kWh 0.36 kWh
$0.16 $0.11 $0.07 $0.07 $0.08 $0.06 $0.03
*Assumes a unit cost of $0.08/kWh for electricity and $0.60/therm for gas.

81
2 kW 3 kW
5 kW
Efficiency = =
Energy utilized Energy supplied to appliance 3 kWh = 0.60 5 kWh
FIGURE 2–56 The efficiency of a cooking appliance represents the fraction of the energy supplied to the appliance that is transferred to the food.
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Thermodynamics 38% Gas Range
73% Electric Range
EXAMPLE 2–15
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. 2–57). Consider a 2kW electric burner at a location where the unit costs of electricity and natural gas are $0.09/kWh and $0.55/therm, 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
FIGURE 2–57 Schematic of the 73 percent efficient electric heating unit and 38 percent efficient gas burner discussed in Example 2–15.
# Qutilized 1Energy input 2 1Efficiency 2 12 kW2 10.732 1.46 kW of useful energy. The unit cost of utilized energy is inversely proportional to the efficiency, and is determined from
Cost of utilized energy
$0.09>kWh Cost of energy input $0.123>kWh Efficiency 0.73
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 (1.46 kW) is
# Qinput, gas
# Qutilized 1.46 kW 3.84 kW¬¬1 13,100 Btu>h 2 Efficiency 0.38
since 1 kW 3412 Btu/h. Therefore, a gas burner should have a rating of at least 13,100 Btu/h to perform as well as the electric unit. Noting that 1 therm 29.3 kWh, the unit cost of utilized energy in the case of a gas burner is determined to be
Cost of utilized energy
$0.55>29.3 kWh Cost of energy input Efficiency 0.38
$0.049>kWh Discussion The cost of utilized gas is less than half of the unit cost of utilized electricity. Therefore, despite its higher efficiency, cooking with an electric burner will cost more than twice as much compared to a gas burner in this case. This explains why costconscious consumers always ask for gas appliances, and it is not wise to use electricity for heating purposes.
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
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Chapter 2 converted entirely from one mechanical form to another, and the mechanical efficiency of a device or process can be defined as (Fig. 2–58) h mech
E mech,out E mech,loss Mechanical energy output 1 Mechanical energy input E mech,in E mech,in
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, velocity, and/or 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 h pump
# # Wpump,u ¢E mech,fluid Mechanical energy increase of the fluid # # Mechanical energy input Wshaft,in Wpump
(2–45)
# # # where ¢E mech,fluid E mech,out E mech,in is the rate of increase in the mechanical # energy of the fluid, which is equivalent to the useful pumping power Wpump,u supplied to the fluid, and # # Wshaft,out Mechanical energy output Wturbine h turbine # (2–46) # Mechanical energy decrease of the fluid 0 ¢E mech,fluid 0 Wturbine,e
# # # where 0 ¢E mech,fluid 0 E mech,in E mech,out is the rate of decrease in the mechanical energy of the fluid, which is equivalent to the mechanical power # extracted from the fluid by the turbine Wturbine,e, 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. 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 hmotor, 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 consumed 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:
h motor
# Wshaft,out Mechanical power output # Electric power input Welect,in
(2–47)
83
Fan 50 W
(2–44)

1
m· = 0.50 kg/s 2
V1 = 0, V2 = 12 m/s z1 = z2 P1 = P2 · · 2/2 ∆Emech,fluid mV 2 hmech, fan = –––––––––– = ––––––– · · Wshaft,in Wshaft,in (0.50 kg/s)(12 m/s)2/2 = ––––––––––––––––– 50 W = 0.72
FIGURE 2–58 The mechanical efficiency of a fan is the ratio of the kinetic energy of air at the fan exit to the mechanical power input.
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Thermodynamics and
hturbine = 0.75
hgenerator = 0.97
# Welect,out Electric power output h generator # Mechanical power input Wshaft,in
Generator: Turbine
Generator
h turbine–gen = hturbineh generator = 0.75 × 0.97 = 0.73
FIGURE 2–59 The overall efficiency of a turbine–generator is the product of the efficiency of the turbine and the efficiency of the generator, and represents the fraction of the mechanical energy of the fluid converted to electric energy.
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 pump–motor and turbine–generator combinations (Fig. 2–59), which are defined as # # Wpump,u ¢E mech,fluid h pumpmotor h pumph motor # # Welect,in Welect,in
Lake h = 50 m Turbine
Generator
m· = 5000 kg/s
FIGURE 2–60 Schematic for Example 2–16.
(2–49)
and h turbinegen
# # Welect,out Welect,out h turbineh generator # # Wturbine,e 0 ¢E mech,fluid
(2–50)
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.
EXAMPLE 2–16 hgenerator = 0.95 1862 kW
(2–48)
Performance of a Hydraulic Turbine–Generator
The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine–generator at a location where the depth of the water is 50 m (Fig. 2–60). Water is to be supplied at a rate of 5000 kg/s. If the electric power generated 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.
Solution A hydraulic turbine–generator is to generate electricity from the water of a lake. The overall efficiency, the turbine efficiency, and the turbine shaft power are to be determined. Assumptions 1 The elevation of the lake remains constant. 2 The mechanical energy of water at the turbine exit is negligible. Properties The density of water can be taken to be r 1000 kg/m3. Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the change in its mechanical energy per unit mass becomes
emech,in emech,out
1 kJ>kg P b 0 gh 19.81 m>s2 2 150 m2 a r 1000 m2>s2
0.491 kJ>kg
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Chapter 2 Then the rate at which mechanical energy is supplied to the turbine by the fluid and the overall efficiency become
0 ¢E mech,fluid 0 m 1emech,in emech,out 2 15000 kg>s2 10.491 kJ>kg2 2455 kW #
#
h overall h turbinegen
# Welect,out 1862 kW 0.76 # 2455 kW 0 ¢E mech,fluid 0
(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from
hturbinegen hturbinehgenerator S hturbine
hturbinegen hgenerator
0.76 0.80 0.95
(c) The shaft power output is determined from the definition of mechanical efficiency,
# # Wshaft,out h turbine 0 ¢E mech,fluid 0 10.802 12455 kW2 1964 kW Discussion Note that the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that drives the generator, which generates 1862 kW of electric power. There are losses associated with each component.
EXAMPLE 2–17
Cost Savings Associated with HighEfficiency Motors
60 hp
h = 89.0% Standard Motor
A 60hp electric motor (a motor that delivers 60 hp of shaft power at full load) that has an efficiency of 89.0 percent is worn out and is to be replaced by a 93.2 percent efficient highefficiency motor (Fig. 2–61). The motor operates 3500 hours a year at full load. Taking the unit cost of electricity to be $0.08/kWh, 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. Analysis The electric power drawn by each motor and their difference can be expressed as
# # Welectric in,standard Wshaft>h st 1Rated power2 1Load factor2 >h st # # Welectric in,efficient Wshaft>h eff 1Rated power2 1Load factor2 >h eff # # Power savings Welectric in,standard Welectric in,efficient 1Rated power2 1Load factor2 11>h st 1>h eff 2
60 hp
h = 93.2% HighEfficiency Motor
FIGURE 2–61 Schematic for Example 2–17.

85
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Thermodynamics where hst is the efficiency of the standard motor, and heff is the efficiency of the comparable highefficiency motor. Then the annual energy and cost savings associated with the installation of the highefficiency motor become
Energy savings 1Power savings 2 1Operating hours2
1Rated power2 1Operating hours 2 1Load factor2 11>h st 1>h eff 2 160 hp2 10.7457 kW>hp2 13500 h>year2 11 2 11>0.89 1>0.93.2 2 7929 kWh>year
Cost savings 1Energy savings 2 1Unit cost of energy 2 17929 kWh>year2 1$0.08> kWh2 $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 $640 1.01 year Annual cost savings $634>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.
2–8
FIGURE 2–62 Energy conversion processes are often accompanied by environmental pollution. © Corbis Royalty Free
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 8 ON THE DVD.
■
ENERGY AND ENVIRONMENT
The conversion of energy from one form to another often affects the environment 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. 2–62). Fossil 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, and global warming and climate change. The environmental pollution has reached such high levels that it became a serious threat to vegetation, wild life, and human health. Air pollution has been the cause of numerous health problems including asthma and cancer. It is estimated that over 60,000 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 fireplaces. 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 pollution is the motor vehicles, and the pollutants released by the vehicles are
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Chapter 2 usually grouped as hydrocarbons (HC), nitrogen oxides (NOx), and carbon monoxide (CO) (Fig. 2–63). The HC emissions are a large component of volatile organic compounds (VOCs) emissions, and the two terms are generally used interchangeably for motor vehicle emissions. A significant portion of the VOC or HC emissions are 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 significant 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 limits on pollutants emitted by large plants and vehicles. These early standards focused on emissions of hydrocarbons, nitrogen oxides, and carbon monoxide. The new cars were required to have catalytic converters in their exhaust systems 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 matter (PM). As a result, today’s 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 0.4 gpm in 1980 and about 0.1 gpm in 1999. 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 pollutants. This becomes apparent when the air pollution levels in their neighborhoods rise to high levels.
Ozone and Smog If you live in a metropolitan area such as Los Angeles, you are probably familiar with urban smog—the 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 numerous other chemicals, including carbon monoxide (CO), particulate matter such as soot and dust, 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 sun’s 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 the motor vehicles. Hydrocarbons and nitrogen oxides react in the presence of sunlight on hot calm days to form groundlevel ozone, which is the primary

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NOx CO HC
FIGURE 2–63 Motor vehicles are the largest source of air pollution.
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SUN
O3 NOx HC
SMOG
FIGURE 2–64 Groundlevel ozone, which is the primary component of smog, forms when HC and NOx react in the presence of sunlight in hot calm days.
component of smog (Fig. 2–64). The 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 of 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, headaches, and nausea, and aggravates respiratory problems such as asthma. Every exposure to ozone does a little damage to the lungs, just like cigarette smoke, eventually reducing the individual’s 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 contains at least 2 percent oxygen was introduced. The use of RFG has resulted in 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 colorless, 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 body’s organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars. 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
FIGURE 2–65 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.
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 sulfur oxides (the cause of the rotten egg smell). The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids (Fig. 2–65). 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 on to the soil by rain or snow. This is known as acid rain. The soil is capable of neutralizing
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a certain amount of acid, but the amounts produced by the power plants using inexpensive highsulfur coal has exceeded this capability, and 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 due to absorbing the acids through their leaves, needles, and roots. Even marble structures deteriorate due to acid rain. The magnitude of the problem was not recognized until the early 1970s, and serious measures have been taken since then to reduce the sulfur dioxide emissions drastically by installing scrubbers in 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 sunlight 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 (nontransparent) to radiation in the longer wavelength infrared regions. Therefore, 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, since it is utilized 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 absorption 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. 2–66). Therefore, they 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 human activities in producing water (such as the burning of fossil fuels) do not make much difference on its concentration in the atmosphere (which is mostly due to evaporation from rivers, lakes, oceans, etc.). CO2 is different, however, in that people’s 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 30°C 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. The global climate change is due to the excessive use of fossil fuels such as coal, petroleum products, and natural gas in electric power generation, transportation, buildings, and manufacturing, and it has been a concern in recent decades. In 1995, a total of 6.5 billion tons of carbon was released to the atmosphere as CO2. The current concentration of CO2 in the atmosphere
SUN Greenhouse gases
Some infrared radiation emitted by earth is absorbed by greenhouse gases and emitted back
Solar radiation passes through and is mostly absorbed by earth’s surface
FIGURE 2–66 The greenhouse effect on earth.
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FIGURE 2–67 The average car produces several times its weight in CO2 every year (it is driven 12,000 miles a year, consumes 600 gallons of gasoline, and produces 20 lbm of CO2 per gallon). © Vol. 39/PhotoDisc
is about 360 ppm (or 0.36 percent). This is 20 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 the photosynthesis process, and thus 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 the CO2 production in recent decades disturbed this balance. In a 1995 report, the world’s leading climate scientists concluded that the earth has already warmed about 0.5°C during the last century, and they estimate that the earth’s temperature will rise another 2°C by the year 2100. A rise of this magnitude is feared to cause severe changes in weather patterns with storms and heavy rains and flooding at some parts and drought in others, major floods due to the melting of ice at the poles, loss of wetlands 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 species 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 world’s 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. This can be done by increasing conservation efforts and improving conversion efficiencies, while meeting new energy demands 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. A major source of greenhouse gas emissions is transportation. Each liter of gasoline burned by a vehicle produces about 2.5 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 12,000 miles a year, and it consumes about 600 gallons of gasoline. Therefore, a car emits about 12,000 lbm of CO2 to the atmosphere a year, which is about four times the weight of a typical car (Fig. 2–67). 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 atmosphere every year while reducing the fuel cost by $400 per year (under average driving conditions of 12,000 miles a year and at a fuel cost of $2.00/gal). It is clear from these discussions that considerable amounts of pollutants are emitted as the chemical energy in fossil fuels is converted to thermal, mechanical, or electrical energy via combustion, and thus power plants, motor vehicles, and even stoves take the blame for air pollution. In contrast, no pollution is emitted as electricity is converted to thermal, chemical, or
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Chapter 2 mechanical energy, and thus electric 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 the electric cars is generated somewhere else mostly by burning fuel and thus emitting pollution. Therefore, each time an electric car consumes 1 kWh of electricity, it bears the responsibility for the pollutions emitted as 1 kWh of electricity (plus the conversion and transmission losses) is generated elsewhere. The electric cars can be claimed to be zero emission vehicles only when the electricity they consume is generated by emissionfree renewable resources such as hydroelectric, solar, wind, and geothermal energy (Fig. 2–68). 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 2–18
Reducing Air Pollution by Geothermal Heating
A geothermal power plant in Nevada is generating electricity using geothermal water extracted at 180°C, and reinjected back to the ground at 85°C. It is proposed to utilize the reinjected brine for heating the residential and commercial buildings in the area, and calculations 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 system will save a year. Take the average NOx and CO2 emissions of gas furnaces to be 0.0047 kg/therm and 6.4 kg/therm, 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. 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,
NOx savings 1NOx emission per therm2 1No. of therms per year2 10.0047 kg>therm2 118 106 therm>year2 8.5 104 kg>year CO2 savings 1CO2 emission per therm2 1No. of therms per year2 16.4 kg>therm2 118 106 therm>year2 1.2 108 kg>year
Discussion A typical car on the road generates about 8.5 kg of NOx and 6000 kg of CO2 a year. Therefore the environmental impact of replacing the gas heating systems in the area by the geothermal heating system is equivalent to taking 10,000 cars off the road for NOx emission and taking 20,000 cars off the road for CO2 emission. The proposed system should have a significant effect on reducing smog in the area.

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FIGURE 2–68 Renewable energies such as wind are called “green energy” since they emit no pollutants or greenhouse gases. © Corbis Royalty Free
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Thermodynamics
TOPIC OF SPECIAL INTEREST*
Heat can be transferred in three different ways: conduction, convection, and radiation. We will give a brief description of each mode to familiarize the reader 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 molecules in a lattice and the 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. 2–69). # It is observed that the rate of heat conduction Qcond through a layer of constant thickness x is proportional to the temperature difference T 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,
AIR Heat COLA
∆T T1
T2 AIR Heat
COLA
Mechanisms of Heat Transfer
∆x Wall of aluminum can
FIGURE 2–69 Heat conduction from warm air to a cold canned drink through the wall of the aluminum can.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 9 ON THE DVD.
# ¢T Qcond kt A¬ ¬¬1W2 ¢x
(2–51)
where the constant of proportionality kt is the thermal conductivity of the material, which is a measure of the ability of a material to conduct heat (Table 2–3). Materials such as copper and silver, which are good electric conductors, are also good heat conductors, and therefore have high kt values. Materials such as rubber, wood, and styrofoam are poor conductors of heat, and therefore have low kt values. In the limiting case of x → 0, the equation above reduces to the differential form # dT Qcond¬¬ kt A¬ ¬¬1W2 dx
(2–52)
which is known as Fourier’s law of heat conduction. It indicates that the rate of heat conduction in a direction is proportional to the temperature gradient in that direction. Heat is conducted in the direction of decreasing temperature, and the temperature gradient becomes negative when temperature decreases with increasing x. Therefore, a negative sign is added in Eq. 2–52 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 molecules 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. *This section can be skipped without a loss in continuity.
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Chapter 2 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 transported 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 2–3. 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 presence 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 of cool air over its top surface (Fig. 2–70). 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 convection; that 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 by the cooler air. 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 variation of temperature in the fluid (Fig. 2–71). For example, in the absence of a fan, heat transfer from the surface of the hot block in Fig. 2–70 will be by natural 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 surrounding 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 considered to be convection because of the fluid motion induced during the process 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 Qconv is determined from Newton’s law of cooling, expressed as # Qconv hA 1Ts Tf 2¬¬1W2
(2–53)
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 temperature equals the surface temperature of the solid.)

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TABLE 2–3 Thermal conductivities of some materials at room conditions Thermal conductivity, W/m · K
Material Diamond Silver Copper Gold Aluminium Iron Mercury () Glass Brick Water () Human skin Wood (oak) Helium (g) Soft rubber Glass fiber Air (g) Urethane, rigid foam
Velocity variation of air
2300 429 401 317 237 80.2 8.54 1.4 0.72 0.613 0.37 0.17 0.152 0.13 0.043 0.026 0.026
Tf V
AIR FLOW
T Temperature variation of air
Qconv A
Ts HOT BLOCK
FIGURE 2–70 Heat transfer from a hot surface to air by convection.
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Forced convection
Natural convection AIR
AIR hot egg
hot egg
FIGURE 2–71 The cooling of a boiled egg by forced and natural convection.
Person 30°C
Air 5°C
Fire 900°C
Radiation
FIGURE 2–72 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.
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 variables 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 W/m2 · K, are in the range of 2–25 for the free convection of gases, 50–1000 for the free convection of liquids, 25–250 for the forced convection of gases, 50–20,000 for the forced convection of liquids, and 2500–100,000 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. 2–72). 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 differs 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 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 Stefan–Boltzmann law as # Q emit,max sAT s4¬¬1W2
108
(2–54)
where A is the surface area and s 5.67 · is the Stefan–Boltzmann constant. The idealized surface that emits radiation at this maximum 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 # Qemit esATs4¬¬1W2
W/m2
K4
(2–55)
where e is the emissivity of the surface. The property emissivity, whose value is in the range 0 e 1, is a measure of how closely a surface approximates a blackbody for which e 1. The emissivities of some surfaces are given in Table 2–4. Another important radiation property of a surface is its absorptivity, a, 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 a 1. A blackbody absorbs the entire radiation incident on it. That is, a blackbody is a perfect absorber (a 1) as well as a perfect emitter.
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Chapter 2 In general, both e and a of a surface depend on the temperature and the wavelength of the radiation. Kirchhoff’s law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength. In most practical applications, the dependence of e and a on the temperature and wavelength is ignored, and the average absorptivity of a surface is taken to be equal to its average emissivity. The rate at which a surface absorbs radiation is determined from (Fig. 2–73) # # Q abs aQ incident¬¬1W2
(2–56)
# where Q incident is the rate at which radiation is incident on the surface and a is the absorptivity of the surface. For opaque (nontransparent) surfaces, the portion 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 transfer 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 e and surface 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 (i.e., 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. 2–74) # Qrad esA 1T 4s T 4surr 2¬¬1W2

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TABLE 2–4 Emissivity of some materials at 300 K Material
Emissivity
Aluminium foil Anodized aluminum Polished copper Polished gold Polished silver Polished stainless steel Black paint White paint White paper Asphalt pavement Red brick Human skin Wood Soil Water Vegetation
0.07 0.82 0.03 0.03 0.02 0.17 0.98 0.90 0.92–0.97 0.85–0.93 0.93–0.96 0.95 0.82–0.92 0.93–0.96 0.96 0.92–0.96
· Qincident · · Qref = (1 – α ) Qincident
(2–57)
In this special case, the emissivity and the surface area of the surrounding surface do not have any effect on the net radiation heat transfer.
· · Qabs = α Qincident
FIGURE 2–73 The absorption of radiation incident on an opaque surface of absorptivity a. EXAMPLE 2–19
Heat Transfer from a Person
Consider a person standing in a breezy room at 20°C. 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 1.6 m2 and 29°C, respectively, and the convection heat transfer coefficient is 6 W/m2 · °C (Fig. 2–75).
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 uniform. 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
LARGE ENCLOSURE
,A,Ts SMALL BODY
Qrad
Tsurr
FIGURE 2–74 Radiation heat transfer between a body and the inner surfaces of a much larger enclosure that completely surrounds it.
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Room air 20°C
Qconv 29°C
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. 2–53,
# Qconv hA 1Ts Tf 2
16 W>m2 # °C2 11.6 m2 2 129 20 2 °C
Qrad
86.4 W
Qcond
FIGURE 2–75 Heat transfer from the person described in Example 2–19.
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 the floor is, from Eq. 2–57, # Qrad esA 1T 4s T 4surr 2
10.95 2 15.67 108 W>m2 # K4 2 11.6 m2 2 3 129 273 2 4 120 273 2 4 4K4 81.7 W
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
# # # Qtotal Qconv Qrad 86.4 81.7 168.1 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 above 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 represents the molecular energy of a system and may exist in sensible, latent, chemical, and nuclear forms.
. Mass flow rate m is defined as the amount of mass flowing through a cross section per unit time. It is related to the vol. ume flow rate V, which is the volume of a fluid flowing through a cross section per unit time, by # # m rV rAcVavg
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Chapter 2
P V gz r 2
Change in internal, kinetic, potential, etc., energies
It can also be expressed in the rate form as . . E in Eout¬ ¬ dE system>dt¬¬1kW2 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
emech
⎫ ⎪ ⎬ ⎪ ⎭
2
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
The efficiencies of various devices are defined as # # Wpump,u ¢E mech,fluid h pump # # Wshaft,in Wpump
and # V2 # # P gz b E mech memech m a r 2 where P/r is the flow energy, V 2/2 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 temperature 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:
h turbine
Wsh 2pnT
Spring work: Wspring
1 k 1x 22 x 21 2 2
The first law of thermodynamics is essentially an expression 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
# # Wshaft,out Wturbine # # 0 ¢E mech,fluid 0 Wturbine,e
# Wshaft,out Mechanical power output h motor # Electric power input Welect,in # Welect,out Electric power output # h generator Mechanical power input Wshaft,in # ¢E mech,fluid h pumpmotor h pumph motor # Welect,in
Electrical work: We VI¬¢t Shaft work:
97
¢E system¬¬1kJ2
Net energy transfer by heat, work, and mass
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
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 in E out¬ ¬
⎫ ⎪ ⎬ ⎪ ⎭
The energy flow rate associated with a fluid flowing at a rate . of m is # # E me

h turbine–gen
# Welect,out h turbineh generator # 0 ¢E mech,fluid 0
The conversion of energy from one form to another is often associated with adverse effects on the environment, and environmental impact should be an important consideration in the conversion and utilization of energy.
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 Teaching Thermodynamics.” ASME International Mechanical Engineering Congress and Exposition, Atlanta, Georgia, AESVol. 36, pp. 251–260, November 17–22, 1996.
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PROBLEMS* Forms of Energy 2–1C Portable electric heaters are commonly used to heat small rooms. Explain the energy transformation involved during this heating process. 2–2C Consider the process of heating water on top of an electric range. What are the forms of energy involved during this process? What are the energy transformations that take place? 2–3C What is the difference between the macroscopic and microscopic forms of energy? 2–4C What is total energy? Identify the different forms of energy that constitute the total energy. 2–5C List the forms of energy that contribute to the internal energy of a system. 2–6C How are heat, internal energy, and thermal energy related to each other? 2–7C What is mechanical energy? How does it differ from thermal energy? What are the forms of mechanical energy of a fluid stream? 2–8 Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s 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. River
3 m/s
2–10 At a certain location, wind is blowing steadily at 10 m/s. 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 1.25 kg/m3. 2–11 A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s 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. 2–12 Two sites are being considered for wind power generation. In the first site, the wind blows steadily at 7 m/s for 3000 hours per year, whereas in the second site the wind blows at 10 m/s for 2000 hours per year. Assuming the wind velocity is negligible at other times for simplicity, determine which is a better site for wind power generation. Hint: Note that the mass flow rate of air is proportional to wind velocity. 2–13 A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam can be built to collect water and release it from an elevation difference of 50 m to generate power. Determine how much power can be generated from this river water after the dam is filled. 2–14 A person gets into an elevator at the lobby level of a hotel together with his 30kg suitcase, and gets out at the 10th floor 35 m above. Determine the amount of energy consumed by the motor of the elevator that is now stored in the suitcase.
Energy Transfer by Heat and Work 90 m
2–15C In what forms can energy cross the boundaries of a closed system? 2–16C When is the energy crossing the boundaries of a closed system heat and when is it work?
FIGURE P2–8
2–17C What is an adiabatic process? What is an adiabatic system?
2–9 Electric power is to be generated by installing a hydraulic turbine–generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. Determine the power generation potential.
2–18C A gas in a piston–cylinder device is compressed, and as a result its temperature rises. Is this a heat or work interaction?
*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 a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
2–19C 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. 2–20C 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? 2–21C An insulated room is heated by burning candles. Is this a heat or work interaction? Take the entire room, including the candles, as the system. 2–22C What are point and path functions? Give some examples.
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Chapter 2 2–23C What is the caloric theory? When and why was it abandoned?
Mechanical Forms of Work 2–24C A car is accelerated from rest to 85 km/h in 10 s. Would the energy transferred to the car be different if it were accelerated to the same speed in 5 s? 2–25C Lifting a weight to a height of 20 m takes 20 s for one crane and 10 s for another. Is there any difference in the amount of work done on the weight by each crane? 2–26 Determine the energy required to accelerate an 800kg car from rest to 100 km/h on a level road. Answer: 309 kJ 2–27 Determine the energy required to accelerate a 1300kg car from 10 to 60 km/h on an uphill road with a vertical rise of 40 m. 2–28E Determine the torque applied to the shaft of a car that transmits 450 hp and rotates at a rate of 3000 rpm. 2–29 Determine the work required to deflect a linear spring with a spring constant of 70 kN/m by 20 cm from its rest position. 2–30 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 km/h at full power on a level road. Is your answer realistic? 2–31 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 km/h. 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.

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a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance. Answers: (a) 98.1 kW, (b) 188 kW, (c) 21.9 kW 2–33 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 km/h on a 30° (from horizontal) uphill road, and (c) to accelerate on a level road from stop to 90 km/h in 12 s. Answers: (a) 0, (b) 81.7 kW, (c) 31.3 kW
The First Law of Thermodynamics 2–34C For a cycle, is the net work necessarily zero? For what kind of systems will this be the case? 2–35C On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will the room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed. 2–36C What are the different mechanisms for transferring energy to or from a control volume? 2–37 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 N · m. Determine the final energy of the system if its initial energy is 10 kJ. Answer: 35.5 kJ 5 kJ
2–32 Determine the power required for a 2000kg car to climb a 100mlong uphill road with a slope of 30° (from horizontal) in 10 s (a) at a constant velocity, (b) from rest to
30 kJ 2000 kg
500 N·m m 100
FIGURE P2–37
30°
FIGURE P2–32
2–38E A vertical piston–cylinder device contains water and is being heated on top of a range. During the process, 65 Btu
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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
2–39 A classroom that normally contains 40 people is to be airconditioned with window airconditioning units of 5kW cooling capacity. A person at rest may be assumed to dissipate heat at a rate of about 360 kJ/h. There are 10 lightbulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window airconditioning units required. Answer: 2 units
2–40 The lighting requirements of an industrial facility are being met by 700 40W standard fluorescent lamps. The lamps are close to completing their service life and are to be replaced by their 34W highefficiency counterparts that operate on the existing standard ballasts. The standard and highefficiency fluorescent lamps can be purchased in quantity at a cost of $1.77 and $2.26 each, respectively. The facility operates 2800 hours a year, and all of the lamps are kept on during operating hours. Taking the unit cost of electricity to be $0.08/kWh and the ballast factor to be 1.1 (i.e., ballasts consume 10 percent of the rated power of the lamps), determine how much energy and money will be saved per year as a result of switching to the highefficiency fluorescent lamps. Also, determine the simple payback period. 2–41 The lighting needs of a storage room are being met by 6 fluorescent light fixtures, each fixture containing four lamps rated at 60 W each. All the lamps are on during operating 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 $0.08/kWh, 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 hour to install it at a cost of $40. 2–42 A university campus has 200 classrooms and 400 faculty 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 classrooms 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 $0.082/kWh, determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods. 2–43 Consider a room that is initially at the outdoor temperature of 20°C. The room contains a 100W lightbulb, a 110W TV set, a 200W refrigerator, and a 1000W 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. 2–44 A fan is to accelerate quiescent air to a velocity of 10 m/s at a rate of 4 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3. Answer: 236 W 2–45E Consider a fan located in a 3 ft 3 ft square duct. Velocities at various points at the outlet are measured, and the average flow velocity is determined to be 22 ft/s. Taking the air density to 0.075 lbm/ft3, estimate the minimum electric power consumption of the fan motor. 2–46 A water pump that consumes 2 kW of electric power when operating is claimed to take in water from a lake and pump it to a pool whose free surface is 30 m above the free surface of the lake at a rate of 50 L/s. Determine if this claim is reasonable. 2–47
The driving force for fluid flow is the pressure difference, and a pump operates by raising the pressure of a fluid (by converting the mechanical shaft work to flow energy). A gasoline pump is measured to consume 5.2 kW of electric power when operating. If the pressure differential between the outlet and inlet of the pump is measured to be 5 kPa and the changes in velocity and elevation are negligible, determine the maximum possible volume flow rate of gasoline. ∆P = 5 kPa
Pump
FIGURE P2–47 2–48 The 60W fan of a central heating system is to circulate air through the ducts. The analysis of the flow shows that the fan needs to raise the pressure of air by 50 Pa to maintain flow. The fan is located in a horizontal flow section whose diameter is 30 cm at both the inlet and the outlet. Determine the highest possible average flow velocity in the duct. 2–49E At winter design conditions, a house is projected to lose heat at a rate of 60,000 Btu/h. The internal heat gain from people, lights, and appliances is estimated to be 6000 Btu/h. If this house is to be heated by electric resistance heaters, determine the required rated power of these heaters in kW to maintain the house at constant temperature. 2–50 An escalator in a shopping center is designed to move 30 people, 75 kg each, at a constant speed of 0.8 m/s at 45° slope. Determine the minimum power input needed to drive
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Chapter 2 this escalator. What would your answer be if the escalator velocity were to be doubled? 2–51 Consider a 1400kg car cruising at constant speed of 70 km/h. Now the car starts to pass another car, by accelerating to 110 km/h 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: 77.8 kW, 38.9 kW
Energy Conversion Efficiencies 2–52C What is mechanical efficiency? What does a mechanical efficiency of 100 percent mean for a hydraulic turbine? 2–53C How is the combined pump–motor efficiency of a pump and motor system defined? Can the combined pump–motor efficiency be greater than either the pump or the motor efficiency? 2–54C Define turbine efficiency, generator efficiency, and combined turbine–generator efficiency. 2–55C Can the combined turbinegenerator efficiency be greater than either the turbine efficiency or the generator efficiency? Explain. 2–56 Consider a 3kW hooded electric open burner in an area where the unit costs of electricity and natural gas are $0.07/kWh and $1.20/therm, respectively. 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. 2–57 A 75hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a highefficiency 75hp motor that has an efficiency of 95.4 percent. Determine the reduction in the heat gain of the room due to higher efficiency under fullload conditions. 2–58 A 90hp (shaft output) electric car is powered by an electric 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. 2–59 A 75hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a highefficiency motor that has an efficiency of 95.4 percent. The motor operates 4368 hours a year at a load factor of 0.75. Taking the cost of electricity to be $0.08/kWh, 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 $5449 and $5520, respectively.

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2–60E The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 3.6 106 Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a handheld flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 1500 hours a year intermittently. Taking the unit cost of energy to be $4.35/106 Btu, determine the annual energy and cost savings as a result of tuning up the boiler. 2–61E
Reconsider Prob. 2–60E. Using EES (or other) software, study the effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings. Let the efficiency vary from 0.6 to 0.9, and the unit cost to vary from $4 to $6 per million Btu. Plot the annual energy used and the cost savings against the efficiency for unit costs of $4, $5, and $6 per million Btu, and discuss the results.
2–62 An exercise room has eight weightlifting machines that have no motors and four treadmills each equipped with a 2.5hp (shaft output) motor. The motors operate at an average load factor of 0.7, at which their efficiency is 0.77. During peak evening hours, all 12 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 525 W, determine the rate of heat gain of the exercise room from people and the equipment at peak load conditions. 2–63 Consider a classroom for 55 students and one instructor, each generating heat at a rate of 100 W. Lighting is provided by 18 fluorescent lightbulbs, 40 W each, and the ballasts consume an additional 10 percent. Determine the rate of internal heat generation in this classroom when it is fully occupied. 2–64 A room is cooled by circulating chilled water through a heat exchanger located in a room. The air is circulated through the heat exchanger by a 0.25hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25hp equipment is 54 percent. Determine the rate of heat supply by the fan–motor assembly to the room. 2–65 Electric power is to be generated by installing a hydraulic turbine–generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. 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 turbine–generator efficiency of this plant. Neglect losses in the pipes. 2–66 At a certain location, wind is blowing steadily at 12 m/s. Determine the mechanical energy of air per unit mass
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and the power generation potential of a wind turbine with 50mdiameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of 30 percent. Take the air density to be 1.25 kg/m3. 2–67
Reconsider Prob. 2–66. Using EES (or other) software, investigate the effect of wind velocity and the blade span diameter on wind power generation. Let the velocity vary from 5 to 20 m/s in increments of 5 m/s, and the diameter vary from 20 to 80 m in increments of 20 m. Tabulate the results, and discuss their significance.
2–68 A wind turbine is rotating at 15 rpm under steady winds flowing through the turbine at a rate of 42,000 kg/s. The tip velocity of the turbine blade is measured to be 250 km/h. 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 1.31 kg/m3. 2–69 Water is pumped from a lake to a storage tank 20 m above at a rate of 70 L/s while consuming 20.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine (a) the overall efficiency of the pump–motor unit and (b) the pressure difference between the inlet and the exit of the pump.
2–72 Large wind turbines with blade span diameters of over 100 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 m/s. Taking the overall efficiency of the wind turbine to be 32 percent and the air density to be 1.25 kg/m3, determine the electric power generated by this wind turbine. Also, assuming steady winds of 8 m/s during a 24hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.06/kWh for electricity. 2–73E A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the pump is measured to be 1.2 psi when the flow rate is 8 ft3/s and the changes in velocity and elevation are negligible, determine the mechanical efficiency of this pump. 2–74 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 0.03 m3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.
2 Storage tank 0.03 m3/s
45 m
20 m 1 z1 = 0
Pump 20 kW Pump Control surface
FIGURE P2–69 FIGURE P2–74 2–70 A geothermal pump is used to pump brine whose density is 1050 kg/m3 at a rate of 0.3 m3/s 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. 2–71 Consider an electric motor with a shaft power output of 20 kW and an efficiency of 88 percent. Determine the rate at which the motor dissipates heat to the room it is in when the motor operates at full load. In winter, this room is normally heated by a 2kW resistance heater. Determine if it is necessary to turn the heater on when the motor runs at full load.
2–75 A 7hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, determine the maximum volume flow rate of water. 2–76 A hydraulic turbine has 85 m of elevation difference available at a flow rate of 0.25 m3/s, and its overall turbine– generator efficiency is 91 percent. Determine the electric power output of this turbine. 2–77 An oil pump is drawing 35 kW of electric power while pumping oil with r 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8 cm and 12 cm,
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Chapter 2 respectively. If the pressure rise of oil in the pump is measured to be 400 kPa and the motor efficiency is 90 percent, determine the mechanical efficiency of the pump.
35 kW 12 cm Motor
Pump 8 cm Oil
∆P = 400 kPa
0.1 m3/s
FIGURE P2–77 2–78E A 73percent efficient pump with a power input of 12 hp is pumping water from a lake to a nearby pool at a rate of 1.2 ft3/s through a constantdiameter pipe. The free surface of the pool is 35 ft above that of the lake. Determine the mechanical power used to overcome frictional effects in piping. Answer: 4.0 hp
Energy and Environment 2–79C How does energy conversion affect the environment? What are the primary chemicals that pollute the air? What is the primary source of these pollutants?
2–86 Repeat Prob. 2–85 assuming the electricity is produced by a power plant that burns coal. The average production of CO2 in this case is 1.1 kg per kWh. 2–87E Consider a household that uses 11,000 kWh of electricity per year and 1500 gallons of fuel oil during a heating season. The average amount of CO2 produced is 26.4 lbm/gallon of fuel oil and 1.54 lbm/kWh of electricity. If this household reduces its oil and electricity usage by 15 percent as a result of implementing some energy conservation measures, determine the reduction in the amount of CO2 emissions by that household per year. 2–88 A typical car driven 12,000 miles 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 4.3 g of NOx per therm, and the electric power plants emit about 7.1 g of NOx per kWh of electricity produced. Consider a household that has two cars and consumes 9000 kWh of electricity and 1200 therms of natural gas. Determine the amount of NOx emission to the atmosphere per year for which this household is responsible.
11 kg NOx per year
2–81C 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?
2–83C Why is carbon monoxide a dangerous air pollutant? How does it affect human health at low and at high levels? 2–84E A Ford Taurus driven 15,000 miles a year will use about 715 gallons of gasoline compared to a Ford Explorer that would use 940 gallons. About 19.7 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 production a man is responsible for during a 5year period if he trades his Taurus for an Explorer. 2–85 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
103
effect and thus global climate change. On average, 0.59 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. Determine the amount of CO2 production that is due to the refrigerators in a city with 200,000 households.
2–80C What is smog? What does it consist of? How does groundlevel ozone form? What are the adverse effects of ozone on human health?
2–82C 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 greenhouse effect? How can we combat this problem?

FIGURE P2–88
Special Topic: Mechanisms of Heat Transfer 2–89C What are the mechanisms of heat transfer? 2–90C Does any of the energy of the sun reach the earth by conduction or convection? 2–91C Which is a better heat conductor, diamond or silver? 2–92C How does forced convection differ from natural convection? 2–93C Define emissivity and absorptivity. What is Kirchhoff’s law of radiation? 2–94C What is blackbody? How do real bodies differ from a blackbody?
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2–95 The inner and outer surfaces of a 5m 6m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m · °C are maintained at temperatures of 20°C and 5°C, respectively. Determine the rate of heat transfer through the wall, in W.
2–102 Hot air at 80°C is blown over a 2m 4m flat surface at 30°C. If the convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of heat transfer from the air to the plate, in kW.
Brick wall
20°C
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 W/m2 · °C. Plot the rate of heat transfer against the convection heat transfer coefficient for the surface emissivities of 0.1, 0.5, 0.8, and 1, and discuss the results.
5°C 30 cm
FIGURE P2–95 2–96 The inner and outer surfaces of a 0.5cmthick 2m 2m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass is 0.78 W/m · °C, determine the amount of heat loss, in kJ, through the glass over a period of 5 h. What would your answer be if the glass were 1cm thick? Reconsider Problem 2–96. Using EES (or other) software, investigate the effect of glass thickness on heat loss for the specified glass surface temperatures. Let the glass thickness vary from 0.2 to 2 cm. Plot the heat loss versus the glass thickness, and discuss the results.
2–103 A 1000W iron is left on the ironing board with its base exposed to the air at 20°C. The convection heat transfer coefficient between the base surface and the surrounding air is 35 W/m2 · °C. If the base has an emissivity of 0.6 and a surface area of 0.02 m2, determine the temperature of the base of the iron. 1000W iron
Air 20°C
2–97
2–98 An aluminum pan whose thermal conductivity is 237 W/m · °C has a flat bottom whose diameter is 20 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 500 W. If the inner surface of the bottom of the pan is 105°C, determine the temperature of the outer surface of the bottom of the pan. 2–99 For heat transfer purposes, a standing man can be modeled as a 30cm diameter, 170cm long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine the rate of heat loss from this man by convection in an environment at 20°C. Answer: 336 W
FIGURE P2–103 2–104 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 0.6 for solar radiation. If solar radiation is incident on the plate at a rate of
700 W/m 2
α = 0.6 25°C
2–100 A 5cmdiameter spherical ball whose surface is maintained at a temperature of 70°C is suspended in the middle of a room at 20°C. If the convection heat transfer coefficient is 15 W/m2 · C and the emissivity of the surface is 0.8, determine the total rate of heat transfer from the ball. 2–101
Reconsider Problem 2–100. Using EES (or other) software, investigate the effect of the
FIGURE P2–104
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Chapter 2 700 W/m2 and the surrounding air temperature is 25°C, determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Assume the convection heat transfer coefficient to be 50 W/m2 · °C, and disregard heat loss by radiation. 2–105
Reconsider Problem 2–104. Using EES (or other) software, investigate the effect of the convection heat transfer coefficient on the surface temperature of the plate. Let the heat transfer coefficient vary from 10 to 90 W/m2 · °C. Plot the surface temperature against the convection heat transfer coefficient, and discuss the results.
2–106 A 5cmexternaldiameter, 10mlong hotwater pipe at 80°C is losing heat to the surrounding air at 5°C by natural convection with a heat transfer coefficient of 25 W/m2 · °C. Determine the rate of heat loss from the pipe by natural convection, in kW. 2–107 The outer surface of a spacecraft in space has an emissivity of 0.8 and an absorptivity of 0.3 for solar radiation. If solar radiation is incident on the spacecraft at a rate of 1000 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed. 2–108
Reconsider Problem 2–107. Using EES (or other) software, investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature. Plot the surface temperature against emissivity for solar absorbtivities of 0.1, 0.5, 0.8, and 1, and discuss the results.
2–109 A hollow spherical iron container whose outer diameter is 20 cm and thickness is 0.4 cm is filled with iced water at 0°C. If the outer surface temperature is 5°C, determine the approximate rate of heat loss from the sphere, and the rate at which ice melts in the container. 5°C
Iced water
0.4 cm
FIGURE P2–109 2–110 The inner and outer glasses of a 2m 2m double pane window are at 18°C and 6°C, respectively. If the 1cm space between the two glasses is filled with still air, determine the rate of heat transfer through the window, in kW.

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2–111 Two surfaces of a 2cmthick plate are maintained at 0°C and 100°C, respectively. If it is determined that heat is transferred through the plate at a rate of 500 W/m2, determine its thermal conductivity.
Review Problems 2–112 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 elevator 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 constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. 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? 2–113 Consider a homeowner who is replacing his 25yearold 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 presently pays $1200 a year for heating, determine if he should buy the conventional or highefficiency model. 2–114 Wind energy has been used since 4000 BC to power sailboats, grind grain, pump water for farms, and, more recently, generate electricity. In the United States alone, more than 6 million small windmills, most of them under 5 hp, have been used since the 1850s to pump water. Small windmills have been used to generate electricity since 1900, but the development of modern wind turbines occurred only recently in response to the energy crises in the early 1970s. The cost of wind power has dropped an order of magnitude from about $0.50/kWh in the early 1980s to about $0.05/kWh in the mid1990s, which is about the price of electricity generated at coalfired power plants. Areas with an average wind speed of 6 m/s (or 14 mph) are potential sites for economical wind power generation. Commercial wind turbines generate from 100 kW to 3.2 MW of electric power each at peak design conditions. The blade span (or rotor) diameter of the 3.2 MW wind turbine built by Boeing Engineering is 320 ft (97.5 m). The rotation speed of rotors of wind turbines is usually under 40 rpm (under 20 rpm for large turbines). Altamont Pass in California is the world’s largest wind farm with 15,000 modern wind turbines. This farm and two others in California produced 2.8 billion kWh of electricity in 1991, which is enough power to meet the electricity needs of San Francisco.
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Thermodynamics and 1 kWh 3600 kJ, gas heating with $1.24/therm and 1 therm 105,500 kJ, and oil heating with $1.25/gal and 1 gal of oil 138,500 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. 2–118 A typical household pays about $1200 a year on energy bills, and the U.S. 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.
FIGURE P2–114 © Vol. 57/PhotoDisc
In 2003, 8133 MW of new wind energy generating capacity were installed worldwide, bringing the world’s total wind energy capacity to 39,294 MW. The United States, Germany, Denmark, and Spain account for over 75 percent of current wind energy generating capacity worldwide. Denmark uses wind turbines to supply 10 percent of its national electricity. Many wind turbines currently in operation have just two blades. This is because at tip speeds of 100 to 200 mph, the efficiency of the twobladed turbine approaches the theoretical maximum, and the increase in the efficiency by adding a third or fourth blade is so little that they do not justify the added cost and weight. Consider a wind turbine with an 80mdiameter rotor that is rotating at 20 rpm under steady winds at an average velocity of 30 km/h. Assuming the turbine has an efficiency of 35 percent (i.e., it converts 35 percent of the kinetic energy of the wind to electricity), determine (a) the power produced, in kW; (b) the tip speed of the blade, in km/h; and (c) the revenue generated by the wind turbine per year if the electric power produced is sold to the utility at $0.06/kWh. Take the density of air to be 1.20 kg/m3. 2–115 Repeat Prob. 2–114 for an average wind velocity of 25 km/h. 2–116E The energy contents, unit costs, and typical conversion efficiencies of various energy sources for use in water heaters are given as follows: 1025 Btu/ft3, $0.012/ft3, and 55 percent for natural gas; 138,700 Btu/gal, $1.15/gal, and 55 percent for heating oil; and 1 kWh/kWh, $0.084/kWh, and 90 percent for electric heaters, respectively. Determine the lowestcost energy source for water heaters. 2–117 A homeowner is considering these heating systems for heating his house: Electric resistance heating with $0.09/kWh
2–119 The U.S. Department of Energy estimates that up to 10 percent of the energy use of a house can be saved by caulking and weatherstripping doors and windows to reduce air leaks at a cost of about $50 for materials for an average home with 12 windows and 2 doors. Caulking and weatherstripping every gasheated home properly would save enough energy to heat about 4 million homes. The savings can be increased by installing storm windows. Determine how long it will take for the caulking and weatherstripping to pay for itself from the energy they save for a house whose annual energy use is $1100. 2–120 The U.S. Department of Energy estimates that 570,000 barrels of oil would be saved per day if every household in the United States lowered the thermostat setting in winter by 6°F (3.3°C). Assuming the average heating season to be 180 days and the cost of oil to be $40/barrel, determine how much money would be saved per year. 2–121 Consider a TV set that consumes 120 W of electric power when it is on and is kept on for an average of 6 hours per day. For a unit electricity cost of 8 cents per kWh, determine the cost of electricity this TV consumes per month (30 days). 2–122 The pump of a water distribution system is powered by a 15kW electric motor whose efficiency is 90 percent.
Water 50 L/s 300 kPa 2
h motor = 90% Motor 15 kW
Pump •
1 100 kPa
Wpump
FIGURE P2–122
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Chapter 2 The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference 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: 74.1 percent 2–123 In a hydroelectric power plant, 100 m3/s of water flows from an elevation of 120 m to a turbine, where electric power is generated. The overall efficiency of the turbine–generator is 80 percent. Disregarding frictional losses in piping, estimate the electric power output of this plant. Answer: 94.2 MW

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efficiencies are expected to be 75 percent each. Disregarding the frictional losses in piping and assuming the system operates for 10 h each in the pump and turbine modes during a typical day, determine the potential revenue this pump–turbine system can generate per year.
Reservoir
40 m 1 Pump– turbine 100 m3/s
120 m Lake 2
Generator
Turbine
FIGURE P2–124
h turbine–gen = 80%
FIGURE P2–123 2–124 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 consumers 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 $0.03/kWh at night and is willing to pay $0.08/kWh 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 reservoir at night using cheap power, and letting the water flow from the reservoir back to the lake during the day, producing power as the pump–motor operates as a turbine–generator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m3/s can be used in either direction. The combined pump–motor and turbine–generator
2–125 A diesel engine with an engine volume of 4.0 L and an engine speed of 2500 rpm operates on an air–fuel ratio of 18 kg air/kg fuel. The engine uses light diesel fuel that contains 750 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 kg/h, 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, one kmol sulfurous acid will be added to the environment. The molar mass of the sulfur is 32 kg/kmol. 2–126 Leaded gasoline contains lead that ends up in the engine exhaust. Lead is a very toxic engine emission. The use of leaded gasoline in the United States has been unlawful for most vehicles since the 1980s. However, leaded gasoline is still used in some parts of the world. Consider a city with 10,000 cars using leaded gasoline. The gasoline contains 0.15 g/L of lead and 35 percent of lead is exhausted to the environment. Assuming that an average car travels 15,000 km per year with a gasoline consumption of 10 L/100 km, determine the amount of lead put into the atmosphere per year in that city. Answer: 788 kg
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Thermodynamics
Fundamentals of Engineering (FE) Exam Problems 2–127 A 2kW electric resistance heater in a room is turned on and kept on for 30 min. The amount of energy transferred to the room by the heater is (a) 1 kJ (e) 7200 kJ
(b) 60 kJ
(c) 1800 kJ
(d) 3600 kJ
2–128 On a hot summer day, the air in a wellsealed room is circulated by a 0.50hp fan driven by a 65 percent efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan.) The rate of energy supply from the fanmotor assembly to the room is (a) 0.769 kJ/s (d) 0.373 kJ/s
(b) 0.325 kJ/s (e) 0.242 kJ/s
(c) 0.574 kJ/s
2–129 A fan is to accelerate quiescent air to a velocity to 12 m/s at a rate of 3 m3/min. If the density of air is 1.15 kg/m3, the minimum power that must be supplied to the fan is (a) 248 W (e) 162 W
(b) 72 W
(c) 497 W
(d) 216 W
2–130 A 900kg car cruising at a constant speed of 60 km/s is to accelerate to 100 km/h in 6 s. The additional power needed to achieve this acceleration is (a) 41 kW (e) 37 kW
(b) 222 kW
(c) 1.7 kW
(d) 26 kW
2–131 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 m/s using an electric motor. Minimum power rating of the motor should be (a) 0 kW (e) 36 kW
(b) 4.8 kW
(c) 47 kW
(d) 12 kW
2–132 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3/s from an elevation of 65 m using a turbine–generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is (a) 3.9 MW (e) 65 MW
(b) 38 MW
(c) 45 MW
(d) 53 MW
2–133 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 88 percent. If the unit cost of electricity is $0.06/kWh, the annual electricity cost of this compressor is (a) $7382 (e) $8389
(b) $9900
(c) $12,780
(d) $9533
2–134 Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is (a) $3.56 (e) $20.74
(b) $5.18
(c) $8.54
(d) $9.28
2–135 A 2kW pump is used to pump kerosene (r 0.820 kg/L) 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) 8.3 L/s (d) 12.1 L/s
(b) 7.2 L/s (e) 17.8 L/s
(c) 6.8 L/s
2–136 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 L/s and the changes in elevation and the flow velocity across the pump are negligible, the overall efficiency of the pump is (a) 69 percent (d) 79 percent
(b) 72 percent (e) 82 percent
(c) 76 percent
The Following Problems Are Based on the Optional Special Topic of Heat Transfer 2–137 A 10cm high and 20cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.08 W and transferring it by convection to the surrounding air at 40°C. 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 W/m2 · °C and radiation heat transfer is negligble, the average surface temperature of the chips is (a) 80°C (e) 60°C
(b) 54°C
(c) 41°C
(d) 72°C
2–138 A 50cmlong, 0.2cmdiameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130°C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is (a) 43,500 W/m2 · °C (c) 68,330 W/m2 · °C (e) 37,540 W/m2 · °C
(b) 137 W/m2 · °C (d) 10,038 W/m2 · °C
2–139 A 3m2 hot black surface at 80°C is losing heat to the surrounding air at 25°C by convection with a convection heat transfer coefficient of 12 W/m2 · °C, and by radiation to the surrounding surfaces at 15°C. The total rate of heat loss from the surface is (a) 1987 W (e) 3811 W
(b) 2239 W
(c) 2348 W
(d) 3451 W
2–140 Heat is transferred steadily through a 0.2m thick 8 m 4 m wall at a rate of 1.6 kW. The inner and outer surface temperatures of the wall are measured to be 15°C to 5°C. The average thermal conductivity of the wall is (a) 0.001 W/m · °C (d) 2.0 W/m · °C
(b) 0.5 W/m · °C (e) 5.0 W/m · °C
(c) 1.0 W/m · °C
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Chapter 2 2–141 The roof of an electrically heated house is 7m long, 10m wide, and 0.25m thick. It is made of a flat layer of concrete whose thermal conductivity is 0.92 W/m · °C. During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be 15°C and 4°C, respectively. The average rate of heat loss through the roof that night was (a) 41 W (e) 2834 W
(b) 177 W
(c) 4894 W
(d) 5567 W
Design and Essay Problems 2–142 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 U.S. government 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 reasonable assumptions, evaluate if this is a reasonable claim or a gross exaggeration. 2–143 Solar energy reaching the earth is about 1350 W/m2 outside the earth’s atmosphere, and 950 W/m2 on earth’s surface normal to the sun on a clear day. Someone is marketing 2 m 3 m photovoltaic cell panels with the claim that a single panel can meet the electricity needs of a house. How do you evaluate this claim? Photovoltaic cells have a conversion efficiency of about 15 percent. 2–144 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. 2–145 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 selection of a heating system? Give some guidelines. Identify

109
the conditions under which each heating system would be the best choice in your area. 2–146 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 performance 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. 2–147 Your neighbor lives in a 2500squarefoot (about 250 m2) older house heated by natural gas. The current gas heater was installed in the early 1970s 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 furnace 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. 2–148 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 article 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 433,000 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 253,000 lbm of coal, 21,000 gallons 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.
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Chapter 3 PROPERTIES OF PURE SUBSTANCES
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 bestknown 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 phasechange 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.

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Thermodynamics INTERACTIVE TUTORIAL
SEE TUTORIAL CH. 3, SEC. 1 ON THE DVD.
N2
AIR
FIGURE 3–1 Nitrogen and gaseous air are pure substances.
VAPOR
LIQUID (a) H2O
VAPOR
LIQUID (b) AIR
FIGURE 3–2 A mixture of liquid and gaseous water is a pure substance, but a mixture of liquid and gaseous air is not.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 2 ON THE DVD.
FIGURE 3–3 The molecules in a solid are kept at their positions by the large springlike intermolecular forces.
3–1
■
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. 3–1). 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. 3–2). 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 gaseous 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.
3–2
■
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 different phase. Even though there are three principal phases—solid, liquid, and gas—a 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 pressures. A phase is identified as having a distinct molecular arrangement that is homogeneous throughout and separated from the others by easily identifiable 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 different 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. One reason is that 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. 3–3). Because of the small distances between molecules in a solid, the attractive forces of molecules on each other are large and keep the molecules at fixed positions (Fig. 3–4). Note that the attractive forces between molecules turn to repulsive forces as the
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Chapter 3 distance between the molecules approaches zero, thus preventing the molecules from piling up on top of each other. Even though the molecules in a solid cannot move relative to each other, they continually oscillate about their equilibrium 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. 3–5). 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 experience 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 molecular order is nonexistent. Gas molecules move about at random, continually colliding with each other and the walls of the container they are in. Particularly 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.
3–3
■
113
FIGURE 3–4 In a solid, the attractive and repulsive forces between the molecules tend to maintain them at relatively constant distances from each other. © Reprinted with special permission of King Features Syndicate.
PHASECHANGE PROCESSES OF PURE SUBSTANCES
INTERACTIVE TUTORIAL
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 home owners consider the freezing of water in underground pipes as the most
(a)

(b)
SEE TUTORIAL CH. 3, SEC. 3 ON THE DVD.
(c)
FIGURE 3–5 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.
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Thermodynamics important phasechange process, attention in this section is focused on the liquid 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.
STATE 1
Compressed Liquid and Saturated Liquid P = 1 atm T = 20°C Heat
FIGURE 3–6 At 1 atm and 20°C, water exists in the liquid phase (compressed liquid).
STATE 2
P = 1 atm T = 100°C
Consider a piston–cylinder device containing liquid water at 20°C and 1 atm pressure (state 1, Fig. 3–6). 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, 40°C. 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 100°C (state 2, Fig. 3–7). At this point water is still a liquid, but any heat addition 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 Heat
FIGURE 3–7 At 1 atm pressure and 100°C, water exists as a liquid that is ready to vaporize (saturated liquid).
STATE 3
P = 1 atm T = 100°C
Saturated vapor Saturated liquid
Heat
FIGURE 3–8 As more heat is transferred, part of the saturated liquid vaporizes (saturated liquid–vapor mixture).
Once boiling starts, the temperature stops rising until the liquid is completely 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 100°C 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. 3–8), the cylinder contains 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. 3–9). 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 liquid–vapor mixture since the liquid and vapor phases coexist in equilibrium at these states. Once the phasechange process is completed, we are back to a singlephase region again (this time vapor), and further transfer of heat results in an increase in both the temperature and the specific volume (Fig. 3–10). At state 5, the temperature of the vapor is, let us say, 300°C; and if we transfer some heat from the vapor, the temperature may drop somewhat but no condensation will take place as long as the temperature remains above 100°C
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Chapter 3 (for P 1 atm). A vapor that is not about to condense (i.e., 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. 3–11. 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 100°C. Strictly speaking, the statement “water boils at 100°C” is incorrect. The correct statement is “water boils at 100°C at 1 atm pressure.” The only reason water started boiling at 100°C was because we held the pressure constant at 1 atm (101.325 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 151.8°C. That is, the temperature at which water starts boiling depends on the pressure; 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 temperature, the pressure at which a pure substance changes phase is called the saturation pressure Psat. At a pressure of 101.325 kPa, Tsat is 99.97°C. Conversely, at a temperature of 99.97°C, Psat is 101.325 kPa. (At 100.00°C, Psat is 101.42 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
P=
1a
tm
T, °C
300
Su
pe rh vap e a t e d or
5
2
Saturated mixture
3 4
Com
pres sed liqu id
100
20
1
v
FIGURE 3–11 Tv diagram for the heating process of water at constant pressure.

115
STATE 4
P = 1 atm T = 100°C
Heat
FIGURE 3–9 At 1 atm pressure, the temperature remains constant at 100°C until the last drop of liquid is vaporized (saturated vapor). STATE 5
P = 1 atm T = 300°C
Heat
FIGURE 3–10 As more heat is transferred, the temperature of the vapor starts to rise (superheated vapor).
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Thermodynamics EXPERIMENT
Use actual data from the experiment shown here to obtain the latent heat of fusion of water. See endofchapter problem 3–146. © Ronald Mullisen
TABLE 3–1 Saturation (boiling) pressure of water at various temperatures Temperature, T, °C 10 5 0 5 10 15 20 25 30 40 50 100 150 200 250 300
Saturation pressure, Psat, kPa 0.26 0.40 0.61 0.87 1.23 1.71 2.34 3.17 4.25 7.39 12.35 101.4 476.2 1555 3976 8588
FIGURE 3–12 The liquid–vapor saturation curve of a pure substance (numerical values are for water).
practically all substances. A partial listing of such a table is given in Table 3–1 for water. This table indicates that the pressure of water changing phase (boiling or condensing) at 25°C must be 3.17 kPa, and the pressure of water must be maintained at 3976 kPa (about 40 atm) to have it boil at 250°C. Also, water can be frozen by dropping its pressure below 0.61 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 333.7 kJ/kg and the latent heat of vaporization is 2256.5 kJ/kg. During a phasechange process, pressure and temperature are obviously dependent properties, and there is a definite relation between them, that is, Tsat f (Psat). A plot of Tsat versus Psat, such as the one given for water in Fig. 3–12, is called a liquid–vapor saturation curve. A curve of this kind is characteristic of all pure substances. It is clear from Fig. 3–12 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: 134°C). 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 79.50 kPa, which corresponds to a boiling temperature of 93.3°C as opposed to 100°C at sea level (zero elevation). The variation of the boiling temperature of water with altitude at standard atmospheric conditions is given in Table 3–2. For each 1000 m increase in elevation, the boiling temperature
Psat, kPa
600
400
200
0 0
50
100
150
200 Tsat,°C
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Chapter 3 drops by a little over 3°C. Note that the atmospheric pressure at a location, and thus the boiling temperature, changes slightly with the weather conditions. But the corresponding change in the boiling temperature is no more than about 1°C.
Some Consequences of Tsat and Psat Dependence We mentioned earlier that a substance at a specified pressure boils at the saturation 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. Below 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 25°C. If the can has been in the room long enough, the temperature of the refrigerant in the can is also 25°C. Now, if the lid is opened slowly and some refrigerant is 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 temperature dropping rapidly, and even ice forming outside the can if the air is humid. A thermometer inserted in the can will register 26°C when the pressure drops to 1 atm, which is the saturation temperature of refrigerant134a at that pressure. The temperature of the liquid refrigerant will remain at 26°C 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 kJ/kg 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 26°C indefinitely. The boiling temperature of nitrogen at atmospheric pressure is 196°C (see Table A–3a). This means the temperature of liquid nitrogen exposed to the atmosphere must be 196°C since some nitrogen will be evaporating. The temperature of liquid nitrogen remains constant at 196°C 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 196°C. 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 196°C (Fig. 3–13). 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 swap 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 saturation pressure at the desired low temperature, and evaporating some water

117
TABLE 3–2 Variation of the standard atmospheric pressure and the boiling (saturation) temperature of water with altitude Elevation, m 0 1,000 2,000 5,000 10,000 20,000
Atmospheric Boiling pressure, temperakPa ture, °C 101.33 89.55 79.50 54.05 26.50 5.53
100.0 96.5 93.3 83.3 66.3 34.7
N2 vapor –196°C
Test chamber –196°C
25°C
Liquid N2 –196°C Insulation
FIGURE 3–13 The temperature of liquid nitrogen exposed to the atmosphere remains constant at 196°C, and thus it maintains the test chamber at 196°C.
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Thermodynamics
Temperature °C Start of cooling (25°C, 100 kPa) 25
End of cooling (0°C, 0.61 kPa) 0
0.61 1
3.17
10
100 Pressure (kPa)
FIGURE 3–14 The variation of the temperature of fruits and vegetables with pressure during vacuum cooling from 25C to 0C. Vacuum pump
Insulation
Air + Vapor Low vapor pressure Evaporation High
vapor
pressure
Ice Water
FIGURE 3–15 In 1775, ice was made by evacuating the air space in a water tank.
from the products to be cooled. The heat of vaporization during evaporation is absorbed from the products, which lowers the product temperature. The saturation pressure of water at 0°C is 0.61 kPa, and the products can be cooled to 0°C by lowering the pressure to this level. The cooling rate can be increased by lowering the pressure below 0.61 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 products at ambient temperature, say at 25°C, are loaded into the chamber, and the operation begins. The temperature in the chamber remains constant until the saturation pressure is reached, which is 3.17 kPa at 25°C. 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. 3–14). Vacuum cooling is usually more expensive than the conventional refrigerated cooling, and its use is limited to applications that result in much faster cooling. Products with large surface area per unit mass and a high tendency to release moisture such as lettuce and spinach are wellsuited for vacuum cooling. Products with low surface area to mass 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 successfully by wetting them first. The vacuum cooling just described becomes vacuum freezing if the vapor pressure in the vacuum chamber is dropped below 0.61 kPa, the saturation pressure of water at 0°C. 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. 3–15). 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.
3–4
■
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. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 4 ON THE DVD.
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. 3–11. 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. 3–16, but there are some noticeable differences. First, water starts boiling at a much higher tempera
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Chapter 3

119
T, °C
Pa
M
Pa
22
Pa
P
=
0.
01
M
P
Pa
=
0.
1
M
Pa
P
=
1
M
Pa
P
=
8
373.95
=
M
M
P
6 .0
15
=
Pa
=
P
M
P
Critical point
25
Saturated liquid
Saturated vapor
v, m3/kg
0.003106
FIGURE 3–16 Tv diagram of constantpressure phasechange processes of a pure substance at various pressures (numerical values are for water).
PO
P
A
>
V
P
cr
T
P
cr
R
Critical point
Pc
r
Tcr
P
P
1
Critical point
v
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Chapter 3

121
P
Critical point
tu ra te
line
Sa
SUPERHEATED VAPOR REGION
d
liquid
r
T2 = cons t.
>T
1
e
SATURATED LIQUID –VAPOR REGION
lin
ted
po
Satura
va
COMPRESSED LIQUID REGION
T1
=c
ons
t.
v
the pressure decreases, the volume of the water increases slightly. When the pressure reaches the saturationpressure value at the specified temperature (0.4762 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 f (Psat)], 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. 3–19.
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 solid–liquid and the solid–vapor saturation regions. The basic principles discussed in conjunction with the liquid–vapor phasechange process apply equally to the solid–liquid and solid–vapor phasechange processes. Most substances contract during a solidification (i.e., freezing) process. Others, like water, expand as they freeze. The Pv diagrams for both groups of substances are given in Figs. 3–21 and 3–22. These two diagrams differ only in
FIGURE 3–19 Pv diagram of a pure substance.
P = 1 MPa T = 150°C
Heat
FIGURE 3–20 The pressure in a piston–cylinder device can be reduced by reducing the weight of the piston.
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Thermodynamics P
VAPOR
LIQUID
SOLID
SOLID + LIQUID
Critical point
LIQUID + VAPOR
Triple line
FIGURE 3–21 Pv diagram of a substance that contracts on freezing.
SOLID + VAPOR
v
P
LIQUID
SOLID + LIQUID
Critical point
VAPOR
SOLID
LIQUID + VAPOR
FIGURE 3–22 Pv diagram of a substance that expands on freezing (such as water).
Triple line
SOLID + VAPOR
v
the solid–liquid 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 sun’s
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Chapter 3 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. 3–23). 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 pressure 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 3–3. For water, the triplepoint temperature and pressure are 0.01°C and 0.6117 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 substances that contract on freezing. However,
TABLE 3–3 Triplepoint temperatures and pressures of various substances Substance
Formula
Acetylene Ammonia Argon Carbon (graphite) Carbon dioxide Carbon monoxide Deuterium Ethane Ethylene Helium 4 (l 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
Ttp, K 192.4 195.40 83.81 3900 216.55 68.10 18.63 89.89 104.0 2.19 13.84 158.96 234.2 90.68 24.57 109.50 63.18 182.34 54.36 1825 2045 197.69 1941 337.17 273.16 161.3 692.65
Source: Data from National Bureau of Standards (U.S.) Circ., 500 (1952).
Ptp, kPa 120 6.076 68.9 10,100 517 15.37 17.1 8 104 0.12 5.1 7.04 13.9 1.65 107 11.7 43.2 21.92 12.6 87.85 0.152 3.5 103 2.0 104 1.67 5.3 103 151.7 0.61 81.5 0.065

VAPOR
LIQUID
SOLID
FIGURE 3–23 At triplepoint pressure and temperature, a substance exists in three phases in equilibrium.
123
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Thermodynamics
VAPOR
SOLID
FIGURE 3–24 At low pressures (below the triplepoint value), solids evaporate without melting first (sublimation).
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 pressure at temperatures below 0°C, but it can exist as a liquid at 20°C 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 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 triplepoint value, since a pure substance cannot exist in the liquid phase at those pressures (Fig. 3–24). Passing from the solid phase directly into the vapor phase is called sublimation. For substances that have a triplepoint pressure above the atmospheric pressure such as solid CO2 (dry ice), sublimation is the only way to change from the solid to vapor phase at atmospheric conditions.
3 The PT Diagram Figure 3–25 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 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.
P
Substances that expand on freezing
Substances that contract on freezing
ing
ltin
LIQUID
g
Melt
Me
Critical point
Va
SOLID
p
iz or
ati
on
Triple point VAPOR
S
FIGURE 3–25 PT diagram of pure substances.
li ub
ma
tio
n
T
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Chapter 3

The PvT Surface
Liquid
Solid–Liquid
Solid
Critical point Ga s
Pressure
The state of a simple compressible substance is fixed by any two independent, 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 z(x, y) represents a surface in space, we can represent the PvT behavior of a substance as a surface in space, as shown in Figs. 3–26 and 3–27. Here T and v may be
Liq Va uid– po r Tri ple lin e So Vo
lum
lid
–V a
po
po
r
r
e
Te
Liquid
e mp
So Vo
lid
lum
e
ple
lin
e Va
–V a
po
FIGURE 3–26 PvT surface of a substance that contracts on freezing.
s
Tri
ure
Critical point
Liq Va uid– po r
Solid
rat
Ga
Pressure
Va
po
r
r
Te
er mp
atu
re
FIGURE 3–27 PvT surface of a substance that expands on freezing (like water).
125
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Thermodynamics 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 bird’seye view of this surface. The PvT surfaces present a great deal of information at once, but in a thermodynamic analysis it is more convenient to work with twodimensional diagrams, such as the Pv and Tv diagrams.
3–5
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 5 ON THE DVD.
u1
■
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 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 A–6 and A–6E, for example, list properties 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.
Enthalpy—A Combination Property
P1v1 Control volume u2 P2v2
FIGURE 3–28 The combination u Pv is frequently encountered in the analysis of control volumes.
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 thermodynamics, 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 generation and refrigeration (Fig. 3–28), we frequently encounter the combination of properties u Pv. For the sake of simplicity and convenience, this combination is defined as a new property, enthalpy, and given the symbol h: h u Pv¬¬1kJ>kg 2
(3–1)
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Chapter 3

127
or, H U PV¬¬1kJ2
(3–2)
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 pressure–volume product may differ from the unit of the internal energy by only a factor (Fig. 3–29). For example, it can be easily shown that 1 kPa · m3 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 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 consistent with the modern thermodynamic terminology and were replaced in the 1930s by the term enthalpy (from the Greek word enthalpien, which means to heat).
kPa · m3 ≡ kPa · m3/kg ≡ bar · m3 ≡ MPa · m3 ≡ psi · ft3 ≡
kJ kJ/kg 100 kJ 1000 kJ 0.18505 Btu
FIGURE 3–29 The product pressure volume has energy units.
1a Saturated Liquid and Saturated Vapor States The properties of saturated liquid and saturated vapor for water are listed in Tables A–4 and A–5. Both tables give the same information. The only difference is that in Table A–4 properties are listed under temperature and in Table A–5 under pressure. Therefore, it is more convenient to use Table A–4 when temperature is given and Table A–5 when pressure is given. The use of Table A–4 is illustrated in Fig. 3–30. 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, vf specific volume of saturated liquid vg specific volume of saturated vapor
vfg difference between vg and vf 1that is, vfg vg vf 2
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.
Sat. Temp. press. kPa °C Psat T 85 90 95
Specific volume m3/kg Sat. liquid vf
57.868 0.001032 70.183 0.001036 84.609 0.001040
Specific temperature
Sat. vapor vg 2.8261 2.3593 1.9808
Specific volume of saturated liquid
Corresponding saturation pressure
FIGURE 3–30 A partial list of Table A–4.
Specific volume of saturated vapor
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Thermodynamics
T,°C
EXAMPLE 3–1 T = 90°C
Pressure of Saturated Liquid in a Tank
A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.
Sat. liquid
kP
a
Solution A rigid tank contains saturated liquid water. The pressure and vol
83
ume of the tank are to be determined. Analysis The state of the saturated liquid water is shown on a Tv diagram in Fig. 3–31. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90°C:
7 P=
90
1 0.
P Psat @ 90°C 70.183 kPa¬¬1Table A–42 vf
The specific volume of the saturated liquid at 90°C is
v
v vf @ 90°C 0.001036 m3>kg¬¬1Table A–42
FIGURE 3–31 Schematic and Tv diagram for Example 3–1.
Then the total volume of the tank becomes
V mv 150 kg2 10.001036 m3>kg 2 0.0518 m3
EXAMPLE 3–2 P, psia
A piston–cylinder device contains 2 ft3 of saturated water vapor at 50psia pressure. Determine the temperature and the mass of the vapor inside the cylinder.
Saturated vapor P = 50 psia V = 2 ft3
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. 3–32. Since the cylinder contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this pressure:
T = 280.99F
50
Temperature of Saturated Vapor in a Cylinder
T Tsat @ 50 psia 280.99°F¬¬1Table A–5E2 The specific volume of the saturated vapor at 50 psia is
vg
FIGURE 3–32 Schematic and PV diagram for Example 3–2.
v vg @ 50 psia 8.5175 ft3>lbm¬1Table A–5E2
v
Then the mass of water vapor inside the cylinder becomes
m
EXAMPLE 3–3
V 2 ft3 0.235 lbm v 8.5175 ft3>lbm
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. 3–33. The volume change per unit mass during a vaporization process is vfg, which
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Chapter 3 is the difference between vg and vf. Reading these values from Table A–5 at 100 kPa and substituting yield

129
P, kPa
vfg vg vf 1.6941 0.001043 1.6931 m3>kg Thus,
¢V mvfg 10.2 kg2 11.6931 m >kg 2 0.3386 m 3
3
Sat. liquid P = 100 kPa
Sat. vapor P = 100 kPa
vf
vg
(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 2257.5 kJ/kg for water at 100 kPa. Thus, the amount of energy transferred is
mhfg 10.2 kg2 12257.5 kJ>kg2 451.5 kJ
100
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 1.694100, which is not necessarily the case. It could very well be that vg 1.694138 since this number, too, would truncate to 1.6941. All the digits in our result (1.6931) are significant. But if we did not truncate the result, we would obtain vfg 1.693057, which falsely implies that our result is accurate to the sixth decimal place.
v
FIGURE 3–33 Schematic and Pv diagram for Example 3–3.
1b Saturated Liquid–Vapor Mixture
t a tes uid s
ted liq Satura
s
te
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
sta
mtotal mliquid mvapor mf mg
or
where
ap
(3–3)
dv
mtotal
ate
mvapor
Critical point tur
x
P or T
Sa
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. 3–34). 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:
Sat. vapor Sat. liquid
v
FIGURE 3–34 The relative amounts of liquid and vapor phases in a saturated mixture are specified by the quality x.
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Thermodynamics
Saturated vapor vg
vf Saturated liquid
≡
vavg Saturated liquid–vapor mixture
to imagine that the two phases are mixed well, forming a homogeneous mixture (Fig. 3–35). Then the properties of this “mixture” will simply be the average properties of the saturated liquid–vapor mixture under consideration. Here is how it is done. Consider a tank that contains a saturated liquid–vapor 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 Vf Vg V mv ¡ mtvavg mf vf mgvg
mf mt m g ¡ mtvavg 1m t mg 2vf mgvg
Dividing by mt yields
vavg 11 x2vf xvg
since x mg /mt. This relation can also be expressed as FIGURE 3–35 A twophase system can be treated as a homogeneous mixture for convenience.
AB AC
v avg – v f
A
B
where vfg vg vf. Solving for quality, we obtain x
vavg vf vfg
(3–5)
vavg
u avg u f xu fg¬¬1kJ>kg2
(3–6)
havg h f xh fg¬¬1kJ>kg2
(3–7)
C
vfg vf
(3–4)
Based on this equation, quality can be related to the horizontal distances on a Pv or Tv diagram (Fig. 3–36). At a given temperature or pressure, the numerator of Eq. 3–5 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:
P or T
x=
vavg vf xvfg¬¬1m3>kg2
vg
FIGURE 3–36 Quality is related to the horizontal distances on Pv and Tv diagrams.
v
All the results are of the same format, and they can be summarized in a single equation as y avg y f xy 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. 3–37). 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 A–4 and A–5 in the case of water).
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Chapter 3 EXAMPLE 3–4
Pressure and Volume of a Saturated Mixture

131
P or T
A rigid tank contains 10 kg of water at 90°C. 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.
Sat. liquid vg Sat. liquid vf
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 liquid–vapor mixture is shown in Fig. 3–38. 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 Psat @ 90°C 70.183 kPa¬¬ 1Table A–4 2
(b) At 90°C, we have vf 0.001036 m3/kg and vg 2.3593 m3/kg (Table A–4). One way of finding the volume of the tank is to determine the volume occupied by each phase and then add them:
V Vf Vg mfvf mgvg 18 kg2 10.001036 m3>kg2 12 kg2 12.3593 m3>kg2 4.73 m3
vf
vf < v < vg
vg
v
FIGURE 3–37 The v value of a saturated liquid–vapor mixture lies between the vf and vg values at the specified T or P.
Another way is to first determine the quality x, then the average specific volume v, and finally the total volume:
x
mg mt
T,°C
2 kg 0.2 10 kg
v vf xvfg
T = 90°C mg = 2 kg
0.001036 m >kg 10.22 3 12.3593 0.001036 2 m >kg4 3
3
0.473 m3>kg
m f = 8 kg
and
V mv 110 kg2 10.473 m >kg 2 4.73 m 3
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.
0. P =7
3 18
kP
a
90
vf = 0.001036 vg = 2.3593 v, m 3/kg
EXAMPLE 3–5
Properties of Saturated Liquid–Vapor 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 liquid–vapor mixture is shown in Fig. 3–39. At this point we do not know whether the refrigerant is in the compressed liquid, superheated vapor, or saturated mixture region. This can
FIGURE 3–38 Schematic and Tv diagram for Example 3–4.
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Thermodynamics
P, kPa
be determined by comparing a suitable property to the saturated liquid and saturated vapor values. From the information given, we can determine the specific volume:
R134a P = 160 kPa m = 4 kg
v
0.080 m3 V 0.02 m3>kg m 4 kg
At 160 kPa, we read
vf 0.0007437 m3>kg
vg 0.12348 m3>kg
T = 15.60C
160
m3/kg
vf = 0.0007437 vg = 0.12348 v, hg = 241.11 h, kJ/kg h f = 31.21
FIGURE 3–39 Schematic and Pv diagram for Example 3–5.
(Table A–12)
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 Tsat @ 160 kPa 15.60°C (b) Quality can be determined from
x
v vf vfg
0.02 0.0007437 0.157 0.12348 0.0007437
(c) At 160 kPa, we also read from Table A–12 that hf 31.21 kJ/kg and hfg 209.90 kJ/kg. Then,
h hf xhfg
31.21 kJ>kg 10.157 2 1209.90 kJ>kg 2 64.2 kJ>kg
(d) The mass of the vapor is
mg xmt 10.1572 14 kg2 0.628 kg
and the volume occupied by the vapor phase is
Vg mgvg 10.628 kg2 10.12348 m3>kg 2 0.0775 m3 1or 77.5 L2 The rest of the volume (2.5 L) is occupied by the liquid. v
1300
u kJ/kg
h kJ/kg
7.2605 4687.2
…
P = 0.1 MPa (99.61°C) 1.6941 2505.6 2675.0 1.6959 2506.2 2675.8 1.9367 2582.9 2776.6
…
…
Sat. 100 150
m3/kg
…
T,°C
5413.3
P = 0.5 MPa (151.83°C) Sat. 0.37483 2560.7 200 0.42503 2643.3 250 0.47443 2723.8
FIGURE 3–40 A partial listing of Table A–6.
2748.1 2855.8 2961.0
Property tables are also available for saturated solid–vapor mixtures. Properties of saturated ice–water vapor mixtures, for example, are listed in Table A–8. Saturated solid–vapor mixtures can be handled just as saturated liquid–vapor 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 superheated vapor tables is illustrated in Fig. 3–40. 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.
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Chapter 3

133
Compared to saturated vapor, superheated vapor is characterized by Lower pressures (P Psat at a given T) Higher tempreatures (T Tsat at a given P) Higher specific volumes (v vg at a given P or T) Higher internal energies (u ug at a given P or T) Higher enthalpies (h hg at a given P or T)
EXAMPLE 3–6
Internal Energy of Superheated Vapor
Determine the internal energy of water at 20 psia and 400°F.
Solution The internal energy of water at a specified state is to be determined. Analysis At 20 psia, the saturation temperature is 227.92°F. Since T Tsat, the water is in the superheated vapor region. Then the internal energy at the given temperature and pressure is determined from the superheated vapor table (Table A–6E) to be
u 1145.1 Btu>lbm
a
Determine the temperature of water at a state of P 0.5 MPa and h 2890 kJ/kg.
T MP
Temperature of Superheated Vapor
0.5
EXAMPLE 3–7
Solution The temperature of water at a specified state is to be determined. Analysis At 0.5 MPa, the enthalpy of saturated water vapor is hg 2748.1 kJ/kg. Since h hg, as shown in Fig. 3–41, we again have superheated vapor. Under 0.5 MPa in Table A–6 we read T, °C
h, kJ/kg
200 250
2855.8 2961.0
Obviously, the temperature is between 200 and 250°C. By linear interpolation it is determined to be
T 216.3°C
3 Compressed Liquid Compressed liquid tables are not as commonly available, and Table A–7 is the only compressed liquid table in this text. The format of Table A–7 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 properties to change less than 1 percent.
hg h > hg
h
FIGURE 3–41 At a specified P, superheated vapor exists at a higher h than the saturated vapor (Example 3–7).
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134

Thermodynamics In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature (Fig. 3–42). This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure. Thus,
Given: P and T = vf @T v ~ u~ = uf @T ~ hf @T h=
y ≅ yf @ T
FIGURE 3–42 A compressed liquid may be approximated as a saturated liquid at the given temperature.
(3–8)
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 above approximation results in negligible error in v and u, the error in h may reach undesirable levels. However, the error in h at low to moderate pressures and temperatures can be reduced significantly by evaluating it from h ≅ hf @ T vf @ T (P Psat @T)
(3–9)
instead of taking it to be just hf . Note, however, that the approximation in Eq. 3–9 does not yield any significant improvement at moderate to high temperatures and pressures, and it may even backfire and result in greater error due to overcorrection at very high temperatures and pressures (see Kostic, Ref. 4). In general, a compressed liquid is characterized by Higher pressures (P Psat at a given T) Lower tempreatures (T Tsat at a given P) Lower specific volumes (v vf at a given P or T) Lower internal energies (u uf at a given P or T) Lower enthalpies (h hf at a given P or T)
T, °C
But unlike superheated vapor, the compressed liquid properties are not much different from the corresponding saturated liquid values. T = 80°C P= 5 MPa
EXAMPLE 3–8
Approximating Compressed Liquid as Saturated Liquid
5M
Pa
Determine the internal energy of compressed liquid water at 80°C 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
80
u ≅ uf @ 80°C
FIGURE 3–43 Schematic and Tu diagram for Example 3–8.
u
water are to be determined. Analysis At 80°C, the saturation pressure of water is 47.416 kPa, and since 5 MPa Psat, we obviously have compressed liquid, as shown in Fig. 3–43. (a) From the compressed liquid table (Table A–7)
P 5 MPa f u 333.82 kJ>kg T 80°C (b) From the saturation table (Table A–4), we read
u uf @ 80°C 334.97 kJ>kg The error involved is
334.97 333.82 100 0.34% 333.82 which is less than 1 percent.
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Chapter 3
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 properties. 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 properties at that state. For water, the state of saturated liquid at 0.01°C is taken 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 40°C 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 3–9
The Use of Steam Tables to Determine Properties
Determine the missing properties and the phase descriptions in the following table for water: T, °C (a) (b) (c) (d) (e)
P, kPa
u, kJ/kg
200 125 75
1000 500 850
x
Phase description
0.6 1600 2950 0.0
Solution Properties and phase descriptions of water are to be determined at various states. Analysis (a) The quality is given to be x 0.6, 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 liquid–vapor mixture at a pressure of 200 kPa. Then the temperature must be the saturation temperature at the given pressure:
T Tsat @ 200 kPa 120.21°C¬¬1Table A–5 2
At 200 kPa, we also read from Table A–5 that uf 504.50 kJ/kg and ufg 2024.6 kJ/kg. Then the average internal energy of the mixture is
u uf xufg
504.50 kJ>kg 10.6 2 12024.6 kJ>kg2 1719.26 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

135
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Thermodynamics saturation table (Table A–4) and determine the uf and ug values at the given temperature. At 125°C, we read uf 524.83 kJ/kg and ug 2534.3 kJ/kg. Next we compare the given u value to these uf and ug values, keeping in mind that
if¬¬u 6 uf¬¬
we have compressed liquid
if¬¬uf u ug¬¬we have saturated mixture if¬¬u 7 ug¬¬
we have superheated vapor
In our case the given u value is 1600, which falls between the uf and ug values at 125°C. Therefore, we have saturated liquid–vapor mixture. Then the pressure must be the saturation pressure at the given temperature:
P Psat @ 125°C 232.23 kPa¬¬1Table A–42 The quality is determined from
x
u uf ufg
1600 524.83 0.535 2009.5
The criteria above 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 761.39 kJ/kg and ug 2582.8 kJ/kg. The specified u value is 2950 kJ/kg, 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 395.2°C¬¬1Table A–6 2
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 A–5) and determine the saturation temperature value at the given pressure. At 500 kPa, we have Tsat 151.83°C. We then compare the given T value to this Tsat value, keeping in mind that
50
kP
a
T, °C
P
=
if¬¬T 6 Tsat @ given P¬¬we have compressed liquid if¬¬T Tsat @ given P ¬¬we have saturated mixture
151.83 75
if¬¬T 7 Tsat @ given P¬¬we have superheated vapor
~u u= f @ 75°C
FIGURE 3–44 At a given P and T, a pure substance will exist as a compressed liquid if T Tsat @ P.
u
In our case, the given T value is 75°C, which is less than the Tsat value at the specified pressure. Therefore, we have compressed liquid (Fig. 3–44), 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 compressed 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 uf @ 75°C 313.99 kJ>kg¬¬1Table A–4 2
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Chapter 3

137
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 Tsat @ 850 kPa 172.94°C
u u f @ 850 kPa 731.00 kJ>kg¬¬1Table A–5 2
3–6
■
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 properties 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, experimentally determined that at low pressures the volume of a gas is proportional to its temperature. That is,
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 6 ON THE DVD.
T P Ra b v
or Pv RT
(3–10)
where the constant of proportionality R is called the gas constant. Equation 3–10 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. 3–45) and is determined from Ru R ¬¬1kJ>kg # K or kPa # m3>kg # K 2 M
where Ru is the universal gas constant and M is the molar mass (also
Substance Air Helium Argon Nitrogen
R,, kJ/kg·K kJ/kg 0.2870 2.0769 0.2081 0.2968
FIGURE 3–45 Different substances have different gas constants.
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Thermodynamics
Ru
Per unit mass m3/kg
v, u, kJ/kg h, kJ/kg
Per unit mole
v, m3/kmol u , kJ/kmol h , kJ/kmol
789
called molecular weight) of the gas. The constant Ru is the same for all substances, and its value is 8.31447 kJ>kmol # K 8.31447 kPa # m3>kmol # K 0.0831447 bar # m3>kmol # K 1.98588 Btu>lbmol # R 10.7316 psia # ft3>lbmol # R 1545.37 ft # lbf>lbmol # R
(3–11)
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 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 kg/kmol 28 lbm/lbmol. The mass of a system is equal to the product of its molar mass M and the mole number N: m MN¬¬1kg2
(3–12)
The values of R and M for several substances are given in Table A–1. The idealgas equation of state can be written in several different forms: FIGURE 3–46 Properties per unit mole are denoted with a bar on the top.
V mv ¡ PV mRT mR 1MN 2R NR u ¡ PV NR uT V Nv ¡ Pv R uT
(3–13) (3–14) (3–15)
where v is the molar specific volume, that is, the volume per unit mole (in m3/kmol or ft3/lbmol). A bar above a property denotes values on a unitmole basis throughout this text (Fig. 3–46). By writing Eq. 3–13 twice for a fixed mass and simplifying, the properties of an ideal gas at two different states are related to each other by P1V1 P2V2 T1 T2
FIGURE 3–47 The idealgas relation often is not applicable to real gases; thus, care should be exercised when using it. © Reprinted with special permission of King Features Syndicate.
(3–16)
An ideal gas is an imaginary substance that obeys the relation Pv RT (Fig. 3–47). 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 temperatures, 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, krypton, and even heavier gases such as carbon dioxide 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.
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Chapter 3 EXAMPLE 3–10
Mass of Air in a Room

139
6m
Determine the mass of the air in a room whose dimensions are 4 m 5 m 6 m at 100 kPa and 25°C.
Solution The mass of air in a room is to be determined. Analysis A sketch of the room is given in Fig. 3–48. Air at specified conditions can be treated as an ideal gas. From Table A–1, the gas constant of air is R 0.287 kPa · m3/kg · K, and the absolute temperature is T 25°C 273 298 K. The volume of the room is
4m AIR P = 100 kPa T = 25°C m=?
5m
V 14 m2 15 m2 16 m2 120 m3
The mass of air in the room is determined from the idealgas relation to be
1100 kPa2 1120 m 2 PV 140.3 kg RT 10.287 kPa # m3>kg # K2 1298 K2 3
m
FIGURE 3–48 Schematic for Example 3–10.
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. 3–49. 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 (less than 0.1 percent). At higher pressures, however, the idealgas assumption yields unacceptable errors, particularly in the vicinity of the critical point and the saturated vapor line (over 100 percent). 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, idealgas relations should not be used.
3–7
■
COMPRESSIBILITY FACTOR—A MEASURE OF DEVIATION FROM IDEALGAS BEHAVIOR
The idealgas equation is very simple and thus very convenient to use. However, as illustrated in Fig. 3–49, gases deviate from idealgas behavior significantly 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
(3–17)
or Pv ZRT
(3–18)
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 7 ON THE DVD.
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Thermodynamics T, °C 10.8 5.0 2.4
17.3
600
500
37.1
0.5
4.1
20.8 8.8
0.8
0.1
0.0
IDEAL 271.0
17.6
56.2
7.4
1.3
0.0
GAS
0.1
0.0
0.0
30
MP
a
400
0.0
0.0
0.0
152.7 20 MPa 10 MPa
49.5
300 5 MPa
200
FIGURE 3–49 Percentage of error ([vtable videal/vtable] 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.
100
16.7
2.6
0.2
0.0
0.0
25.7
6.0 7.6
1 MPa
0.5
100 kPa
0.0
1.6
0.0
0.0
10 kPa
0.0
0.1
0.8 kPa 0 0.001
0.01
0.0 0.1
1
10
100
v, m3/kg
It can also be expressed as Z
IDEAL GAS
REAL GASES
Z=1
>1 Z = 1 kg # K
v
FIGURE 3–53 Gases deviate from the idealgas behavior the most in the neighborhood of the critical point.
Pcr 4.059 MPa Tcr 374.2 K
(a) The specific volume of refrigerant134a under the idealgas assumption is
v
10.0815 kPa # m3>kg # K 2 1323 K2 RT 0.026325 m3>kg P 1000 kPa
Therefore, treating the refrigerant134a vapor as an ideal gas would result in an error of (0.026325 0.021796)/0.021796 0.208, or 20.8 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 1 MPa 0.246 Pcr 4.059 MPa ∂ ¬Z 0.84 T 323 K TR 0.863 Tcr 374.2 K
PR
Thus
v Zvideal 10.84 2 10.026325 m3>kg2 0.022113 m3>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.
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Chapter 3 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 property called the pseudoreduced specific volume vR as vactual vR RTcr>Pcr
(3–21)
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. 3–54).
EXAMPLE 3–12
PR = P Pcr
Using Generalized Charts to Determine Pressure
Determine the pressure of water vapor at 600°F and 0.51431 ft3/lbm, using (a) the steam tables, (b) the idealgas equation, and (c) the generalized compressibility chart.
vR =
v RTcr /P cr

Z =… (Fig. A–15)
FIGURE 3–54 The compressibility factor can also be determined from a knowledge of PR and vR.
Solution The pressure of water vapor is to be determined in three different ways. Analysis A sketch of the system is given in Fig. 3–55. The gas constant, the critical pressure, and the critical temperature of steam are determined from Table A–1E to be
R 0.5956 psia # ft3>lbm # R
Pcr 3200 psia Tcr 1164.8 R
(a) The pressure at the specified state is determined from Table A–6E to be
v 0.51431 ft >lbm f ¬P 1000 psia T 600°F 3
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
10.5956 psia # ft3>lbm # R 2 11060 R 2 RT 1228 psia v 0.51431 ft3>lbm
Therefore, treating the steam as an ideal gas would result in an error of (1228 1000)/1000 0.228, or 22.8 percent in this case. (c) To determine the correction factor Z from the compressibility chart (Fig. A–15), we first need to calculate the pseudoreduced specific volume and the reduced temperature:
10.51431 ft3>lbm2 13200 psia2 vactual 2.372 RTcr>Pcr 10.5956 psia # ft3>lbm # R2 11164.8 R 2 ∂ ¬PR 0.33 T 1060 R TR 0.91 Tcr 1164.8 R vR
143
H 2O T = 600°F v = 0.51431 ft3/lbm P=?
FIGURE 3–55 Schematic for Example 3–12.
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Thermodynamics Thus,
P, psia Exact Z chart Ideal gas
1000 1056 1228
P PRPcr 10.33 2 13200 psia2 1056 psia
Discussion Using the compressibility chart reduced the error from 22.8 to 5.6 percent, which is acceptable for most engineering purposes (Fig. 3–56). 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.
(from Example 312)
FIGURE 3–56 Results obtained by using the compressibility chart are usually within a few percent of actual values.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 8 ON THE DVD.
3–8
■
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. 3–57), but we shall discuss only three: the van der Waals equation because it is one of the earliest, the BeattieBridgeman equation of state because it is one of the best known and is reasonably accurate, 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 aP
van der Waals Berthelet RedlichKwang BeattieBridgeman BenedictWebbRubin Strobridge Virial
FIGURE 3–57 Several equations of state have been proposed throughout history.
a b 1v b 2 RT v2
(3–22)
Van der Waals intended to improve the idealgas equation of state by including two of the effects not considered in the idealgas model: the intermolecular attraction forces and the volume occupied by the molecules themselves. The term a/v 2 accounts for the intermolecular forces, and b accounts for the volume occupied by the gas molecules. In a room at atmospheric pressure and temperature, the volume actually occupied by molecules is only about onethousandth 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. 3–58). Thus, the first and the second derivatives of P with respect to v at the critical point must be zero. That is, a
0P 0 2P 0¬and¬a 2 b 0 b 0v TTcrconst 0v TTcrconst
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Chapter 3 By performing the differentiations and eliminating vcr, the constants a and b are determined to be a
27R 2 T 2cr RTcr ¬and¬b 64Pcr 8Pcr
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 R uT v2
a1
c A b 1v B2 2 v T3 v
(3–24)
where A A0 a 1
a b b ¬and¬B B0 a 1 b v v
(3–25)
The constants appearing in the above equation are given in Table 3–4 for various substances. The BeattieBridgeman equation is known to be reasonably accurate for densities up to about 0.8rcr, where rcr 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
C0 1 R uT bR uT a g aa c 2 a B0R uT A 0 2 b 2 6 3 2 a 1 2 b e g>v 3 v T v v v T v v (3–26)
The values of the constants appearing in this equation are given in Table 3–4. This equation can handle substances at densities up to about 2.5rcr. In 1962, Strobridge further extended this equation by raising the number of constants to 16 (Fig. 3–59).
Virial Equation of State The equation of state of a substance can also be expressed in a series form as P
a 1T2 b 1T2 c 1T2 d 1T2 RT 2 3 4 5 ... v v v v v
145
P
(3–23)
The constants a and b can be determined for any substance from the criticalpoint data alone (Table A–1). 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 replacing the v in Eq. 3–22 by v and the R in Eqs. 3–22 and 3–23 by Ru.
P

(3–27)
Critical point T
cr
=c
on
sta
nt
v
FIGURE 3–58 Critical isotherm of a pure substance has an inflection point at the critical state.
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Thermodynamics
TABLE 3–4 Constants that appear in the BeattieBridgeman and the BenedictWebbRubin equations of state (a) When P is in kPa, v– is in m3/kmol, T is in K, and Ru 8.314 kPa · m3/kmol · K, the five constants in the BeattieBridgeman equation are as follows: Gas
A0
Air Argon, Ar Carbon dioxide, CO2 Helium, He Hydrogen, H2 Nitrogen, N2 Oxygen, O2
a
131.8441 130.7802 507.2836 2.1886 20.0117 136.2315 151.0857
B0
0.01931 0.02328 0.07132 0.05984 0.00506 0.02617 0.02562
0.04611 0.03931 0.10476 0.01400 0.02096 0.05046 0.04624
b
c
0.001101 0.0 0.07235 0.0 0.04359 0.00691 0.004208
4.34 104 5.99 104 6.60 105 40 504 4.20 104 4.80 104
Source: Gordon J. Van Wylen and Richard E. Sonntag, Fundamentals of Classical Thermodynamics, English/SI Version, 3rd ed. (New York: John Wiley & Sons, 1986), p. 46, table 3.3.
(b) When P is in kPa, v– is in m3/kmol, T is in K, and Ru 8.314 kPa · m3/kmol · K, the eight constants in the BenedictWebbRubin equation are as follows: Gas
a
nButane, 190.68 C4H10 Carbon dioxide, CO2 13.86 Carbon monoxide, CO 3.71 Methane, CH4 5.00 Nitrogen, N2 2.54
A0
b
B0
c
0.039998
0.12436
3.205 107
1.006 108 1.101 103
0.0340
277.30
0.007210
0.04991
1.511 106
1.404 107 8.470 105
0.00539
135.87 187.91 106.73
0.002632 0.003380 0.002328
0.05454 0.04260 0.04074
1.054 105 2.578 105 7.379 104
8.673 105 1.350 104 2.286 106 1.244 104 8.164 105 1.272 104
0.0060 0.0060 0.0053
1021.6
C0
a
g
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.
van der Waals: 2 constants. Accurate over a limited range. BeattieBridgeman: 5 constants. Accurate for ρ < 0.8rcr. BenedictWebbRubin: 8 constants. Accurate for ρ < 2.5rcr. Strobridge: 16 constants. More suitable for computer calculations. Virial: may vary. Accuracy depends on the number of terms used.
FIGURE 3–59 Complex equations of state represent the PvT behavior of gases more accurately over a wider range.
This and similar equations are called the virial equations of state, and the coefficients a(T), b(T), c(T), 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 equation 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 liquid–vapor mixtures. Complex equations represent the PvT behavior of substances reasonably well and are very suitable for digital computer applications. For hand calculations, 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,
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Chapter 3

147
4.7% 0.2% 0.2%
300
3.7% 0.1% 0.4%
2.9% 0.3% 0.7%
200
4.2% 0.1% 0.2%
5.3% 0.1% 0.1%
6.7% 0.7% 0.1%
11.6% 6.3% 1.2%
2.3% 0.1% 0.0%
2.8% 0.1% 0.1%
3.2% 0.1% 1.0%
1.0% 0.1% 0.1%
1.1% 0.1% 0.0%
1.2% 0.1% 0.1%
0.4% 0.1% 0.4%
a
a
0.2 MP
0.1 MP
1 MPa
2 MPa
1.9% 0.1% 0.1%
0.5% 0.0% 0.0%
0.1% 0.0% 0.0%
0.5% 0.1% 0.0%
0.1% 0.0% 0.0%
0.5% 0.0% 0.0%
0.1% 0.0% 0.0%
0.1% 0.0% 0.2%
0.0% 0.0% 0.0%
0.0% 0.0% 0.0%
Van der Waals (top) BeattieBridgeman (middle) BenedictWebbRubin (bottom)
0.0% 0.0% 0.0%
0.0% 0.0% 0.0%
0.0% 0.0% 0.0%
>100% >100% >100%
5.7% 59.3% 18.7%
0 0.01
0.1
15.2% 74.5% 51.0%
7.9% 0.7% 5.2%
5.2% 0.6% 3.7%
0.9% 0.1% 0.1% 3.3% 0.4% 2.5%
0.4% 0.1% 0.1%
1.6% 0.2% 1.3%
1
10
0.0
100
125
MP
a
20.7% 14.1% 2.1%
4 MPa
20 MPa
10 MPa
T, K
0.8% 0.4% 0.1% 0.1% 0.8% 0.3%
100
v, m3/kmol
FIGURE 3–60 Percentage of error involved in various equations of state for nitrogen (% error [(vtable vequation)/vtable] 100).
BeattieBridgeman, and BenedictWebbRubin equations of state is illustrated in Fig. 3–60. It is apparent from this figure that the BenedictWebbRubin equation of state is usually the most accurate.
EXAMPLE 3–13
Different Methods of Evaluating Gas Pressure
Predict the pressure of nitrogen gas at T 175 K and v 0.00375 m3/kg 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 10,000 kPa.
Solution The pressure of nitrogen gas is to be determined using four different equations of state.
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Thermodynamics Properties The gas constant of nitrogen gas is 0.2968 kPa m3/kg K (Table A–1). Analysis (a) Using the idealgas equation of state, the pressure is found to be
P
10.2968 kPa # m3>kg # K 2 1175 K2 RT 13,851 kPa v 0.00375 m3>kg
which is in error by 38.5 percent. (b) The van der Waals constants for nitrogen are determined from Eq. 3–23 to be
a 0.175 m6 # kPa>kg2
b 0.00138 m3>kg From Eq. 3–22,
P
RT a 2 9471 kPa vb v
which is in error by 5.3 percent. (c) The constants in the BeattieBridgeman equation are determined from Table 3–4 to be
A 102.29 B 0.05378 c 4.2 104 Also, v– Mv (28.013 kg/mol)(0.00375 m3/kg) 0.10505 m3/kmol. Substituting these values into Eq. 3–24, we obtain
P
RuT v2
a1
c A b 1v B2 2 10,110 kPa vT3 v
which is in error by 1.1 percent. (d) The constants in the BenedictWebbRubin equation are determined from Table 3–4 to be
a 2.54
A0 106.73
b 0.002328 c 7.379 10
B0 0.04074 4
C0 8.164 105
a 1.272 104 g 0.0053 Substituting these values into Eq. 3–26 gives
P
C0 1 R uT bR uT a a B0R uT A 0 2 b 2 v T v v3
g 2 c aa 3 2 a 1 2 b e g>v 6 v T v v
10,009 kPa which is in error by only 0.09 percent. Thus, the accuracy of the BenedictWebbRubin equation of state is rather impressive in this case.
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Chapter 3 TOPIC OF SPECIAL INTEREST*

149
Vapor Pressure and Phase Equilibrium
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 proportional to the average velocity of the molecules and the number of molecules per unit volume of the container (i.e., molar density). Therefore, the pressure 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 moisture content) and water vapor (also referred to as moisture), and the atmospheric pressure is the sum of the pressure of dry air Pa and the pressure of water vapor, called the vapor pressure Pv (Fig. 3–61). That is, Patm Pa Pv
(3–29)
where Psat @ T is the saturation pressure of water at the specified temperature. For example, the vapor pressure of air at 25°C and 60 percent relative humidity is Pv fPsat @ 25°C 0.6 13.17 kPa2 1.90 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 saturation 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 *This section can be skipped without a loss in continuity.
Air
Water vapor
(3–28)
(Note that in some applications, the phrase “vapor pressure” is used to indicate saturation pressure.) The vapor pressure constitutes a small fraction (usually 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. 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 f. 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 25°C is 3.17 kPa. The amount of moisture in the air is completely specified by the temperature and the relative humidity, and the vapor pressure is related to relative humidity f by Pv fPsat @ T
Patm = Pa + Pv
FIGURE 3–61 Atmospheric pressure is the sum of the dry air pressure Pa and the vapor pressure Pv.
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Thermodynamics
Water
Salty water
Salt (a) Before
(b) After
FIGURE 3–62 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 low concentration region.
Pv
Water vapor
Liquid water T
FIGURE 3–63 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.
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 electronic devices such as camcorders come with warnings against bringing them into moist indoors when the devices are cold to avoid moisture condensation on the sensitive electronics of the devices. It is a common observation 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 commodity, 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. 3–62). 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 eventually 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 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 saturation pressure of water at the water temperature. That is (Fig. 3–63), Phase equilibrium criterion for water exposed to air: Pv Psat @ T
(3–30)
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 pressure 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 equilibrium with the water vapor in the surrounding air forms the basis for the
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Chapter 3

operation of the 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 consumes 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 liquid–vapor interface when the vapor pressure is less than the saturation pressure of the liquid at a given temperature. Water in a lake at 20°C, for example, evaporates to air at 20°C and 60 percent relative humidity since the saturation pressure of water at 20°C is 2.34 kPa, and the vapor pressure of air at 20°C and 60 percent relative humidity is 1.4 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 evaporation involves no bubble formation or bubble motion (Fig. 3–64). Boiling, on the other hand, occurs at the solid–liquid 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 110°C boils since the saturation temperature of water at 1 atm is 100°C. The boiling process is characterized by the rapid motion of vapor bubbles that form at the solid– liquid 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.
© Vol. 30/PhotoDisc
© Vol. 93/PhotoDisc
FIGURE 3–64 A liquidtovapor phase change process is called evaporation if it occurs at a liquid–vapor interface, and boiling if it occurs at a solid–liquid interface.
151
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Thermodynamics Air 25°C φ = 10% Pv
EXAMPLE 3–14
Pv = Psat @ T
T Lake
FIGURE 3–65 Schematic for Example 3–14.
Temperature Drop of a Lake Due to Evaporation
On a summer day, the air temperature over a lake is measured to be 25°C. Determine 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. 3–65).
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 25°C, from Table 3–1, is 3.17 kPa. Then the vapor pressures at relative humidities of 10, 80, and 100 percent are determined from Eq. 3–29 to be
Relative humidity 10%:¬¬Pv1 f1Psat @ 25°C 0.1 13.17 kPa2 0.317 kPa
Relative humidity 80%:¬¬Pv 2 f2Psat @ 25°C 0.8 13.17 kPa2 2.536 kPa
Relative humidity 100%:¬¬Pv3 f3Psat @25°C 1.0 13.17 kPa2 3.17 kPa
The saturation temperatures corresponding to these pressures are determined from Table 3–1 (or Table A–5) by interpolation to be
T1 8.0°C¬T2 21.2°C¬and¬T3 25°C 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 25°C, and you are right. The water temperature drops to 8°C 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 8°C 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 relatively cool.
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153
SUMMARY A substance that has a fixed chemical composition throughout 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 pressure, 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 liquid–vapor mixture, the mass fraction of vapor is called the quality and is expressed as x
mvapor mtotal
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 determined from
Real gases exhibit idealgas behavior at relatively low pressures and high temperatures. The deviation from idealgas behavior can be properly accounted for by using the compressibility factor Z, defined as Z
The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure, which are defined as TR
vR
Pv RT where R is the gas constant. Caution should be exercised in using this relation since an ideal gas is a fictitious substance.
vactual RTcr>Pcr
The PvT behavior of substances can be represented more accurately by more complex equations of state. Three of the best known are aP
van der Waals:
a b 1v b 2 RT v2
where a
y yf @ 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 liquid 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 simplest and bestknown equation of state is the idealgas equation of state, given as
T P ¬and¬PR Tcr Pcr
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 pseudoreduced specific volume, defined as
y yf xyfg where f stands for saturated liquid and g for saturated vapor. In the absence of compressed liquid data, a general approximation is to treat a compressed liquid as a saturated liquid at the given temperature,
vactual Pv ¬or¬Z RT videal
27R 2 T 2cr RTcr ¬and¬b 64Pcr 8Pcr
BeattieBridgeman: P
RuT v2
a1
c A b 1v B2 2 vT3 v
where A A0 a 1
a b b ¬and¬B B0 a 1 b v v
BenedictWebbRubin: C0 1 RuT bRuT a aa P a B0RuT A0 2 b 2 6 v v T v v3
g c 2 a 1 2 b e g>v 2 v T v 3
where Ru is the universal gas constant and v– is the molar specific volume.
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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. 2nd ed. New York: Wiley, 1997. 4. M. Kostic. Analysis of Enthalpy Approximation for Compressed Liquid Water. IMECE 2004, ASME Proceedings, ASME, New York, 2004.
PROBLEMS* Pure Substances, PhaseChange Processes, Property Diagrams
3–12C How does the boiling process at supercritical pressures differ from the boiling process at subcritical pressures?
3–1C Is iced water a pure substance? Why?
Property Tables
3–2C What is the difference between saturated liquid and compressed liquid?
3–13C 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.
3–3C What is the difference between saturated vapor and superheated vapor? 3–4C Is there any difference between the intensive properties of saturated vapor at a given temperature and the vapor of a saturated mixture at the same temperature? 3–5C Is there any difference between the intensive properties of saturated liquid at a given temperature and the liquid of a saturated mixture at the same temperature? 3–6C Is it true that water boils at higher temperatures at higher pressures? Explain. 3–7C If the pressure of a substance is increased during a boiling process, will the temperature also increase or will it remain constant? Why? 3–8C Why are the temperature and pressure dependent properties in the saturated mixture region? 3–9C What is the difference between the critical point and the triple point? 3–10C Is it possible to have water vapor at 10°C? 3–11C 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?
*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 a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
3–14C 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. 3–15C It is well known that warm air in a cooler environment 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? 3–16C In 1775, Dr. William Cullen made ice in Scotland by evacuating the air in a water tank. Explain how that device works, and discuss how the process can be made more efficient. 3–17C Does the amount of heat absorbed as 1 kg of saturated liquid water boils at 100°C have to be equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100°C? 3–18C Does the reference point selected for the properties of a substance have any effect on thermodynamic analysis? Why? 3–19C What is the physical significance of hfg? Can it be obtained from a knowledge of hf and hg? How? 3–20C Is it true that it takes more energy to vaporize 1 kg of saturated liquid water at 100°C than it would at 120°C? 3–21C What is quality? Does it have any meaning in the superheated vapor region? 3–22C 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? 3–23C Does hfg change with pressure? How?
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Chapter 3 3–24C Can quality be expressed as the ratio of the volume occupied by the vapor phase to the total volume? Explain. 3–25C In the absence of compressed liquid tables, how is the specific volume of a compressed liquid at a given P and T determined? 3–26
Complete this table for H2O:
3–33E
Complete this table for refrigerant134a:
T, °F
P, psia 80
15 10
h, Btu/lbm
P, kPa
50 250 110
v, m3/kg
Phase description
4.16
3–34 T, °C
200 400 600
Saturated vapor
3–27
Reconsider Prob. 3–26. Using EES (or other) software, determine the missing properties of water. Repeat the solution for refrigerant134a, refrigerant22, and ammonia.
3–35 T, °C
3–28E Complete this table for H2O:
129.46
T, °F
P, psia
300 500 400
u, Btu/lbm
Phase description
782 40 120 400
Saturated liquid
3–29E
Reconsider Prob. 3–28E. Using EES (or other) software, determine the missing properties of water. Repeat the solution for refrigerant134a, refrigerant22, and ammonia.
3–30 T, °C
P, kPa
h, kJ/kg
200
x
Phase description 0.7
950 500 800
T, °C
P, kPa
8 30
320
v,
0.05 550 750
Saturated liquid 0.140
Complete this table for H2O: P, kPa
u, kJ/kg
Phase description
1450 Saturated vapor
2500 4000
3040
3–36 A 1.8m3 rigid tank contains steam at 220°C. 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.
Steam 1.8 m3 220°C
Phase description
3–37 A piston–cylinder device contains 0.85 kg of refrigerant134a at 10°C. 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
0.015 180 600
Saturated vapor
Complete this table for refrigerant134a: P, kPa
20 12 8
Phase description
FIGURE P3–36
3162.2
m3/kg
T, °C
v, m3/kg
0.0
Complete this table for refrigerant134a:
3–32
P, kPa
1800
3–31
80
Complete this table for H2O:
Complete this table for H2O:
140 80
1.0
400 220 190
Phase description
0.6 70 180
140 125 500
155
78
110 T, °C
x

u, kJ/kg
Phase description
95
Q R134a 0.85 kg –10°C
Saturated liquid 400 600
300
FIGURE P3–37
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until the temperature is 15°C. Determine (a) the final pressure, (b) the change in the volume of the cylinder, and (c) the change in the enthalpy of the refrigerant134a.
60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water.
3–38E The temperature in a pressure cooker during cooking at sea level is measured to be 250°F. Determine the absolute pressure inside the cooker in psia and in atm. Would you modify your answer if the place were at a higher elevation?
3–42 Repeat Prob. 3–41 for a location at an elevation of 1500 m where the atmospheric pressure is 84.5 kPa and thus the boiling temperature of water is 95°C. 3–43 Water is boiled at 1 atm pressure in a 25cminternaldiameter 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. 3–44 Repeat Prob. 3–43 for a location at 2000m elevation where the standard atmospheric pressure is 79.5 kPa.
Pressure cooker 250°F
3–45 Saturated steam coming off the turbine of a steam power plant at 30°C condenses on the outside of a 3cmouterdiameter, 35mlong tube at a rate of 45 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe.
FIGURE P3–38E 3–39E The atmospheric pressure at a location is usually specified at standard conditions, but it changes with the weather conditions. As the weather forecasters frequently state, the atmospheric pressure drops during stormy weather and it rises during clear and sunny days. If the pressure difference between the two extreme conditions is given to be 0.3 in of mercury, determine how much the boiling temperatures of water will vary as the weather changes from one extreme to the other. 3–40 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 20°C. 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 during cooling, determine if the lid will open or the pan will move up together with the lid.
3–46 The average atmospheric pressure in Denver (elevation 1610 m) is 83.4 kPa. Determine the temperature at which water in an uncovered pan boils in Denver. Answer: 94.6°C. 3–47 Water in a 5cmdeep pan is observed to boil at 98°C. At what temperature will the water in a 40cmdeep pan boil? Assume both pans are full of water. 3–48 A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4kg lid. If the local atmospheric pressure is 101 kPa, determine the temperature at which the water starts boiling when it is heated. Answer: 100.2°C
Patm = 101 kPa m lid = 4 kg
3–41 Water is to be boiled at sea level in a 30cmdiameter stainless steel pan placed on top of a 3–kW electric burner. If H2O Vapor
FIGURE P3–48 3–49
Reconsider Prob. 3–48. Using EES (or other) software, investigate the effect of the mass of the lid on the boiling temperature of water in the pan. Let the mass vary from 1 kg to 10 kg. Plot the boiling temperature against the mass of the lid, and discuss the results.
60%
3 kW
FIGURE P3–41
40%
3–50 Water is being heated in a vertical piston–cylinder device. The piston has a mass of 20 kg and a crosssectional area of 100 cm2. If the local atmospheric pressure is 100 kPa, determine the temperature at which the water starts boiling.
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Chapter 3 3–51 A rigid tank with a volume of 2.5 m3 contains 15 kg of saturated liquid–vapor mixture of water at 75°C. 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: 187.0°C
3–52 A rigid vessel contains 2 kg of refrigerant134a at 800 kPa and 120°C. Determine the volume of the vessel and the total internal energy. Answers: 0.0753 m3, 655.7 kJ 3–53E A 5ft3 rigid tank contains 5 lbm of water at 20 psia. Determine (a) the temperature, (b) the total enthalpy, and (c) the mass of each phase of water. 3–54 A 0.5m3 vessel contains 10 kg of refrigerant134a at 20°C. Determine (a) the pressure, (b) the total internal energy, and (c) the volume occupied by the liquid phase. Answers: (a) 132.82 kPa, (b) 904.2 kJ, (c) 0.00489 m3
A piston–cylinder device contains 0.1 m3 of liquid water and 0.9 m3 of water vapor in equilibrium at 800 kPa. Heat is transferred at constant pressure until the temperature reaches 350°C.
3–55
(a) (b) (c) (d)
What is the initial temperature of the water? Determine the total mass of the water. Calculate the final volume. Show the process on a Pv diagram with respect to saturation lines.

157
3–58E
Reconsider Prob. 3–57E. Using EES (or other) software, investigate the effect of initial pressure on the quality of water at the final state. Let the pressure vary from 100 psi to 300 psi. Plot the quality against initial pressure, and discuss the results. Also, show the process in Prob. 3–57E on a Tv diagram using the property plot feature of EES.
3–59 A piston–cylinder device initially contains 50 L of liquid water at 40°C and 200 kPa. Heat is transferred to the water at constant pressure until the entire liquid is vaporized. (a) (b) (c) (d)
What is the mass of the water? What is the final temperature? Determine the total enthalpy change. Show the process on a Tv diagram with respect to saturation lines.
Answers: (a) 49.61 kg, (b) 120.21°C, (c) 125,943 kJ
3–60 A 0.3m3 rigid vessel initially contains saturated liquid– vapor mixture of water at 150°C. The water is now heated until it reaches the critical state. Determine the mass of the liquid water and the volume occupied by the liquid at the initial state. Answers: 96.10 kg, 0.105 m3 3–61 Determine the specific volume, internal energy, and enthalpy of compressed liquid water at 100°C and 15 MPa using the saturated liquid approximation. Compare these values to the ones obtained from the compressed liquid tables. 3–62
H 2O P = 800 kPa
Reconsider Prob. 3–61. Using EES (or other) software, determine the indicated properties of compressed liquid, and compare them to those obtained using the saturated liquid approximation.
3–63E A 15ft3 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. 3–64 A piston–cylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled at constant pressure until onehalf of the mass condenses.
FIGURE P3–55 3–56
Reconsider Prob. 3–55. Using EES (or other) software, investigate the effect of pressure on the total mass of water in the tank. Let the pressure vary from 0.1 MPa to 1 MPa. Plot the total mass of water against pressure, and discuss the results. Also, show the process in Prob. 3–55 on a Pv diagram using the property plot feature of EES.
3–57E Superheated water vapor at 180 psia and 500°F is allowed to cool at constant volume until the temperature drops to 250°F. 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) 29.84 psia, (b) 0.219, (c) 426.0 Btu/lbm
(a) Show the process on a Tv diagram. (b) Find the final temperature. (c) Determine the volume change. 3–65 A rigid tank contains water vapor at 250°C and an unknown pressure. When the tank is cooled to 150°C, the vapor starts condensing. Estimate the initial pressure in the tank. Answer: 0.60 MPa 3–66 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 1.19 kg in 30 minutes. If it is estimated that 75 percent of electricity consumed by the heater is transferred to the water as heat, determine the local atmospheric pressure in that location. Answer: 85.4 kPa
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3–67 A rigid tank initially contains 1.4kg saturated liquid water at 200°C. At this state, 25 percent of the volume is occupied 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 pressure, and (c) the internal energy change of the water. Q Water 1.4 kg 200°C
3–74
Reconsider Prob. 3–73. Using EES (or other) software, 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.
3–75 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 50°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure is 100 kPa.
FIGURE P3–67 3–68 A piston–cylinder device initially contains steam at 3.5 MPa, superheated by 5°C. 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 200°C. 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).
V = 0.025 m3 T = 25°C Pg = 210 kPa AIR
Steam 3.5 MPa
FIGURE P3–75
Q
FIGURE P3–68
3–76E The air in an automobile tire with a volume of 0.53 ft3 is at 90°F and 20 psig. Determine the amount of air that must be added to raise the pressure to the recommended value of 30 psig. Assume the atmospheric pressure to be 14.6 psia and the temperature and the volume to remain constant.
Ideal Gas
Answer: 0.0260 lbm
3–69C Propane and methane are commonly used for heating 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.
3–77 The pressure gage on a 2.5m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C and the atmospheric pressure is 97 kPa.
3–70C Under what conditions is the idealgas assumption suitable for real gases? 3–71C What is the difference between R and Ru? How are these two related? 3–72C What is the difference between mass and molar mass? How are these two related? 3–73 A spherical balloon with a diameter of 6 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of the helium in the balloon. Answers: 9.28 kmol, 37.15 kg
Pg = 500 kPa
O2
V = 2.5 m 3 T = 28˚C
FIGURE P3–77
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159
3–78E A rigid tank contains 20 lbm of air at 20 psia and 70°F. More air is added to the tank until the pressure and temperature rise to 35 psia and 90°F, respectively. Determine the amount of air added to the tank. Answer: 13.73 lbm
3–89E Refrigerant134a at 400 psia has a specific volume of 0.13853 ft3/lbm. Determine the temperature of the refrigerant based on (a) the idealgas equation, (b) the generalized compressibility chart, and (c) the refrigerant tables.
3–79 A 400L rigid tank contains 5 kg of air at 25°C. Determine the reading on the pressure gage if the atmospheric pressure is 97 kPa.
3–90 A 0.016773m3 tank contains 1 kg of refrigerant134a at 110°C. Determine the pressure of the refrigerant, using (a) the idealgas equation, (b) the generalized compressibility chart, and (c) the refrigerant tables. Answers: (a) 1.861 MPa,
3–80 A 1m3 tank containing air at 25°C and 500 kPa is connected through a valve to another tank containing 5 kg of air at 35°C and 200 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20°C. Determine the volume of the second tank and the final equilibrium pressure of air.
(b) 1.583 MPa, (c) 1.6 MPa
3–91 Somebody claims that oxygen gas at 160 K and 3 MPa can be treated as an ideal gas with an error of less than 10 percent. Is this claim valid?
Answers: 2.21 m3, 284.1 kPa
3–92 What is the percentage of error involved in treating carbon dioxide at 3 MPa and 10°C as an ideal gas?
Compressibility Factor
Answer: 25 percent
3–81C What is the physical significance of the compressibility factor Z?
3–93 What is the percentage of error involved in treating carbon dioxide at 7 MPa and 380 K as an ideal gas?
3–82C What is the principle of corresponding states?
3–94 Carbon dioxide gas enters a pipe at 3 MPa and 500 K at a rate of 2 kg/s. CO2 is cooled at constant pressure as it flows in the pipe and the temperature 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 each case.
3–83C How are the reduced pressure and reduced temperature defined? 3–84 Determine the specific volume of superheated water vapor at 10 MPa and 400°C, 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) 0.03106 m3/kg, 17.6 percent; (b) 0.02609 m3/kg, 1.2 percent; (c) 0.02644 m3/kg
Reconsider Prob. 3–84. Solve the problem using the generalized compressibility factor feature of the EES software. Again using EES, compare the specific volume of water for the three cases at 10 MPa over the temperature range of 325 to 600°C in 25°C intervals. Plot the percent error involved in the idealgas approximation against temperature, and discuss the results.
3 MPa 500 K 2 kg/s
3–85
3–86 Determine the specific volume of refrigerant134a vapor at 0.9 MPa and 70°C 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. 3–87 Determine the specific volume of nitrogen gas at 10 MPa and 150 K based on (a) the idealgas equation and (b) the generalized compressibility chart. Compare these results with the experimental value of 0.002388 m3/kg, and determine the error involved in each case. Answers: (a) 0.004452 m3/kg, 86.4 percent; (b) 0.002404
m3/kg,
0.7 percent
3–88 Determine the specific volume of superheated water vapor at 3.5 MPa and 450°C based on (a) the idealgas equation, (b) the generalized compressibility chart, and (c) the steam tables. Determine the error involved in the first two cases.
CO2
450 K
FIGURE P3–94
Other Equations of State 3–95C 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? 3–96 A 3.27m3 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 BeattieBridgeman equation. Compare your results with the actual value of 1505 kPa. 3–97 A 1m3 tank contains 2.841 kg of steam at 0.6 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) 457.6 K, (b) 465.9 K, (c) 473 K 3–98
Reconsider Prob. 3–97. Solve the problem using EES (or other) software. Again using the EES, compare the temperature of water for the three cases at constant specific volume over the pressure range of 0.1 MPa to
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1 MPa in 0.1 MPa increments. Plot the percent error involved in the idealgas approximation against pressure, and discuss the results. 3–99E Refrigerant134a at 100 psia has a specific volume of 0.54022 ft3/lbm. Determine the temperature of the refrigerant based on (a) the idealgas equation, (b) the van der Waals equation, and (c) the refrigerant tables. 3–100
Nitrogen at 150 K has a specific volume of 0.041884 m3/kg. 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) 1000.4 kPa
since you notice no condensation forming outside the can. Can the store owner be telling the truth?
Review Problems 3–108 The combustion in a gasoline engine may be approximated by a constant volume heat addition process. There exists the air–fuel mixture in the cylinder before the 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 1.8 MPa and 450°C before the combustion and 1300°C after it. Determine the pressure at the end of the combustion process. Answer: 3916 kPa
3–101
Reconsider Prob. 3–100. Using EES (or other) software, compare the pressure results of the idealgas and BeattieBridgeman equations with nitrogen data supplied by EES. Plot temperature versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K T 150 K.
Combustion chamber 1.8 MPa 450°C
Special Topic: Vapor Pressure and Phase Equilibrium 3–102 Consider a glass of water in a room that is at 20°C and 60 percent relative humidity. If the water temperature is 15°C, determine the vapor pressure (a) at the free surface of the water and (b) at a location in the room far from the glass. 3–103 During a hot summer day at the beach when the air temperature is 30°C, someone claims the vapor pressure in the air to be 5.2 kPa. Is this claim reasonable? 3–104 On a certain day, the temperature and relative humidity of air over a large swimming pool are measured to be 20°C and 40 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.
FIGURE P3–108 3–109 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 temperature was reached at the end of the process.
Ideal gas 300 kPa 600 K
3–105 Consider two rooms that are identical except that one is maintained at 30°C and 40 percent relative humidity while the other is maintained at 20°C and 70 percent relative humidity. Noting that the amount of moisture is proportional to the vapor pressure, determine which room contains more moisture. 3–106E A thermos bottle is halffilled with water and is left open to the atmospheric air at 70°F and 35 percent relative humidity. If heat transfer to the water through the thermos walls and the free surface is negligible, determine the temperature of water when phase equilibrium is established. 3–107 During a hot summer day when the air temperature is 35°C and the relative humidity is 70 percent, you buy a supposedly “cold” canned drink from a store. The store owner claims that the temperature of the drink is below 10°C. Yet the drink does not feel so cold and you are skeptical
FIGURE P3–109 3–110 Carbondioxide gas at 3 MPa and 500 K flows steadily in a pipe at a rate of 0.4 kmol/s. 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 3 MPa 500 K 0.4 kmol/s
CO2
FIGURE P3–110
450 K
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Chapter 3 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. 3–111 A piston–cylinder device initially contains 0.2 kg of steam at 200 kPa and 300°C. Now, the steam is cooled at constant pressure until it is at 150°C. Determine the volume change of the cylinder during this process using the compressibility factor and compare the result to the actual value.
Steam 0.2 kg 200 kPa 300°C
Q
FIGURE P3–111 3–112 Combustion in a diesel engine may be modeled as a constantpressure heat addition process with air in the cylinder before and after combustion. Consider a diesel engine with cylinder conditions of 950 K and 75 cm3 before combustion, and 150 cm3 after it. The engine operates with an air–fuel ratio of 22 kg air/kg fuel (the mass of the air divided by the mass of the fuel). Determine the temperature after the combustion process.
161
(b) On the Tv diagram sketch the constant specific volume process through the state T 120°C, v 0.7163 m3/kg 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. 3–114 The gage pressure of an automobile tire is measured to be 200 kPa before a trip and 220 kPa after the trip at a location where the atmospheric pressure is 90 kPa. Assuming the volume of the tire remains constant at 0.035 m3, determine the percent increase in the absolute temperature of the air in the tire. 3–115 Although balloons have been around since 1783 when the first balloon took to the skies in France, a real breakthrough in ballooning occurred in 1960 with the design of the modern hotair balloon fueled by inexpensive propane and constructed of lightweight nylon fabric. Over the years, ballooning has become a sport and a hobby for many people around the world. Unlike balloons filled with the light helium gas, hotair balloons are open to the atmosphere. Therefore, the pressure in the balloon is always the same as the local atmospheric pressure, and the balloon is never in danger of exploding. Hotair balloons range from about 15 to 25 m in diameter. The air in the balloon cavity is heated by a propane burner located at the top of the passenger cage. The flames from the burner that shoot into the balloon heat the air in the balloon cavity, raising the air temperature at the top of the balloon from 65°C to over 120°C. The air temperature is maintained at the desired levels by periodically firing the propane burner.
Combustion chamber 950 K 75 cm3
FIGURE P3–112 3–113 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 constant temperature process through the state P 300 kPa, v 0.525 m3/kg 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.

FIGURE P3–115 © Vol. 1/PhotoDisc
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Thermodynamics
The buoyancy force that pushes the balloon upward is proportional to the density of the cooler air outside the balloon and the volume of the balloon, and can be expressed as
H 2O V= 4L m = 2 kg T = 50°C
FB rcool air gVballoon where g is the gravitational acceleration. When air resistance is negligible, the buoyancy force is opposed by (1) the weight of the hot air in the balloon, (2) the weight of the cage, the ropes, and the balloon material, and (3) the weight of the people and other load in the cage. The operator of the balloon can control the height and the vertical motion of the balloon by firing the burner or by letting some hot air in the balloon escape, to be replaced by cooler air. The forward motion of the balloon is provided by the winds. Consider a 20mdiameter hotair balloon that, together with its cage, has a mass of 80 kg when empty. This balloon is hanging still in the air at a location where the atmospheric pressure and temperature are 90 kPa and 15°C, respectively, while carrying three 65kg people. Determine the average temperature of the air in the balloon. What would your response be if the atmospheric air temperature were 30°C? 3–116
Reconsider Prob. 3–115. Using EES (or other) software, investigate the effect of the environment temperature on the average air temperature in the balloon when the balloon is suspended in the air. Assume the environment temperature varies from 10 to 30°C. Plot the average air temperature in the balloon versus the environment temperature, and discuss the results. Investigate how the number of people carried affects the temperature of the air in the balloon.
3–117 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 70kg people, is heated by propane burners at a location where the atmospheric pressure and temperature are 93 kPa and 12°C, 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 25°C? 3–118E Water in a pressure cooker is observed to boil at 260°F. What is the absolute pressure in the pressure cooker, in psia? 3–119 A rigid tank with a volume of 0.117 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 diagram with respect to saturation lines. 3–120 A 4L rigid tank contains 2 kg of saturated liquid–vapor mixture of water at 50°C. 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 P3–120 3–121 A 10kg mass of superheated refrigerant134a at 1.2 MPa and 70°C is cooled at constant pressure until it exists as a compressed liquid at 20°C. (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) 0.187 m3, (c) 1984 kJ
3–122 A 0.5m3 rigid tank containing hydrogen at 20°C and 600 kPa is connected by a valve to another 0.5m3 rigid tank that holds hydrogen at 30°C and 150 kPa. Now the valve is opened and the system is allowed to reach thermal equilibrium with the surroundings, which are at 15°C. Determine the final pressure in the tank.
H2
H2 m3
V = 0.5 T = 20°C P = 600 kPa
V = 0.5 m3 T = 30°C P = 150 kPa
FIGURE P3–122 3–123
Reconsider Prob. 3–122. Using EES (or other) software, investigate the effect of the surroundings temperature on the final equilibrium pressure in the tanks. Assume the surroundings temperature to vary from 10 to 30°C. Plot the final pressure in the tanks versus the surroundings temperature, and discuss the results.
3–124 A 20m3 tank contains nitrogen at 23°C 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 20°C, determine the amount of nitrogen that has escaped. Answer: 44.6 kg
3–125 Steam at 400°C has a specific volume of 0.02 m3/kg. Determine the pressure of the steam based on (a) the idealgas equation, (b) the generalized compressibility chart, and (c) the steam tables. Answers: (a) 15,529 kPa, (b) 12,576 kPa, (c) 12,500 kPa
3–126 A tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains 0.01 m3
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Chapter 3 of refrigerant134a that is a saturated liquid at 0.8 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 refrigerant is 20°C and 400 kPa, determine the volume of the tank.
R134a V = 0.01 m3 P = 0.8 MPa
Evacuated
FIGURE P3–126 Reconsider Prob. 3–126. Using EES (or other) software, investigate the effect of the initial pressure of refrigerant134a on the volume of the tank. Let the initial pressure vary from 0.5 to 1.5 MPa. Plot the volume of the tank versus the initial pressure, and discuss the results.
Leak
Propane
FIGURE P3–128 3–129
Repeat Prob. 3–128 for isobutane.
3–130 A tank contains helium at 100°C and 10 kPa gage. The helium is heated in a process by heat transfer from the surroundings such that the helium reaches a final equilibrium state at 300°C. Determine the final gage pressure of the helium. Assume atmospheric pressure is 100 kPa.
163
3–13l A tank contains argon at 600°C and 200 kPa gage. The argon is cooled in a process by heat transfer to the surroundings such that the argon reaches a final equilibrium state at 300°C. Determine the final gage pressure of the argon. Assume atmospheric pressure is 100 kPa. 3–132 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
200 270.3
30 130 400
3–127
3–128 Liquid propane is commonly used as a fuel for heating homes, powering vehicles such as forklifts, and filling portable picnic tanks. Consider a propane tank that initially contains 5 L of liquid propane at the environment temperature of 20°C. If a hole develops in the connecting tube of a propane tank and the propane starts to leak out, determine the temperature of propane when the pressure in the tank drops to 1 atm. Also, determine the total amount of heat transfer from the environment to the tank to vaporize the entire propane in the tank.

300 500
v, m3/kg
u, kJ/kg
Condition description and quality (if applicable)
1.5493 0.500 3084
3–133 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 mixture, superheated vapor, or insufficient information; and, if applicable, give the quality.
P, kPa
T, °C
v, m3/kg u, kJ/kg
320 12 1000 39.37 40 0.17794 180 0.0700 200
Condition description and quality (if applicable)
249
3–134 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 constant temperature process through the state P 280 kPa, v 0.06 m3/kg as pressure 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 constant specific volume process through the state T 20°C, v 0.02 m3/kg 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.
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Thermodynamics
Fundamentals of Engineering (FE) Exam Problems 3–135 A rigid tank contains 6 kg of an ideal gas at 3 atm and 40°C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 186°C (b) 59° (c) 43°C (d) 20°C (e) 230°C 3–136 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 25°C, the air temperature after the trip is (a) 51.1°C (b) 64.2°C (c) 27.2°C (d) 28.3°C (e) 25.0°C 3–137 A 300m3 rigid tank is filled with saturated liquid– vapor mixture of water at 200 kPa. If 25 percent of the mass is liquid and 75 percent of the mass is vapor, the total mass in the tank is (a) 451 kg (b) 556 kg (c) 300 kg (d) 331 kg (e) 195 kg 3–138 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersiontype electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 18 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 0.90 kW (d) 1.05 kW (b) 1.52 kW (e) 1.24 kW (c) 2.09 kW 3–139 A 1m3 rigid tank contains 10 kg of water (in any phase or phases) at 160°C. The pressure in the tank is (a) 738 kPa (d) 2000 MPa (b) 618 kPa (e) 1618 kPa (c) 370 kPa 3–140 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid water evaporates in 30 min. The rate of heat transfer to the water is (a) 2.51 kW (d) 0.47 kW (b) 2.32 kW (e) 3.12 kW (c) 2.97 kW 3–141 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 evaporated. Then the rate of heat transfer to the water is (a) 0.84 kJ/min (d) 53.5 kJ/min (b) 45.1 kJ/min (e) 225.7 kJ/min (c) 41.8 kJ/min
3–142 A 3m3 rigid vessel contains steam at 10 MPa and 500°C. The mass of the steam is (a) 3.0 kg (b) 19 kg (c) 84 kg (d) 91 kg (e) 130 kg 3–143 Consider a sealed can that is filled with refrigerant134a. The contents of the can are at the room temperature of 25°C. 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) 0°C (b) 29°C (c) 16°C (d) 5°C (e) 25°C
Design, Essay, and Experiment Problems 3–144 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. 3–145 It is well known that water freezes at 0°C at atmospheric pressure. The mixture of liquid water and ice at 0°C 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 2°C or even lower without any formation of ice at atmospheric pressure. But at that state even a small disturbance can initiate the formation of ice abruptly, and the water temperature stabilizes at 0°C following this sudden change. The water at 2°C is said to be in a metastable state. Write an essay on metastable states and discuss how they differ from stable equilibrium states. 3–146 Enthalpy of Fusion for Water Experiment. The enthalpy of fusion for water (also known as latent heat of fusion) is obtained with an ice calorimeter that is constructed from a copper tube with closed ends and two access ports. Inside the calorimeter is coiled thermocouple wire that serves as electric heater wire. The calorimeter is filled with water, placed in a freezer and removed after the water is frozen. The calorimeter is insulated with Styrofoam and placed in a chamber with double walls that hold crushed ice and water providing a 0°C air environment. Electrical power input into the heater causes the solid ice at 0°C to melt to liquid water at 0°C the energy supplied for this phasechange is the enthalpy of fusion. Obtain the enthalpy of fusion for water using the video clip, the complete writeup, and the data provided on the DVD accompanying this book.
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Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS
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 thermodynamics properties of substances. In this chapter, we apply the energy balance relation to systems that do not involve any mass flow across their boundaries; that 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 Esystem, 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.
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 (fixed mass) 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 (fixed mass) systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

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Thermodynamics INTERACTIVE TUTORIAL
SEE TUTORIAL CH. 4, SEC. 1 ON THE DVD.
The moving boundary
GAS
FIGURE 4–1 The work associated with a moving boundary is called boundary work.
F
4–1
MOVING BOUNDARY WORK
One form of mechanical work frequently encountered in practice is associated with the expansion or compression of a gas in a piston–cylinder 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. 4–1). 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 expansion, 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 without 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 quasiequilibrium process, a process during which the system remains nearly in equilibrium at all times. A quasiequilibrium process, also called a quasistatic 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 compressors to be a minimum when quasiequilibrium processes are used in place of nonquasiequilibrium processes. Below, the work associated with a moving boundary is evaluated for a quasiequilibrium process. Consider the gas enclosed in the piston–cylinder device shown in Fig. 4–2. The initial pressure of the gas is P, the total volume is V, and the crosssectional area of the piston is A. If the piston is allowed to move a distance ds in a quasiequilibrium manner, the differential work done during this process is dWb F ds PA ds P¬dV
A
ds P GAS
FIGURE 4–2 A gas does a differential amount of work dWb as it forces the piston to move by a differential amount ds.
(4–1)
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. 4–1 that P is the absolute pressure, which is always positive. However, the volume change dV is positive during an expansion process (volume increasing) and negative during a compression process (volume decreasing). Thus, the boundary work is positive during an expansion process and negative during a compression process. Therefore, Eq. 4–1 can be viewed as an expression for boundary work output, Wb,out. 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: Wb
2
1
¬P dV¬¬1kJ2
(4–2)
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Chapter 4 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 diagram in Fig. 4 –3. 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 1–2 is obtained by adding these differential areas: Area A
2
dA
¬
1
P dV
1 Process path
2 dA = P dV
V1
V
(4–3)
2
Pi dV
V2
dV
1
A comparison of this equation with Eq. 4–2 reveals that the area under the process curve on a PV diagram is equal, in magnitude, to the work done during 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. 4 –4). This is expected, since work is a path function (i.e., 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. 4–5 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. But we can always 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. 4–2 is the pressure at the inner surface 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 cylinder is at the same pressure at any given time. Equation 4–2 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 Wb
167
P
2 ¬

(4–4)
1
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
P
FIGURE 4–3 The area under the process curve on a PV diagram represents the boundary work. P
WA = 10 kJ 1
WB = 8 kJ WC = 5 kJ A B C 2
V1
V2
V
FIGURE 4–4 The boundary work done during a process depends on the path followed as well as the end states. P 2 A Wnet B 1
V2
V1
V
FIGURE 4–5 The net work done during a cycle is the difference between the work done by the system and the work done on the system.
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Thermodynamics compression process). Therefore, it has to appear somewhere else and we must be able 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, Wb Wfriction Watm Wcrank
2
1
1Ffriction Patm¬A Fcrank 2 dx
(4–5)
Of course the work used to overcome friction appears as frictional heat and the energy transmitted through the crankshaft is transmitted to other components (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 processes of gases only. It can also be used for solids and liquids.
EXAMPLE 4–1
Boundary Work for a ConstantVolume Process
A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C 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. 4–6. The boundary work can be determined from Eq. 4–2 to be
Wb
2
0 P dV ¬ 0 ¡ ˛
1
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).
P, kPa
AIR P1 = 500 kPa T1 = 150°C
FIGURE 4–6 Schematic and PV diagram for Example 4–1.
P2 = 400 kPa T2 = 65°C
Heat
500
1
400
2
V
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Chapter 4 EXAMPLE 4–2

Boundary Work for a ConstantPressure Process
A frictionless piston–cylinder 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 piston cylinder device is heated and the temperature rises at constant pressure. 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. 4–7. Assumption The expansion process is quasiequilibrium. Analysis 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 and the weight of the piston remain constant. Therefore, this is a constantpressure process, and, from Eq. 4–2
Wb
2
P dV P0
1
1
or
2
dV P0 1V2 V1 2
(4–6)
Wb mP0 1v2 v1 2
since V mv. From the superheated vapor table (Table A–6E), the specific volumes are determined to be v1 7.4863 ft3/lbm at state 1 (60 psia, 320F) and v2 8.3548 ft3/lbm at state 2 (60 psia, 400F). Substituting these values yields
Wb 110 lbm2 160 psia 2 3 18.3548 7.48632 ft3>lbm4 a
1 Btu b 5.404 psia # ft3
96.4 Btu Discussion The positive sign indicates that the work is done by the system. That is, the steam used 96.4 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 V for this case.
P, psia
1
P0 = 60 psia
2
60 H2O m = 10 lbm P = 60 psia
Heat
Area = wb
v1 = 7.4863
v2 = 8.3548
v, ft3/lbm
FIGURE 4–7 Schematic and Pv diagram for Example 4–2.
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Thermodynamics EXAMPLE 4–3
Isothermal Compression of an Ideal Gas
A piston–cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.
Solution Air in a piston–cylinder device is compressed isothermally. 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. 4–8. Assumptions 1 The compression process is quasiequilibrium. 2 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 For an ideal gas at constant temperature T0,
PV mRT0 C¬or¬P
C V
where C is a constant. Substituting this into Eq. 4–2, we have
P dV 2
Wb
1
2
1
C dV C V
2
1
V2 V2 dV C ln¬ P1V1 ln¬ V V1 V1
(4–7)
In Eq. 4–7, P1V1 can be replaced by P2V2 or mRT0. Also, V2/V1 can be replaced by P1/P2 for this case since P1V1 P2V2. Substituting the numerical values into Eq. 4–7 yields
Wb 1100 kPa2 10.4 m3 2 a ln
0.1 1 kJ ba b 0.4 1 kPa # m3
55.5 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.
P 2 T0 = 80°C = const. AIR V1 = 0.4 m3 P1 = 100 kPa T0 = 80°C = const.
1
0.1
FIGURE 4–8 Schematic and PV diagram for Example 4–3.
0.4
V, m3
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Chapter 4

171
P 1
P1
P1V1n = P2V 2n PV n = const.
GAS P2
PV n = C = const.
2
V1
V2
FIGURE 4–9 Schematic and PV diagram for a polytropic process.
V
Polytropic Process During actual expansion and compression processes of gases, pressure and volume are often related by PV n C, where n and C are constants. A process of this kind is called a polytropic process (Fig. 4–9). Below we develop a general expression for the work done during a polytropic process. The pressure for a polytropic process can be expressed as P CV n
EXPERIMENT
(4–8)
Substituting this relation into Eq. 4–2, we obtain 2
Wb
2
P dV CV 1
n ¬dV
C
1
V 2n1 V 1n1 P2V2 P1V1 n 1 1n
(4–9)
since C P1V1n P2V2n. For an ideal gas (PV mRT), this equation can also be written as Wb
mR 1T2 T1 2 1n
¬¬n 1¬¬1kJ2
(4–10)
For the special case of n 1 the boundary work becomes Wb
1
2
P dV
1
2
CV 1 dV PV ln a
V2 b V1
For an ideal gas this result is equivalent to the isothermal process discussed in the previous example.
EXAMPLE 4–4
Expansion of a Gas against a Spring
A piston–cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m 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 crosssectional area of the piston is 0.25 m2, determine (a) the final pressure inside the cylinder, (b) the total work done by
Use actual data from the experiment shown here to find the polytropic exponent for expanding air. See endofchapter problem 4–174. © Ronald Mullisen
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Thermodynamics the gas, and (c) the fraction of this work done against the spring to compress it.
Solution A gas in a piston–cylinder 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. 4–10. (a) The enclosed volume at the final state is
V2 2V1 122 10.05 m3 2 0.1 m3 Then the displacement of the piston (and of the spring) becomes
10.1 0.052 m3 ¢V 0.2 m A 0.25 m2
x
The force applied by the linear spring at the final state is
F kx 1150 kN>m2 10.2 m2 30 kN The additional pressure applied by the spring on the gas at this state is
P
F 30 kN 120 kPa A 0.25 m2
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. 4–10 the area under the process curve (a trapezoid) is determined to be
W area
1200 3202 kPa 2
¬3
10.1 0.052 m3 4 a
1 kJ b 13 kJ 1 kPa # m3
k = 150 kN/m P, kPa
320 II 200
FIGURE 4–10 Schematic and PV diagram for Example 4–4.
A = 0.25 m2 P1 = 200 kPa V1 = 0.05 m3
I 0.05
Heat
0.1 V, m3
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Chapter 4

173
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,
Wspring 12 3 1320 200 2 kPa4 10.05 m3 2 a Discussion
1 kJ b 3 kJ 1 kPa # m3
This result could also be obtained from
Wspring 12k 1x22 x21 2 12 1150 kN>m2 3 10.2 m2 2 02 4 a
4–2
1 kJ b 3 kJ 1 kN # m
ENERGY BALANCE FOR CLOSED SYSTEMS
Energy balance for any system undergoing any kind of process was expressed as (see Chap. 2)
SEE TUTORIAL CH. 4, SEC. 2 ON THE DVD.
(4–11)
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬ ¢E system¬¬1kJ2
Net energy transfer by heat, work, and mass
INTERACTIVE TUTORIAL
Change in internal, kinetic, potential, etc., energies
or, in the rate form, as
Rate of net energy transfer by heat, work, and mass
(4–12)
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
. . E in E out¬ ¬dE system>dt¬¬1kW2 Rate of change in internal, kinetic, potential, etc., energies
For constant rates, the total quantities during a time interval t are related to the quantities per unit time as # # Q Q ¢t,¬W W ¢t,¬and¬¢E 1dE>dt2 ¢t¬¬1kJ2
(4–13)
The energy balance can be expressed on a per unit mass basis as ein eout ¢esystem¬¬1kJ>kg 2
(4–14)
which is obtained by dividing all the quantities in Eq. 4–11 by the mass m of the system. Energy balance can also be expressed in the differential form as dE in dE out dE system¬or¬dein deout desystem
(4–15)
For a closed system undergoing a cycle, the initial and final states are identical, and thus Esystem 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 # # Wnet,out Q net,in¬or¬Wnet,out Q net,in¬¬1for a cycle 2
P
(4–16)
That is, the net work output during a cycle is equal to net heat input (Fig. 4–11).
Qnet = Wnet
V
FIGURE 4–11 For a cycle E 0, thus Q W.
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Thermodynamics
General Q – W = ∆E Stationary systems Q – W = ∆U Per unit mass q – w = ∆e Differential form δq – δw = de
FIGURE 4–12 Various forms of the firstlaw relation for closed systems.
EXPERIMENT
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 net,in Wnet,out ¢E system¬or¬Q W ¢E
(4–17)
where Q Qnet,in Qin Qout is the net heat input and W Wnet,out 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. 4–12. 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 principles instead of being a fundamental physical law itself. As energy quantities, heat and work are not that different, and you probably 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 relations 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. EXAMPLE 4–5
Electric Heating of a Gas at Constant Pressure
A piston–cylinder 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 0.2 A for 5 min from a 120V source. At the same time, a heat loss of 3.7 kJ occurs. (a) Show that for a closed system the boundary work Wb and the change in internal energy U in the firstlaw relation can be combined into one term, H, for a constantpressure process. (b) Determine the final temperature of the steam.
Solution Saturated water vapor in a piston–cylinder device expands at con
Use actual data from the experiment shown here to verify the first law of thermodynamics. See endofchapter problem 4–175. © Ronald Mullisen
stant pressure as a result of heating. It is to be shown that U Wb H, and the final temperature is to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero, KE PE 0. Therefore, E U 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.
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Chapter 4

175
P, kPa
H 2O
0.2 A
m = 25 g P1 = P2 = 300 kPa
1
300
2
120 V
EXPERIMENT
Sat. vapor 5 min
v
Qout = 3.7 kJ
FIGURE 4–13 Schematic and Pv diagram for Example 4–5.
Analysis We take the contents of the cylinder, including the resistance wires, as the system (Fig. 4–13). This is a closed system since no mass crosses the system boundary during the process. We observe that a piston–cylinder 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 undergoing 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¬ ¬
© Ronald Mullisen
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭ Net energy transfer by heat, work, and mass
Use actual data from the experiment shown here to verify the first law of thermodynamics. See endofchapter problem 4–176.
Change in internal, kinetic, potential, etc., energies
EXPERIMENT
0 0 Q W ¢U ¢KE ¢PE ¡ ¡ Q Wother Wb U2 U1 For a constantpressure process, the boundary work is given as Wb P0(V2 V1). Substituting this into the preceding relation gives
Q Wother P0 1V2 V1 2 U2 U1
However,
P0 P2 P1¬ S ¬Q Wother 1U2 P2V2 2 1U1 P1V1 2
Also H U PV, and thus
Q Wother H2 H1¬¬1kJ2
(4–18)
which is the desired relation (Fig. 4–14). This equation is very convenient to use in the analysis of closed systems undergoing a constantpressure quasiequilibrium process since the boundary work is automatically taken care of by the enthalpy terms, and one no longer needs to determine it separately.
Use actual data from the experiment shown here to verify the first law of thermodynamics. See endofchapter problem 4–177. © Ronald Mullisen
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Thermodynamics EXPERIMENT
(b) The only other form of work in this case is the electrical work, which can be determined from
We VI¬¢t 1120 V2 10.2 A2 1300 s2 a State 1:
1 kJ>s 1000 VA
b 7.2 kJ
P1 300 kPa f ¬h 1 h g @ 300 kPa 2724.9 kJ>kg¬¬1Table A–52 sat. vapor
The enthalpy at the final state can be determined directly from Eq. 4–18 by expressing 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 U by H for a constantpressure expansion or compression process:
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬
⎫ ⎪ ⎬ ⎪ ⎭
Use actual data from the experiment shown here to verify the first law of thermodynamics. See endofchapter problem 4–178.
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
We,in Q out Wb ¢U
We,in Q out ¢H m 1h 2 h 1 2¬¬1since P constant 2
© Ronald Mullisen
7.2 kJ 3.7 kJ 10.025 kg2 1h 2 2724.9 2 kJ>kg h 2 2864.9 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 = const.
∆H Q – W other – Wb = ∆U
P2 300 kPa f ¬T2 200°C¬¬1Table A–6 2 h 2 2864.9 kJ>kg
Therefore, the steam will be at 200°C 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 0.0002 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.
Q – W other = ∆H
FIGURE 4–14 For a closed system undergoing a quasiequilibrium, P constant process, U Wb H.
EXAMPLE 4–6
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 25°C, 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 25°C. Determine (a) the volume of the tank, (b) the final pressure, and (c) the heat transfer for this process.
Solution One half 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 to expand and fill the entire tank while the temperature is maintained constant. The volume of tank, the final pressure, and the heat transfer are to be to determined.
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Chapter 4 Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 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 The water temperature remains constant during the process. 5 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. 4–15). 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 liquid–vapor 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 25°C (3.1698 kPa). Approximating the compressed liquid as a saturated liquid at the given temperature, we find
v1 vf @ 25°C 0.001003 m3>kg 0.001 m3>kg¬¬1Table A–4 2
Then the initial volume of the water is
V1 mv1 15 kg2 10.001 m3>kg2 0.005 m3 The total volume of the tank is twice this amount:
Vtank 12 2 10.005 m3 2 0.01 m3 (b) At the final state, the specific volume of the water is
v2
V2 0.01 m3 0.002 m3>kg m 5 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:¬vf 0.001003 m3>kg¬and¬vg 43.340 m3>kg¬1Table A–4 2 Since vf v2 vg, the water is a saturated liquid–vapor mixture at the final state, and thus the pressure is the saturation pressure at 25°C:
P2 Psat @ 25°C 3.1698 kPa¬¬1Table A–4 2
P, kPa System boundary
Evacuated space
Partition
200
1
H2 O m = 5 kg P1 = 200 kPa T1 = 25 °C
3.17
2
Qin
v
FIGURE 4–15 Schematic and Pv diagram for Example 4–6.

177
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Thermodynamics (c) Under stated assumptions and observations, the energy balance on the system can be expressed as
Vacuum P=0 W=0
E in E out¬ ¬
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
H 2O
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Qin ¢U m 1u2 u1 2
Heat
FIGURE 4–16 Expansion against a vacuum involves no work and thus no energy transfer.
¢E system
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 boundary work is zero (Fig. 4–16). 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,
u1 uf @ 25°C 104.83 kJ>kg The quality at the final state is determined from the specific volume information:
1 kg IRON
1 kg WATER
20 ← 30°C
20 ← 30°C
4.5 kJ
x2
m = 1 kg ∆T = 1°C Specific heat = 5 kJ/kg ·°C
5 kJ
FIGURE 4–18 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 3 ON THE DVD.
vfg
0.002 0.001 2.3 105 43.34 0.001
Then
u2 uf x2ufg 104.83 kJ>kg 12.3 105 2 12304.3 kJ>kg2
41.8 kJ
FIGURE 4–17 It takes different amounts of energy to raise the temperature of different substances by the same amount.
v2 vf
104.88 kJ>kg Substituting yields
Qin 15 kg2 3 1104.88 104.83 2 kJkg4 0.25 kJ
Discussion The positive sign indicates that the assumed direction is correct, and heat is transferred to the water.
4–3
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 4.5 kJ of energy to raise the temperature of 1 kg of iron from 20 to 30°C, whereas it takes about 9 times this energy (41.8 kJ to be exact) to raise the temperature of 1 kg of liquid water by the same amount (Fig. 4–17). Therefore, it is desirable 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 temperature of a unit mass of a substance by one degree (Fig. 4–18). 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
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Chapter 4 do the same as the pressure is maintained constant is the specific heat at constant pressure cp. This is illustrated in Fig. 4–19. The specific heat at constant pressure 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 supplied to the system. Now we attempt to express the specific heats in terms of other thermodynamic properties. First, consider a fixed mass in a stationary closed system undergoing a constantvolume process (and thus no expansion or compression work is involved). The conservation of energy principle ein eout esystem for this process can be expressed in the differential form as dein deout du
cv¬d T du¬¬at constant volume ˛
179 (2)
(1)
V = constant
P = constant
m = 1 kg ∆T = 1°C cv = 3.12
m = 1 kg ∆ T = 1 °C kJ kg. °C
3.12 kJ
The lefthand side of this equation represents the net amount of energy transferred 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,

cp = 5.19
kJ kg.°C
5.19 kJ
FIGURE 4–19 Constantvolume and constantpressure specific heats cv and cp (values given are for helium gas).
or cv a
0u b 0T v
(4–19)
Similarly, an expression for the specific heat at constant pressure cp can be obtained by considering a constantpressure expansion or compression process. It yields cp a
0h b 0T p
(4–20)
Equations 4 –19 and 4 –20 are the defining equations for cv and cp, and their interpretation is given in Fig. 4 –20. 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 independent, intensive properties. That is, the energy required to raise the temperature of a substance by one degree is different at different temperatures and pressures (Fig. 4 –21). But this difference is usually not very large. A few observations can be made from Eqs. 4–19 and 4–20. 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 originated. 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 also explains the origin of the name specific heat at constant pressure. Another observation that can be made from Eqs. 4–19 and 4–20 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
( (
cv = ∂u
∂T
v
= the change in internal energy with temperature at constant volume
( (
cp = ∂h ∂T p = the change in enthalpy with temperature at constant pressure
FIGURE 4–20 Formal definitions of cv and cp.
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Thermodynamics
AIR
AIR
m = 1 kg
m = 1 kg
300 ← 301 K
1000 ← 1001 K
0.718 kJ
0.855 kJ
FIGURE 4–21 The specific heat of a substance changes with temperature.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 4 ON THE DVD.
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 measure 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 kJ/kg · °C or kJ/kg · K. Notice that these two units are identical since T(°C) T(K), and 1°C change in temperature is equivalent to a change of 1 K. The specific heats are sometimes given on a molar basis. They are then denoted by c–v and c–p and have the unit kJ/kmol · °C or kJ/kmol · K.
4–4
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 1T2
Thermometer
WATER
AIR (high pressure)
Evacuated
FIGURE 4–22 Schematic of the experimental apparatus used by Joule.
(4–21)
In his classical experiment, Joule submerged two tanks connected with a pipe and a valve in a water bath, as shown in Fig. 4–22. Initially, one tank contained air at a high pressure and the other tank was evacuated. When thermal 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 f ¬h u RT Pv RT
Since R is constant and u u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only: h h 1T2
(4–22)
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Chapter 4 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 temperature, u, h, cv, and cp of an ideal gas have fixed values regardless of the specific volume or pressure (Fig. 4 –23). Thus, for ideal gases, the partial derivatives in Eqs. 4 –19 and 4–20 can be replaced by ordinary derivatives. Then the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as and
du cv 1T2 dT
(4–23)
dh cp 1T2 dT
(4–24)
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: ¢u u 2 u 1
2
2
1
cv 1T2 dT¬¬1kJ>kg2
(4–25)
cp 1T2 dT¬¬1kJ>kg2
(4–26)
u= h= cv = cp =

181
u(T ) h(T ) cv (T ) c p(T )
FIGURE 4–23 For ideal gases, u, h, cv, and cp vary with temperature only.
and ¢h h2 h 1
1
To carry out these integrations, we need to have relations for cv and cp as functions 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 are often denoted cp0 and cv 0. Accurate analytical expressions for idealgas specific heats, based on direct measurements or calculations from statistical behavior of molecules, are available and are given as thirddegree polynomials in the appendix (Table A–2c) for several gases. A plot of c–p0(T) data for some common gases is given in Fig. 4 –24. 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. 4 –25 and 4–26 are straightforward but rather timeconsuming and thus impractical. To avoid these laborious calculations, u and h data for a number of gases have been tabulated over small temperature intervals. These tables are obtained by choosing an arbitrary reference point and performing the integrations in Eqs. 4–25 and 4–26 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. 4–25). The choice of the reference state has no effect on u or h calculations. The u and h data are given in kJ/kg for air (Table A–17) and usually in kJ/kmol for other gases. The unit kJ/kmol is very convenient in the thermodynamic analysis of chemical reactions. Some observations can be made from Fig. 4 –24. 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
Cp0 kJ/kmol · K CO 2
60
H2 O 50
O2
40
H2 Air
30
Ar, He, Ne, Kr, Xe, Rn 20 1000
2000 Temperature, K
3000
FIGURE 4–24 Idealgas constantpressure specific heats for some gases (see Table A–2c for cp equations).
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Thermodynamics 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. 4 –25 and 4 –26 can be replaced by the constant average specific heat values. Then the integrations in these equations can be performed, yielding
AIR T, K
u, kJ/kg
0 . . 300 310 . .
0 . . 214.07 221.25 . .
h, kJ/kg 0 . . 300.19 310.24 . .
cp Approximation 2
cp,avg 1
T1
T avg
T2
(4–27)
h 2 h 1 cp,avg 1T2 T1 2¬¬1kJ>kg 2
(4–28)
and
FIGURE 4–25 In the preparation of idealgas tables, 0 K is chosen as the reference temperature.
Actual
u 2 u 1 cv,avg 1T2 T1 2¬¬1kJ>kg2
T
FIGURE 4–26 For small temperature intervals, the specific heats may be assumed to vary linearly with temperature.
The specific heat values for some common gases are listed as a function of temperature in Table A–2b. The average specific heats cp,avg and cv,avg are evaluated from this table at the average temperature (T1 + T2)/2, as shown in Fig. 4–26. 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 reasonably good results, and one is not necessarily better than the other. Another observation that can be made from Fig. 4 –24 is that the idealgas specific heats of monatomic gases such as argon, neon, and helium remain constant over the entire temperature range. Thus, u and h of monatomic gases can easily be evaluated from Eqs. 4 –27 and 4 –28. Note that the u and h 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 u cv,avg T is valid for any ideal gas undergoing any process (Fig. 4–27). A similar argument can be given for cp and h. To summarize, there are three ways to determine the internal energy and enthalpy changes of ideal gases (Fig. 4 –28): 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
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Chapter 4

183
Replacing dh by cp dT and du by cv dT and dividing the resulting expression by dT, we obtain cp cv R¬¬1kJ>kg # K2
(4–29)
This is an important relationship for ideal gases since it enables us to determine cv from a knowledge of cp and the gas constant R. When the specific heats are given on a molar basis, R in the above equation should be replaced by the universal gas constant Ru (Fig. 4–29). cp cv Ru¬¬1kJ>kmol # K2
cp cv
(4–31)
The specific ratio also varies with temperature, but this variation is very mild. For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature.
EXAMPLE 4–7
AIR P = constant T1 = 20 °C T2 = 30 °C
Q2
(4–30)
At this point, we introduce another idealgas property called the specific heat ratio k, defined as k
Q1
AIR V = constant T1 = 20 °C T2 = 30 °C
∆u = cv ∆T
∆u = cv ∆T = 7.18 kJ/kg
= 7.18 kJ/kg
FIGURE 4–27 The relation u cv T is valid for any kind of process, constantvolume or not.
Evaluation of the u 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 A–17), (b) the functional form of the specific heat (Table A–2c), and (c) the average specific heat value (Table A–2b).
∆u = u 2 – u1 (table) 2
∆u =
c
v
(T ) dT
1
∆u ≅ cv,avg ∆T
Solution The internal energy change of air is to be determined in three different 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 u 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 A–17 and take the difference:
u1 u @ 300 K 214.07 kJ>kg u2 u @ 600 K 434.78 kJ>kg Thus,
¢u u2 u1 1434.78 214.072 kJ>kg 220.71 kJ>kg
(b) The c–p(T) of air is given in Table A–2c in the form of a thirddegree polynomial expressed as
cp 1T 2 a bT cT 2 dT 3
FIGURE 4–28 Three ways of calculating u.
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184

Thermodynamics where a 28.11, b 0.1967 102, c 0.4802 105, and d 1.966 109. From Eq. 4–30,
AIR at 300 K cv = 0.718 kJ/kg · K R = 0.287 kJ/kg . K
{
From Eq. 4–25,
or cv = 20.80 kJ/kmol . K Ru = 8.314 kJ/kmol . K
cv 1T2 cp Ru 1a Ru 2 bT cT 2 dT 3
cp = 1.005 kJ/kg . K
{
cp = 29.114 kJ/kmol . K
¢u
1
2
¬cv 1T2 ¬dT
T2
T1
¬
3 1a R u 2 bT cT 2 dT 3 4¬ dT
Performing the integration and substituting the values, we obtain
FIGURE 4–29 The cp of an ideal gas can be determined from a knowledge of cv and R.
¢u 6447 kJ>kmol The change in the internal energy on a unitmass basis is determined by dividing this value by the molar mass of air (Table A–1):
¢u
6447 kJ>kmol ¢u 222.5 kJ>kg M 28.97 kg>kmol
which differs from the tabulated value by 0.8 percent. (c) The average value of the constantvolume specific heat cv,avg is determined from Table A–2b at the average temperature of (T1 T2)/2 450 K to be
cv,avg cv @ 450 K 0.733 kJ>kg # K Thus,
¢u cv,avg 1T2 T1 2 10.733 kJ>kg # K2 3 1600 3002K4 220 kJ>kg
Discussion This answer differs from the tabulated value (220.71 kJ/kg) by only 0.4 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 215.4 kJ/kg, which is in error by about 2 percent. Errors of this magnitude are acceptable for most engineering purposes.
EXAMPLE 4–8
Heating of a Gas in a Tank by Stirring
An insulated rigid tank initially contains 1.5 lbm of helium at 80°F and 50 psia. A paddle wheel with a power rating of 0.02 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 451°F. 2 Constant specific heats can be used for helium. 3 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 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.
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Chapter 4

Analysis We take the contents of the tank as the system (Fig. 4–30). 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
2545 Btu>h # b 25.45 Btu Wsh Wsh¬¢t 10.02 hp 2 10.5 h 2 a 1 hp Under the stated assumptions and observations, the energy balance on the system can be expressed as ⎫ ⎪ ⎬ ⎪ ⎭
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Wsh,in ¢U m 1u 2 u 1 2 mcv,avg 1T2 T1 2 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 A–2Ea to be cv 0.753 Btu/lbm · °F. Substituting this and other known quantities into the above equation, we obtain
25.45 Btu 11.5 lbm2 10.753 Btu>lbm # °F 2 1T2 80°F2 T2 102.5°F
(b) The final pressure is determined from the idealgas relation
P1V1 P2V2 T1 T2 where V1 and V2 are identical and cancel out. Then the final pressure becomes
50 psia P2 180 4602 R 1102.5 460 2R P2 52.1 psia
Discussion Note that the pressure in the idealgas relation is always the absolute pressure.
P, psia
He m = 1.5 lbm T1 = 80°F P1 = 50 psia
W sh
P2
2
50
1
V2 = V1
V
FIGURE 4–30 Schematic and PV diagram for Example 4–8.
185
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Thermodynamics EXAMPLE 4–9
Heating of a Gas by a Resistance Heater
A piston–cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27°C. An electric heater within the device is turned on and is allowed to pass a current 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 piston–cylinder 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 147°C, and 3.39 MPa. 2 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 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. 4–31). This is a closed system since no mass crosses the system boundary during the process. We observe that a piston–cylinder 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:
We VI ¢t 1120 V2 12 A 2 15 60 s2 a
1 kJ>s 1000 VA
b 72 kJ
The mass of nitrogen is determined from the idealgas relation:
m
1400 kPa2 10.5 m3 2 P1V1 2.245 kg RT1 10.297 kPa # m3>kg # K 2 1300 K2
Under the stated assumptions and observations, the energy balance on the system can be expressed as
E in E out¬ ¬
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
We,in Q out Wb,out ¢U
We,in Q out ¢H m 1h 2 h 1 2 mcp 1T2 T1 2
since U Wb H for a closed system undergoing a quasiequilibrium expansion or compression process at constant pressure. From Table A–2a, cp 1.039 kJ/kg · K for nitrogen at room temperature. The only unknown quantity in the previous equation is T2, and it is found to be
72 kJ 2.8 kJ 12.245 kg2 11.039 kJ>kg # K 2 1T2 27°C2 T2 56.7°C
Discussion Note that we could also solve this problem by determining the boundary work and the internal energy change rather than the enthalpy change.
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Chapter 4 P, kPa
2A 120 V
N2 P = const.
2800 J
1
400
2
V 1 = 0.5 m 3 P 1 = 400 kPa T1 = 27°C 0.5
V, m3
FIGURE 4–31 Schematic and PV diagram for Example 4–9.
EXAMPLE 4–10
Heating of a Gas at Constant Pressure
A piston–cylinder device initially contains air at 150 kPa and 27°C. At this state, the piston is resting on a pair of stops, as shown in Fig. 4–32, 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 temperature, (b) the work done by the air, and (c) the total heat transferred to the air.
Solution Air in a piston–cylinder 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 pressure relative to its criticalpoint values. 2 The system is stationary and thus the kinetic and potential energy changes are zero, KE PE 0 and E U. 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. 4–32). 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 transferred 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:
1150 kPa2 1V1 2 1350 kPa2 12V1 2 P3V3 P1V1 ¡ T1 T3 300 K T3 T3 1400 K

187
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Thermodynamics P, kPa 2
350
3
AIR A
V 1 = 400 L P 1 = 150 kPa
FIGURE 4–32 Schematic and PV diagram for Example 4–10.
EXPERIMENT
T1 = 27°C
1
150 Q
0.4
0.8
V, m3
(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. 4–32:
A 1V2 V1 2P2 10.4 m3 2 1350 kPa2 140 m3 # kPa Therefore,
W13 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 1–3) can be expressed as
© Ronald Mullisen
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
E in E out¬ ¬
⎫ ⎪ ⎬ ⎪ ⎭
Use actual data from the experiment shown here to obtain the specific heat of aluminum. See endofchapter problem 4–179.
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Q in Wb,out ¢U m 1u 3 u 1 2 EXPERIMENT
The mass of the system can be determined from the idealgas relation:
m
1150 kPa2 10.4 m3 2 P1V1 0.697 kg RT1 10.287 kPa # m3>kg # K2 1300 K2
The internal energies are determined from the air table (Table A–17) to be
u1 u @ 300 K 214.07 kJ>kg u3 u @ 1400 K 1113.52 kJ>kg Thus,
Use actual data from the experiment shown here to obtain the specific heat of aluminum. See endofchapter problem 4–180. © Ronald Mullisen
Q in 140 kJ 10.697 kg2 3 11113.52 214.072 kJ>kg4 Q in 767 kJ
Discussion
The positive sign verifies that heat is transferred to the system.
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Chapter 4
4–5
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 essentially remain constant during a process (Fig. 4–33). 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 that (see Chap. 12) the constantvolume and constantpressure specific heats are identical for incompressible substances (Fig. 4–34). 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, cp cv c
(4–32)

189
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 5 ON THE DVD.
LIQUID vl = constant
SOLID vs = constant
FIGURE 4–33 The specific volumes of incompressible substances remain constant during a process.
This result could also be deduced from the physical definitions of constantvolume and constantpressure specific heats. Specific heat values for several common liquids and solids are given in Table A–3.
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 cv¬dT c 1T 2 dT ˛˛
˛
1
2
c 1T 2 dT¬¬1kJ>kg 2
(4–34)
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 ¢u cavg 1T2 T1 2 ¬¬1kJ>kg2
(4–35)
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 0 →
= 0.45 kJ/kg . °C
(4–33)
The change in internal energy between states 1 and 2 is then obtained by integration: ¢u u 2 u 1
IRON 25°C c = cv = cp
dh du v dP P dv du v dP
(4–36)
¢h ¢u v ¢P cavg ¢T v ¢P¬¬1kJ>kg2
(4–37)
Integrating,
FIGURE 4–34 The cv and cp values of incompressible substances are identical and are denoted by c.
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Thermodynamics For solids, the term v P is insignificant and thus h u ≅ cavgT. For liquids, two special cases are commonly encountered: 1. Constantpressure processes, as in heaters (P 0): h u ≅ cavgT 2. Constanttemperature processes, as in pumps (T 0): h v P For a process between states 1 and 2, the last relation can be expressed as h2 h1 v(P2 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 temperature, the enthalpy of the compressed liquid can be expressed as h @P,T h f @ T vf @ T 1P Psat @ T 2
(4–38)
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 temperature (that is, h@ P,T ≅ hf @ T). However, the contribution of the last term is often very small, and is neglected. (Note that at high temperature and pressures, Eq. 4–38 may overcorrect the enthalpy and result in a larger error than the approximation h ≅ hf @ T.)
EXAMPLE 4–11
Enthalpy of Compressed Liquid
Determine the enthalpy of liquid water at 100°C and 15 MPa (a) by using compressed liquid tables, (b) by approximating it as a saturated liquid, and (c) by using the correction given by Eq. 4–38.
Solution The enthalpy of liquid water is to be determined exactly and approximately. Analysis At 100°C, the saturation pressure of water is 101.42 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 f ¬h 430.39 kJ>kg¬¬1Table A–7 2 T 100°C This is the exact value. (b) Approximating the compressed liquid as a saturated liquid at 100°C, as is commonly done, we obtain
h hf @ 100°C 419.17 kJ>kg This value is in error by about 2.6 percent. (c) From Eq. 4–38,
h@P,T h f @ T vf @ T 1P Psat @ T 2 1419.17 kJ>kg2 10.001 m3 kg2 3 115,000 101.422 kPa4 a
1 kJ b 1 kPa # m3
434.07 kJ>kg Discussion Note that the correction term reduced the error from 2.6 to about 1 percent in this case. However, this improvement in accuracy is often not worth the extra effort involved.
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Chapter 4 EXAMPLE 4–12

Cooling of an Iron Block by Water
A 50kg iron block at 80°C is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25°C. 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, KE PE 0 and E U. 4 There are no electrical, shaft, or other forms of work involved. 5 The system is wellinsulated and thus there is no heat transfer. Analysis We take the entire contents of the tank as the system (Fig. 4–35). This is a closed system since no mass crosses the system boundary during the process. We 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¬ ¬
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
0 ¢U 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
¢Usys ¢Uiron ¢Uwater 0
3mc 1T2 T1 2 4 iron 3mc 1T2 T1 2 4 water 0 The specific volume of liquid water at or about room temperature can be taken to be 0.001 m3/kg. Then the mass of the water is
mwater
V 0.5 m3 500 kg v 0.001 m3>kg
The specific heats of iron and liquid water are determined from Table A–3 to be ciron 0.45 kJ/kg · °C and cwater 4.18 kJ/kg · °C. Substituting these values into the energy equation, we obtain
150 kg 2 10.45 kJ>kg # °C2 1T2 80°C2 1500 kg2 14.18 kJ>kg # °C2 1T2 25°C2 0 T2 25.6°C
Therefore, when thermal equilibrium is established, both the water and iron will be at 25.6°C. Discussion The small rise in water temperature is due to its large mass and large specific heat.
WATER 25°C IRON m = 50 kg 80°C 0.5 m 3
FIGURE 4–35 Schematic for Example 4–12.
191
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192

Thermodynamics EXAMPLE 4–13
Temperature Rise due to Slapping
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 person, which caused the temperature of the affected area of your face to rise by 1.8°C (ouch!). Assuming the slapping hand has a mass of 1.2 kg and about 0.150 kg of the tissue on the face and the hand is affected by the incident, estimate the velocity of the hand just before impact. Take the specific heat of the tissue to be 3.8 kJ/kg · °C.
Solution The face of a person is slapped. For the specified temperature rise of the affected part, the impact velocity of the hand is to be determined. Assumptions 1 The hand is brought to a complete stop after the impact. 2 The face takes the blow without significant movement. 3 No heat is transferred from the affected area to the surroundings, and thus the process is adiabatic. 4 No work is done on or by the system. 5 The potential energy change is zero, PE 0 and E U KE. Analysis We analyze this incident in a professional manner without involving any emotions. First, we identify the system, draw a sketch of it, and state our observations about the specifics of the problem. We take the hand and the affected portion of the face as the system (Fig. 4–36). This is a closed system since it involves a fixed amount of mass (no mass transfer). We observe that the kinetic energy of the hand decreases during the process, as evidenced by a decrease in velocity from initial value to zero, while the internal energy of the affected area increases, as evidenced by an increase in the temperature. There seems to be no significant energy transfer between the system and its surroundings during this process. Under the stated assumptions and observations, the energy balance on the system can be expressed as
FIGURE 4–36 Schematic for Example 4–13.
E in E out¬ ¬ Net energy transfer by heat, work, and mass
¢E system
Change in internal, kinetic, potential, etc., energies
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
0 ¢Uaffected tissue ¢KEhand
0 1mc¢T2 affected tissue 3m 10 V 2 2>2 4 hand
That is, the decrease in the kinetic energy of the hand must be equal to the increase in the internal energy of the affected area. Solving for the velocity and substituting the given quantities, the impact velocity of the hand is determined to be
Vhand
B
2 1mc¢T2 affected tissue mhand
2 10.15 kg2 13.8 kJ>kg # °C2 11.8°C2 1000 m2>s2 a b B 1.2 kg 1 kJ>kg
41.4 m>s 1or 149 km>h2 Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering.
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Chapter 4 TOPIC OF SPECIAL INTEREST*

193
Thermodynamic Aspects of Biological Systems
An important and exciting application area of thermodynamics is biological systems, which are the sites of rather complex and intriguing energy transfer and transformation 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 25.5 percent, carbon 9.5 percent, and nitrogen 1.4 percent of all the atoms. The remaining 0.6 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 0.01 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 during 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 37°C while performing the necessary bodily tasks, is called metabolism. In simple terms, metabolism 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 functions (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, 1.8m2 body surface area), the basal metabolic rate is 84 W. That is, the body dissipates energy to the environment at a rate of 84 W, which means that the body is converting chemical energy of the food (or of the body fat if the person has not eaten) into thermal energy at a rate of 84 W (Fig. 4–37). The metabolic rate increases with the level of activity, 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. 4–38). 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 conditions, 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 remarkable that about half of the basal metabolic activity occurs in the brain alone. *This section can be skipped without a loss in continuity.
FIGURE 4–37 An average person dissipates energy to the surroundings at a rate of 84 W when resting. © Vol. 124/PhotoDisc
1.2 kJ/s
1 kJ/s
FIGURE 4–38 Two fastdancing people supply more energy to a room than a 1kW electric resistance heater.
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Thermodynamics
A 300W refrigerator
A 300W resistance heater
A 300W fan
A 300W TV
Two people, each dissipating 150 W
Three light bulbs, 100 W each A 100W computer with a 200W monitor
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 eventually 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 moving 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 electricity 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 light bulbs will produce the same heating effect in a room as a 300W resistance heater (Fig. 4–39). 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 system remains constant during a process.
Food and Exercise Solar energy 300 W
FIGURE 4–39 Some arrangements that supply a room the same amount of energy as a 300W electric resistance heater. Mixer and motor Electrical switch Thermometer Water Bomb (combustion chamber) Insulation
Food sample
FIGURE 4–40 Schematic of a bomb calorimeter used to determine the energy content of food samples.
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 2:1 ratio in their molecules. The molecules of carbohydrates range from very simple (as in plain sugar) to very complex or large (as in starch). Bread and plain 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 proteins 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 carbohydrate, 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 wellinsulated rigid tank (Fig. 4–40). The tank contains a small combustion chamber surrounded by water. The food is ignited and burned in the combustion chamber in the presence of excess oxygen, and the energy released is transferred 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
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Chapter 4 18.0 MJ/kg for carbohydrates, 22.2 MJ/kg for proteins, and 39.8 MJ/kg for fats. These food groups are not entirely metabolized in the human body, however. The fraction of metabolizable energy contents are 95.5 percent for carbohydrates, 77.5 percent for proteins, and 97.7 percent for fats. That is, the fats we eat are almost entirely metabolized in the body, but close to one quarter of the protein we eat is discarded from the body unburned. This corresponds to 4.1 Calories/g for proteins and carbohydrates and 9.3 Calories/g for fats (Fig. 4–41) commonly seen in nutrition books and on food labels. The energy contents of the foods we normally eat are much lower than the values above 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 4.2 MJ/kg for carbohydrates, 8.4 MJ/kg for proteins, and 33.1 MJ/kg 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 nutritionists in terms of the capitalized Calories. One Calorie is equivalent to one kilocalorie (1000 calories), which is equivalent to 4.1868 kJ. That is, 1 Cal 1Calorie2 1000 calories 1 kcal 1kilocalorie2 4.1868 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. 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 2.205 times the body weight in kg) by 11 for a sedentary person, 13 for a moderately active person, 15 for a moderate exerciser 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 the spare energy of the body 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 33.1 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 requirements by burning only 9211/33,100 0.28 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

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FIGURE 4–41 Evaluating the calorie content of one serving of chocolate chip cookies (values are for Chips Ahoy cookies made by Nabisco). © Vol. 30/PhotoDisc
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Thermodynamics 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 a reasonable amount of calorie intake. The average metabolizable energy contents of various foods and the energy consumption during various activities are given in Tables 4–1 and 4–2. Considering 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. The rates of energy consumption listed in Table 4–2 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 4–2 to be 639 Calories/h. Then the rate of energy consumption by a 50kg bicyclist is 150 kg2
639 Cal>h 68 kg
470 Cal>h
For a 100kg person, it would be 940 Cal/h. The thermodynamic analysis of the human body is rather complicated since it involves mass transfer (during breathing, perspiring, etc.) as well as energy transfer. As such, 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
TABLE 4–1 Approximate metabolizable energy content of some common foods (1 Calorie 4.1868 kJ 3.968 Btu) Food 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)
Calories 70 250 550 70 35 325 105 87 80
Food Fish sandwich French fries (regular) Hamburger Hot dog Ice cream (100 ml, 10% fat) Lettuce salad with French dressing
Calories 450 250 275 300 110 150
Food Milk (skim, 200 ml) Milk (whole, 200 ml) Peach (one, medium) Pie (one 18– slice, 23 cm diameter) Pizza (large, cheese, one 18– slice)
Calories 76 136 65 300 350
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Chapter 4 will gain weight whereas a person who consumes less 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 fuelefficient 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 exercise 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. 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 burning calories for several hours after the workout, raising the overall metabolic rate considerably. Another theory suggests that people with too many fat cells developed during childhood or adolescence are much more likely to gain weight. Some people believe that the fat content of the 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 Rockefeller University have identified a gene, called the RIIbeta, that seems to control the rate of metabolism. The body tries to keep the body fat at a particular level, called the set point, that differs from person to person (Fig. 4–42). This is done by speeding up the metabolism and thus burning extra calories much faster when a person tends to gain weight and by slowing down the metabolism and thus burning calories at a slower rate 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

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TABLE 4–2 Approximate energy consumption of a 68kg adult during some activities (1 Calorie 4.1868 kJ 3.968 Btu) Activity Basal metabolism Basketball Bicycling (21 km/h) Crosscountry skiing (13 km/h) Driving a car Eating Fast dancing Fast running (13 km/h) Jogging (8 km/h) Swimming (fast) Swimming (slow) Tennis (advanced) Tennis (beginner) Walking (7.2 km/h) Watching TV
Calories/h 72 550 639 936 180 99 600 936 540 860 288 480 288 432 72
Body fat level
Set point New set point
FIGURE 4–42 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.
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TABLE 4–3 The range of healthy weight for adults of various heights (Source: National Institute of Health) English Units Healthy Height, weight, in. lbm* 58 60 62 64 66 68 70 72 74 76
91–119 97–127 103–136 111–146 118–156 125–165 133–175 140–185 148–195 156–205
SI Units Height, m
Healthy weight, kg*
1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90
40–53 43–56 46–60 49–64 52–68 55–72 58–77 62–81 65–86 69–90
*The upper and lower limits of healthy range correspond to mass body indexes of 19 and 25, respectively.
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. However, 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 don’t impose on it. Don’t 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. 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 weightrelated 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 don’t 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
W 1kg2
H 2 1m2 2
¬with
BMI 6 19¬¬ underweight healthy weight 19 BMI 25¬¬ overweight BMI 7 25¬¬
(4–39)
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 above is equivalent in English units to BMI 705 W/H 2 where W is in pounds and H is in inches. The proper range of weight for adults of various heights is given in Table 4–3 in both SI and English units.
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Chapter 4 EXAMPLE 4–14

Burning Off Lunch Calories
A 90kg man had two hamburgers, a regular serving of french fries, and a 200ml Coke for lunch (Fig. 4–43). 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 times 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 4–1 and 4–2 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 4–1 to be
FIGURE 4–43 A typical lunch discussed in Example 4–14. © Vol. 30/PhotoDisc
Ein 2 Ehamburger Efries Ecola 2 275 250 87 887 Cal The rate of energy output for a 68kg man watching TV is given in Table 4–2 to be 72 Calories/h. For a 90kg man it becomes
72 Cal>h 95.3 Cal>h E out 190 kg2 ¬ 68 kg Therefore, it will take
¢t
887 Cal 9.3 h 95.3 Cal>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: 18.6 h by watching TV and 94 min by fast swimming.
EXAMPLE 4–15
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 (28.3g) serving of regular potato chips has 10 g of fat and 150 Calories, whereas 1 oz of the socalled 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. 4–44).
FIGURE 4–44 Schematic for Example 4–15.
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Thermodynamics 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
Ereduced 175 Cal>day2 1365 day>year2 27,375 Cal>year
The metabolizable energy content of 1 kg of body fat is 33,100 kJ. Therefore, assuming the deficit in the calorie intake is made up by burning body fat, the person who switches to fatfree chips will lose
mfat
Ereduced 27,375 Cal 4.1868 kJ a b 3.46 kg Energy content of fat 33,100 kJ>kg 1 Cal
(about 7.6 pounds) of body fat that year.
SUMMARY
(2) Isobaric process
Wb P0 1V2 V1 2
Net energy transfer by heat, work, and mass
It can also be expressed in the rate form as . . E in Eout¬ ¬ dE system>dt¬¬1kW2 Rate of net energy transfer by heat, work, and mass
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
(P1 P2 P0 constant)
(3) Polytropic process
Q W ¢U ¢KE ¢PE¬¬1kJ2 where
P2V2 P1V1 Wb ¬1n 1 2 (PV n constant) 1n (4) Isothernal process of an ideal gas V2 V2 Wb P1V1 ln¬ mRT0 ln¬ V1 V1
Change in internal, kinetic, potential, etc., energies
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
P dV
¬
1
⎫ ⎪ ⎬ ⎪ ⎭
¢E system¬¬1kJ2
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
Wb
(1) General
2
E in E out¬ ¬
⎫ ⎪ ⎬ ⎪ ⎭
Work is the energy transferred as a force acts on a system through a distance. The most common form of mechanical work is the boundary work, which is the work associated 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:
(PV mRT0 constant)
The first law of thermodynamics is essentially an expression of the conservation of energy principle, also called the energy balance. The general energy balances for any system undergoing any process can be expressed as
W Wother Wb
¢U m 1u 2 u 1 2
¢KE 12m 1V 22 V 21 2 ¢PE mg 1z 2 z 1 2
For a constantpressure process, Wb U H. Thus, Q Wother ¢H ¢KE ¢PE¬¬1kJ2
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Chapter 4 The amount of energy needed to raise the temperature of a unit mass of a substance by one degree is called the specific heat at constant volume cv for a constantvolume process and the specific heat at constant pressure cp for a constantpressure process. They are defined as cv a
0u 0h b ¬and¬cp a b 0T v 0T p
For ideal gases u, h, cv, and cp are functions of temperature alone. The u and h of ideal gases are expressed as ¢u u 2 u 1
2
2
c 1T2 dT cv,avg 1T2 T1 2 ¬ v
1
¢h h 2 h 1
1
cp 1T2 dT cp,avg 1T2 T1 2
¬

201
where R is the gas constant. The specific heat ratio k is defined as k
cp cv
For incompressible substances (liquids and solids), both the constantpressure and constantvolume specific heats are identical and denoted by c: cp cv c¬¬1kJ>kg # K2 The u and h of imcompressible substances are given by ¢u
1
2
c 1T2 dT cavg 1T2 T1 2¬¬1kJ>kg2
¬
¢h ¢u v¢P¬¬1kJ>kg2
For ideal gases, cv and cp are related by cp cv R¬¬1kJ>kg # K2
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.
PROBLEMS* Moving Boundary Work 4 –1C On a Pv diagram, what does the area under the process curve represent? 4 –2C Is the boundary work associated with constantvolume systems always zero? 4 –3C An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. For which case is the work done greater? 4 –4C Show that 1 kPa · m3 1 kJ.
*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 a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
4–5 A piston–cylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120°C. The nitrogen is now expanded polytropically to a state of 100 kPa and 100°C. Determine the boundary work done during this process. 4–6 A piston–cylinder device with a set of stops initially contains 0.3 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 60 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is (a) 1.0 MPa and 250°C and (b) 500 kPa. (c) Also determine the temperature at the final state in part (b).
Steam 0.3 kg 1 MPa 400°C
FIGURE P4–6
Q
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Thermodynamics
4 –7 A piston–cylinder device initially contains 0.07 m3 of nitrogen gas at 130 kPa and 120°C. The nitrogen is now expanded to a pressure of 100 kPa polytropically with a polytropic exponent whose value is equal to the specific heat ratio (called isentropic expansion). Determine the final temperature and the boundary work done during this process.
process, the pressure changes with volume according to the relation P aV b, where a 1200 kPa/m3 and b 600 kPa. Calculate the work done during this process (a) by plotting the process on a PV diagram and finding the area under the process curve and (b) by performing the necessary integrations.
4 –8 A mass of 5 kg of saturated water vapor at 300 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate the work done by the steam during this process. Answer: 165.9 kJ GAS
4 –9 A frictionless piston–cylinder device initially contains 200 L of saturated liquid refrigerant134a. The piston is free to move, and its mass is such that it maintains a pressure of 900 kPa on the refrigerant. The refrigerant is now heated until its temperature rises to 70°C. Calculate the work done during this process. Answer: 5571 kJ
R134a P = const.
FIGURE P4–9
P = aV + b
FIGURE P4–14 4–15E 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 psia/ft3 and b is a constant. If the initial volume of the gas is 7 ft3, calculate the work done during the process. Answer: 181 Btu 4–16
During some actual expansion and compression processes in piston–cylinder devices, the gases have been observed to satisfy the relationship PV n C, where n and C are constants. Calculate the work done when a gas expands from 150 kPa and 0.03 m3 to a final volume of 0.2 m3 for the case of n 1.3.
4 –10
Reconsider Prob. 4 –9. Using EES (or other) software, investigate the effect of pressure on the work done. Let the pressure vary from 400 kPa to 1200 kPa. Plot the work done versus the pressure, and discuss the results. Explain why the plot is not linear. Also plot the process described in Prob. 4 –9 on the Pv diagram.
4–17
4 –11E A frictionless piston–cylinder device contains 16 lbm of superheated water vapor at 40 psia and 600°F. Steam is now cooled at constant pressure until 70 percent of it, by mass, condenses. Determine the work done during this process.
4–18 A frictionless piston–cylinder device contains 2 kg of nitrogen at 100 kPa and 300 K. Nitrogen is now compressed slowly according to the relation PV1.4 constant until it reaches a final temperature of 360 K. Calculate the work input during this process. Answer: 89 kJ
4 –12 A mass of 2.4 kg of air at 150 kPa and 12°C is contained in a gastight, frictionless piston–cylinder 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 temperature inside the cylinder remains constant. Calculate the work input during this process. Answer: 272 kJ 4 –13 Nitrogen at an initial state of 300 K, 150 kPa, and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 800 kPa. Determine the work done during this process. 4 –14 A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasiequilibrium
Reconsider Prob. 4–16. Using the EES (or other) 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. Let the polytropic exponent vary from 1.1 to 1.6. Plot the boundary work versus the polytropic exponent, and discuss the results.
N2 PV1.4 = const.
FIGURE P4–18
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Chapter 4 The equation of state of a gas is given as v (P 10/ v 2) RuT, where the units of v and P are 3 m /kmol and kPa, respectively. Now 0.5 kmol of this gas is expanded in a quasiequilibrium manner from 2 to 4 m3 at a constant temperature of 300 K. Determine (a) the unit of the quantity 10 in the equation and (b) the work done during this isothermal expansion process.
4 –19
4 –20
Reconsider Prob. 4 –19. Using the integration feature of the EES software, calculate the work done, and compare your result with the “handcalculated” result obtained in Prob. 4 –19. Plot the process described in the problem on a Pv diagram.
4 –21 Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volume are related by P aV2, where a 8 kPa · m6. Calculate the work done on the carbon dioxide during this process. Answer: 53.3 kJ 4 –22E Hydrogen is contained in a piston–cylinder device at 14.7 psia and 15 ft3. At this state, a linear spring (F ∝ x) with a spring constant of 15,000 lbf/ft is touching the piston but exerts no force on it. The crosssectional area of the piston is 3 ft2. Heat is transferred to the hydrogen, causing it to expand until its volume doubles. Determine (a) the final pressure, (b) the total work done by the hydrogen, and (c) the fraction of this work done against the spring. Also, show the process on a PV diagram. 4 –23 A piston–cylinder device contains 50 kg of water at 250 kPa and 25°C. The crosssectional area of the piston is 0.1 m2. Heat is now transferred to the water, causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the piston reaches a linear spring whose spring constant is 100 kN/m. More heat is transferred to the water until the piston rises 20 cm more. Determine (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, 147.9°C, (b) 44.5 kJ

203
4–24
Reconsider Prob. 4–23. Using the EES software, 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 kN/m to 500 kN/m. Plot the final pressure and the boundary work against the spring constant, and discuss the results.
4–25 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, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L. 4–26 A piston–cylinder device initially contains 0.25 kg of nitrogen gas at 130 kPa and 120°C. The nitrogen is now expanded isothermally to a pressure of 100 kPa. Determine the boundary work done during this process. Answer: 7.65 kJ
N2 130 kPa 120°C
FIGURE P4–26 4–27 A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the boundary work for each process and the net work of the cycle.
Closed System Energy Analysis 4–28 A 0.5m3 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. 2
A = 0.1 m H 2O m = 50 kg
FIGURE P4–23
4–29E A 20ft3 rigid tank initially contains saturated refrigerant134a 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 refrigerant that has condensed, and (c) the heat transfer.
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4 –30 A wellinsulated rigid tank contains 5 kg of a saturated liquid–vapor mixture of water at l00 kPa. Initially, threequarters of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 110V source, and a current of 8 A flows through the resistor when the switch is turned on. Determine how long it will take to vaporize all the liquid in the tank. Also, show the process on a Tv diagram with respect to saturation lines.
H 2O
V = constant
We
FIGURE P4–30
4–34 A piston–cylinder device contains 5 kg of refrigerant134a at 800 kPa and 70°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 15°C. Determine the amount of heat loss and show the process on a Tv diagram with respect to saturation lines. Answer: 1173 kJ 4–35E A piston–cylinder device contains 0.5 lbm of water initially at 120 psia and 2 ft3. Now 200 Btu of heat is transferred to the water while its pressure is held constant. Determine the final temperature of the water. Also, show the process on a Tv diagram with respect to saturation lines. 4–36 An insulated piston–cylinder 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 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
4 –31
Reconsider Prob. 4 –30. Using EES (or other) software, investigate the effect of the initial mass of water on the length of time required to completely vaporize the liquid. Let the initial mass vary from 1 to 10 kg. Plot the vaporization time against the initial mass, and discuss the results.
H2O P = constant
4 –32 An insulated tank is divided into two parts by a partition. One part of the tank contains 2.5 kg of compressed liquid water at 60°C and 600 kPa while the other part is evacuated. The partition is now removed, and the water expands to fill the entire tank. Determine the final temperature of the water and the volume of the tank for a final pressure of 10 kPa.
Evacuated Partition
H2O
FIGURE P4–32 4 –33
Reconsider Prob. 4 –32. Using EES (or other) software, 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.
Wsh
We
FIGURE P4–36 4–37 A piston–cylinder device contains steam initially at 1 MPa, 450°C, and 2.5 m3. Steam is allowed to cool at constant pressure until it first starts condensing. Show the process on a Tv diagram with respect to saturation lines and determine (a) the mass of the steam, (b) the final temperature, and (c) the amount of heat transfer. 4–38
A piston–cylinder device initially contains steam at 200 kPa, 200°C, and 0.5 m3. At this state, a linear spring (F x) is touching the piston but exerts no force on it. Heat is now slowly transferred to the steam, causing the pressure and the volume to rise to 500 kPa and 0.6 m3, respectively. Show the process on a Pv diagram with respect to saturation lines and determine (a) the final temperature, (b) the work done by the steam, and (c) the total heat transferred. Answers: (a) 1132°C, (b) 35 kJ, (c) 808 kJ
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4–42 A 30L electrical radiator containing heating oil is placed in a 50m3 room. Both the room and the oil in the radiator are initially at 10°C. The radiator with a rating of 1.8 kW is now turned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s. After some time, the average temperature is measured to be 20°C for the air in the room, and 50°C for the oil in the radiator. Taking the density and the specific heat of the oil to be 950 kg/m3 and 2.2 kJ/kg °C, respectively, determine how long the heater is kept on. Assume the room is wellsealed so that there are no air leaks.
H2 O 200 kPa 200°C
Q

FIGURE P4–38 4 –39
Reconsider Prob. 4 –38. Using EES (or other) software, investigate the effect of the initial temperature of steam on the final temperature, the work done, and the total heat transfer. Let the initial temperature vary from 150 to 250°C. Plot the final results against the initial temperature, and discuss the results.
4 –40 A piston–cylinder device initially contains 0.8 m3 of saturated 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) 662°C, (b) 240 kJ, (c) 1213 kJ
4 –41 Two tanks (Tank A and Tank B) are separated by a partition. Initially Tank A contains 2kg steam at 1 MPa and 300°C while Tank B contains 3kg saturated liquid–vapor mixture with a vapor mass fraction of 50 percent. Now the partition is removed and the two sides are allowed to mix until the 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.
TANK A 2 kg 1 MPa 300°C
TANK B 3 kg 150°C x = 0.5
Q
FIGURE P4–41
Room 10°C
Q
Radiator
FIGURE P4–42 Specific Heats, u, and h of Ideal Gases 4–43C Is the relation u mcv,avgT restricted to constantvolume processes only, or can it be used for any kind of process of an ideal gas? 4–44C Is the relation h mcp,avgT restricted to constantpressure processes only, or can it be used for any kind of process of an ideal gas? _ _ 4–45C Show that for an ideal gas cp cv Ru. 4–46C 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. 4–47C In the relation u mcv T, what is the correct unit of cv — kJ/kg · °C or kJ/kg · K? 4–48C A fixed mass of an ideal gas is heated from 50 to 80°C 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? 4–49C A fixed mass of an ideal gas is heated from 50 to 80°C 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? 4–50C A fixed mass of an ideal gas is heated from 50 to 80°C (a) at constant volume and (b) at constant pressure. For which case do you think the energy required will be greater? Why? 4–51 Determine the kJ/kg, as it is heated empirical specific heat (Table A–2c), (b) the
enthalpy change h of nitrogen, in from 600 to 1000 K, using (a) the equation as a function of temperature cp value at the average temperature
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(Table A–2b), and (c) the cp value at room temperature (Table A–2a). Answers: (b) 447.8 kJ/kg, (b) 448.4 kJ/kg, (c) 415.6 kJ/kg
4 –52E Determine the enthalpy change h 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 A–2Ec), (b) the cp value at the average temperature (Table A–2Eb), and (c) the cp value at room temperature (Table A–2Ea). Answers: (a) 170.1 Btu/lbm, (b) 178.5 Btu/lbm, (c) 153.3 Btu/lbm
4 –53 Determine the internal energy change u of hydrogen, in kJ/kg, as it is heated from 200 to 800 K, using (a) the empirical specific heat equation as a function of temperature (Table A–2c), (b) the cv value at the average temperature (Table A–2b), and (c) the cv value at room temperature (Table A–2a).
at a rate of 10,000 kJ/h, 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 kJ/h. If the initial temperature of the room air is 10°C, determine how long it will take for the air temperature to rise to 20°C. Assume constant specific heats at room temperature. 4–59 A student living in a 4m 6m 6m dormitory room turns on her 150W fan before she leaves the room on a summer day, hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 h later. Use specific heat values at room temperature, and assume the room to be at 100 kPa and 15°C in the morning when she leaves. Answer: 58.2°C
ClosedSystem Energy Analysis: Ideal Gases 4 –54C Is it possible to compress an ideal gas isothermally in an adiabatic piston–cylinder device? Explain. 4 –55E A rigid tank contains 20 lbm of air at 50 psia and 80°F. The air is now heated until its pressure doubles. Determine (a) the volume of the tank and (b) the amount of heat transfer. Answers: (a) 80 ft3, (b) 1898 Btu
ROOM 4m×6m×6m
Fan
4 –56 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.
FIGURE P4–59
4 –57 A 4m 5m 6m room is to be heated by a baseboard resistance heater. It is desired that the resistance heater be able to raise the air temperature in the room from 7 to 23°C within 15 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: 1.91 kW
4–60E A 10ft3 tank contains oxygen initially at 14.7 psia and 80°F. 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.
4 –58 A 4m 5m 7m room is heated by the radiator of a steamheating system. The steam radiator transfers heat 5000 kJ/h
4–61 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 50°C, and the other part is evacuated. The partition is now removed, and the gas expands into the entire tank. Determine the final temperature and pressure in the tank.
ROOM 4m×5m×7m Steam
IDEAL GAS
· Wpw
800 kPa 50°C
10,000 kJ/h
FIGURE P4–58
Evacuated
FIGURE P4–61
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Chapter 4 4 –62 A piston–cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa and 25°C. The mass of the piston is such that 500 kPa of pressure is required to raise it. How much heat must be transferred to the helium before the piston starts rising? Answer: 1857 kJ
4 –63 An insulated piston–cylinder device contains 100 L of air at 400 kPa and 25°C. 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.

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transfer. Let the polytropic exponent vary from 1.1 to 1.6. Plot the boundary work and the heat transfer versus the polytropic exponent, and discuss the results. 4–69 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 6500 kJ/h, the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in kW.
ROOM
4 –64E A piston–cylinder device contains 25 ft3 of nitrogen at 40 psia and 700°F. Nitrogen is now allowed to cool at constant pressure until the temperature drops to 200°F. Using specific heats at the average temperature, determine the amount of heat loss.
Q
Tair = constant We
4 –65 A mass of 15 kg of air in a piston–cylinder device is heated from 25 to 77°C 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.
FIGURE P4–69
Answer: 0.235 kWh
4–70E A piston–cylinder device contains 3 ft3 of air at 60 psia and 150°F. Heat is transferred to the air in the amount of 40 Btu as the air expands isothermally. Determine the amount of boundary work done during this process. AIR P = constant We
Q
FIGURE P4–65 4 –66 An insulated piston–cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa and 27°C. 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. 4 –67 A piston–cylinder device contains 0.8 kg of nitrogen initially at 100 kPa and 27°C. The nitrogen is now compressed slowly in a polytropic process during which PV1.3 constant until the volume is reduced by onehalf. Determine the work done and the heat transfer for this process. 4 –68
Reconsider Prob. 4 –67. Using EES (or other) 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
4–71 A piston–cylinder device contains 4 kg of argon at 250 kPa and 35°C. During a quasiequilibrium, isothermal expansion process, 15 kJ of boundary work is done by the system, and 3 kJ of paddlewheel work is done on the system. Determine the heat transfer for this process. 4–72 A piston–cylinder device, whose piston is resting on a set of stops, initially contains 3 kg of air at 200 kPa and 27°C. 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
4–73
A piston–cylinder device, with a set of stops on the top, initially contains 3 kg of air at 200 kPa and 27°C. Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a Pv diagram.
ClosedSystem Energy Analysis: Solids and Liquids 4–74 In a manufacturing facility, 5cmdiameter brass balls (r 8522 kg/m3 and cp 0.385 kJ/kg · °C) initially at 120°C are quenched in a water bath at 50°C for a period of 2 min at
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a rate of 100 balls per minute. If the temperature of the balls after quenching is 74°C, determine the rate at which heat needs to be removed from the water in order to keep its temperature constant at 50°C.
120°C
Brass balls
50°C
Water bath
4–78 Stainless steel ball bearings (r 8085 kg/m3 and cp 0.480 kJ/kg · °C) having a diameter of 1.2 cm are to be quenched in water at a rate of 800 per minute. The balls leave the oven at a uniform temperature of 900°C and are exposed to air at 25°C for a while before they are dropped into the water. If the temperature of the balls drops to 850°C prior to quenching, determine the rate of heat transfer from the balls to the air. 4–79 Carbon steel balls (r 7833 kg/m3 and cp 0.465 kJ/kg · °C) 8 mm in diameter are annealed by heating them first to 900°C in a furnace, and then allowing them to cool slowly to 100°C in ambient air at 35°C. If 2500 balls are to be annealed per hour, determine the total rate of heat transfer from the balls to the ambient air. Answer: 542 W
FIGURE P4–74 4 –75
4 –76E During a picnic on a hot summer day, all the cold drinks disappeared quickly, and the only available drinks were those at the ambient temperature of 75°F. 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 32°F. Using the properties of water for the drink, determine the mass of ice that will melt by the time the canned drink cools to 45°F. 4 –77 Consider a 1000W iron whose base plate is made of 0.5cmthick aluminum alloy 2024T6 (r 2770 kg/m3 and cp 875 J/kg · °C). The base plate has a surface area of 0.03 m2. Initially, the iron is in thermal equilibrium with the ambient air at 22°C. Assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine the minimum time needed for the plate temperature to reach 140°C.
Air, 35°C
Furnace
Repeat Prob. 4 –74 for aluminum balls.
900°C
Steel ball
100°C
FIGURE P4–79 4–80 An electronic device dissipating 30 W has a mass of 20 g and a specific heat of 850 J/kg · °C. The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of 25°C. Determine 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 0.2kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal. 4–81
Reconsider Prob. 4–80. Using EES (or other) software, investigate the effect of the mass of the heat sink on the maximum device temperature. Let the mass of heat sink vary from 0 to 1 kg. Plot the maximum temperature against the mass of heat sink, and discuss the results.
4–82 An ordinary egg can be approximated as a 5.5cmdiameter sphere. The egg is initially at a uniform temperature of 8°C and is dropped into boiling water at 97°C. Taking the properties of the egg to be r 1020 kg/m3 and cp 3.32 kJ/kg · °C, determine how much heat is transferred to the egg by the time the average temperature of the egg rises to 80°C.
FIGURE P4–77 © Vol. 58/PhotoDisc
4–83E ln a production facility, 1.2inthick 2ft 2ft square brass plates (r 532.5 lbm/ft3 and cp 0.091 Btu/lbm · °F) that are initially at a uniform temperature of 75°F are heated by passing them through an oven at 1300°F at a rate of 300 per minute. If the plates remain in the oven until their average temperature rises to 1000°F, determine the rate of heat transfer to the plates in the furnace.
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4–94 A 55kg 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: 2.52 h
4–95 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.
1.2 in.
Brass plate, 75°F
FIGURE P4–83E 4 –84 Long cylindrical steel rods (r 7833 kg/m3 and cp 0.465 kJ/kg · °C) of 10cm diameter are heattreated by drawing them at a velocity of 3 m/min through an oven maintained at 900°C. If the rods enter the oven at 30°C and leave at a mean temperature of 700°C, determine the rate of heat transfer to the rods in the oven.
Special Topic: Biological Systems 4 –85C What is metabolism? What is basal metabolic rate? What is the value of basal metabolic rate for an average man? 4 –86C For what is the energy released during metabolism in humans used? 4 –87C Is the metabolizable energy content of a food the same as the energy released when it is burned in a bomb calorimeter? If not, how does it differ? 4 –88C Is the number of prospective occupants an important consideration in the design of heating and cooling systems of classrooms? Explain. 4 –89C 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? 4 –90 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 faster? 4 –91 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: 1.045 kg 4 –92 Consider a classroom that is losing heat to the outdoors at a rate of 20,000 kJ/h. If there are 30 students in class, each dissipating sensible heat at a rate of 100 W, determine if it is necessary to turn the heater in the classroom on to prevent the room temperature from dropping. 4 –93 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.
4–96 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 teaspoons of butter while Wendy eats hers plain every evening. Determine the difference in the weights of Candy and Wendy after one year. Answer: 6.5 kg 4–97 A woman who used to drink about one liter of regular cola every day switches to diet cola (zero calorie) and starts eating two slices of apple pie every day. Is she now consuming fewer or more calories? 4–98 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: 0.087kg gain 4–99 The average specific heat of the human body is 3.6 kJ/kg · °C. If the body temperature of an 80kg man rises from 37°C to 39°C during strenuous exercise, determine the increase in the thermal energy of the body as a result of this rise in body temperature. 4–100E Alcohol provides 7 Calories per gram, but it provides no essential nutrients. A 1.5 ounce serving of 80proof liquor contains 100 Calories in alcohol alone. Sweet wines and beer provide additional calories since they also contain carbohydrates. About 75 percent of American adults drink some sort of alcoholic beverage, which adds an average of 210 Calories a day to their diet. Determine how many pounds less an average American adult will weigh per year if he or she quit drinking alcoholic beverages and started drinking diet soda. 4–101 A 12oz serving of a regular beer contains 13 g of alcohol and 13 g of carbohydrates, and thus 150 Calories. A 12oz serving of a light beer contains 11 g of alcohol and 5 g Regular beer
Light beer
12 oz. 150 cal
12 oz. 100 cal
FIGURE P4–101
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of carbohydrates, and thus 100 Calories. An average person burns 700 Calories per hour while exercising on a treadmill. Determine how long it will take to burn the calories from a 12oz can of (a) regular beer and (b) light beer on a treadmill. 4 –102 A 5oz serving of a Bloody Mary contains 14 g of alcohol and 5 g of carbohydrates, and thus 116 Calories. A 2.5oz 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 exercising 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. 4 –103E A 176pound man and a 132pound woman went to Burger King for lunch. The man had a BK Big Fish sandwich (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 360 Cal/h for the woman and 480 Cal/h for the man. Determine how long each one of them needs to shovel snow to burn off the lunch calories. 4 –104 Consider two friends who go to Burger King every day for lunch. One of them orders a Double Whopper sandwich, 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. 4 –105E A 150pound person goes to Hardee’s for dinner and orders a regular roast beef (270 Cal) and a big roast beef (410 Cal) sandwich together with a 12oz can of Pepsi (150 Cal). A 150pound person burns 400 Calories per hour while climbing stairs. Determine how long this person needs to climb stairs to burn off the dinner calories. 4 –106 A person eats a McDonald’s Big Mac sandwich (530 Cal), a second person eats a Burger King Whopper sandwich (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.
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. 4–109 The body mass index (BMI) of a 1.7m tall 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 25. Use the data in the text for calories and take the metabolizable energy content of 1 kg of body fat to be 33,100 kJ. Answer: 262 days
Review Problems 4–110 Consider a piston–cylinder device that contains 0.5 kg air. Now, heat is transferred to the air at constant pressure and the air temperature increases by 5°C. Determine the expansion work done during this process. 4–111 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 phase change process, and thus materials that change phase at about room temperature such as glaubers salt (sodium sulfate decahydrate), which has a melting point of 32°C and a heat of fusion of 329 kJ/L, are very suitable for this purpose. Determine 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 2.32 kJ/kg · °C and a temperature change of 20°C, and (c) water with a heat capacity of 4.00 kJ/kg · °C and a temperature change of 20°C. 4–112 A piston–cylinder device contains 0.8 kg of an ideal gas. Now, the gas is cooled at constant pressure until its temperature decreases by 10°C. If 16.6 kJ of compression work
4 –107 A 100kg man decides to lose 5 kg without cutting down his intake of 3000 Calories a day. Instead, he starts fast swimming, fast dancing, jogging, and biking each for an hour every day. He sleeps or relaxes the rest of the day. Determine how long it will take him to lose 5 kg. 4 –108E The range of healthy weight for adults is usually expressed in terms of the body mass index (BMI), defined, in SI units, as BMI
Ideal gas 0.8 kg ∆T = 10°C
W 1kg2
H 2 1m2 2
FIGURE P4–112
Q
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Chapter 4 is done during this process, determine the gas constant and the molar mass of the gas. Also, determine the constantvolume and constantpressure specific heats of the gas if its specific heat ratio is 1.667. 4 –113 The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation 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: 119.8 m/s, 731.9 m
4 –114 A frictionless piston–cylinder device initially contains air at 200 kPa and 0.2 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 0.5 m3 and 800 kPa. Determine (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) 150 kJ, (b) 90 kJ

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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 300 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. 4–116E A spherical balloon contains 10 lbm of air at 30 psia and 800 R. The balloon material is such that the pressure inside is always proportional to the square of the diameter. Determine the work done when the volume of the balloon doubles as a result of heat transfer. Answer: 715 Btu 4–117E
Reconsider Prob. 4–116E. Using the integration feature of the EES software, determine the work done. Compare the result with your “handcalculated” result.
4–118 A mass of 12 kg of saturated refrigerant134a vapor is contained in a piston–cylinder 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 70°C. Also, show the process on a Tv diagram with respect to the saturation lines. Answer: 12.8 A
AIR P1 = 200 kPa
V1 = 0.2 m3 R134a P = constant
FIGURE P4–114 We
4 –115 A mass of 5 kg of saturated liquid–vapor mixture of water is contained in a piston–cylinder device at 125 kPa. Initially, 2 kg of the water is in the liquid phase and the rest is in
H 2O m = 5 kg
FIGURE P4–115
Q
FIGURE P4–118 4–119 A mass of 0.2 kg of saturated refrigerant134a is contained in a piston–cylinder device at 200 kPa. Initially, 75 percent of the mass is in the liquid phase. Now heat is transferred 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. 4–120 A piston–cylinder device contains helium gas initially at 150 kPa, 20°C, and 0.5 m3. The helium is now compressed in a polytropic process (PV n constant) to 400 kPa and 140°C. Determine the heat loss or gain during this process. Answer: 11.2 kJ loss
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Thermodynamics 4–124 One ton (1000 kg) of liquid water at 80°C is brought into a wellinsulated and wellsealed 4m 5m 6m room initially at 22°C and 100 kPa. Assuming constant specific heats for both air and water at room temperature, determine the final equilibrium temperature in the room. Answer: 78.6°C He PV n = constant Q
FIGURE P4–120 4 –121 A frictionless piston–cylinder device and a rigid tank initially contain 12 kg of an ideal gas each at the same temperature, pressure, and volume. It is desired to raise the temperatures of both systems by 15°C. 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. 4 –122 A passive solar house that is losing heat to the outdoors at an average rate of 50,000 kJ/h is maintained at 22°C 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 80°C during the day by absorbing solar energy. A thermostatcontrolled 15kW backup electric resistance heater turns on whenever necessary to keep the house at 22°C. (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) 4.77 h,
4–125 A 4m 5m 6m room is to be heated by one ton (1000 kg) of liquid water contained in a tank that is placed in the room. The room is losing heat to the outside at an average rate of 8000 kJ/h. The room is initially at 20°C and 100 kPa and is maintained at an average temperature of 20°C at all times. If the hot water is to meet the heating requirements of this room for a 24h period, determine the minimum temperature of the water when it is first brought into the room. Assume constant specific heats for both air and water at room temperature. 4–126 The energy content of a certain food is to be determined 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 3.2°C when equilibrium is established, determine the energy content of the food, in kJ/kg, 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: 20,060 kJ/kg
Reaction chamber
(b) 9.26 h
Food
∆T = 3.2°C
FIGURE P4–126 22ºC Water 80°C
Pump
FIGURE P4–122 4 –123 An 1800W electric resistance heating element is immersed in 40 kg of water initially at 20°C. Determine how long it will take for this heater to raise the water temperature to 80°C.
4–127 A 68kg man whose average body temperature is 39°C drinks 1 L of cold water at 3°C in an effort to cool down. Taking the average specific heat of the human body to be 3.6 kJ/kg · °C, determine the drop in the average body temperature of this person under the influence of this cold water. 4–128 A 0.2L glass of water at 20°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at (a) 0°C and (b) 8°C. Also determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are
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Chapter 4 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L. 4 –129
Reconsider Prob. 4 –128. Using EES (or other) software, investigate the effect of the initial temperature of the ice on the final mass required. Let the ice temperature vary from –20 to 0°C. Plot the mass of ice against the initial temperature of ice, and discuss the results.

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evaporates in 25 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 18°C to the boiling temperature. 1 atm
4 –130 In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 80 kg of ice at 5°C into the water. Determine the final equilibrium temperature in the tank. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively.
Coffee maker 1L
Answer: 12.4°C
4 –131 An insulated piston–cylinder device initially contains 0.01 m3 of saturated liquid–vapor mixture with a quality of 0.2 at 120°C. Now some ice at 0°C is added to the cylinder. If the cylinder contains saturated liquid at 120°C when thermal equilibrium is established, determine the amount of ice added. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively. 4 –132 The early steam engines were driven by the atmospheric pressure acting on the piston fitted into a cylinder filled with saturated steam. A vacuum was created in the cylinder by cooling the cylinder externally with cold water, and thus condensing the steam. Consider a piston–cylinder device with a piston surface area of 0.1 m2 initially filled with 0.05 m3 of saturated water vapor at the atmospheric pressure of 100 kPa. Now cold water is poured outside the cylinder, and the steam inside starts condensing as a result of heat transfer to the cooling water outside. If the piston is stuck at its initial position, determine the friction force acting on the piston and the amount of heat transfer when the temperature inside the cylinder drops to 30°C.
FIGURE P4–133 4–134 Two rigid tanks are connected by a valve. Tank A contains 0.2 m3 of water at 400 kPa and 80 percent quality. Tank B contains 0.5 m3 of water at 200 kPa and 250°C. 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 25°C. Answers: 3.17 kPa, 2170 kJ
H 2O
H2 O
400 kPa
200 kPa
Q A
Cold water
B
FIGURE P4–134 4–135
0.05 m 3 100 kPa Steam
FIGURE P4–132 4 –133 Water is boiled at sea level in a coffee maker 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 coffee maker
Reconsider Prob. 4–134. Using EES (or other) software, investigate the effect of the environment temperature on the final pressure and the heat transfer. Let the environment temperature vary from 0 to 50°C. Plot the final results against the environment temperature, and discuss the results.
4–136 A rigid tank containing 0.4 m3 of air at 400 kPa and 30°C is connected by a valve to a piston–cylinder device with zero clearance. The mass of the piston is such that a pressure of 200 kPa is required to raise the piston. The valve is now opened slightly, and air is allowed to flow into the cylinder until the pressure in the tank drops to 200 kPa. During this process, heat is exchanged with the surroundings such that
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the entire air remains at 30°C at all times. Determine the heat transfer for this process.
N2 1 m3 500 kPa 80°C
AIR T = const. Q
He 1 m3 500 kPa 25°C
FIGURE P4–138 4–139 Repeat Prob. 4–138 by assuming the piston is made of 5 kg of copper initially at the average temperature of the two gases on both sides. Answer: 56°C
FIGURE P4–136 4 –137 A wellinsulated 4m 4m 5m room initially at 10°C 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 200°C. 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 the average temperature of air in 30 min. Assume the air pressure in the room remains constant at 100 kPa.
4–140
Reconsider Prob. 4–139. Using EES (or other) software, investigate the effect of the mass of the copper piston on the final equilibrium temperature. Let the mass of piston vary from 1 to 10 kg. Plot the final temperature against the mass of piston, and discuss the results.
4–141 An insulated rigid tank initially contains 1.4kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. 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, and (c) the electric power rating of the resistor. Answers: (a ) 0.00648 m3, (b) 371°C, (c) 1.58 kW
10°C 4m×4m×5m
Fan
Steam radiator
We
Water 1.4 kg, 200°C
FIGURE P4–141
FIGURE P4–137
4 –138 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 80°C while the other side contains 1 m3 of He gas at 500 kPa and 25°C. 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 the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move?
4–142 A vertical 12cm diameter piston–cylinder device contains an ideal gas at the ambient conditons of 1 bar and 24°C. 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 0.1 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 piston is displaced. 4–143 A piston–cylinder device initially contains 0.15kg steam at 3.5 MPa, superheated by 5°C. 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 200°C. Determine (a) the final pressure and the quality (if mix
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Chapter 4 ture), (b) the boundary work, (c) the amount of heat transfer when the piston first hits the stops, (d) and the total heat transfer.
Q
FIGURE P4–143
P2 T2
Steam boiler
where m3 and T3 are the mass and temperature of the final mixture, respectively.
SIDE 2 Mass = m 2 Temperature = T2
FIGURE P4–144 4 –145 Catastrophic explosions of steam boilers in the 1800s and early 1900s resulted in hundreds of deaths, which prompted the development of the ASME Boiler and Pressure Vessel Code in 1915. Considering that the pressurized fluid in a vessel eventually reaches equilibrium with its surroundings shortly after the explosion, the work that a pressurized fluid would do if allowed to expand adiabatically to the state of the surroundings can be viewed as the explosive energy of the pressurized fluid. Because of the very short time period of the explosion and the apparent stability afterward, the explosion process can be considered to be adiabatic with no changes in kinetic and potential energies. The closedsystem conservation of energy relation in this case reduces to Wout m(u1 – u2). Then the explosive energy Eexp becomes Eexp m 1u1 u2 2
where the subscripts 1 and 2 refer to the state of the fluid before and after the explosion, respectively. The specific
P1 T2 a1 b k1 T1
Also, determine the total explosion energy of 20 m3 of air at 5 MPa and 100°C when the surroundings are at 20°C.
m1 m2 ,T,T b T3 f a , m3 m3 1 2
SIDE 1 Mass = m1 Temperature = T1
u1 u2 v1
where v1 is the specific volume of the fluid before the explosion. Show that the specific explosion energy of an ideal gas with constant specific heat is eexp
4 –144 An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal gas at identical pressure but at different temperatures and masses. The wall separating the two compartments 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
215
explosion energy eexp is usually expressed per unit volume, and it is obtained by dividing the quantity above by the total V of the vessel: eexp
Steam 0.15 kg 3.5 MPa

P1 T1
FIGURE P4–145 4–146 Using the relations in Prob. 4–145, determine the explosive energy of 20 m3 of steam at 10 MPa and 500°C assuming the steam condenses and becomes a liquid at 25°C after the explosion. To how many kilograms of TNT is this explosive energy equivalent? The explosive energy of TNT is about 3250 kJ/kg.
Fundamentals of Engineering (FE) Exam Problems 4–147 A room is filled with saturated steam at 100°C. Now a 5kg bowling ball at 25°C is brought to the room. Heat is transferred to the ball from the steam, and the temperature of the ball rises to 100°C while some steam condenses on the ball as it loses heat (but it still remains at 100°C). The specific heat of the ball can be taken to be 1.8 kJ/kg · C. 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 4–148 A frictionless piston–cylinder device and a rigid tank contain 2 kmol of an ideal gas at the same temperature, pressure, and volume. Now heat is transferred, and the temperature of both systems is raised by 10°C. The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure is (a) 0 kJ (b) 42 kJ (c) 83 kJ
(d ) 102 kJ (e) 166 kJ
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4 –149 The specific heat of a material is given in a strange unit to be c 3.60 kJ/kg °F. The specific heat of this material in the SI units of kJ/kg °C is (a) 2.00 kJ/kg · °C (b) 3.20 kJ/kg · °C (c) 3.60 kJ/kg · °C
(d ) 4.80 kJ/kg · °C (e) 6.48 kJ/kg · °C
4 –150 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
4 –151 A 0.8m3 rigid tank contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a volume of 0.1 m3. The work done on the gas during this compression process is (a) 746 kJ (b) 0 kJ (c) 420 kJ
(d ) 998 kJ (e) 1890 kJ
4 –152 A wellsealed room contains 60 kg of air at 200 kPa and 25°C. Now solar energy enters the room at an average rate of 0.8 kJ/s 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) 25.6°C (b) 49.8°C (c) 53.4°C
(d ) 52.5°C (e) 63.4°C
4 –153 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) 8.5°C (b) 12.4°C (c) 24.0°C
(d) 33.4°C (e) 54.8°C
4 –154 A room contains 60 kg of air at 100 kPa and 15°C. The room has a 250W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120W TV, a 1kW 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) 3312 kJ/h (b) 4752 kJ/h (c) 5112 kJ/h
(d) 2952 kJ/h (e) 4680 kJ/h
4 –155 A piston–cylinder device contains 5 kg of air at 400 kPa and 30°C. During a quasiequilibium isothermal expansion process, 15 kJ of boundary work is done by the system,
and 3 kJ of paddlewheel work is done on the system. The heat transfer during this process is (a) 12 kJ (b) 18 kJ (c) 2.4 kJ
(d) 3.5 kJ (e) 60 kJ
4–156 A container equipped with a resistance heater and a mixer is initially filled with 3.6 kg of saturated water vapor at 120°C. Now the heater and the mixer are turned on; the steam is compressed, and there is heat loss to the surrounding air. At the end of the process, the temperature and pressure of steam in the container are measured to be 300°C and 0.5 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
4–157 A 6pack canned drink is to be cooled from 25°C to 3°C. The mass of each canned drink is 0.355 kg. The drinks can be treated as water, and the energy stored in the aluminum can itself is negligible. The amount of heat transfer from the 6 canned drinks is (a) 33 kJ (b) 37 kJ (c) 47 kJ
(d) 196 kJ (e) 223 kJ
4–158 A glass of water with a mass of 0.45 kg at 20°C is to be cooled to 0°C by dropping ice cubes at 0°C into it. The latent heat of fusion of ice is 334 kJ/kg, and the specific heat of water is 4.18 kJ/kg · °C. The amount of ice that needs to be added is (a) 56 g (b) 113 g (c) 124 g
(d) 224 g (e) 450 g
4–159 A 2kW electric resistance heater submerged in 5kg water is turned on and kept on for 10 min. During the process, 300 kJ of heat is lost from the water. The temperature rise of water is (a) 0.4°C (b) 43.1°C (c) 57.4°C
(d ) 71.8°C (e) 180.0°C
4–160 3 kg of liquid water initially at 12°C is to be heated at 95°C in a teapot equipped with a 1200W electric heating element inside. The specific heat of water can be taken to be 4.18 kJ/kg · °C, 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) 4.8 min (b) 14.5 min (c) 6.7 min
(d ) 9.0 min (e) 18.6 min
4–161 An ordinary egg with a mass of 0.1 kg and a specific heat of 3.32 kJ/kg · °C is dropped into boiling water at 95°C.
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If the initial temperature of the egg is 5°C, the maximum amount of heat transfer to the egg is
until the temperature rises to 1200°C, and the piston moves to maintain a constant pressure.
(a) 12 kJ (b) 30 kJ (c) 24 kJ
1. The initial pressure of the steam is most nearly (a) 250 kPa (d ) 1000 kPa (b) 500 kPa (e) 1250 kPa (c) 750 kPa
(d ) 18 kJ (e) infinity
4 –162 An apple with an average mass of 0.18 kg and average specific heat of 3.65 kJ/kg · °C is cooled from 22°C to 5°C. The amount of heat transferred from the apple is (a) 0.85 kJ (b) 62.1 kJ (c) 17.7 kJ
(d ) 11.2 kJ (e) 7.1 kJ
4 –163 The specific heat at constant pressure for an ideal gas is given by cp 0.9 (2.7 104)T (kJ/kg · K) where T is in kelvin. The change in the enthalpy for this ideal gas undergoing a process in which the temperature changes from 27 to 127°C is most nearly (a) 90 kJ/kg (d ) 108.9 kJ/kg (b) 92.1 kJ/kg (e) 105.2 kJ/kg (c) 99.5 kJ/kg 4 –164 The specific heat at constant volume for an ideal gas is given by cv 0.7 (2.7 104)T (kJ/kg · K) where T is in kelvin. The change in the internal energy for this ideal gas undergoing a process in which the temperature changes from 27 to 127°C is most nearly (a) 70 kJ/kg (d ) 82.1 kJ/kg (b) 72.1 kJ/kg (e) 84.0 kJ/kg (c) 79.5 kJ/kg 4 –165 A piston–cylinder 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 –34 T1. Then the piston is held stationary while the gas is further cooled to T3 –12 T1, where all temperatures are in K. 1. The ratio of the final volume to the initial volume of the gas is (a) 0.25 (d ) 0.75 (b) 0.50 (e) 1.0 (c) 0.67 2. The work done on the gas by the piston is (a) RT1/4 (d ) (cv cp)T1/4 (e) cv (T1 T2)/2 (b) cvT1/2 (c) cpT1/2 3. The total heat transferred from the gas is (a) RT1/4 (d ) (cv cp)T1/4 (e) cv (T1 T3)/2 (b) cvT1/2 (c) cpT1/2 4 –166 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 1.2 MPa and 700°C, respectively. Additional heat is added to the steam
2. The work done by the steam on the piston is most nearly (a) 230 kJ/kg (d ) 2340 kJ/kg (b) 1100 kJ/kg (e) 840 kJ/kg (c) 2140 kJ/kg 3. The total heat transferred to the steam is most nearly (a) 230 kJ/kg (d ) 2340 kJ/kg (b) 1100 kJ/kg (e) 840 kJ/kg (c) 2140 kJ/kg
Design, Essay, and Experiment Problems 4–167 Using a thermometer, measure the boiling temperature of water and calculate the corresponding saturation pressure. From this information, estimate the altitude of your town and compare it with the actual altitude value. 4–168 Find out how the specific heats of gases, liquids, and solids are determined in national laboratories. Describe the experimental apparatus and the procedures used. 4–169 Design an experiment complete with instrumentation to determine the specific heats of a gas 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? 4–170 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? 4–171 You are asked to design a heating system for a swimming pool that is 2 m deep, 25 m long, and 25 m wide. Your client desires that the heating system be large enough to raise the water temperature from 20 to 30°C in 3 h. The rate of heat loss from the water to the air at the outdoor design conditions is determined to be 960 W/m2, and the heater must also be able to maintain the pool at 30°C at those conditions. Heat losses to the ground are expected to be small and can be disregarded. 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? 4–172 It is claimed that fruits and vegetables are cooled by 6°C for each percentage point of weight loss as moisture
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during vacuum cooling. Using calculations, demonstrate if this claim is reasonable. 4 –173 A 1982 U.S. Department of Energy article (FS #204) states that a leak of one drip of hot water per second can cost $1.00 per month. Making reasonable assumptions about the drop size and the unit cost of energy, determine if this claim is reasonable. 4 –174 Polytropic Expansion of Air Experiment The expansion on compression of a gas can be described by the polytropic relation pv n c, where p is pressure, v is specific volume, c is a constant and the exponent n depends on the thermodynamic process. In our experiment compressed air in a steel pressure vessel is discharged to the atmosphere while temperature and pressure measurements of the air inside the vessel are recorded. There measurements, along with the first law of thermodynamics, are used to produce the polytropic exponent n for the process. Obtain the polytropic exponent n for the process using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4 –175 First Law of Thermodynamics—Lead Smashing Experiment The first law of thermodynamics is verified with a lead smashing experiment. A small piece of lead, instrumented with a thermocouple, is smashed with two steel cylinders. The cylinders are suspended by nylon chords and swing as pendulums from opposite directions, simultaneously striking the lead. The loss in gravitational potential energy of the cylinders is equated to the rise in internal energy of the lead. Verify the first law of thermodynamics using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4 –176 First Law of Thermodynamics—Friction Bearing Experiment The first law of thermodynamics is verified with a friction bearing experiment. A copper friction bearing is attached to one end of a wood shaft that is driven in rotation with a falling weight turning a pulley attached to the shaft. Friction causes the bearing to heat up. Data reduction analysis accounts for gravitational potential energy, elastic potential energy, translational and rotational kinetic energy, internal energy, and heat loss from the bearing. Verify the first law of thermodynamics using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4 –177 First Law of Thermodynamics—Copper Cold Working Experiment The first law of thermodynamics is verified again, but this time with a copper hinge calorimeter that is “worked” by a swinging pendulum, which causes a rise in the hinge temperature. The loss in potential energy of the pendulum is equated to the rise in internal energy of the hinge, plus
the heat unavoidably transferred into the hinge clamps. Verify the first law of thermodynamics using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4–178 First Law of Thermodynamics—Bicycle Braking Experiment The first law of thermodynamics is verified yet again—this time with a bicycle. A bicycle front caliper brake is removed and replaced with a levermounted, copper calorimeter friction pad. The calorimeter friction pad rubs on the front tire, heats up, brings the bicycle to a stop, and verifies the first law of thermodynamics. Used in the data reduction analysis are aerodynamics drag and rolling friction, which are obtained using bicycle coastdown data read into a cassette audio recorder by the bicycle rider. Verify the first law of thermodynamics using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4–179 Specific Heat of Aluminum—Electric Calorimeter Experiment The specific heat of aluminum is obtained with an electric calorimeter. The design consists of two individual calorimeters—each an assembly of 13 aluminum plates with electric resistance heater wires laced inbetween the plates. The exterior surfaces of both calorimeters and the surrounding insulation are identical. However, the interior plates are different—one calorimeter has solid interior plates and the other has perforated interior plates. By initially adjusting the electrical power into each calorimeter the temperatureversustime curves for each calorimeter are matched. This curve match allows cancellation of the unknown heat loss from each calorimeter and cancellation of the unknown heater thermal capacity to deliver an accurate specific heat value. Obtain the specific heat of aluminum using the video clip, the complete writeup, and the data provided on the DVD accompanying this book. 4–180 Specific Heat of Aluminum—Transient Cooling Experiment The specific heat of aluminum is obtained with an entirely different experiment than the one described in Prob. 4–179. In the present experiment a hollow, aluminum cylinder calorimeter is fitted with a plug forming a watertight cavity. The calorimeter is heated with a hair drier and then allowed to cool in still air. Two tests are performed: one with water in the cavity and one without water in the cavity. Transient temperature measurements from the two tests give different cooling rates characterized with Trendlines in EXCEL. These Trendlines are used to compute the aluminum specific heat. Obtain the specific heat of aluminum using the video clip, the complete writeup, and the data provided on the DVD accompanying this book.
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Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES
I
n Chap. 4, we applied the general energy balance relation expressed as Ein Eout Esystem to closed systems. In this chapter, we extend the energy analysis to systems that involve mass flow across their boundaries i.e., control volumes, with particular emphasis to 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 balance to systems that involve steadyflow processes and analyze the common steadyflow devices such as nozzles, diffusers, compressors, turbines, throttling devices, mixing chambers, and heat exchangers. Finally, we apply the energy balance to general unsteadyflow processes such as the charging and discharging of vessels.
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 unsteadyflow control volumes. • Apply the first law of thermodynamics 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, mixers, heaters, and heat exchangers. • Apply the energy balance to general unsteadyflow processes with particular emphasis on the uniformflow process as the model for commonly encountered charging and discharging processes.

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Thermodynamics
5–1
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 1 ON THE DVD.
2 kg H2
16 kg O2
18 kg H2O
FIGURE 5–1 Mass is conserved even during chemical reactions.
■
CONSERVATION OF MASS
Conservation of mass is one of the most fundamental principles in nature. We are all familiar with this principle, and it is not difficult to understand. As the saying goes, You cannot have your cake and eat it too! A person does not have to be a scientist to figure out how much vinegarandoil dressing is 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. 5–1). In an electrolysis process, the water separates back to 2 kg of hydrogen and 16 kg of oxygen. Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. However, mass m and energy E can be converted to each other according to the wellknown formula proposed by Albert Einstein (1879–1955): E mc2
(5–1)
where c is the speed of light in a vacuum, which is c 2.9979 m/s. This equation suggests that the mass of a system changes when its energy changes. However, for all energy interactions encountered in practice, with the exception of nuclear reactions, the change in mass is extremely small and cannot be detected by even the most sensitive devices. For example, when 1 kg of water is formed from oxygen and hydrogen, the amount of energy released is 15,879 kJ, which corresponds to a mass of 1.76 1010 kg. A mass of this magnitude is beyond the accuracy required by practically all engineering calculations and thus can be disregarded. 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. 108
Mass and Volume Flow Rates →
V
dAc
Vn →
n
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, as explained in Chap. 2. A fluid usually flows into or out of a control volume through pipes or ducts. The differential mass flow rate of fluid flowing across a small area element dAc on a flow cross section is proportional to dAc itself, the fluid density r, and the component of the flow velocity normal to dAc, which we denote as Vn, and is expressed as (Fig. 5–2) # dm rVn dA c
Control surface
FIGURE 5–2 The normal velocity Vn for a surface is the component of velocity perpendicular to the surface.
(5–2)
Note that both d and d are used to indicate differential quantities, but d 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, 2 2 # # dA c A c2 A c1 p 1r 22 r 21 2 but dm mtotal (total mass flow rate 1 1 . . through the annulus), not m2 m1. For specified values of r1 and r2, the value of the integral of dAc is fixed (thus the names point function and exact
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221
. differential), but this is not the case for the integral of dm (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
dm# rV
n
Ac
dA c¬¬1kg>s2
(5–3)
Ac
While Eq. 5–3 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 r 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 r outside the integral of Eq. 5–3. Velocity, however, is never uniform over a cross section of a pipe because of the fluid sticking to the surface and thus having zero velocity at the wall (the noslip condition). 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 (Fig. 5–3), Vavg
Average velocity:
1 Ac
V
n
dA c
(5–4)
Ac
where Ac is the area of the cross section normal to the flow direction. Note that if the velocity were Vavg all through the cross section, the mass flow rate would be identical to that obtained by integrating the actual velocity profile. Thus for incompressible flow or even for compressible flow where r is uniform across Ac, Eq. 5–3 becomes # m rVavg A c¬¬1kg>s2
(5–5)
For compressible flow, we can think of r as the bulk average density over the cross section, and then Eq. 5–5 can still be used as a reasonable approximation. For simplicity, we drop the subscript on the average velocity. Unless otherwise 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. 5–4) and is given by # V
V
n
Ac
dA c Vavg A c VA c¬¬1m3>s2
(5–6)
An early form of Eq. 5–6 was published in 1628 by the Italian monk Benedetto Castelli (circa. 1577–1644). Note that most fluid . mechanics textbooks use Q instead of V for volume flow rate. We use V to avoid confusion with heat transfer. The mass and volume flow rates are related by # # V # m rV v
(5–7)
where v is the specific volume. This relation is analogous to m rV V/v, which is the relation between the mass and the volume of a fluid in a container.
Vavg
FIGURE 5–3 The average velocity Vavg is defined as the average speed through a cross section.
Ac
Vavg
V = Vavg Ac
Cross section
FIGURE 5–4 The volume flow rate is the volume of fluid flowing through a cross section per unit time.
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Thermodynamics
Conservation of Mass Principle
min = 50 kg
thtub
ter Wa mout – = min
= 20
kg
∆ mba
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 t is equal to the net change (increase or decrease) in the total mass within the control volume during t. That is, a
Total mass entering Total mass leaving Net change in mass b a b a b the CV during ¢t the CV during ¢t within the CV during ¢t
or min mout ¢mCV¬¬1kg2
FIGURE 5–5 Conservation of mass principle for an ordinary bathtub.
(5–8)
where mCV mfinal minitial is the change in the mass of the control volume during the process (Fig. 5–5). It can also be expressed in rate form as # # min mout dmCV>dt¬¬1kg>s2
(5–9)
(5–10)
. . where min and mout are the total rates of mass flow into and out of the control volume, and dmCV/dt is the time rate of change of mass within the control volume boundaries. Equations 5–8 and 5–9 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. 5–6. The mass of a differential volume dV within the control volume is dm r 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
r dV
CV
Then the time rate of change of the amount of mass within the control volume can be expressed as Rate of change of mass within the CV:
dV →
n
dm dA Control volume (CV)
u →
V
Control surface (CS)
FIGURE 5–6 The differential control volume dV and the differential control surface dA used in the derivation of the conservation of mass relation.
dmCV d dt dt
r dV
(5–11)
CV
For the special case of no mass crossing the control surface (i.e., the control volume resembles a closed system), the conservation of mass principle reduces to that of a system that can be expressed as dmCV/dt 0. This relation is valid whether the control volume is fixed, moving, or deforming. Now consider mass flow into or out of the control volume through a differ→ ential 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. 5–6. In general, the velocity may cross dA at an angle u off the normal of dA,→and the mass flow → rate is proportional to the normal→component of velocity Vn V cos u ranging from a maximum outflow of V for u 0 (flow is normal to dA) to a min→ imum of zero for u 90° (flow is tangent to dA) to a maximum inflow of V for u 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 can be expressed as S
Normal component of velocity:
Vn V cos u V # n
S
(5–12)
The mass flow rate through dA is proportional to the fluid density r, normal velocity Vn, and the flow area dA, and can be expressed as S S # Differential mass flow rate: dm rVn dA r 1V cos u2 dA r 1 V # n 2 dA (5–13)
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223
The net flow rate into or out of the control volume through the entire con. trol surface is obtained by integrating dm over the entire control surface, # mnet
Net mass flow rate: →
# dm
CS
rVn dA
CS
r 1V # n 2 dA S
CS
S
(5–14)
→
Note that V · n V cos u is positive for u 90° (outflow) and negative for u 90° (inflow). Therefore, the direction of flow is automatically accounted for, and the surface integral in Eq. 5–14 directly gives the net . mass flow rate. A positive value for mnet indicates net outflow, and a negative value indicates a net inflow of mass. . . Rearranging Eq. 5–9 as dmCV/dt mout min 0, the conservation of mass relation for a fixed control volume can then be expressed as General conservation of mass:
d dt
r dV
CV
CS
r 1 V # n 2 dA 0 S
S
(5–15)
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. 5–15 into two parts—one for the outgoing flow streams (positive) and one for the incoming streams (negative)— the general conservation of mass relation can also be expressed as d dt
r dV a out
CV
rV dA a rV dA 0 n
n
in
A
(5–16)
A
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. 5–16 can also be expressed as d dt
dmCV # # # # r dV a m a m¬or¬ amam dt in out in out CV
(5–17)
Equations 5–15 and 5–16 are also valid for moving or deforming control vol→ is replaced by the relative velocity umes provided that the absolute velocity V → Vr , which is the fluid velocity relative to the control surface.
Mass Balance for SteadyFlow Processes
˙ 1 = 2 kg/s m
During a steadyflow process, the total amount of mass contained within a control volume does not change with time (mCV constant). Then the conservation 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 can be expressed in rate form as (Fig. 5–7) Steady flow:
# # a m a m¬¬1kg>s2 in
out
(5–18)
˙ 2 = 3 kg/s m
CV
˙ 3 = m˙ 1 + ˙m2 = 5 kg/s m FIGURE 5–7 Conservation of mass principle for a twoinlet–oneoutlet steadyflow system.
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Thermodynamics 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, compressors, and pumps involve a single stream (only one inlet and one outlet). For these cases, we denote the inlet state by the subscript 1 and the outlet state by the subscript 2, and drop the summation signs. Then Eq. 5–18 reduces, for singlestream steadyflow systems, to Steady flow (single stream):
# # m1 m 2¬S ¬r 1V1A 1 r 2V2 A 2
(5–19)
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
˙ 2 = 2 kg/s m ˙ 2 = 0.8 m3/s V
Steady, incompressible flow:
# # 3 a V a V¬¬1m >s2 in
For singlestream steadyflow systems it becomes Steady, incompressible flow (single stream):
Air compressor
˙ 1 = 2 kg/s m ˙ 1 = 1.4 m3/s V FIGURE 5–8 During a steadyflow process, volume flow rates are not necessarily conserved although mass flow rates are.
(5–20)
out
# # V1 V2 S V1A 1 V2 A 2
(5–21)
It should always be kept in mind that there is no such thing as a “conservation 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. 5–8). This is due to the higher density of air at the compressor exit. For steady flow of liquids, however, the volume flow rates, as well as the mass flow rates, remain constant since liquids are essentially incompressible (constantdensity) substances. Water flow through the nozzle of a garden hose is an example of the latter case. The conservation of mass principle is based on experimental observations and requires every bit of mass to be accounted for during a process. If you can balance your checkbook (by keeping track of deposits and withdrawals, or by simply observing the “conservation of money” principle), you should have no difficulty applying the conservation of mass principle to engineering systems.
EXAMPLE 5–1
Water Flow through a Garden Hose Nozzle
Nozzle
Garden hose
Bucket
A garden hose attached with a nozzle is used to fill a 10gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit (Fig. 5–9). 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 FIGURE 5–9 Schematic for Example 5–1.
mass flow rates of water and the exit velocity are to be determined. Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing.
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225
Properties We take the density of water to be 1000 kg/m3 1 kg/L. Analysis (a) Noting that 10 gal of water are discharged in 50 s, the volume and mass flow rates of water are
# 10 gal 3.7854 L V a b 0.757 L/s V ¢t 50 s 1 gal # # m rV 11 kg>L2 10.757 L>s2 0.757 kg/s (b) The crosssectional area of the nozzle exit is
Ae pr 2e p 10.4 cm2 2 0.5027 cm2 0.5027 104 m2 The volume flow rate through the hose and the nozzle is constant. Then the average velocity of water at the nozzle exit becomes
Ve
# V 0.757 L/s 1 m3 ¬a b 15.1 m/s 4 2 Ae 1000 L 0.5027 10 m
Discussion It can be shown that the average velocity in the hose is 2.4 m/s. Therefore, the nozzle increases the water velocity by over six times.
EXAMPLE 5–2
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 0.5 in streams out (Fig. 5–10). The average velocity of the jet is given by V 12gh, 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 will take 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 an 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 32.2 ft/s2. 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 the rate form as
dmCV # # min mout dt
(1)
. During this process no mass enters the control volume (min 0), and the mass flow rate of discharged water can be expressed as
# mout (rVA)out r22ghAjet
(2)
Air Water
h0 h2
h 0
Djet
Dtank
FIGURE 5–10 Schematic for Example 5–2.
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Thermodynamics where Ajet pD 2jet /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
mCV rV rAtankh
(3)
where Atank p is the base area of the cylindrical tank. Substituting Eqs. 2 and 3 into the mass balance relation (Eq. 1) gives D 2tank /4
r22ghAjet
d(rAtankh) r(pD2tank/4) dh → r22gh(pD2jet/4) dt dt
Canceling the densities and other common terms and separating the variables give
dt
D2tank
dh
D2jet
22gh
Integrating from t 0 at which h h0 to t t at which h h2 gives
t
D 2tank dt ¬ D 2jet 22g
h2
h0
dh 2h
S t
2h0 2h 2 2g/2
¬a
D tank 2 b D jet
Substituting, the time of discharge is
t
24 ft 22 ft 232.2/2 ft/s
2
¬
a
3 12 in 2 b 757 s 12.6 min 0.5 in
Therefore, half of the tank is emptied in 12.6 min after the discharge hole is unplugged. Discussion Using the same relation with h2 0 gives t 43.1 min for the discharge 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.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 2 ON THE DVD.
A
F
V P m
CV
L Imaginary piston
FIGURE 5–11 Schematic for flow work.
5–2
■
FLOW WORK AND THE ENERGY OF A FLOWING FLUID
Unlike closed systems, control volumes involve mass flow across their boundaries, 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 necessary 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. 5–11. 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. 5–12), the force applied on the fluid element by the imaginary piston is F PA
(5–22)
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Chapter 5 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 (i.e., the flow work) is Wflow FL PAL PV¬¬1kJ2
wflow Pv¬¬1kJ>kg2
(5–24)
The flow work relation is the same whether the fluid is pushed into or out of the control volume (Fig. 5–13). 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 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.
As we discussed in Chap. 2, the total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies (Fig. 5–14). On a unitmass basis, it is expressed as e u ke pe u
V2 gz¬¬1kJ>kg2 2
(5–25)
where V is the velocity and z is the elevation of the system relative to some external reference point.
Flow energy
Kinetic energy
Non lowing Nonflowing fluid
e=u+ Internal energy
V2 + gz 2 Potential energy
Flowing fluid
F
P
FIGURE 5–12 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 v
wflow
CV
(a) Before entering
wflow
Total Energy of a Flowing Fluid
227
A
(5–23)
The flow work per unit mass is obtained by dividing both sides of this equation by the mass of the fluid element:

P v CV
(b) After entering
FIGURE 5–13 Flow work is the energy needed to push a fluid into or out of a control volume, and it is equal to Pv.
Kinetic energy
2
θ = Pv + u + V + gz 2 Internal energy
Potential energy
FIGURE 5–14 The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid.
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Thermodynamics The fluid entering or leaving a control volume possesses an additional form of energy—the flow energy Pv, as already discussed. Then the total energy of a flowing fluid on a unitmass basis (denoted by u) becomes u Pv e Pv 1u ke pe2
(5–26)
But the combination Pv u has been previously defined as the enthalpy h. So the relation in Eq. 5–26 reduces to u h ke pe h
V2 gz¬¬1kJ>kg2 2
(5–27)
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. 5–27, and no reference will be made to flow work or flow energy.
Energy Transport by Mass
˙ i,kg/s m θ i,kJ/kg
CV
˙ i θi m (kW)
Noting that u is total energy per unit mass, the total energy of a flowing fluid of mass m is simply mu, 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 mu (Fig. 5–15). That is,
Rate of energy transport:
FIGURE 5–15 . The product miui is the energy transported into control volume by mass per unit time.
E mass mu m a h
Amount of energy transport:
V2 gz b ¬¬1kJ2 2
# V2 # # E mass m u m a h gz b ¬¬1kW2 2
(5–28)
(5–29)
When the kinetic and potential energies of a fluid stream are. negligible, as . is often the case, these relations simplify to Emass mh and Emass mh. 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 opening as a result of mass flow is to consider sufficiently small differential masses dm that have uniform properties and to add their total energies during flow. Again noting that u is total energy per unit mass, the total energy of a flowing fluid of mass dm is u dm. Then the total energy transported by mass through an inlet or exit (miui and meue) is obtained by integration. At an inlet, for example, it becomes E in,mass
mi
u i dmi
mi
a hi
V i2 gz i b dmi 2
(5–30)
Most flows encountered in practice can be approximated as being steady and onedimensional, and thus the simple relations in Eqs. 5–28 and 5–29 can be used to represent the energy transported by a fluid stream.
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Chapter 5 EXAMPLE 5–3
Energy Transport by Mass
Steam is leaving a 4L pressure cooker whose operating pressure is 150 kPa (Fig. 5–16). It is observed that the amount of liquid in the cooker has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the crosssectional area of the exit opening is 8 mm2. 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 leaves the cooker by steam.
Steam
150 kPa
Solution Steam leaves a pressure cooker at a specified pressure. The velocity, flow rate, the total and flow energies, and the rate of energy transfer by mass are to be determined. Assumptions 1 The flow is steady, and the initial startup period is disregarded. 2 The kinetic and potential energies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at the cooker pressure. Properties The properties of saturated liquid water and water vapor at 150 kPa are vf 0.001053 m3/kg, vg 1.1594 m3/kg, ug 2519.2 kJ/kg, and hg 2693.1 kJ/kg (Table A–5). Analysis (a) Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity are
m
¢V liquid vf
0.6 L 1 m3 ¬ a b 0.570 kg 0.001053 m3>kg 1000 L
0.570 kg m # m 0.0142 kg>min 2.37 104 kg/s ¢t 40 min # # m vg 12.37 10 4 kg>s2 11.1594 m3>kg2 m 34.3 m/s V r gA c Ac 8 10 6 m2 (b) Noting that h u Pv and that the kinetic and potential energies are disregarded, the flow and total energies of the exiting steam are
eflow Pv h u 2693.1 2519.2 173.9 kJ/kg u h ke pe h 2693.1 kJ/kg Note that the kinetic energy in this case is ke V 2/2 (34.3 m/s)2/2 588 m2/s2 0.588 kJ/kg, which is small compared to enthalpy. (c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass,
# # E mass m u 12.37 10 4 kg/s2 12693.1 kJ/kg2 0.638 kJ/s 0.638 kW Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside (which is hfg) since it relates directly to the amount of energy supplied to the cooker.
Pressure Cooker
FIGURE 5–16 Schematic for Example 5–3.

229
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Thermodynamics
5–3
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 3 ON THE DVD.
FIGURE 5–17 Many engineering systems such as power plants operate under steady conditions. © Vol. 57/PhotoDisc
Mass in
Control volume mCV = constant ECV = constant Mass out
FIGURE 5–18 Under steadyflow conditions, the mass and energy contents of a control volume remain constant.
■
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. 5–17). 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 (Fig. 5–18). As a result, the boundary work is zero for steadyflow systems (since VCV constant), and the total mass or energy entering the control volume 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. 5–19). 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 steadyflow 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. 5–1 as # # a m a m¬¬1kg>s2 in
˙2 m h2
˙1 m h1 Control volume
˙3 m h3 FIGURE 5–19 Under steadyflow conditions, the fluid properties at an inlet or exit remain constant (do not change with time).
(5–31)
out
The mass balance for a singlestream (oneinlet and oneoutlet) steadyflow system was given as # # m1 m2¬S ¬r 1V1A 1 r 2V2 A 2
(5–32)
where the subscripts 1 and 2 denote the inlet and the exit states, respectively, r 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 (ECV 0). Therefore, the amount of energy entering 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 balance reduces for a steadyflow process to
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# # E in E out
0 (steady) dE system>dt ¡ 0
Chapter 5

231
(5–33)
⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
or Energy balance:
. E in
1kW2
. E out
⎫ ⎬ ⎭
⎫ ⎬ ⎭
Rate of net energy transfer in by heat, work, and mass
Rate of net energy transfer out by heat, work, and mass
(5–34)
Noting that energy can be transferred by heat, work, and mass only, the energy balance in Eq. 5–34 for a general steadyflow system can also be written more explicitly as # # # # # # Q in Win a mu Q out Wout a mu in
(5–35)
out
or # # # # V2 V2 # # Q in Win a m a h gz b Q out Wout a m a h gz b 2 2 in out 123 123 for each inlet
˙ 2 = m˙ 1 m
for each exit
since the energy of a flowing fluid per unit mass is u h ke pe h V 2/2 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. 5–20. 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 Qout, and the electric heating element is supplying electrical work (heating) to the . water at a rate of Win. 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 practice 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 # # V2 V2 # # Q W a mah gz b a m a h gz b 2 2 out in 123
123
for each exit
for each inlet
(5–37)
. . Obtaining a negative quantity for Q or W simply means that the assumed direction is wrong and should be reversed. For singlestream devices, the steadyflow energy balance equation becomes # # V 22 V 21 # Q W m c h 2 h1 g 1z2 z1 2 d 2
Heat ˙out loss Q
(5–36)
(5–38)
Electric heating element ˙ in W
Hot water out CV (Hotwater tank)
˙1 m Cold water in
FIGURE 5–20 A water heater in steady operation.
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Thermodynamics . Dividing Eq. 5–38 by m gives the energy balance on a unitmass basis as q w h 2 h1
V 22 V 21 g 1z2 z1 2 2
(5–39)
. . . . 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 negligible changes in its kinetic and potential energies (that is, ke 0, pe 0), the energy balance equation is reduced further to q w h2 h 1
˙e W
CV
˙ sh W FIGURE 5–21 Under steady operation, shaft work and electrical work are the only forms of work a simple compressible system may involve.
J N. m ≡ ≡ kg m2 kg kg s
(
( kgm ≡ ms
2
2
2
Btu ft ≡ 25,037 (Also, lbm ( s 2
FIGURE 5–22 The units m2/s2 and J/kg are equivalent.
(5–40)
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 (i.e., 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. 5–21). 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. h 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 h cp,avg(T2 T1). Note that (kg/s)(kJ/kg) kW. ke (V 22 V 21)/2. The unit of kinetic energy is m2/s2, which is equivalent to J/kg (Fig. 5–22). The enthalpy is usually given in kJ/kg. To add these two quantities, the kinetic energy should be expressed in kJ/kg. This is easily accomplished by dividing it by 1000. A velocity of 45 m/s corresponds to a kinetic energy of only 1 kJ/kg, 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. 5–23). pe g(z2 z1). A similar argument can be given for the potential energy term. A potential energy change of 1 kJ/kg 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.
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Chapter 5
5–4
■
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. 5–24). Therefore, these devices can be conveniently analyzed as steadyflow devices. In this section, some common steadyflow devices are described, and the thermodynamic aspects of the flow through them are analyzed. The conservation of mass and the conservation of energy principles for these devices are illustrated with examples.
5Stage LPC Bleed Low Pressure Air Collector Compressor (LPC) Cold End Drive Flange
14Stage High Pressure Compressor
Combustor Fuel System Manifolds
2Stage High Pressure Turbine
SEE TUTORIAL CH. 5, SEC. 4 ON THE DVD.
V1 m/s
V2 m/s
∆ke kJ/kg
0 50 100 200 500
45 67 110 205 502
1 1 1 1 1
FIGURE 5–23 At very high velocities, even small changes in velocities can cause significant changes in the kinetic energy of the fluid.
5Stage Low Pressure Turbine
Hot End Drive Flange
FIGURE 5–24 A modern landbased gas turbine used for electric power production. This is a General Electric LM5000 turbine. It has a length of 6.2 m, it weighs 12.5 tons, and produces 55.2 MW at 3600 rpm with steam injection. Courtesy of GE Power Systems
233
INTERACTIVE TUTORIAL
1 Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, 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 (pe 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. 5–25). Therefore, the kinetic energy changes must be accounted for in analyzing the flow through these devices (ke 0).

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Thermodynamics EXAMPLE 5–4
V1
Nozzle
V2
>> V1
Deceleration of Air in a Diffuser
Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 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 velocV1
Diffuser
V2
>> V1
FIGURE 5–25 Nozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies.
ity. 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 mCV 0 and ECV 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, pe 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. 5–26). 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 m1 m2 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:
v1 P1 = 80 kPa T1 = 10°C V1 = 200 m/s A1 = 0.4 m2
0.287 kPa # m3/kg # K2 1283 K 2 RT1 1.015 m3/kg P1 80 kPa
Then, AIR m=?
T2 = ?
1 1 # 2 ¬ 1200 m/s2 10.4 m 2 78.8 kg/s m ¬V1A 1 v1 1.015 m3/kg Since the flow is steady, the mass flow rate through the entire diffuser remains constant at this value.
FIGURE 5–26 Schematic for Example 5–4.
(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
dE system>dt ¡
0 (steady) 0
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # V 12 V 22 # # b m a h2 b ¬¬1since Q 0, W 0, and ¢pe 02 m a h1 2 2 h 2 h1
V 22 V 21 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 A–17) to be
h1 h @ 283 K 283.14 kJ/kg
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235
Substituting, we get
h 2 283.14 kJ/kg
0 1200 m/s 2 2 2
a
1 kJ/kg 1000 m2/s2
b
303.14 kJ/kg From Table A–17, the temperature corresponding to this enthalpy value is
T2 303 K Discussion This result shows that the temperature of the air increases by about 20°C 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 5–5
Acceleration of Steam in a Nozzle
Steam at 250 psia and 700°F steadily enters a nozzle whose inlet area is 0.2 ft2. The mass flow rate of steam through the nozzle is 10 lbm/s. Steam leaves the nozzle at 200 psia with a velocity of 900 ft/s. Heat losses from the nozzle per unit mass of the steam are estimated to be 1.2 Btu/lbm. 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 mCV 0 and ECV 0. 2 There are no work interactions. 3 The potential energy change is zero, pe 0. Analysis We take the nozzle as the system (Fig. 5–26A). 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 m1 m2 m. (a) The specific volume and enthalpy of steam at the nozzle inlet are
P1 250 psia v 2.6883 ft3/lbm f¬ 1 ¬¬1Table A–6E 2 T1 700°F h 1 1371.4 Btu/lbm Then,
1 # m V1A 1 v1 10 lbm>s
1 2 ¬ 1V1 2 10.2 ft 2 2.6883 ft3>lbm
V1 134.4 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
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
0 (steady) dE system>dt ¡ 0
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # V 12 V 22 # # m a h1 b Qout m a h2 b ¬¬1since W 0, and ¢pe 02 2 2
qout = 1.2 Btu/lbm
STEAM m = 10 lbm/s
P1 = 250 psia T1 = 700°F A1 = 0.2 ft2
P2 = 200 psia V2 = 900 ft/s
FIGURE 5–26A Schematic for Example 5–5.
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Thermodynamics . Dividing by the mass flow rate m and substituting, h2 is determined to be
h 2 h 1 qout
V 22 V 21 2
11371.4 1.22 Btu/lbm
1900 ft/s 2 2 1134.4 ft/s 2 2 2
a
1 Btu/lbm b 25,037 ft2/s2
1354.4 Btu/lbm Then,
P2 200 psia f ¬T2 662.0°F¬¬1Table A–6E 2 h 2 1354.4 Btu/lbm Discussion Note that the temperature of steam drops by 38.0°F 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.)
2 Turbines and Compressors In steam, gas, or hydroelectric power plants, the device that drives the electric 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. 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 (pe 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 (ke 0). The fluid velocities 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. EXAMPLE 5–6
Compressing Air by 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 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor.
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Chapter 5 Solution Air is compressed steadily by a compressor to a specified temperature 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 mCV 0 and ECV 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, ke pe 0. Analysis We take the compressor as the system (Fig. 5–27). 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 m1 m2 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
dE system>dt ¡
0 (steady) 0
237
qout = 16 kJ/kg P2 = 600 kPa T2 = 400 K AIR m ˙ = 0.02 kg/s
˙ in = ? W P1 = 100 kPa T1 = 280 K
FIGURE 5–27 Schematic for Example 5–6.
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Rate of net energy transfer by heat, work, and mass

Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # # # Win mh1 Qout mh2¬¬1since ¢ke ¢pe 02 # # # Win mqout m 1h2 h1 2 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 A–17) to be
h1 h @ 280 K 280.13 kJ/kg h2 h@ 400 K 400.98 kJ/kg Substituting, the power input to the compressor is determined to be
# Win 10.02 kg/s2 116 kJ/kg2 10.02 kg/s2 1400.98 280.132 kJ/kg 2.74 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 5–7
Power Generation by a Steam Turbine
P1 = 2 MPa T1 = 400°C V1 = 50 m/s z1 = 10 m
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. 5–28. (a) Compare the magnitudes of h, ke, and pe. (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 mCV 0 and ECV 0. 2 The system is adiabatic and thus there is no heat transfer.
STEAM TURBINE Wout = 5 MW
P2 = 15 kPa x2 = 90% V2 = 180 m/s z2 = 6 m
FIGURE 5–28 Schematic for Example 5–7.
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Thermodynamics 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 m1 m2 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
P1 2 MPa f T1 400 C
h1 3248.4 kJ/kg
(Table A–6)
At the turbine exit, we obviously have a saturated liquid–vapor mixture at 15kPa pressure. The enthalpy at this state is
h2 hf x2hfg [225.94 (0.9)(2372.3)] kJ/kg 2361.01 kJ/kg Then
¢h h 2 h 1 12361.01 3248.42 kJ>kg 887.39 kJ/kg
¢ke
1 kJ>kg 1180 m>s2 2 150 m>s2 2 V 22 V 21 ¬a b 14.95 kJ/kg 2 2 1000 m2>s2
¢pe g 1z 2 z 1 2 19.81 m>s2 2 3 16 10 2 m4 a
1 kJ>kg
1000 m2>s2
b 0.04 kJ/kg
(b) The energy balance for this steadyflow system can be expressed in the rate form as
# # E in E out
0 (steady) 0
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
dE system>dt ¡
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # V 12 V 22 # # gz1 b Wout m a h2 gz2 b ¬¬1since Q 0 2 m a h1 2 2 . 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
wout c 1h 2 h 1 2
V 22 V 21 g 1z 2 z 1 2 d 1¢h ¢ke ¢pe2 2
[887.39 14.95 0.04] kJ/kg 872.48 kJ/kg (c) The required mass flow rate for a 5MW power output is
# Wout 5000 kJ/s # m 5.73 kg/s wout 872.48 kJ/kg 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 pressure 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.
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239
3 Throttling Valves Throttling valves are any kind of flowrestricting devices that cause a significant pressure drop in the fluid. Some familiar examples are ordinary adjustable valves, capillary tubes, and porous plugs (Fig. 5–29). 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 temperature, and for that reason throttling devices are commonly used in refrigeration 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 (pe 0). Even though the exit velocity is often considerably higher than the inlet velocity, in many cases, the increase in kinetic energy is insignificant (ke 0). Then the conservation of energy equation for this singlestream steadyflow device reduces to h2 h 1¬¬1kJ>kg2
(a) An adjustable valve
(b) A porous plug
(c) A capillary tube
FIGURE 5–29 Throttling valves are devices that cause large pressure drops in the fluid.
(5–41)
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. 5–41 as follows: u 1 P1v1 u 2 P2v 2
Throttling valve
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 h(T ), and thus the temperature has to remain constant during a throttling process (Fig. 5–30). EXAMPLE 5–8
Expansion of Refrigerant134a in a Refrigerator
Refrigerant134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.
Solution Refrigerant134a that enters a capillary tube as saturated liquid is throttled to a specified pressure. The exit quality of the refrigerant and the temperature drop are to be determined.
IDEAL GAS
T1
T2 = T1
h1
h2 = h1
FIGURE 5–30 The temperature of an ideal gas does not change during a throttling (h constant) process since h h(T).
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Thermodynamics Throttling valve
u1 = 94.79 kJ/kg P1v1 = 0.68 kJ/kg (h 1 = 95.47 kJ/kg)
u2 = 88.79 kJ/kg P2v2 = 6.68 kJ/kg (h 2 = 95.47 kJ/kg)
FIGURE 5–31 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.
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. 5–31).
At inlet:
At exit:
P1 0.8 MPa f sat. liquid P2 0.12 MPa (h2 h1)
T1 Tsat @ 0.8 MPa 31.31 C h1 hf @ 0.8 MPa 95.47 kJ/kg ¡
(Table A–12)
hf 22.49 kJ/kg Tsat 22.32 C hg 236.97 kJ/kg
Obviously hf h2 hg; thus, the refrigerant exists as a saturated mixture at the exit state. The quality at this state is
x2
h2 hf h fg
95.47 22.49 0.340 236.97 22.49
Since the exit state is a saturated mixture at 0.12 MPa, the exit temperature must be the saturation temperature at this pressure, which is 22.32°C. Then the temperature change for this process becomes
¢T T2 T1 122.32 31.312°C 53.63°C
Discussion Note that the temperature of the refrigerant drops by 53.63°C during this throttling process. Also note that 34.0 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
Cold water
Hot water
Telbow
FIGURE 5–32 The Telbow of an ordinary shower serves as the mixing chamber for the hot and the coldwater streams.
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. 5–32). 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 energies 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.
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Chapter 5 EXAMPLE 5–9
Mixing of Hot and Cold Waters in a Shower
m· 1
Mixing chamber P = 20 psia
Solution In a shower, cold water is mixed with hot water at a specified
Mass balance:
Energy balance:
241
T1 = 140°F
Consider an ordinary shower where hot water at 140°F is mixed with cold water at 50°F. If it is desired that a steady stream of warm water at 110°F 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.
temperature. 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 mCV 0 and ECV 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. 5–33). 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:

T3 = 110°F m· 3
T2 = 50°F m· 2
FIGURE 5–33 Schematic for Example 5–9.
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
0 (steady) # # min mout dmsystem>dt ¡ 0 # # # # # min mout S m1 m2 m3 0 (steady) # # E in E out dE system>dt ¡ 0
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # # # # m1h1 m2h2 m3h3 1since Q 0, W 0, ke pe 02
T
Combining the mass and energy balances, t.
# # # # m1h1 m2h2 (m1 m2)h3
h 1 h f @ 140°F 107.99 Btu/lbm h 2 h f @ 50°F 18.07 Btu/lbm h 3 h f @ 110°F 78.02 Btu/lbm Solving for y and substituting yields
y
h3 h2 78.02 18.07 2.0 h1 h3 107.99 78.02
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 110°F.
co = P
yh1 h2 (y 1)h3 . . where y m1/m2 is the desired mass flow rate ratio. The saturation temperature of water at 20 psia is 227.92°F. 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. 5–34). A compressed liquid can be approximated as a saturated liquid at the given temperature. Thus,
ns
. Dividing this equation by m2 yields Tsat
Compressed liquid states
v
FIGURE 5–34 A substance exists as a compressed liquid at temperatures below the saturation temperatures at the given pressure.
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4b Heat Exchangers
Fluid B 70°C
Heat Fluid A 20°C
50°C Heat
35°C
FIGURE 5–35 A heat exchanger can be as simple as two concentric pipes.
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 (also called tubeandshell) heat exchanger, shown in Fig. 5–35. It is composed of two concentric pipes of different diameters. 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 outbound 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 (ke 0, pe 0) for each fluid stream. The heat transfer rate associated with heat exchangers 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 surrounding 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 insulation and little or no heat crosses the boundary (Fig. 5–36). 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. EXAMPLE 5–10
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 kg/min at 1 MPa and 70°C and leaves at 35°C. The cooling water enters at 300 kPa and 15°C and leaves
Fluid B
Fluid B
CV boundary
Fluid A
FIGURE 5–36 The heat transfer associated with a heat exchanger may be zero or nonzero depending on how the control volume is selected.
CV boundary
Heat
Heat
(a) System: Entire heat exchanger (QCV = 0)
(b) System: Fluid A (QCV ≠ 0)
Fluid A
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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 mCV 0 and ECV 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. 5–37). This is a control volume since mass crosses the system boundary during the process. In general, there are several possibilities for selecting the control volume for multiplestream 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:
# # min mout
Mass balance:
for each fluid stream since there is no mixing. Thus,
# # # m1 m2 mw # # # m3 m4 mR # # Energy balance: E in E out
0 (steady) 0
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
dE system>dt ¡
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # # # # # m1h1 m3h3 m2h2 m4h4¬¬1since Q 0, W 0, ke pe 02 Combining the mass and energy balances and rearranging give
# # mw(h1 h2) mR(h4 h3) 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 (133.52°C). Approximating the compressed liquid as a saturated liquid at the given temperatures, we have
h 1 h f @ 15° C 62.982 kJ/kg h 2 h f @ 25° C 104.83 kJ/kg
(Table A–4)
The refrigerant enters the condenser as a superheated vapor and leaves as a compressed liquid at 35°C. From refrigerant134a tables,
P3 1 MPa f T3 70 C
h3 303.85 kJ/kg
P4 1 MPa f T4 35 C
h4 hf @ 35 C 100.87 kJ/kg
(Table A–13) (Table A–11)
243
Water 15°C 300 kPa 1
at 25°C. 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

R134a 3 70°C 1MPa 4 35°C
2 25°C
FIGURE 5–37 Schematic for Example 5–10.
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Thermodynamics Substituting, we find
# mw 162.982 104.832 kJ/kg 16 kg/min 2[ 1100.87 303.85 2 kJ/kg] # mw 29.1 kg/min
.
.
Qw,in = QR,out R134a
(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
0 (steady) 0
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
FIGURE 5–38 In a heat exchanger, the heat transfer depends on the choice of the control volume.
dE system>dt ¡
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
Control volume boundary
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
# # Ein Eout # # # Qw, in mwh1 mwh2 Rearranging and substituting,
# # Q w, in mw 1h2 h 1 2 129.1 kg/min 2[ 1104.83 62.9822 kJ/kg] ¬ 1218 kJ/min
.
Qout
Surroundings 20°C
Discussion Had we chosen the volume occupied by the refrigerant as . the control volume (Fig. 5–38), we would have obtained the same result for QR,out since the heat gained by the water is equal to the heat lost by the refrigerant.
70°C Hot fluid
FIGURE 5–39 Heat losses from a hot fluid flowing through an uninsulated pipe or duct to the cooler environment may be very significant.
˙e W
Control volume
˙ sh W FIGURE 5–40 Pipe or duct flow may involve more than one form of work at the same time.
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. 5–39). 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 examples 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 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. 5–40). Of these, fan work is usually small and often neglected in energy analysis.
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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 negligible. Kinetic energy changes may be significant, however, for gas flow in ducts with variable crosssectional areas especially when the compressibility 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. EXAMPLE 5–11
Electric Heating of Air in a House
The electric heating systems used in many houses consist of a simple duct with resistance heaters. Air is heated as it flows over resistance wires. Consider a 15kW electric heating system. Air enters the heating section at 100 kPa and 17°C with a volume flow rate of 150 m3/min. 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 specified 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 mCV 0 and ECV 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, ke pe 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. 5–41). 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 m1 m2 m. Also, heat is lost from the system and electrical work is supplied to the system. At temperatures encountered in heating and airconditioning applications, h can be replaced by cp T where cp 1.005 kJ/kg · °C—the value at room temperature—with negligible error (Fig. 5–42). Then the energy balance for this steadyflow system can be expressed in the rate form as
# # E in E out
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
0 (steady) dE system>dt ¡ 0
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
Q˙ out = 200 W T2 = ?
˙e, in = 15 kW W T1 = 17°C P1 = 100 kPa
˙ 1 = 150 m3/min V FIGURE 5–41 Schematic for Example 5–11.
# # E in E out # # # # We,in mh1 Q out mh 2 ¬1since ¢ke ¢pe 0 2 # # # We,in Q out mcp 1T2 T1 2
AIR –20 to 70°C
From the idealgas relation, the specific volume of air at the inlet of the duct is
∆h = 1.005 ∆T (kJ/kg)
v1
RT1 P1
10.287 kPa # m3/kg # K 2 1290 K2 100 kPa
0.832 m3/kg
The mass flow rate of the air through the duct is determined from
# V1 150 m3/min 1 min # b 3.0 kg/s m ¬a v1 60 s 0.832 m3/kg
FIGURE 5–42 The error involved in h cp T, where cp 1.005 kJ/kg · °C, is less than 0.5 percent for air in the temperature range 20 to 70°C.
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Thermodynamics Substituting the known quantities, the exit temperature of the air is determined to be
115 kJ/s 2 10.2 kJ/s 2 13 kg/s2 11.005 kJ/kg # °C 2 1T2 17 2°C T2 21.9°C
Discussion of air.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 5 ON THE DVD.
Supply line
Control volume
CV boundary
FIGURE 5–43 Charging of a rigid tank from a supply line is an unsteadyflow process since it involves changes within the control volume.
CV boundary
Control volume
5–5
■
Note that heat loss from the duct reduces the exit temperature
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 transientflow, 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 volume as well as the energy interactions across the boundary. Some familiar unsteadyflow processes are the charging of rigid vessels from supply lines (Fig. 5–43), 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 t 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 steadyflow systems are fixed in space, size, and shape. Unsteadyflow systems, however, are not (Fig. 5–44). 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. 5–1) min mout ¢msystem¬¬1kg2
(5–42)
where msystem mfinal minitial is the change in the mass of the system. For control volumes, it can also be expressed more explicitly as FIGURE 5–44 The shape and size of a control volume may change during an unsteadyflow process.
mi m e 1m2 m1 2 CV
(5–43)
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 exam
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ple, 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 E in E out
Energy balance:
¢E system
⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭
Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
1kJ2
(5–44)
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 specified by two independent intensive properties for simple compressible systems. Then the energy balance for a uniformflow system can be expressed explicitly as a Q in
Win a mu b a Q out Wout a mu b 1m2e2 m1e1 2 system in
Q
W
Closed system
Closed
Q – W = ∆U Closed
FIGURE 5–45 The energy equation of a uniformflow system reduces to that of a closed system when all the inlets and exits are closed.
(5–45)
out
where u 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 a mh a mh 1m2u 2 m 1u 1 2 system out
Wb Moving boundary
We
(5–46)
in
where Q Qnet,in Qin Qout is the net heat input and W Wnet,out Wout Win is the net work output. Note that if no mass enters or leaves the control volume during a process (mi me 0, and m1 m2 m), this equation reduces to the energy balance relation for closed systems (Fig. 5–45). Also note that an unsteadyflow system may involve boundary work as well as electrical and shaft work (Fig. 5–46). Although both the steadyflow and uniformflow processes are somewhat idealized, many actual processes can be approximated reasonably well by
Wsh
FIGURE 5–46 A uniformflow system may involve electrical, shaft, and boundary work all at once.
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Thermodynamics one of these with satisfactory results. The degree of satisfaction depends on the desired accuracy and the degree of validity of the assumptions made. EXAMPLE 5–12
Charging of a Rigid Tank by Steam
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300°C. 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, KE PE 0 and Esystem Usystem. 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. 5–47). 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
Imaginary piston Pi = 1 MPa Ti = 300°C
Steam
Pi = 1 MPa (constant) mi = m2
m1 = 0 P2 = 1 MPa T2 = ?
(a) Flow of steam into an evacuated tank
FIGURE 5–47 Schematic for Example 5–12.
P2 = 1 MPa
(b) The closedsystem equivalence
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0 Q min mout ¢msystem¬S ¬mi m2 m1 m2
Mass balance:
E in E out
Energy balance:
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭ Net energy transfer by heat, work, and mass
mihi m2u2
Change in internal, kinetic, potential, etc., energies
(since W Q 0, ke pe 0, m1 0)
Combining the mass and energy balances gives
u2 hi 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
Pi 1 MPa f Ti 300 C
hi 3051.6 kJ/kg
(Table A–6)
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
P2 1 MPa f ¬T2 456.1°C u 2 3051.6 kJ/kg Discussion Note that the temperature of the steam in the tank has increased by 156.1°C. 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 temperature (Fig. 5–48). 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. 5–47b. 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
Wb,in
2
1
Pi dV Pi 1V2 V1 2 Pi 3Vtank 1Vtank Vi 2 4 PiVi
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 system
⎫ ⎪ ⎬ ⎪ ⎭
E in E out Net energy transfer by heat, work, and mass
Change in internal, kinetic, potential, etc., energies
Wb,in ¢U miPivi m2u 2 miu i u 2 u i Pivi h i
Steam Ti = 300°C
T2 = 456.1°C
FIGURE 5–48 The temperature of steam rises from 300 to 456.1°C as it enters a tank as a result of flow energy being converted to internal energy.
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Thermodynamics 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.
EXAMPLE 5–13
System boundary
H2O m1 = 1 kg V=6L P = 75 kPa (gage) Vapor Liquid
˙ in = 500 W Q FIGURE 5–49 Schematic for Example 5–13.
Cooking with a Pressure Cooker
A pressure cooker is a pot that cooks food much faster than ordinary pots by maintaining a higher pressure and temperature 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 absolute) inside. Therefore, pressure cookers cook at a temperature of about 133°C (or 271°F) instead of 100°C (or 212°F), cutting the cooking time by as much as 70 percent while minimizing the loss of nutrients. 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 operating 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 temperature at which cooking takes place and (b) the amount of water left in the pressure cooker at the end of the process.
Solution Heat is transferred to a pressure cooker at a specified rate for a
P = 175 kPa T = Tsat@P = 116°C
specified time period. The cooking temperature and the water remaining in the cooker are to be determined. Assumptions 1 This process can be analyzed as a uniformflow process since the properties of the steam leaving the control volume remain constant during the entire cooking process. 2 The kinetic and potential energies of the streams are negligible, ke pe 0. 3 The pressure cooker is stationary and thus its kinetic and potential energy changes are zero; that is, KE PE 0 and Esystem Usystem. 4 The pressure (and thus temperature) in the pressure cooker remains constant. 5 Steam leaves as a saturated vapor at the cooker pressure. 6 There are no boundary, electrical, or shaft work interactions involved. 7 Heat is transferred to the cooker at a constant rate. Analysis We take the pressure cooker as the system (Fig. 5–49). This is a control volume since mass crosses the system boundary during the process. We observe that this is an unsteadyflow process since changes occur within the control volume. Also, there is one exit and no inlets for mass flow. (a) The absolute pressure within the cooker is
Pabs Pgage Patm 75 100 175 kPa
FIGURE 5–50 As long as there is liquid in a pressure cooker, the saturation conditions exist and the temperature remains constant at the saturation temperature.
Since saturation conditions exist in the cooker at all times (Fig. 5–50), the cooking temperature must be the saturation temperature corresponding to this pressure. From Table A–5, it is
T Tsat @ 175 kPa 116.04°C which is about 16°C higher than the ordinary cooking temperature.
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(b) Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as
Mass balance: min mout ¢msystem S me 1m 2 m 1 2 CV¬or¬m e 1m1 m2 2 CV E in E out
Energy balance:
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
⎫ ⎪ ⎬ ⎪ ⎭ Net energy transfer by heat, work, and mass
Qin mehe (m2u2 m1u1)CV
Change in internal, kinetic, potential, etc., energies
(since W 0, ke pe 0)
Combining the mass and energy balances gives
Qin (m1 m2)he (m2u2 m1u1)CV
Sat. vapor he = hg@175 kPa
The amount of heat transfer during this process is found from
# Q in Q in ¢t 10.5 kJ/s 2 130 60 s 2 900 kJ Steam leaves the pressure cooker as saturated vapor at 175 kPa at all times (Fig. 5–51). Thus,
he hg @ 175 kPa 2700.2 kJ/kg
P Sat. vapor
The initial internal energy is found after the quality is determined:
v1 x1 Thus,
V 0.006 m3 0.006 m3/kg m1 1 kg v1 vf vfg
0.006 0.001 0.00499 1.004 0.001
u 1 u f x1u fg 486.82 10.004992 12037.72 kJ/kg 497 kJ/kg
and
U1 m1u1 (1 kg)(497 kJ/kg) 497 kJ The mass of the system at the final state is m2 V/v2. Substituting this into the energy equation yields
Q in a m 1
Sat. liquid
V V b h a ¬u 2 m1u 1 b v2 e v2
There are two unknowns in this equation, u2 and v2. Thus we need to relate them to a single unknown before we can determine these unknowns. Assuming there is still some liquid water left in the cooker at the final state (i.e., saturation conditions exist), v2 and u2 can be expressed as
v2 vf x2vfg 0.001 x2 11.004 0.0012 m3/kg
u 2 u f x2u fg 486.82 x2 12037.72 kJ/kg
Recall that during a boiling process at constant pressure, the properties of each phase remain constant (only the amounts change). When these expressions are substituted into the above energy equation, x2 becomes the only unknown, and it is determined to be
x2 0.009
FIGURE 5–51 In a pressure cooker, the enthalpy of the exiting steam is hg @ 175 kPa (enthalpy of the saturated vapor at the given pressure).
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v2 0.001 10.0092 11.004 0.0012 m3/kg 0.010 m3/kg
and
m2
V 0.006 m3 0.6 kg v2 0.01 m3/kg
Therefore, after 30 min there is 0.6 kg water (liquid vapor) left in the pressure cooker. Discussion Note that almost half of the water in the pressure cooker has evaporated during cooking.
TOPIC OF SPECIAL INTEREST*
General Energy Equation 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 conservation of energy for a fixed quantity of mass can be expressed in rate form as dE sys # # # # d ¬or¬Q W QW dt dt
re¬dV
(5–47)
sys
. . . . where Q Qnet,in Qin Qout is. the net to the system . rate of. heat transfer . (negative, if from the system), W Wnet,out Wout Win is the net power output from the system in all forms (negative, if power input) and dEsys /dt 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
V2 gz 2
(5–48)
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 difference, 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 Wtotal Wshaft Wpressure Wviscous Wother
(5–49)
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 *This section can be skipped without a loss in continuity.
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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 piston–cylinder device shown in Fig. 5–52a. 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 piston, the boundary work done on the system is dWboundary PA ds. Dividing both sides of this relation by the differential time interval dt gives the time rate of boundary work (i.e., power), # # dWpressure dWboundary PAVpiston
where Vpiston ds/dt 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. 5–52b. 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 # S S dWpressure P dA Vn P dA 1 V # n 2
(5–50)
since the normal component of velocity through the differential area dA is → → → Vn V cos u V · n . Note that n is the outer normal of dA, and thus the → → quantity V · n is positive for expansion and negative for compression. . The total rate of work done by pressure forces is obtained by integrating dWpressure over the entire surface A, # Wpressure,net out
S S P P 1V # n 2 dA r 1V # n 2 dA r A A S
S
(5–51)
In light of these discussions, the net power transfer can be expressed as # # # # Wnet,out Wshaft,net out Wpressure,net out Wshaft,net out
P 1 V # n 2 dA (5–52) S
S
A
Then the rate form of the conservation of energy relation for a closed system becomes dE sys # # # Q net,in Wshaft,net out Wpressure,net out dt
(5–53)
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
P
ds
A Vpiston System (gas in cylinder)
(a)
dV
→
P dm dA System
n
u →
V
System boundary, A (b)
FIGURE 5–52 The pressure force acting on (a) the moving boundary of a system in a piston–cylinder device, and (b) the differential surface area of a system of arbitrary shape.
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Thermodynamics energy per unit mass e, which is e u ke pe u V 2/2 gz (Fig. 5–53). This yields
dBsys dt
d =
dt
br dV +
CV
B=E
dE sys
→
br(Vr · n ) dA
dt
d dt
er¬dV
CV
dt
d dt
er dV +
CV
b=e
→
→
er(Vr · n ) dA
CS
⋅
Qnet,in In
m⋅ out , energyout
Fixed control volume
In
⋅ in , m
energyin
Out
⋅
Out
⋅ out , m energyout
Out
Wshaft, net,in
⋅ out , m energyout
FIGURE 5–54 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 transfer and net shaft work.
er¬dV
CV
CS
er 1 Vr # n 2 dA S
S
(5–55)
which can be stated as The net rate of energy The time rate of The net flow rate of ° transfer into a CV by ¢ °change of the energy¢ °energy out of the control¢ heat and work transfer content of the CV surface by mass flow →
→
→
Here V r V → V CS is the fluid velocity relative to the control surface, and → the product r(V r · 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. 5–51 into Eq. 5–55 and combining it with the surface integral on the right give # # d Q net,in Wshaft,net out dt
energyin
(5–54)
Substituting the lefthand side of Eq. 5–53 into Eq. 5–54, the general form of the energy equation that applies to fixed, moving, or deforming control volumes becomes
FIGURE 5–53 The conservation of energy equation is obtained by replacing an extensive property B in the Reynolds transport theorem by energy E and its associated intensive property b by e (Ref. 3).
m⋅ in ,
S
CS
b=e
=
er 1 Vr # n 2A S
CS
# # # d Q net,in Wshaft,net out Wpressure,net out dt dEsys
CV
er dV
CS
a
S S P e b r 1 Vr # n 2 dA (5–56) r
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 P/r Pv wflow is the flow work, which is the work associated 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 (i.e., 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 P/r e is nearly uniform across an inlet or outlet, we can simply take it outside the S S # integral. Noting that m r 1 Vr # n 2 dAc is the mass flow rate across an inlet
Ac
or outlet, the rate of inflow or outflow of energy through the inlet or outlet can . be approximated as m(P/r e). Then the energy equation becomes (Fig. 5–54) # # d Q net,in Wshaft,net out dt
# P # P er dV a m a e b a m a e b r r out in CV
(5–57)
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where e u V 2/2 gz is the total energy per unit mass for both the control volume and flow streams. Then, # # d Q net,in Wshaft,net out dt
V2 # P er¬dV a m a u gz b r 2 out CV
V2 # P a ma u gz b r 2 in
(5–58)
or # # d Q net,in Wshaft,net out dt
V2 # er¬dV a m a h gz b 2 out CV
V2 # a mah gz b 2 in
(5–59)
where we used the definition of enthalpy h u Pv u P/r. 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 negative 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 system during that process, and is expressed as # # min mout ¢msystem¬and¬min mout dmsystem>dt
where msystem mfinal minitial is the change in the mass of . . the system during the process, min and mout are the total rates of mass flow into and out of the system, and dmsystem/dt is the rate of change of mass within the system boundaries. The relations above 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 rVA
V2 gz 2
The total energy transported by a flowing fluid of mass m with uniform properties is mu. The rate of energy transport . . by a fluid with a mass flow rate of m is mu. When the kinetic and potential energies of a fluid stream are negligible, the amount and rate of energy transport become Emass mh and . . Emass mh, respectively. The first law of thermodynamics is essentially an expression 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
¢E system ⎫ ⎪ ⎬ ⎪ ⎭
Net energy transfer by heat, work, and mass
Changes in internal, kinetic, potential, etc., energies
⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
It can also be expressed in the rate form as # # E in E out dE system>dt ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭
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
u h ke pe h
⎫ ⎪ ⎬ ⎪ ⎭
where r 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>r
energy into enthalpy. Then the total energy of a flowing fluid is expressed as
Rate of net energy transfer by heat, work, and mass
Rate of change in internal, kinetic, potential, etc., energies
Thermodynamic processes involving control volumes can be considered in two groups: steadyflow processes and
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unsteadyflow 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 process. 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 # # amam in
out
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭
# # V2 V2 # # Q W a mah gz b a m a h gz b 2 2 out in for each exit
for each inlet
These are the most general forms of the equations for steadyflow processes. For singlestream (oneinlet–oneexit) systems such as nozzles, diffusers, turbines, compressors, and pumps, they simplify to 1 1 # # m1 m2 S VA VA v1 1 1 v2 2 2 # # V 22 V 21 # Q W m c h2 h1 g 1z 2 z 1 2 d 2
In these relations, subscripts 1 and 2 denote the inlet and exit states, respectively. Most unsteadyflow processes can be modeled as a uniformflow 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 relations for a uniformflow system are expressed as min mout ¢msystem Q W a mh a mh 1m 2u 2 m1u 1 2 system out
in
where Q Qnet,in Qin Qout is the net heat input and W Wnet,out Wout Win is the net work output. When solving thermodynamic problems, it is recommended that the general form of the energy balance Ein Eout Esystem be used for all problems, and 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.
3. Y. A. Çengel and J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications. New York: McGrawHill, 2006.
2. ASHRAE Handbook of Refrigeration. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and AirConditioning Engineers, Inc., 1994.
PROBLEMS* Conservation of Mass 5–1C Name four physical quantities that are conserved and two quantities that are not conserved during a process. *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 a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
5–2C Define mass and volume flow rates. How are they related to each other? 5–3C Does the amount of mass entering a control volume have to be equal to the amount of mass leaving during an unsteadyflow process? 5–4C When is the flow through a control volume steady? 5–5C 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? 5–6E 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
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reduces to 0.5 in at the nozzle exit. If the average velocity in the hose is 8 ft/s, 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. 5–7 Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 180 m/s. 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:
Air outlet Air inlet
(a) 0.796 kg/s, (b) 58 cm2
5–8 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 1.20 kg/m3 at the inlet and 1.05 kg/m3 at the exit, determine the percent increase in the velocity of air as it flows through the dryer.
Exhaust fan
FIGURE P5–12 1.05 kg/m3
1.20 kg/m3
FIGURE P5–8
5–13 A smoking lounge is to accommodate 15 heavy smokers. The minimum fresh air requirement for smoking lounges is specified to be 30 L/s per person (ASHRAE, Standard 62, 1989). Determine the minimum required flow rate of fresh air that needs to be supplied to the lounge, and the diameter of the duct if the air velocity is not to exceed 8 m/s.
5–9E Air whose density is 0.078 lbm/ft3 enters the duct of an airconditioning system at a volume flow rate of 450 ft3/min. If the diameter of the duct is 10 in, determine the velocity of the air at the duct inlet and the mass flow rate of air. 5–10 A 1m3 rigid tank initially contains air whose density is 1.18 kg/m3. 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 7.20 kg/m3. Determine the mass of air that has entered the tank.
Smoking lounge Fan 15 smokers
Answer: 6.02 kg
5–11 The ventilating fan of the bathroom of a building has a volume flow rate of 30 L/s and runs continuously. If the density of air inside is 1.20 kg/m3, determine the mass of air vented out in one day. 5–12 A desktop computer is to be cooled by a fan whose flow rate is 0.34 m3/min. Determine the mass flow rate of air through the fan at an elevation of 3400 m where the air density is 0.7 kg/m3. Also, if the average velocity of air is not to exceed 110 m/min, determine the diameter of the casing of the fan. Answers: 0.238 kg/min, 0.063 m
FIGURE P5–13 5–14 The minimum fresh air requirement of a residential building is specified to be 0.35 air change per hour (ASHRAE, Standard 62, 1989). That is, 35 percent of the entire air contained in a residence should be replaced by fresh outdoor air every hour. If the ventilation requirement of a 2.7mhigh, 200m2 residence is to be met entirely by a fan, determine the flow capacity in L/min of the fan that needs to be installed. Also determine the diameter of the duct if the air velocity is not to exceed 6 m/s.
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5–15 Air enters a 28cm diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. Air is heated as it flows, and leaves the pipe at 180 kPa and 40°C. 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. Q Air 200 kPa 20°C 5 m/s
180 kPa 40°C
FIGURE P5–15 5–16 Refrigerant134a enters a 28cm diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat as it flows and leaves the pipe at 180 kPa and 40°C. 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. 5–17 Consider a 300L storage tank of a solar water heating system initially filled with warm water at 45°C. Warm water is withdrawn from the tank through a 2cm diameter hose at an average velocity of 0.5 m/s while cold water enters the tank at 20°C at a rate of 5 L/min. Determine the amount of water in the tank after a 20minute period. Assume the pressure in the tank remains constant at 1 atm. Answer: 212 kg Cold water 20°C 5 L/min
(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. 5–22 Refrigerant134a enters the compressor of a refrigeration system as saturated vapor at 0.14 MPa, and leaves as superheated vapor at 0.8 MPa and 60°C at a rate of 0.06 kg/s. Determine the rates of energy transfers by mass into and out of the compressor. Assume the kinetic and potential energies to be negligible. 5–23 A house is maintained at 1 atm and 24°C, and warm air inside a house is forced to leave the house at a rate of 150 m3/h as a result of outdoor air at 5°C infiltrating into the house through the cracks. Determine the rate of net energy loss of the house due to mass transfer. Answer: 0.945 kW
5–24 Air flows steadily in a pipe at 300 kPa, 77°C, and 25 m/s at a rate of 18 kg/min. Determine (a) the diameter of the pipe, (b) the rate of flow energy, (c) the rate of energy transport by mass, and (d) also determine the error involved in part (c) if the kinetic energy is neglected.
SteadyFlow Energy Balance: Nozzles and Diffusers 5–25C How is a steadyflow system characterized? 5–26C Can a steadyflow system involve boundary work? 5–27C 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? 5–28C The kinetic energy of a fluid increases as it is accelerated in an adiabatic nozzle. Where does this energy come from?
300 L 45°C Warm water 45°C 0.5 m/s
FIGURE P5–17 Flow Work and Energy Transfer by Mass 5–18C What are the different mechanisms for transferring energy to or from a control volume?
5–29C Is heat transfer to or from the fluid desirable as it flows through a nozzle? How will heat transfer affect the fluid velocity at the nozzle exit? 5–30 Air enters an adiabatic nozzle steadily at 300 kPa, 200°C, and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the nozzle is 80 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. Answers: (a) 0.5304 kg/s, (b) 184.6°C, (c) 38.7 cm2
5–19C What is flow energy? Do fluids at rest possess any flow energy? 5–20C How do the energies of a flowing fluid and a fluid at rest compare? Name the specific forms of energy associated with each case. 5–21E Steam is leaving a pressure cooker whose operating pressure is 30 psia. It is observed that the amount of liquid in the cooker has decreased by 0.4 gal in 45 minutes after the steady operating conditions are established, and the crosssectional area of the exit opening is 0.15 in2. Determine (a) the mass flow rate of the steam and the exit velocity,
P1 = 300 kPa T 1 = 200°C V1 = 30 m/s A1 = 80 cm 2
AIR
P 2 = 100 kPa V2 = 180 m/s
FIGURE P5–30
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Chapter 5 Reconsider Prob. 5–30. Using EES (or other) software, 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.

259
5–31
P1 = 13 psia T1 = 20°F V1 = 600 ft/s
5–32 Steam at 5 MPa and 400°C enters a nozzle steadily with a velocity of 80 m/s, and it leaves at 2 MPa and 300°C. The inlet area of the nozzle is 50 cm2, and heat is being lost at a rate of 120 kJ/s. Determine (a) the mass flow rate of the steam, (b) the exit velocity of the steam, and (c) the exit area of the nozzle. 5–33E Air enters a nozzle steadily at 50 psia, 140°F, and 150 ft/s and leaves at 14.7 psia and 900 ft/s. The heat loss from the nozzle is estimated to be 6.5 Btu/lbm of air flowing. The inlet area of the nozzle is 0.1 ft2. Determine (a) the exit temperature of air and (b) the exit area of the nozzle. Answers: (a) 507 R, (b) 0.048 ft2
5–34
Steam at 3 MPa and 400°C enters an adiabatic nozzle steadily with a velocity of 40 m/s and leaves at 2.5 MPa and 300 m/s. Determine (a) the exit temperature and (b) the ratio of the inlet to exit area A1/A2. 5–35 Air at 600 kPa and 500 K enters an adiabatic nozzle that has an inlettoexit area ratio of 2:1 with a velocity of
AIR
P2 = 14.5 psia V2 D 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. 5–146 Underground water is being pumped into a pool whose cross section is 3 m 4 m while water is discharged through a 5cmdiameter orifice at a constant average velocity of 5 m/s. If the water level in the pool rises at a rate of 1.5 cm/min, determine the rate at which water is supplied to the pool, in m3/s. 5–147 The velocity of a liquid flowing in a circular pipe of radius R varies from zero at the wall to a maximum at the pipe center. The velocity distribution in the pipe can be represented as V(r), where r is the radial distance from the pipe . center. Based on the definition of mass flow rate m, obtain a relation for the average velocity in terms of V(r), R, and r. 5–148 Air at 4.18 kg/m3 enters a nozzle that has an inlettoexit area ratio of 2:1 with a velocity of 120 m/s and leaves with a velocity of 380 m/s. Determine the density of air at the exit. Answer: 2.64 kg/m3
Sat. vapor 2 m3 1 MPa
FIGURE P5–143 5–144 A piston–cylinder device initially contains 0.6 kg of steam with a volume of 0.1 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 500°C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 250°C, at which point the valve is closed. Determine (a) the mass of steam that has entered and (b) the amount of heat transfer.
5–149 The air in a 6m 5m 4m hospital room is to be completely replaced by conditioned air every 15 min. If the average air velocity in the circular air duct leading to the room is not to exceed 5 m/s, determine the minimum diameter of the duct. 5–150 A long roll of 1mwide and 0.5cmthick 1Mn manganese steel plate (r 7854 kg/m3) 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 m/min, determine the mass flow rate of the steel plate through the oil bath. Furnace Steel plate
10 m/min Q Steam 0.6 kg 0.1 m3 800 kPa
Oil bath Steam 5 kPa 500°C
FIGURE P5–144 Review Problems 5–145 A D0 10mdiameter tank is initially filled with water 2 m above the center of a D 10cmdiameter valve near the bottom. 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
FIGURE P5–150 5–151E It is well established that indoor air quality (IAQ) has a significant effect on general health and productivity of employees at a workplace. A recent study showed that enhancing IAQ by increasing the building ventilation from 5 cfm (cubic feet per minute) to 20 cfm increased the productivity by 0.25 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
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Chapter 5 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: $14,280/yr 5–152 Air enters a pipe at 50°C and 200 kPa and leaves at 40°C and 150 kPa. It is estimated that heat is lost from the pipe in the amount of 3.3 kJ per kg of air flowing in the pipe. The diameter ratio for the pipe is D1/D2 1.8. Using constant specific heats for air, determine the inlet and exit velocities of the air. Answers: 28.6 m/s, 120 m/s 5–153 In a singleflash geothermal power plant, geothermal water enters the flash chamber (a throttling valve) at 230°C as a saturated liquid at a rate of 50 kg/s. The steam resulting 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 at the exit of the flash chamber is (a) 1 MPa, (b) 500 kPa, (c) 100 kPa, (d) 50 kPa.
5–156 Cold water enters a steam generator at 20°C and leaves as saturated vapor at 150°C. Determine the fraction of heat used in the steam generator to preheat the liquid water from 20°C to the saturation temperature of 150°C. 5–157 Cold water enters a steam generator at 20°C and leaves as saturated vapor at the boiler pressure. At what pressure will the amount of heat needed to preheat the water to saturation temperature be equal to the heat needed to vaporize the liquid at the boiler pressure? 5–158 Saturated steam at 1 atm condenses on a vertical plate that is maintained at 90°C by circulating cooling water through the other side. If the rate of heat transfer by condensation to the plate is 180 kJ/s, determine the rate at which the condensate drips off the plate at the bottom. 1 atm Steam
Separator Flash chamber
230°C sat. liq.
2
3
271
water. If the rate of heat transfer from the hot gases to water is 74 kJ/s, determine the rate of evaporation of water.
90°C
1

Steam turbine 4 20 kPa x = 0.95
Liquid
FIGURE P5–153 5–154 The hotwater needs of a household are met by a 60L electric water heater whose heaters are rated at 1.6 kW. The hotwater tank is initially full with hot water at 80°C. Somebody takes a shower by mixing a constant flow of hot water from the tank with cold water at 20°C at a rate of 0.06 kg/s. After a shower period of 8 min, the water temperature in the tank is measured to drop to 60°C. The heater remained on during the shower and hot water withdrawn from the tank is replaced by cold water at the same flow rate. Determine the mass flow rate of hot water withdrawn from the tank during the shower and the average temperature of mixed water used for the shower.
20°C
FIGURE P5–158 5–159 Water is boiled at 100°C electrically by a 3kW resistance wire. Determine the rate of evaporation of water. Steam
Water 100°C
Tank T1 = 80°C T2 = 60°C
m· 20°C 0.06 kg/s
m·
Mixing chamber
Tmix
FIGURE P5–154 5–155 In a gasfired boiler, water is boiled at 150°C by hot gases flowing through a stainless steel pipe submerged in
FIGURE P5–159 5–160 Two streams of the same ideal gas having different mass flow rates and temperatures are mixed in a steadyflow, adiabatic mixing device. Assuming constant specific heats,
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find the simplest expression for the mixture temperature written in the form # # m1 m2 T3 f a # , # , T1, T2 b m3 m3 m• 1, T1 m• 2, T2
Mixing device
m• 3, T3
FIGURE P5–160 5–161 An ideal gas expands in an adiabatic turbine from 1200 K, 600 kPa to 700 K. Determine the turbine inlet volume flow rate of the gas, in m3/s, required to produce turbine work output at the rate of 200 kW. The average values of the specific heats for this gas over the temperature range are cp 1.13 kJ/kg · K and cv 0.83 kJ/kg · K. R 0.30 kJ/kg · K. 5–162 Consider two identical buildings: one in Los Angeles, California, where the atmospheric pressure is 101 kPa and the other in Denver, Colorado, where the atmospheric pressure is 83 kPa. Both buildings are maintained at 21°C, and the infiltration rate for both buildings is 1.2 air changes per hour (ACH). That is, the entire air in the building is replaced completely by the outdoor air 1.2 times per hour on a day when the outdoor temperature at both locations is 10°C. Disregarding latent heat, determine the ratio of the heat losses by infiltration at the two cities. 5–163 The ventilating fan of the bathroom of a building has a volume flow rate of 30 L/s and runs continuously. The 30 L/s 12.2°C
Fan
Bathroom 22°C
building is located in San Francisco, California, where the average winter temperature is 12.2°C, and is maintained at 22°C at all times. The building is heated by electricity whose unit cost is $0.09/kWh. Determine the amount and cost of the heat “vented out” per month in winter. 5–164 Consider a large classroom on a hot summer day with 150 students, each dissipating 60 W of sensible heat. All the lights, with 6.0 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 15°C, and the temperature of the return air is not to exceed 25°C. Determine the required flow rate of air, in kg/s, that needs to be supplied to the room to keep the average temperature of the room constant. Answer: 1.49 kg/s 5–165 Chickens with an average mass of 2.2 kg and average specific heat of 3.54 kJ/kg · °C are to be cooled by chilled water that enters a continuousflowtype immersion chiller at 0.5°C. Chickens are dropped into the chiller at a uniform temperature of 15°C at a rate of 500 chickens per hour and are cooled to an average temperature of 3°C before they are taken out. The chiller gains heat from the surroundings at a rate of 200 kJ/h. Determine (a) the rate of heat removal from the chickens, in kW, and (b) the mass flow rate of water, in kg/s, if the temperature rise of water is not to exceed 2°C. 5–166 Repeat Prob. 5–165 assuming heat gain of the chiller is negligible. 5–167 In a dairy plant, milk at 4°C is pasteurized continuously at 72°C at a rate of 12 L/s 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 82 percent. The pasteurized milk is then cooled by cold water at 18°C before it is finally refrigerated back to 4°C. To save energy and money, the plant installs a regenerator that has an effectiveness of 82 percent. If the cost of natural gas is $1.10/therm (1 therm 105,500 kJ), determine how much energy and money the regenerator will save this company per year.
72°C
Heat (Pasteurizing section)
72°C Hot milk
4°C Regenerator
Cold milk
FIGURE P5–167
FIGURE P5–163
5–168E A refrigeration system is being designed to cool eggs (r 67.4 lbm/ft3 and cp 0.80 Btu/lbm · °F) with an average mass of 0.14 lbm from an initial temperature of 90°F
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Chapter 5 to a final average temperature of 50°F by air at 34°F at a rate of 10,000 eggs per hour. Determine (a) the rate of heat removal from the eggs, in Btu/h and (b) the required volume flow rate of air, in ft3/h, if the temperature rise of air is not to exceed 10°F. 5–169 The heat of hydration of dough, which is 15 kJ/kg, will raise its temperature to undesirable levels unless some cooling mechanism is utilized. A practical way of absorbing the heat of hydration is to use refrigerated water when kneading the dough. If a recipe calls for mixing 2 kg of flour with 1 kg of water, and the temperature of the city water is 15°C, determine the temperature to which the city water must be cooled before mixing in order for the water to absorb the entire heat of hydration when the water temperature rises to 15°C. Take the specific heats of the flour and the water to be 1.76 and 4.18 kJ/kg · °C, respectively. Answer: 4.2°C
Flour
Water Cooling section
Dough
FIGURE P5–169 5–170 A glass bottle washing facility uses a wellagitated hotwater bath at 55°C that is placed on the ground. The bottles enter at a rate of 800 per minute at an ambient temperature of 20°C and leave at the water temperature. Each bottle has a mass of 150 g and removes 0.2 g of water as it leaves the bath wet. Makeup water is supplied at 15°C. 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. Repeat Prob. 5–170 for a water bath temperature of
5–172 Long aluminum wires of diameter 3 mm (r 2702 kg/m3 and cp 0.896 kJ/kg · °C) are extruded at a tem350°C
perature of 350°C and are cooled to 50°C in atmospheric air at 30°C. If the wire is extruded at a velocity of 10 m/min, determine the rate of heat transfer from the wire to the extrusion room. 5–173 Repeat Prob. 5–172 for a copper wire (r 8950 kg/m3 and cp 0.383 kJ/kg · °C). 5–174 Steam at 40°C condenses on the outside of a 5mlong, 3cmdiameter thin horizontal copper tube by cooling water that enters the tube at 25°C at an average velocity of 2 m/s and leaves at 35°C. Determine the rate of condensation of steam. Answer: 0.0245 kg/s Steam 40°C Cooling water
35°C
FIGURE P5–174
15 kJ/kg
Dough
5–171 50°C.
273
25°C
Qout
15°C

Tair = 30°C 10 m/min
Aluminum wire
FIGURE P5–172
5–175E The condenser of a steam power plant operates at a pressure of 0.95 psia. The condenser consists of 144 horizontal tubes arranged in a 12 12 square array. Steam condenses on the outer surfaces of the tubes whose inner and outer diameters are 1 in and 1.2 in, respectively. If steam is to be condensed at a rate of 6800 lbm/h and the temperature rise of the cooling water is limited to 8°F, determine (a) the rate of heat transfer from the steam to the cooling water and (b) the average velocity of the cooling water through the tubes. 5–176 Saturated refrigerant134a vapor at 34°C is to be condensed as it flows in a 1cmdiameter tube at a rate of 0.1 kg/min. Determine the rate of heat transfer from the refrigerant. What would your answer be if the condensed refrigerant is cooled to 20°C? 5–177E The average atmospheric pressure in Spokane, Washington (elevation 2350 ft), is 13.5 psia, and the average winter temperature is 36.5°F. The pressurization test of a 9fthigh, 3000ft2 older home revealed that the seasonal average infiltration rate of the house is 2.2 air changes per hour (ACH). That is, the entire air in the house is replaced completely 2.2 times per hour by the outdoor air. It is suggested that the infiltration rate of the house can be reduced by half to 1.1 ACH by winterizing the doors and the windows. If the house is heated by natural gas whose unit cost is $1.24/therm and the heating season can be taken to be six months, determine how much the home owner will save from the heating costs per year by this winterization project. Assume the house is maintained at 72°F at all times and the efficiency of the furnace is 0.65. Also assume the latent heat load during the heating season to be negligible. 5–178 Determine the rate of sensible heat loss from a building due to infiltration if the outdoor air at 5°C and 90 kPa
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enters the building at a rate of 35 L/s when the indoors is maintained at 20°C. 5–179 The maximum flow rate of standard shower heads is about 3.5 gpm (13.3 L/min) and can be reduced to 2.75 gpm (10.5 L/min) by switching to lowflow shower heads that are equipped with flow controllers. Consider a family of four, with each person taking a 5 min shower every morning. City water at 15°C is heated to 55°C in an electric water heater and tempered to 42°C by cold water at the Telbow of the shower before being routed to the shower heads. Assuming a constant specific heat of 4.18 kJ/kg · °C 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 by the lowflow ones. 5–180
Reconsider Prob. 5–179. Using EES (or other) software, investigate the effect of the inlet temperature of cold water on the energy saved by using the lowflow shower head. Let the inlet temperature vary from 10°C to 20°C. Plot the electric energy savings against the water inlet temperature, and discuss the results.
5–181 A fan is powered by a 0.5hp motor and delivers air at a rate of 85 m3/min. Determine the highest value for the average velocity of air mobilized by the fan. Take the density of air to be 1.18 kg/m3. 5–182 An airconditioning system requires airflow at the main supply duct at a rate of 180 m3/min. The average velocity of air in the circular duct is not to exceed 10 m/s to avoid excessive vibration and pressure drops. Assuming the fan converts 70 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 1.20 kg/m3.
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. 5–184 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 12.5 MPa and 500°C at a rate of 25 kg/s and exits at 10 kPa and a quality of 0.92. Air enters the compressor at 98 kPa and 295 K at a rate of 10 kg/s and exits at 1 MPa and 620 K. Determine the net power delivered to the generator by the turbine. 1 MPa 620 K
12.5 MPa 500°C
Air comp.
Steam turbine
98 kPa 295 K
10 kPa
FIGURE P5–184 5–185 Water flows through a shower head steadily at a rate of 10 L/min. An electric resistance heater placed in the water pipe heats the water from 16 to 43°C. Taking the density of water to be 1 kg/L, determine the electric power input to the heater, in kW. In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of 39°C through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers
180 m3/min 10 m /s
Resistance heater
FIGURE P5–182 5–183 Consider an evacuated rigid bottle of volume V that is surrounded by the atmosphere at pressure P0 and temperature 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
FIGURE P5–185
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Chapter 5 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. If the price of the electric energy is 8.5 ¢/kWh, determine how much money is saved during a 10min shower as a result of installing this heat exchanger. 5–186
Reconsider Prob. 5–185. Using EES (or other) software, investigate the effect of the heat exchanger effectiveness on the money saved. Let effectiveness range from 20 to 90 percent. Plot the money saved against the effectiveness, and discuss the results.
5–187
Steam enters a turbine steadily at 10 MPa and 550°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent. A heat loss of 30 kJ/kg occurs during the process. 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. 5–188
Reconsider Prob. 5–187. Using EES (or other) 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 the exit area to 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.
5–189E Refrigerant134a enters an adiabatic compressor at 15 psia and 20°F with a volume flow rate of 10 ft3/s 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.

275
gases enter the regenerator at 140 kPa and 800 K and leave at 130 kPa and 600 K. Treating the exhaust gases as air, determine (a) the exit temperature of the air and (b) the mass flow rate of exhaust gases. Answers: (a) 775 K, (b) 14.9 kg/s 5–191 It is proposed to have a water heater that consists of an insulated pipe of 5cm diameter and an electric resistor inside. Cold water at 20°C enters the heating section steadily at a rate of 30 L/min. If water is to be heated to 55°C, determine (a) the power rating of the resistance heater and (b) the average velocity of the water in the pipe. 5–192 In large steam power plants, the feedwater is frequently 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 200°C and leaves as saturated liquid at the same pressure. Feedwater enters the heater at 2.5 MPa and 50°C and leaves at 10°C below the exit temperature of the steam. Determine the ratio of the mass flow rates of the extracted steam and the feedwater. 5–193 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 14°C, and the local atmospheric pressure is 95 kPa. The building is losing heat to the surroundings at a steady rate of 450 kJ/min. Air is forced to flow through the duct and the heater steadily by a 250W fan, and it experiences a temperature rise of 5°C 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 24°C? (b) Determine the average mass flow rate of air through the duct. Answers: (a) 146 s, (b) 6.02 kg/s
P2 = 100 psia
450 kJ/min
T2 = T 1+ 5°C R134a
V = 400 m3 P = 95 kPa
45 hp
·
We,in = 30 kW 14°C ← 24°C
· m
P1 = 15 psia T1 = 20°F · V1 = 10 ft3/s
T1
FIGURE P5–193
FIGURE P5–189E 5–190 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 regenerator at 1 MPa and 550 K at a mass flow rate of 800 kg/min. Heat is transferred to the air at a rate of 3200 kJ/s. Exhaust
250 W
5–194
An insulated vertical piston–cylinder device initially contains 0.2 m3 of air at 200 kPa and 22°C. At this state, a linear spring touches the piston but exerts no force on it. The cylinder is connected by a valve to a line that supplies air at 800 kPa and 22°C. The valve is
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opened, and air from the highpressure line is allowed to enter the cylinder. The valve is turned off when the pressure inside the cylinder reaches 600 kPa. If the enclosed volume inside the cylinder doubles during this process, determine (a) the mass of air that entered the cylinder, and (b) the final temperature of the air inside the cylinder.
has escaped from the cylinder, and (c) the work done. Use constant specific heats at the average temperature. 5–197 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 L/s. The diameters of the inlet and outlet pipes are the same, and the elevation difference 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 (a) the mechanical efficiency of the pump and (b) the temperature rise of water as it flows through the pump due to the mechanical inefficiency. Answers: (a) 74.1 percent, (b) 0.017°C Water 50 L/s
AIR
V1 = 0.2 m3 P1 = 200 kPa T1 = 22°C
300 kPa
h motor = 90%
2
Motor 15 kW
Pi = 800 kPa T i = 22°C
⋅ Wpump
100 kPa
1
FIGURE P5–194 5–195 A piston–cylinder device initially contains 2 kg of refrigerant134a at 800 kPa and 80°C. 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 tank is opened, and R134a is withdrawn 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 tank and the temperature in the tank drops to 20°C. Determine (a) the work done and (b) the heat transfer. Answers: (a) 11.6 kJ, (b) 60.7 kJ 5–196 A piston–cylinder device initially contains 1.2 kg of air at 700 kPa and 200°C. 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 temperature of the air in the cylinder, (b) the amount of mass that
FIGURE P5–197 5–198 Steam enters a nozzle with a low velocity at 150°C 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 0.001 m2. 5–199 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 purpose of a turbocharger is to increase the pressure of air so that
350°C
Q Air 1.2 kg 700 kPa 200°C
FIGURE P5–196
Turbine Exhaust gases 400°C 120 kPa
50°C 100 kPa Air Compressor 130 kPa 30°C Cold air
Aftercooler
FIGURE P5–199
40°C
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Chapter 5 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 400°C and 120 kPa at a rate of 0.02 kg/s and leave at 350°C. Air enters the compressor at 50°C and 100 kPa and leaves at 130 kPa at a rate of 0.018 kg/s. The compressor increases the air pressure with a side effect: It also increases the air temperature, which increases the possibility of a gasoline engine to experience an engine knock. To avoid this, an aftercooler is placed after the compressor to cool the warm air by cold ambient air before it enters the engine cylinders. It is estimated that the aftercooler must decrease the air temperature below 80°C if knock is to be avoided. The cold ambient air enters the aftercooler at 30°C and leaves at 40°C. 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.
Fundamentals of Engineering (FE) Exam Problems 5–200 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 210 m/s at a rate of 3.2 kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, the exit area of the nozzle is (a) 24.0 cm2 (d ) 152 cm2 2 (b) 8.4 cm (e) 23.0 cm2 2 (c) 10.2 cm 5–201 Steam enters a diffuser steadily at 0.5 MPa, 300°C, and 122 m/s at a rate of 3.5 kg/s. The inlet area of the diffuser is (a) 15 cm2 (d ) 150 cm2 2 (b) 50 cm (e) 190 cm2 2 (c) 105 cm 5–202 An adiabatic heat exchanger is used to heat cold water at 15°C entering at a rate of 5 kg/s by hot air at 90°C entering also at a rate of 5 kg/s. If the exit temperature of hot air is 20°C, the exit temperature of cold water is (a) 27°C (d ) 85°C (b) 32°C (e) 90°C (c) 52°C 5–203 A heat exchanger is used to heat cold water at 15°C entering at a rate of 2 kg/s by hot air at 100°C entering at a rate of 3 kg/s. The heat exchanger is not insulated and is losing heat at a rate of 40 kJ/s. If the exit temperature of hot air is 20°C, the exit temperature of cold water is (a) 44°C (d ) 72°C (b) 49°C (e) 95°C (c) 39°C 5–204 An adiabatic heat exchanger is used to heat cold water at 15°C entering at a rate of 5 kg/s by hot water at 90°C entering at a rate of 4 kg/s. If the exit temperature of hot water is 50°C, the exit temperature of cold water is
(a) 42°C (b) 47°C (c) 55°C

277
(d ) 78°C (e) 90°C
5–205 In a shower, cold water at 10°C flowing at a rate of 5 kg/min is mixed with hot water at 60°C flowing at a rate of 2 kg/min. The exit temperature of the mixture is (a) 24.3°C (d ) 44.3°C (b) 35.0°C (e) 55.2°C (c) 40.0°C 5–206 In a heating system, cold outdoor air at 10°C flowing at a rate of 6 kg/min is mixed adiabatically with heated air at 70°C flowing at a rate of 3 kg/min. The exit temperature of the mixture is (a) 30°C (d ) 55°C (b) 40°C (e) 85°C (c) 45°C 5–207 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 0.1 kg/s, and exit at 0.2 MPa and 900 K. If heat is lost from the turbine to the surroundings at a rate of 15 kJ/s, the power output of the gas turbine is (a) 15 kW (d) 60 kW (b) 30 kW (e) 75 kW (c) 45 kW 5–208 Steam expands in a turbine from 4 MPa and 500°C to 0.5 MPa and 250°C at a rate of 1350 kg/h. Heat is lost from the turbine at a rate of 25 kJ/s during the process. The power output of the turbine is (a) 157 kW (d ) 287 kW (b) 207 kW (e) 246 kW (c) 182 kW 5–209 Steam is compressed by an adiabatic compressor from 0.2 MPa and 150°C to 2.5 MPa and 250°C at a rate of 1.30 kg/s. The power input to the compressor is (a) 144 kW (d ) 717 kW (b) 234 kW (e) 901 kW (c) 438 kW 5–210 Refrigerant134a is compressed by a compressor from the saturated vapor state at 0.14 MPa to 1.2 MPa and 70°C at a rate of 0.108 kg/s. The refrigerant is cooled at a rate of 1.10 kJ/s during compression. The power input to the compressor is (a) 5.54 kW (d ) 7.74 kW (b) 7.33 kW (e) 8.13 kW (c) 6.64 kW 5–211 Refrigerant134a expands in an adiabatic turbine from 1.2 MPa and 100°C to 0.18 MPa and 50°C at a rate of 1.25 kg/s. The power output of the turbine is (a) 46.3 kW (d ) 89.2 kW (b) 66.4 kW (e) 112.0 kW (c) 72.7 kW
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5–212 Refrigerant134a at 1.4 MPa and 90°C is throttled to a pressure of 0.6 MPa. The temperature of the refrigerant after throttling is (a) 22°C (d) 80°C (b) 56°C (e) 90°C (c) 82°C 5–213 Air at 20°C and 5 atm is throttled by a valve to 2 atm. If the valve is adiabatic and the change in kinetic energy is negligible, the exit temperature of air will be (a) 10°C (d) 20°C (b) 14°C (e) 24°C (c) 17°C 5–214 Steam at 1 MPa and 300°C is throttled adiabatically to a pressure of 0.4 MPa. If the change in kinetic energy is negligible, the specific volume of the steam after throttling is (a) 0.358 m3/kg (d ) 0.646 m3/kg (e) 0.655 m3/kg (b) 0.233 m3/kg 3 (c) 0.375 m /kg 5–215 Air is to be heated steadily by an 8kW electric resistance heater as it flows through an insulated duct. If the air enters at 50°C at a rate of 2 kg/s, the exit temperature of air is (a) 46.0°C (d ) 55.4°C (b) 50.0°C (e) 58.0°C (c) 54.0°C 5–216 Saturated water vapor at 50°C is to be condensed as it flows through a tube at a rate of 0.35 kg/s. The condensate leaves the tube as a saturated liquid at 50°C. The rate of heat transfer from the tube is (a) 73 kJ/s (d ) 834 kJ/s (b) 980 kJ/s (e) 907 kJ/s (c) 2380 kJ/s
Design and Essay Problems 5–217 Design a 1200W electric hair dryer such that the air temperature and velocity in the dryer will not exceed 50°C and 3 m/s, respectively. 5–218 Design a scalding unit for slaughtered chickens to loosen their feathers before they are routed to featherpicking machines with a capacity of 1200 chickens per hour under the following conditions: The unit will be of an immersion type filled with hot water at an average temperature of 53°C at all times. Chicken with an average mass of 2.2 kg and an average temperature of 36°C will be dipped into the tank, held in the water for 1.5 min, and taken out by a slowmoving conveyor. The chicken is expected to leave the tank 15 percent heavier as a result of the water that sticks to its surface. The centertocenter distance between chickens in any direction will be at least 30 cm. The tank can be as wide as 3 m and as high as 60 cm. The water is to be circulated through and heated by a natural gas furnace, but the temperature rise of water will not exceed 5°C as it passes through the furnace. The water loss is to be made up by the city water at an average temperature of 16°C. The walls and the floor of the tank are wellinsulated. The unit operates 24 h a day and 6 days a week. Assuming reasonable values for the average properties, recommend reasonable values for (a) the mass flow rate of the makeup water that must be supplied to the tank, (b) the rate of heat transfer from the water to the chicken, in kW, (c) the size of the heating system in kJ/h, and (d ) the operating cost of the scalding unit per month for a unit cost of $1.12/therm of natural gas.
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Chapter 6 THE SECOND LAW OF THERMODYNAMICS
T
o this point, we have focused our attention on the first law of thermodynamics, 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 thermodynamic temperature scale. The Carnot cycle is introduced next, and the Carnot principles are discussed. Finally, the idealized Carnot heat engines, refrigerators, and heat pumps are examined.
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 Kelvin–Planck 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.

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SEE TUTORIAL CH. 6, SEC. 1 ON THE DVD.
HOT COFFEE
Heat
FIGURE 6–1 A cup of hot coffee does not get hotter in a cooler room. Heat I=0
FIGURE 6–2 Transferring heat to a wire will not generate electricity.
Heat
FIGURE 6–3 Transferring heat to a paddle wheel will not cause it to rotate.
6–1
■
INTRODUCTION TO THE SECOND LAW
In Chaps. 4 and 5, we applied the first law of thermodynamics, or the conservation 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 eventually cools off (Fig. 6–1). This process satisfies the first law of thermodynamics 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 process—the 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. 6–2). 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 attempt 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 generated in the wires. Finally, consider a paddlewheel mechanism that is operated by the fall of a mass (Fig. 6–3). 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. 6–4). 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 discussed above violate the second law of thermodynamics. 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. 6–5). There are numerous 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, however. 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
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Chapter 6 to engineers, and the second law provides the necessary means to determine the quality as well 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.
6–2
■

281
ONE WAY FIGURE 6–4 Processes occur in a certain direction, and not in the reverse direction. PROCESS
1st law
2nd law
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 capacity (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. 6–6). 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 in large rivers by power plants do not cause any significant change in water temperature. A twophase system can be modeled as a reservoir also 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 furnace. The temperatures of most furnaces are carefully controlled, and they are capable of supplying large quantities of thermal energy as heat in an essentially isothermal manner. Therefore, they can be modeled as reservoirs. A body does not actually have to be very large to be considered a reservoir. 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. 6–7). 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 concern 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 rivers. 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 6–5 A process must satisfy both the first and second laws of thermodynamics to proceed.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 2 ON THE DVD.
ATMOSPHERE RIVER LAKE
OCEAN
FIGURE 6–6 Bodies with relatively large thermal masses can be modeled as thermal energy reservoirs. Thermal energy SOURCE HEAT HEAT Thermal energy SINK
FIGURE 6–7 A source supplies energy in the form of heat, and a sink absorbs it.
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SEE TUTORIAL CH. 6, SEC. 3 ON THE DVD.
Work
No work
Heat
Heat WATER
WATER
FIGURE 6–8 Work can always be converted to heat directly and completely, but the reverse is not true.
Hightemperature SOURCE
Qin
HEAT ENGINE
Wnet,out
Qout
Lowtemperature SINK
FIGURE 6–9 Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink.
6–3
■
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. 6–8, 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 characterized by the following (Fig. 6–9): 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 workproducing 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. 6–10. 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: Qin amount of heat supplied to steam in boiler from a hightemperature source (furnace) Qout amount of heat rejected from steam in condenser to a lowtemperature sink (the atmosphere, a river, etc.) Wout amount of work delivered by steam as it expands in turbine Win 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.
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Energy source (such as a furnace) System boundary
Qin Boiler
Wout
Win Pump
Turbine
Condenser Qout Energy sink (such as the atmosphere) Wout HEAT ENGINE
FIGURE 6–10 Schematic of a steam power plant.
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. 6–11): Wnet,out Wout Win¬¬1kJ2
(6–1)
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 they 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. 6–10; thus, it can be analyzed as a closed system. Recall that for a closed system undergoing a cycle, the change in internal energy U is zero, and therefore the net work output of the system is also equal to the net heat transfer to the system: Wnet,out Qin Qout¬¬1kJ2
Win
FIGURE 6–11 A portion of the work output of a heat engine is consumed internally to maintain continuous operation.
SOURCE Heat input 100 kJ 100 kJ
(6–2) 1
Thermal Efficiency In Eq. 6–2, 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 hth (Fig. 6–12). 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
Wnet,out
(6–3)
Waste heat 80 kJ
ηth,1 = 20%
Net work output 20 kJ
SINK
2
Net work output 30 kJ
Waste heat 70 kJ
ηth,2 = 30%
FIGURE 6–12 Some heat engines perform better than others (convert more of the heat they receive to work).
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Thermodynamics or hth
Wnet,out Qin
(6–4)
It can also be expressed as h th 1
Hightemperature reservoir at TH
HE
QL
(6–5)
since Wnet,out 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: QH magnitude of heat transfer between the cyclic device and the hightemperature medium at temperature TH QL magnitude of heat transfer between the cyclic device and the lowtemperature medium at temperature TL
QH
Wnet,out
Q out Q in
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. 6–13) can also be expressed as Wnet,out QH QL
Lowtemperature reservoir at TL
FIGURE 6–13 Schematic of a heat engine.
and hth
Wnet,out QH
or h th 1
Furnace QH = 100 MJ
Wnet,out = 55 MJ HE
QL = 45 MJ
The atmosphere
FIGURE 6–14 Even the most efficient heat engines reject almost onehalf of the energy they receive as waste heat.
QL QH
(6–6)
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 consumption 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 as 60 percent for large combined gassteam power plants. Thus, even with the most efficient heat engines available today, almost onehalf of the energy supplied ends up in the rivers, lakes, or the atmosphere as waste or useless energy (Fig. 6–14).
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Can We Save Qout? In a steam power plant, the condenser is the device where large quantities of waste heat is 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 continuously 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. 6–15 that is used to lift weights. It consists of a piston–cylinder device with two sets of stops. The working fluid is the gas contained within the cylinder. Initially, the gas temperature is 30°C. 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 100°C, 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 90°C. 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 (weightless 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 90°C back to the reservoir at 100°C 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 medium to a lowtemperature one, and never the other way around. Therefore, we cannot cool this gas from 90 to 30°C by transferring heat to a reservoir at 100°C. Instead, we have to bring the system into contact with a lowtemperature reservoir, say at 20°C, 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 (15 kJ) LOAD LOAD GAS 90°C GAS 30°C
Reservoir at 100°C
GAS 30°C
Heat in (100 kJ)
Heat out (85 kJ)
Reservoir at 20°C
FIGURE 6–15 A heatengine cycle cannot be completed without rejecting some heat to a lowtemperature sink.
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Thermodynamics 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 Kelvin–Planck expression of the second law of thermodynamics discussed later in this section.
EXAMPLE 6–1
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.
FURNACE · QH = 80 MW · Wnet,out HE
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. 6–16. 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
# # QH 80 MW¬and¬QL 50 MW
· QL = 50 MW
The net power output of this heat engine is
# # # Wnet,out QH QL 180 502 MW 30 MW
RIVER
FIGURE 6–16 Schematic for Example 6–1.
Then the thermal efficiency is easily determined to be
# Wnet,out 30 MW hth 0.375 1or 37.5% 2 # 80 MW QH Discussion Note that the heat engine converts 37.5 percent of the heat it receives to work.
m· fuel Combustion chamber · QH
CAR ENGINE (idealized)
· Wnet,out = 65 hp
· QL
Atmosphere
FIGURE 6–17 Schematic for Example 6–2.
EXAMPLE 6–2
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 19,000 Btu/lbm (that is, 19,000 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. Analysis A schematic of the car engine is given in Fig. 6–17. 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
Wnet,out # 65 hp 2545 Btu>h QH a b 689,270 Btu>h h th 0.24 1 hp
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To supply energy at this rate, the engine must burn fuel at a rate of
689,270 Btu>h # m 36.3 lbm/h 19,000 Btu>lbm since 19,000 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.
The Second Law of Thermodynamics: Kelvin–Planck Statement We have demonstrated earlier with reference to the heat engine shown in Fig. 6–15 that, even under ideal conditions, a heat engine must reject some heat to a lowtemperature 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 thermal efficiency of heat engines forms the basis for the Kelvin–Planck statement of the second law of thermodynamics, which is expressed as follows:
Thermal energy reservoir · QH = 100 kW
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 Kelvin–Planck statement can also be expressed as no heat engine can have a thermal efficiency of 100 percent (Fig. 6–18), 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 temperatures only.
6–4
■
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 lowtemperature ones. This heat transfer process occurs in nature without requiring 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. 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. 6–19.
· Wnet,out = 100 kW HEAT ENGINE · QL = 0
FIGURE 6–18 A heat engine that violates the Kelvin–Planck statement of the second law.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 4 ON THE DVD.
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Thermodynamics Surrounding medium such as the kitchen air QH CONDENSER 800 kPa 30°C
800 kPa 60°C Wnet,in
EXPANSION VALVE
COMPRESSOR
120 kPa –25°C
120 kPa –20°C EVAPORATOR
FIGURE 6–19 Basic components of a refrigeration system and typical operating conditions.
Warm environment at TH > TL
QH
Required input Wnet,in
R Desired output QL Cold refrigerated space at TL
FIGURE 6–20 The objective of a refrigerator is to remove QL from the cooled space.
QL Refrigerated space
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 evaporates 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. 6–20. Here QL is the magnitude 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 Wnet,in 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 objective, it requires a work input of Wnet,in. Then the COP of a refrigerator can be expressed as COPR
Desired output QL Required input Wnet,in
(6–7)
. This relation . can also be expressed in rate form by replacing QL by QL and Wnet,in by Wnet,in. The conservation of energy principle for a cyclic device requires that Wnet,in QH QL¬¬1kJ2
(6–8)
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Chapter 6 Then the COP relation becomes COPR
QL 1 QH QL Q H>Q L 1

289
Warm heated space at TH > TL (6–9) QH
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 refrigerator by another term—the coefficient of performance—is the desire to avoid the oddity of having efficiencies greater than unity.
Desired output Wnet,in
HP Required input QL
Heat Pumps Another device that transfers heat from a lowtemperature medium to a hightemperature one is the heat pump, shown schematically in Fig. 6–21. Refrigerators 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 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 the hightemperature medium such as a house (Fig. 6–22). 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 (Fig. 6–23). The measure of performance of a heat pump is also expressed in terms of the coefficient of performance COPHP, defined as Desired output QH COPHP Required input Wnet,in
Cold environment at TL
FIGURE 6–21 The objective of a heat pump is to supply heat QH into the warmer space.
Warm indoors at 20°C
QH = 7 kJ Wnet,in = 2 kJ
(6–10)
COP = 3.5
HP
which can also be expressed as COPHP
QH 1 QH QL 1 Q L >Q H
(6–11)
Cold outdoors at 4°C
A comparison of Eqs. 6–7 and 6–10 reveals that COPHP COPR 1
QL = 5 kJ
(6–12)
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 positive 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.
FIGURE 6–22 The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors.
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FIGURE 6–23 When installed backward, an air conditioner functions as a heat pump. © Reprinted with special permission of King Features Syndicate.
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 3.0 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 is about 4.0. Air conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment. A window airconditioning 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 as shown in Fig. 6–23. 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. The performance of refrigerators and air conditioners in the United States is often expressed in terms of the energy efficiency rating (EER), which is the amount of heat removed from the cooled space in Btu’s for 1 Wh (watthour) of electricity consumed. Considering that 1 kWh 3412 Btu and thus 1 Wh 3.412 Btu, a unit that removes 1 kWh of heat from the cooled space for each kWh of electricity it consumes (COP 1) will have an EER of 3.412. Therefore, the relation between EER and COP is EER 3.412 COPR
Most air conditioners have an EER between 8 and 12 (a COP of 2.3 to 3.5). A highefficiency heat pump manufactured by the Trane Company using a reciprocating variablespeed compressor is reported to have a COP of 3.3 in the heating mode and an EER of 16.9 (COP of 5.0) in the airconditioning mode. Variablespeed compressors and fans allow the unit to operate at maximum efficiency for varying heating/cooling 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 temperature than needed. The COPs of refrigerators are in the range of 2.6–3.0 for cutting and preparation rooms; 2.3–2.6 for meat, deli, dairy, and produce; 1.2–1.5 for frozen foods; and 1.0–1.2 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.
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Chapter 6 EXAMPLE 6–3
Heat Rejection by a Refrigerator

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Kitchen
The food compartment of a refrigerator, shown in Fig. 6–24, is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator.
· QH · Wnet,in = 2 kW R
Solution The power consumption of a refrigerator is given. The COP and the rate of heat rejection are to be determined. Assumptions Steady operating conditions exist. Analysis (a) The coefficient of performance of the refrigerator is
# QL
COPR # Wnet,in
360 kJ>min 2 kW
¬a
1 kW b 3 60 kJ>min
That is, 3 kJ of heat is removed from the refrigerated space for each kJ of work supplied. (b) The rate at which heat is rejected to the room that houses the refrigerator is determined from the conservation of energy relation for cyclic devices,
· QL = 360 kJ/min
Food compartment 4°C
FIGURE 6–24 Schematic for Example 6–3.
60 kJ>min # # # QH QL Wnet,in 360 kJ>min 12 kW2 a b 480 kJ/min 1 kW 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, can move from one place to another, but is never destroyed during a process.
EXAMPLE 6–4
Heating a House by a Heat Pump 80,000 kJ/h
House 20°C
A heat pump is used to meet the heating requirements of a house and maintain it at 20°C. On a day when the outdoor air temperature drops to 2°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air.
Heat loss
·
QH
·
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. 6–25, is determined from the definition of the coefficient of performance to be
# Wnet,in
# 80,000 kJ>h QH 32,000 kJ/h 1or 8.9 kW2 COPHP 2.5
(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be maintained at a constant temperature of 20°C, the heat pump must deliver
Wnet,in = ? COP = 2.5
HP
Q· L = ? Outdoor air at – 2°C
FIGURE 6–25 Schematic for Example 6–4.
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Thermodynamics heat to the house at the same rate, that is, at a rate of 80,000 kJ/h. Then the rate of heat transfer from the outdoor becomes
# # # QL QH Wnet,in 180,000 32,000 2 kJ>h 48,000 kJ/h Discussion Note that 48,000 of the 80,000 kJ/h heat delivered to the house is actually extracted from the cold outdoor air. Therefore, we are paying only for the 32,000kJ/h 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 80,000 kJ/h to the resistance heater as electric energy. This would mean a heating bill that is 2.5 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.
The Second Law of Thermodynamics: Clausius Statement There are two classical statements of the second law—the Kelvin–Planck statement, 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:
Warm environment
QH = 5 kJ Wnet,in = 0 R
QL = 5 kJ Cold refrigerated space
FIGURE 6–26 A refrigerator that violates the Clausius statement of the second law.
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. 6–26). This way, the net effect on the surroundings involves the consumption 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 Kelvin–Planck and the Clausius statements of the second law are negative statements, and a negative statement cannot be proved. Like any other physical law, the second law of thermodynamics is based on experimental 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 Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the second law of thermodynamics. Any device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa. This can be demonstrated as follows.
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Chapter 6 Hightemperature reservoir at TH QH
HEAT ENGINE η th = 100%
= QH
293
Hightemperature reservoir at TH QH + QL
Wnet

QL
REFRIGERATOR
REFRIGERATOR
QL
QL
Lowtemperature reservoir at TL
Lowtemperature reservoir at TL
(a) A refrigerator that is powered by a 100 percent efficient heat engine
(b) The equivalent refrigerator
FIGURE 6–27 Proof that the violation of the Kelvin–Planck statement leads to the violation of the Clausius statement.
Consider the heatenginerefrigerator combination shown in Fig. 6–27a, operating between the same two reservoirs. The heat engine is assumed to have, in violation of the Kelvin–Planck 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 hightemperature 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. 6–27b, 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 Kelvin–Planck statement results in the violation of the Clausius statement. It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the Kelvin–Planck statement. Therefore, the Clausius and the Kelvin–Planck statements are two equivalent expressions of the second law of thermodynamics.
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■
PERPETUALMOTION MACHINES
We have repeatedly stated that a process cannot take place unless it satisfies both the first and second laws of thermodynamics. Any device that violates 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).
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 5 ON THE DVD.
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Thermodynamics · Wnet,out System boundary BOILER
Resistance heater TURBINE
PUMP
FIGURE 6–28 A perpetualmotion machine that violates the first law of thermodynamics (PMM1).
GENERATOR
CONDENSER · Qout
Consider the steam power plant shown in Fig. 6–28. 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 world’s energy problem—if it works, of course. A careful examination of this invention reveals that the system enclosed by the . area is continuously supplying energy to the . shaded any energy. That is, this outside at a rate of Qout Wnet,out without . . receiving system is creating energy at a rate of Qout Wnet,out, which is clearly a violation 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 modification 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. 6–29. 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 efficiency 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
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Chapter 6 System boundary

· Qin BOILER
· Wnet,out PUMP
TURBINE
FIGURE 6–29 A perpetualmotion machine that violates the second law of thermodynamics (PMM2).
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 history, and many more are 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 fundraising. For example, a Philadelphia carpenter named J. W. Kelly collected millions of dollars between 1874 and 1898 from investors in his hydropneumaticpulsatingvacuengine, 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. Recently a group of investors was set to invest $2.5 million into 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 attempting to run his demo machine. Tired of applications for perpetualmotion machines, the U.S. 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 U.S. 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. The proposers of perpetualmotion machines generally have innovative minds, but they usually lack formal engineering training, which is very unfortunate. No one is immune from being deceived by an innovative perpetualmotion machine. As the saying goes, however, if something sounds too good to be true, it probably is.
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Thermodynamics INTERACTIVE TUTORIAL
SEE TUTORIAL CH. 6, SEC. 6 ON THE DVD.
(a) Frictionless pendulum
(b) Quasiequilibrium expansion and compression of a gas
FIGURE 6–30 Two familiar reversible processes.
6–6
■
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 without leaving any trace on the surroundings (Fig. 6–30). 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 following a process, regardless of whether the process is reversible or irreversible. But for reversible processes, this restoration is made without leaving any net change on the surroundings, whereas for irreversible processes, the surroundings usually do some work on the system and therefore does not return to their original state. Reversible processes actually do not occur in nature. They are merely idealizations 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 are bound to remain Mr. or Ms. Single for the rest of their lives. The possibility of finding the perfect prospective 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. 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. 6–31). Reversible processes can be viewed as theoretical limits for the corresponding irreversible ones. Some processes are more irreversible than others. We may never be able to have a reversible process, but we can certainly
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Compression
Expansion

297
Compression
Pressure distribution Water
Water
(a) Slow (reversible) process
Water
Water
(b) Fast (irreversible) process
FIGURE 6–31 Reversible processes deliver the most and consume the least work.
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 workconsuming device. The concept of reversible processes leads to the definition of the secondlaw efficiency for actual processes, which is the degree of approximation 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 irreversibilities 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. 6–32), 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 overcoming 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 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
Friction
GAS
FIGURE 6–32 Friction renders a process irreversible.
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(a) Fast compression
(b) Fast expansion
700 kPa
50 kPa
(c) Unrestrained expansion
FIGURE 6–33 Irreversible compression and expansion processes.
20°C Heat
20°C 5°C
(a) An irreversible heat transfer process
20°C Heat
5°C 2°C
this process and recover that lost power, even though doing so would not violate the conservation of energy principle. Another example of irreversibility is the unrestrained expansion of a gas separated from a vacuum by a membrane, as shown in Fig. 6–33. 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. 6–34). 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 surroundings 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 difference 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.
(b) An impossible heat transfer process
Internally and Externally Reversible Processes FIGURE 6–34 (a) Heat transfer through a temperature difference is irreversible, and (b) the reverse process is impossible.
A typical process involves interactions between a system and its surroundings, 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 reversible process, a system proceeds through a series of equilibrium states,
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Chapter 6 and when the process is reversed, the system passes through exactly the same equilibrium states while returning to its initial state. That is, the paths of the forward and reverse processes coincide for an internally reversible process. 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. 6–35). A totally reversible process involves no heat transfer through a finite temperature difference, no nonquasiequilibrium changes, and no friction or other dissipative 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. 6–36. Both processes are internally reversible, since both take place isothermally and both pass through exactly the same equilibrium states. The first process shown is externally reversible also, since heat transfer for this process takes place through an infinitesimal temperature difference dT. The second process, however, is externally irreversible, since it involves heat transfer through a finite temperature difference T.
6–7
■
THE CARNOT CYCLE
We mentioned earlier that heat engines are cyclic devices and that the working 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,
20°C
20°C
Heat
Heat
Thermal energy reservoir at 20.000 ...1°C
Thermal energy reservoir at 30°C
(a) Totally reversible
(b) Internally reversible
No irreversibilities outside the system

299
No irreversibilities inside the system
FIGURE 6–35 A reversible process involves no internal and external irreversibilities.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 7 ON THE DVD.
Boundary at 20°C
FIGURE 6–36 Totally and interally reversible heat transfer processes.
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Thermodynamics (2)
TH = const.
(1) Energy source at TH QH
(a) Process 12
Insulation
(2)
(3)
TH TL (b) Process 23
(3)
TL = const.
(4) Energy sink at TL QL
(c) Process 34
Insulation
(1)
(4)
TH TL (d) Process 41
FIGURE 6–37 Execution of the Carnot cycle in a closed system.
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 irreversibilities 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 processes—two isothermal and two adiabatic—and it can be executed either in a closed or a steadyflow system. Consider a closed system that consists of a gas contained in an adiabatic piston–cylinder device, as shown in Fig. 6–37. 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 temperature 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
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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, the gas returns to its initial state (state 1). The temperature rises from TL to TH during this reversible adiabatic compression process, which completes the cycle. The PV diagram of this cycle is shown in Fig. 6–38. 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. Notice that if we acted stingily and compressed the gas at state 3 adiabatically 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 discussed in later chapters in conjunction with other power cycles. Being a reversible cycle, the Carnot cycle is the most efficient cycle operating 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.
P 1
QH 2 TH = const. Wnet,out 4
TL = con
st.
QL
V
FIGURE 6–38 PV diagram of the Carnot cycle.
P 1
QH
The Reversed Carnot Cycle The Carnot heatengine cycle just described is a totally reversible cycle. Therefore, 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 Wnet,in 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. 6–39.
6–8
■
THE CARNOT PRINCIPLES
The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the Kelvin–Planck and Clausius statements. A heat engine cannot operate by exchanging heat with a single reservoir, and a refrigerator cannot operate without a net energy input from an external source.
3
4 TH = const. Wnet,in 2
TL = con
st.
QL
3
V
FIGURE 6–39 PV diagram of the reversed Carnot cycle.
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 8 ON THE DVD.
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Hightemperature reservoir at TH
1 Irrev. HE
2 Rev. HE
ηth,1 < ηth,2
3 Rev. HE
ηth,2 = ηth,3
Lowtemperature reservoir at TL
FIGURE 6–40 The Carnot principles.
We can draw valuable conclusions from these statements. Two conclusions pertain to the thermal efficiency of reversible and irreversible (i.e., actual) heat engines, and they are known as the Carnot principles (Fig. 6–40), 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 thermodynamics. To prove the first statement, consider two heat engines operating between the same reservoirs, as shown in Fig. 6–41. 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. In violation of the first Carnot principle, we assume that the irreversible heat engine is more efficient than the reversible one (that is, hth,irrev hth,rev) 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 reservoir. Since the refrigerator is rejecting heat in the amount of QH to the hightemperature 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
Hightemperature reservoir at TH QH
QH
Wirrev Irreversible HE
Reversible HE (or R)
QL,irrev < QL,rev (assumed)
QL,rev
Lowtemperature reservoir at TL
FIGURE 6–41 Proof of the first Carnot principle.
(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)
Wrev
Combined HE + R
Wirrev – Wrev
QL,rev – QL,irrev
Lowtemperature reservoir at TL (b) The equivalent combined system
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Chapter 6 while exchanging heat with a single reservoir—a violation of the Kelvin– Planck statement of the second law. Therefore, our initial assumption that hth,irrev hth,rev 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 by 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 reversible 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.
6–9
■
303
Hightemperature reservoir at TH = 1000 K
Another reversible HE ηth,B
A reversible HE ηth,A
ηth,A = ηth,B = 70% Lowtemperature reservoir at TL = 300 K
FIGURE 6–42 All reversible heat engines operating between the same two reservoirs have the same efficiency (the second Carnot principle).
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 calculations, and its derivation is given below using some reversible heat engines. The second Carnot principle discussed in Section 6–8 states that all reversible heat engines have the same thermal efficiency when operating between the same two reservoirs (Fig. 6–42). 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 reservoirs are characterized by their temperatures, the thermal efficiency of reversible heat engines is a function of the reservoir temperatures only. That is, hth,rev g 1TH, TL 2
INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 9 ON THE DVD.
Thermal energy reservoir at T1 Q1
Q1
Rev. HE A
WA
Q2 Rev. HE C
T2
or
Q2
QH f 1TH, TL 2 QL

WC
(6–13)
since hth 1 QL/QH. 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. 6–43. 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
Rev. HE B
WB Q3
Q3 Thermal energy reservoir at T3
FIGURE 6–43 The arrangement of heat engines used to develop the thermodynamic temperature scale.
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Thermodynamics 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. 6–13 to all three engines separately, we obtain Q1 Q2 Q1 f 1T1, T2 2,¬ f 1T2, T3 2,¬and¬ f 1T1, T3 2 Q2 Q3 Q3
Now consider the identity Q1 Q1 Q2 ¬ Q3 Q2 Q3
which corresponds to f 1T1, T3 2 f 1T1, T2 2 # f 1T2, T3 2
A careful examination of this equation reveals that the lefthand side is a function of T1 and T3, and therefore the righthand side must also be a function of 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 1T1, T2 2
f 1T1 2 f 1T2 2
¬and¬f 1T2, T3 2
f 1T2 2 f 1T3 2
so that f(T2) will cancel from the product of f(T1, T2) and f(T2, T3), yielding f 1T1 2 Q1 f 1T1, T3 2 Q3 f 1T3 2
(6–14)
This relation is much more specific than Eq. 6–13 for the functional form of Q1/Q3 in terms of T1 and T3. For a reversible heat engine operating between two reservoirs at temperatures TH and TL, Eq. 6–14 can be written as Hightemperature reservoir at TH
QL
Wnet
QH QL
(6–15)
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 f(T) satisfy this equation, and the choice is completely arbitrary. Lord Kelvin first proposed taking f(T) T to define a thermodynamic temperature scale as (Fig. 6–44)
QH
Reversible heat engine or refrigerator
f 1TH 2 QH QL f 1TL 2
=
TH TL
Lowtemperature reservoir at TL
FIGURE 6–44 For reversible cycles, the heat transfer ratio QH /QL can be replaced by the absolute temperature ratio TH /TL.
a
QH TH b QL rev TL
(6–16)
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 substance. On this scale, temperatures vary between zero and infinity. The thermodynamic temperature scale is not completely defined by Eq. 6–16 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
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Chapter 6 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 273.16 K (Fig. 6–45). The magnitude of a kelvin is defined as 1/273.16 of the temperature interval between absolute zero and the triplepoint temperature of water. The magnitudes of temperature units on the Kelvin and Celsius scales are identical (1 K 1°C). The temperatures on these two scales differ by a constant 273.15: T 1°C2 T 1K2 273.15
(6–17)
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 temperature scale. Absolute temperatures can be measured accurately by other means, such as the constantvolume idealgas thermometer together with extrapolation techniques as discussed in Chap. 1. The validity of Eq. 6–16 can be demonstrated from physical considerations for a reversible cycle using an ideal gas as the working fluid.
6–10
■
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. 6–6 as hth 1
QL QH
where QH is heat transferred to the heat engine from a hightemperature reservoir at TH, and QL is heat rejected to a lowtemperature reservoir at TL. For reversible heat engines, the heat transfer ratio in the above relation can be replaced by the ratio of the absolute temperatures of the two reservoirs, as given by Eq. 6–16. Then the efficiency of a Carnot engine, or any reversible heat engine, becomes hth,rev 1
TL TH
(6–18)
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. 6–46). All irreversible (i.e., actual) heat engines operating between these temperature limits (TL and TH) have lower efficiencies. An actual heat engine cannot reach this maximum theoretical efficiency value because it is impossible to completely eliminate all the irreversibilities associated with the actual cycle. Note that TL and TH in Eq. 6–18 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. 6–47): 6 hth,rev¬irreversible heat engine hth• hth,rev¬reversible heat engine 7 hth,rev¬impossible heat engine
(6–19)

305
Heat reservoir T QH
W Carnot HE
QL 273.16 K (assigned) Water at triple point QH T = 273.16 ––– QL
FIGURE 6–45 A conceptual experimental setup to determine thermodynamic temperatures on the Kelvin scale by measuring heat transfers QH and QL. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 10 ON THE DVD. Hightemperature reservoir at TH = 1000 K QH
Carnot HE ηth = 70%
Wnet,out
QL Lowtemperature reservoir at TL = 300 K
FIGURE 6–46 The Carnot heat engine is the most efficient of all heat engines operating between the same high and lowtemperature reservoirs.
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Thermodynamics Hightemperature reservoir at TH = 1000 K
Rev. HE ηth = 70%
FIGURE 6–47 No heat engine can have a higher efficiency than a reversible heat engine operating between the same high and lowtemperature reservoirs.
Hightemperature reservoir at TH = 652°C
Irrev. HE ηth = 45%
Impossible HE ηth = 80%
Lowtemperature reservoir at TL = 300 K
Most workproducing devices (heat engines) in operation today have efficiencies under 40 percent, which appear low relative to 100 percent. However, 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 temperature limits—because 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. 6–18. 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. It is obvious from Eq. 6–18 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).
QH = 500 kJ Wnet,out Carnot HE
EXAMPLE 6–5
Analysis of a Carnot Heat Engine
A Carnot heat engine, shown in Fig. 6–48, receives 500 kJ of heat per cycle from a hightemperature source at 652°C and rejects heat to a lowtemperature sink at 30°C. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle. QL
Solution The heat supplied to a Carnot heat engine is given. The thermal Lowtemperature reservoir at TL = 30°C
FIGURE 6–48 Schematic for Example 6–5.
efficiency 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. 6–18 to be
hth,C hth,rev 1
130 2732 K TL 1 0.672 TH 1652 2732 K
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That is, this Carnot heat engine converts 67.2 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. 6–16 to be
QL,rev
130 2732 K TL Q 1500 kJ2 164 kJ TH H,rev 1652 2732 K
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 6–5 receives heat from a source at 925 K and converts 67.2 percent of it to work while rejecting the rest (32.8 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. 6–18 and is listed in Fig. 6–49. 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 67.2 to 39.4 percent. That is, the fraction of heat that can be converted to work drops to 39.4 percent when the temperature of the source drops to 500 K. When the source temperature is 350 K, this fraction becomes a mere 13.4 percent. These efficiency values show that energy has quality as well as quantity. It is clear from the thermal efficiency values in Fig. 6–49 that more of the hightemperature thermal energy can be converted to work. Therefore, the higher the temperature, the higher the quality of the energy (Fig. 6–50). Large quantities of solar energy, for example, can be stored in large bodies 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 concentrating 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 1 300/1000 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.
Hightemperature reservoir at TH
Rev. HE ηth
TH, K
ηth, %
925 800 700 500 350
67.2 62.1 56.7 39.4 13.4
Lowtemperature reservoir at TL = 303 K
FIGURE 6–49 The fraction of heat that can be converted to work as a function of source temperature (for TL 303 K). T, K Quality
2000 1500
Thermal energy
1000 500
FIGURE 6–50 The higher the temperature of the thermal energy, the higher its quality.
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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 to 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 at 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 evaluating 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 obvious 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 multibilliondollar diet industry, which is primarily based on the first law of thermodynamics. However, considering 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. This is also confirmed by studies that show that calories that come from fat are more likely to be stored as fat than the calories that come from carbohydrates and protein. A Stanford study found that body weight was related to fat calories consumed and not calories per se. A Harvard study found no correlation between calories eaten and degree of obesity. A major Cornell University survey involving 6500 people in nearly all provinces of China found that the Chinese eat more—gram for gram, calorie for calorie—than Americans do, but they weigh less, with less body fat. Studies indicate that the metabolism rates and hormone levels change noticeably in the mid30s. Some researchers concluded that prolonged dieting teaches a body to survive on fewer calories, making it more fuel efficient. This probably explains why the dieters gain more weight than they lost once they go back to their normal eating levels. 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 weightloss process.
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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.
6–11
■
THE CARNOT REFRIGERATOR AND HEAT PUMP
INTERACTIVE TUTORIAL
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. 6–9 and 6–11 as COPR
SEE TUTORIAL CH. 6, SEC. 11 ON THE DVD.
1 1 ¬and¬COPHP QH>QL 1 1 QL >QH
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 above relations by the ratios of the absolute temperatures of the high and lowtemperature reservoirs, as expressed by Eq. 6–16. Then the COP relations for reversible refrigerators and heat pumps become COPR,rev
1 TH>TL 1
(6–20)
COPHP,rev
1 1 TL >TH
(6–21)
and
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. 6–51).
Warm environment at TH = 300 K
Reversible refrigerator COPR = 11
Irreversible refrigerator COPR = 7
Cool refrigerated space at TL = 275 K
Impossible refrigerator COPR = 13
FIGURE 6–51 No refrigerator can have a higher COP than a reversible refrigerator operating between the same temperature limits.
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Thermodynamics The coefficients of performance of actual and reversible refrigerators operating between the same temperature limits can be compared as follows: 6 COPR,rev¬¬irreversible refrigerator COPR• COPR,rev¬¬reversible refrigerator 7 COPR,rev¬¬impossible refrigerator
Warm environment at TH = 75°F
Refrigerator COP = 13.5
A similar relation can be obtained for heat pumps by replacing all COPR’s in Eq. 6–22 by COPHP. The COP of a reversible refrigerator or heat pump is the maximum theoretical 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. EXAMPLE 6–6
Cool refrigerated space at TL = 35°F
FIGURE 6–52 Schematic for Example 6–6. 135,000 kJ/h Heat loss
House
A Questionable Claim for a Refrigerator
An inventor claims to have developed a refrigerator that maintains the refrigerated space at 35°F while operating in a room where the temperature is 75°F and that has a COP of 13.5. Is this claim reasonable?
Solution An extraordinary claim made for the performance of a refrigerator is to be evaluated. Assumptions Steady operating conditions exist. Analysis The performance of this refrigerator (shown in Fig. 6–52) can be evaluated by comparing it with a reversible refrigerator operating between the same temperature limits:
COPR,max COPR,rev
TH = 21°C
Q·H
·
Wnet,in = ? HP
Q·L Cold outside air TL = –5°C
FIGURE 6–53 Schematic for Example 6–7.
(6–22)
1 TH>TL 1
1 12.4 175 460 R2 > 135 460 R2 1
Discussion This is the highest COP a refrigerator can have when absorbing heat from a cool medium at 35°F and rejecting it to a warmer medium at 75°F. Since the COP claimed by the inventor is above this maximum value, the claim is false.
EXAMPLE 6–7
Heating a House by a Carnot Heat Pump
A heat pump is to be used to heat a house during the winter, as shown in Fig. 6–53. The house is to be maintained at 21°C at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to 5°C. Determine the minimum power required to drive this heat pump.
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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 QH 135,000 kJ/h 37.5 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
COPHP,rev
1 1 11.3 1 TL >TH 1 15 273 K2> 121 273 K2
Then the required power input to this reversible heat pump becomes
# Wnet,in
QH 37.5 kW 3.32 kW COPHP 11.3
Discussion This reversible heat pump can meet the heating requirements of this house by consuming electric power at a rate of 3.32 kW only. If this house were to be heated by electric resistance heaters instead, the power consumption would jump up 11.3 times to 37.5 kW. This is because in resistance heaters the electric energy is converted to heat at 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 3.32 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).
TOPIC OF SPECIAL INTEREST*
Household Refrigerators
Refrigerators to preserve perishable foods have long been one of the essential appliances in a household. They have proven to be highly durable and reliable, providing satisfactory service for over 15 years. A typical household refrigerator is actually a combination refrigeratorfreezer since it has a freezer compartment to make ice and to store frozen food. Today’s refrigerators use much less energy as a result of using smaller and higherefficiency motors and compressors, better insulation materials, larger coil surface areas, and better door seals (Fig. 6–54). At an average electricity rate of 8.3 cents per kWh, an average refrigerator costs about $72 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 energyefficient model will save over 1000 kWh of electricity per year. For the environment, 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 thermoelectric refrigeration systems are currently more expensive and less
*This section can be skipped without a loss in continuity.
Better door seals
Better insulation materials Refrigerator
More efficient motors and compressors
FIGURE 6–54 Today’s refrigerators are much more efficient because of the improvements in technology and manufacturing.
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TABLE 6–1 Typical operating efficiencies of some refrigeration systems for a freezer temperature of 18°C and ambient temperature of 32°C Type of refrigeration system
Coefficient of performance
Vaporcompression Absorption refrigeration Thermoelectric refrigeration
1.3 0.4 0.1
efficient, and they have found limited use in some specialized applications (Table 6–1). A household refrigerator is designed to maintain the freezer section at 18°C (0°F) and the refrigerator section at 3°C (37°F). Lower freezer temperatures 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 vegetables 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 the 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 0.5°C (31°F), which keeps meat at the lowest safe temperature without freezing it, and thus extending its storage life. The more expensive models come with an automatic icemaker located in the freezer section that is connected to the water line, as well as automatic ice and chilledwater dispensers. 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 environments at up to 43°C (110°F). Refrigerators run intermittently, as you may have noticed, running about 30 percent of the time under normal use in a house at 25°C (77°F). 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 conductivity k 0.032–0.040 W/m · °C) to expandedinplace urethane foam insulation (k 0.019 W/m · °C) made it possible to reduce the wall thickness of the refrigerator 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 refrigeration 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. 6–55), and the unpredictable part, which depends on the user habits such as opening the door, making ice, and loading the refrigerator. The amount of energy
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313
Steel or plastic liner Thermal Insulation 6% Defrost heater
6% Fan motor 6% External heater
52% Wall insulation
FIGURE 6–55 The cross section of a refrigerator showing the relative magnitudes of various effects that constitute the predictable heat load.
30% Door gasket region
Plastic breaker strips
Plastic door liner
consumed by the refrigerator can be minimized by practicing good conservation measures as discussed below. 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 environment in its immediate surroundings (Fig. 6–56). 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 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 the 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 18°C (0°F) and 3°C (37°F), respectively. Setting the freezer temperature below 18°C adds significantly to the energy consumption but does not add much to the storage life of frozen foods. Keeping temperatures 6°C (or 10°F)
From ASHRAE Handbook of Refrigeration, Chap. 48, Fig. 2.
Warm air 30°C
6°C Hot food 80°C
5°C
FIGURE 6–56 Putting hot foods into the refrigerator without cooling them first not only wastes energy but also could spoil the foods nearby.
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Cabinet
Warm air
Refrigerator
Coils
Cool air
FIGURE 6–57 The condenser coils of a refrigerator must be cleaned periodically, and the airflow passages must not be blocked to maintain high performance. Light bulb 40 W
below recommended levels can increase the 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 by a 300W to 1000W resistance heater or by 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 air flow 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 lowwattage 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 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 airconditioning. 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 air flow 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. 6–57). 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.
EXAMPLE 6–8
FIGURE 6–58 Schematic for Example 6–8.
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. 6–58). If the refrigerator has a coefficient of performance of 1.3 and the cost of electricity is 8 cents per kWh, determine the increase in the energy consumption of the refrigerator and its cost per year if the switch is not fixed.
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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 1.3, the power consumed by the refrigerator to remove the heat generated by the lightbulb is
# Qrefrig # 40 W 30.8 W Wrefrig COPR 1.3 Therefore, the total additional power consumed by the refrigerator is
# # # Wtotal,additional Wlight Wrefrig 40 30.8 70.8 W The total number of hours in a year is
Annual hours 1365 days>yr2 124 h>day 2 8760 h>yr 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 120 times>day 2 130 s>time 2 11 h>3600 s2 1365 days>yr2 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 Wtotal,additional 1Additional operating hours 2 10.0708 kW2 18699 h>yr2 616 kWh/yr
and
Additional power cost 1Additional power consumption 2 1Unit cost2 1616 kWh>yr2 1$0.08>kWh 2 $49.3/yr
Discussion Note that not repairing the switch will cost the homeowner about $50 a year. This is alarming when we consider that at $0.08/kWh, a typical refrigerator consumes about $70 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 second 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 converted to work only by some devices called heat engines. The thermal efficiency of a heat engine is defined as hth
Wnet,out QH
1
QL QH
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where Wnet,out 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 highertemperature ones. The performance of a refrigerator or a heat pump is expressed in terms of the coefficient of performance, which is defined as COPR
QL 1 Wnet,in QH>QL 1
COPHP
QH 1 Wnet,in 1 QL >QH
The Kelvin–Planck statement of the second law of thermodynamics 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 thermodynamics is called a perpetualmotion machine. A process is said to be reversible if both the system and the surroundings can be restored to their original conditions. Any other process is irreversible. The effects such as friction, nonquasiequilibrium expansion or compression, and heat transfer through a finite temperature difference render a process irreversible and are called irreversibilities. The Carnot cycle is a reversible cycle that is composed of four reversible processes, two isothermal and two adiabatic. The Carnot principles state that the thermal efficiencies 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 establishing a thermodynamic temperature scale related to the heat transfers between a reversible device and the high and lowtemperature reservoirs by a
QH TH b QL rev TL
Therefore, the QH /QL ratio can be replaced by TH /TL for reversible 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 hth,rev 1
TL TH
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 COPR,rev
1 TH>TL 1
COPHP,rev
1 1 TL >TH
and
Again, these are the highest COPs a refrigerator or a heat pump operating between the temperature limits of TH and TL can have.
REFERENCES AND SUGGESTED READINGS 1. ASHRAE Handbook of Refrigeration, SI version. Atlanta, GA: American Society of Heating, Refrigerating, and AirConditioning Engineers, Inc. 1994. 2. W. Z. Black and J. G. Hartley. Thermodynamics. New York: Harper & Row, 1985.
3. D. Stewart. “Wheels Go Round and Round, but Always Run Down.” November 1986, Smithsonian, pp. 193–208. 4. K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGrawHill, 1999.
PROBLEMS* Second Law of Thermodynamics and Thermal Energy Reservoirs 6–1C A mechanic claims to have developed a car engine that runs on water instead of gasoline. What is your response to this claim? 6–2C Describe an imaginary process that satisfies the first law but violates the second law of thermodynamics.
* 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 a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text.
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Chapter 6 6–3C Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics. 6–4C Describe an imaginary process that violates both the first and the second laws of thermodynamics. 6–5C An experimentalist claims to have raised the temperature of a small amount of water to 150°C by transferring heat from highpressure steam at 120°C. Is this a reasonable claim? Why? Assume no refrigerator or heat pump is used in the process. 6–6C What is a thermal energy reservoir? Give some examples. 6–7C Consider the process of baking potatoes in a conventional oven. Can the hot air in the oven be treated as a thermal energy reservoir? Explain. 6–8C Consider the energy generated by a TV set. What is a suitable choice for a thermal energy reservoir?
Heat Engines and Thermal Efficiency 6–9C Is it possible for a heat engine to operate without rejecting any waste heat to a lowtemperature reservoir? Explain. 6–10C What are the characteristics of all heat engines? 6–11C 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. 6–12C Baseboard heaters are basically electric resistance heaters and are frequently used in space heating. A home owner claims that her 5yearold baseboard heaters have a conversion efficiency of 100 percent. Is this claim in violation of any thermodynamic laws? Explain. 6–13C What is the Kelvin–Planck expression of the second law of thermodynamics? 6–14C 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. 6–15C In the absence of any friction and other irreversibilities, can a heat engine have an efficiency of 100 percent? Explain. 6–16C Are the efficiencies of all the workproducing devices, including the hydroelectric power plants, limited by the Kelvin–Planck statement of the second law? Explain.

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ferred to the cooling water at a rate of 145 GJ/h, determine (a) net power output and (b) the thermal efficiency of this power plant. Answers: (a) 35.3 MW, (b) 45.4 percent 6–19E A car engine with a power output of 110 hp has a thermal efficiency of 28 percent. Determine the rate of fuel consumption if the heating value of the fuel is 19,000 Btu/lbm. 6–20 A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,000 kJ/kg, determine the overall efficiency of this plant. Answer: 30.0 percent 6–21 An automobile engine consumes fuel at a rate of 28 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine. Answer: 21.9 percent 6–22E 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 4 percent and a net power output of 350 kW, determine the average value of the required solar energy collection rate, in Btu/h. 6–23 In 2001, the United States produced 51 percent of its electricity in the amount of 1.878 1012 kWh from coalfired power plants. Taking the average thermal efficiency to be 34 percent, determine the amount of thermal energy rejected by the coalfired power plants in the United States that year. 6–24 The Department of Energy projects that between the years 1995 and 2010, the United States will need to build new power plants to generate an additional 150,000 MW of electricity 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 34 percent. Another possibility is to use the cleanburning Integrated Gasification Combined Cycle (IGCC) plants where the coal is subjected to heat and pressure to gasify it while removing sulfur and particulate 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. Currently the construction of IGCC plants costs about $1500 per kW, but their efficiency is about 45 percent. The average heating value of the coal is about 28,000,000 kJ per ton (that is, 28,000,000 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.
6–17 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?
6–25
Reconsider Prob. 6–24. Using EES (or other) software, investigate the price of coal for varying simple payback periods, plant construction costs, and operating efficiency.
6–18 A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJ/h. If the waste heat is trans
6–26 Repeat Prob. 6–24 for a simple payback period of three years instead of five years. 6–27E An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate
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between the temperature limits of 86°F at the ocean surface and 41°F at a depth of 2100 ft. About 13,300 gpm of cold seawater was to be pumped from deep ocean through a 40indiameter pipe to serve as the cooling medium or heat sink. If the cooling water experiences a temperature rise of 6°F and the thermal efficiency is 2.5 percent, determine the amount of power generated. Take the density of seawater to be 64 lbm/ft3. 6–28 A coalburning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent. The actual gravimetric air–fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24hour period and (b) the rate of air flowing through the furnace. Answers: (a) 2.89 106 kg, (b) 402 kg/s
6–40 An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air. Answers: (a) 2.08, (b) 1110 kJ/min
6–41 A household refrigerator runs onefourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h. If the COP of the refrigerator is 2.2, determine the power the refrigerator draws when running. · Win
800 kJ/h
Refrigerators and Heat Pumps
COP = 2.2
REFRIG.
6–29C What is the difference between a refrigerator and a heat pump? 6–30C What is the difference between a refrigerator and an air conditioner? 6–31C In a refrigerator, heat is transferred from a lowertemperature medium (the refrigerated space) to a highertemperature one (the kitchen air). Is this a violation of the second law of thermodynamics? Explain. 6–32C 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. 6–33C Define the coefficient of performance of a refrigerator in words. Can it be greater than unity? 6–34C Define the coefficient of performance of a heat pump in words. Can it be greater than unity? 6–35C A heat pump that is used to heat a house has a COP of 2.5. That is, the heat pump delivers 2.5 kWh of energy to the house for each 1 kWh of electricity it consumes. Is this a violation of the first law of thermodynamics? Explain. 6–36C A refrigerator has a COP of 1.5. That is, the refrigerator removes 1.5 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. 6–37C What is the Clausius expression of the second law of thermodynamics?
FIGURE P6–41 6–42E Water enters an ice machine at 55°F and leaves as ice at 25°F. If the COP of the ice machine is 2.4 during this operation, determine the required power input for an ice production rate of 28 lbm/h. (169 Btu of energy needs to be removed from each lbm of water at 55°F to turn it into ice at 25°F.) 6–43 A household refrigerator that has a power input of 450 W and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8°C. If the watermelons are initially at 20°C, determine how long it will take for the refrigerator to cool them. The watermelons can be treated as water whose specific heat is 4.2 kJ/kg · °C. Is your answer realistic or optimistic? Explain. Answer: 2240 s 6–44
When a man returns to his wellsealed house on a summer day, he finds that the house is at 32°C. He turns on the air conditioner, which cools the entire house to 20°C in 15 min. If the COP of the airconditioning system is 2.5, 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 0.72 kJ/kg · °C and cp 1.0 kJ/kg · °C.
6–38C Show that the Kelvin–Planck and the Clausius expressions of the second law are equivalent. 6–39 A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air. Answers: (a) 0.83 kW, (b) 110 kJ/min
· Win 32°C 20°C
A/C
FIGURE P6–44
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6–45
Reconsider Prob. 6–44. Using EES (or other) software, determine the power input required by the air conditioner to cool the house as a function for airconditioner EER ratings in the range 9 to 16. Discuss your results and include representative costs of airconditioning units in the EER rating range.
meet the additional cooling requirements. Assuming a usage factor of 0.4 (i.e., only 40 percent of the rated power will be consumed at any given time) and additional occupancy of four people, each generating heat at a rate of 100 W, determine how many of these air conditioners need to be installed to the room.
6–46 Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.
6–53 Consider a building whose annual airconditioning load is estimated to be 120,000 kWh in an area where the unit cost of electricity is $0.10/kWh. Two air