Unformatted text preview: Chapter 1 Introduction and Basic Concepts 1-1 Thermodynamics and Energy Application Areas of Thermodynamics 1-2 Importance of Dimensions and Units Some SI and English Units Dimensional Homogeneity Unity Conversion Ratios 1-3 Systems and Control Volumes 1-4 Properties of a System Continuum 1-5 Density and Specific Gravity 1-6 State and Equilibrium The State Postulate 1-7 Processes and Cycles The Steady-Flow Process 1-8 Temperature and the Zeroth Law of Thermodynamics Temperature Scales The International Temperature Scale of 1990 (ITS-90) 1-9 Pressure Variation of Pressure with Depth 1-10 The Manometer Other Pressure Measurement Devices 1-11 The Barometer and Atmospheric Pressure 1-12 Problem-Solving 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 2-1 Introduction 2-2 Forms of Energy Some Physical Insight to Internal Energy Mechanical Energy More on Nuclear Energy 2-3 Energy Transfer by Heat Historical Background on Heat 2-4 Energy Transfer by Work Electrical Work 2-5 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 2-6 The First Law of Thermodynamics Energy Balance Energy Change of a System, Esystem Mechanisms of Energy Transfer, Ein and Eout 2-7 Energy Conversion Efficiencies 2-8 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 3-1 Pure Substance 3-2 Phases of a Pure Substance 3-3 Phase-Change 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 3-4 Property Diagrams for Phase-Change Processes 1 The T-v Diagram 2 The P-v Diagram Extending the Diagrams to Include the Solid Phase 3 The P-T Diagram The P-v-T Surface 3-5 Property Tables EnthalpyA Combination Property 1a Saturated Liquid and Saturated Vapor States 1b Saturated LiquidVapor Mixture 2 Superheated Vapor 3 Compressed Liquid Reference State and Reference Values 3-6 The Ideal-Gas Equation of State Is Water Vapor an Ideal Gas? 3-7 Compressibility FactorA Measure of Deviation from Ideal-Gas Behavior 3-8 Other Equations of State Van der Waals Equation of State Beattie-Bridgeman Equation of State Benedict-Webb-Rubin 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 4-1 Moving Boundary Work Polytropic Process 4-2 Energy Balance for Closed Systems 4-3 Specific Heats 4-4 Internal Energy, Enthalpy, and Specific Heats of Ideal Gases Specific Heat Relations of Ideal Gases 4-5 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 5-1 Conservation of Mass Mass and Volume Flow Rates Conservation of Mass Principle Mass Balance for Steady-Flow Processes Special Case: Incompressible Flow 5-2 Flow Work and the Energy of a Flowing Fluid Total Energy of a Flowing Fluid Energy Transport by Mass 5-3 Energy Analysis of Steady-Flow Systems Energy Balance 5-4 Some Steady-Flow Engineering Devices 1 Nozzles and Diffusers 2 Turbines and Compressors 3 Throttling Valves 4a Mixing Chambers 4b Heat Exchangers 5 Pipe and Duct Flow 5-5 Energy Analysis of Unsteady-Flow 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 6-1 Introduction to the Second Law 6-2 Thermal Energy Reservoirs 6-3 Heat Engines Thermal Efficiency Can We Save Qout ? The Second Law of Thermodynamics: KelvinPlanck Statement 6-5 Refrigerators and Heat Pumps Coefficient of Performance Heat Pumps The Second Law of Thermodynamics: Clausius Statement Equivalence of the Two Statements 6-6 Perpetual-Motion Machines 6-7 Reversible and Irreversible Processes Irreversibilities Internally and Externally Reversible Processes 6-8 The Carnot Cycle The Reversed Carnot Cycle 6-9 The Carnot Principles 6-10 The Thermodynamic Temperature Scale 6-11 The Carnot Heat Engine The Quality of Energy Quantity versus Quality in Daily Life 6-12 The Carnot Refrigerator and Heat Pump Topics of Special Interest: Household Refrigerators Summary References and Suggested Reading Problems Chapter 7 Entropy 7-1 Entropy A Special Case: Internally Reversible Isothermal Heat Transfer Processes 7-2 The Increase of Entropy Principle Some Remarks about Entropy 7-3 Entropy Change of Pure Substances 7-4 Isentropic Processes 7-5 Property Diagrams Involving Entropy 7-6 What Is Entropy? Entropy and Entropy Generation in Daily Life 7-7 The T ds Relations 7-8 Entropy Change of Liquids and Solids 7-9 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 7-10 Reversible Steady-Flow Work Proof that Steady-Flow Devices Deliver the Most and Consume the Least Work when the Process Is Reversible 7-11 Minimizing the Compressor Work Multistage Compression with Intercooling 7-12 Isentropic Efficiencies of Steady-Flow Devices Isentropic Efficiency of Turbines Isentropic Efficiencies of Compressors and Pumps Isentropic Efficiency of Nozzles 7-13 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 8-1 Exergy: Work Potential of Energy Exergy (Work Potential) Associated with Kinetic and Potential Energy 8-2 Reversible Work and Irreversibility 8-3 Second-Law Efficiency, ηII 8-4 Exergy Change of a System Exergy of a Fixed Mass: Nonflow (or Closed System) Exergy Exergy of a Flow Stream: Flow (or Stream) Exergy 8-5 Exergy Transfer by Heat, Work, and Mass Exergy Transfer by Heat Transfer, Q Exergy Transfer by Work, W Exergy Transfer by Mass, m 8-6 The Decrease of Exergy Principle and Exergy Destruction Exergy Destruction 8-7 Exergy Balance: Closed Systems 8-8 Exergy Balance: Control Volumes Exergy Balance for Steady-Flow Systems Reversible Work, W rev Second-Law Efficiency of Steady-Flow Devices, ηII Topics of Special Interest: Second-Law Aspects of Daily Life Summary References and Suggested Reading Problems Chapter 9 Gas Power Cycles 9-1 Basic Considerations in the Analysis of Power Cycles 9-2 The Carnot Cycle and Its Value in Engineering 9-3 Air-Standard Assumptions 9-4 An Overview of Reciprocating Engines 9-5 Otto Cycle: The Ideal Cycle for Spark-Ignition Engines 9-6 Diesel Cycle: The Ideal Cycle for Compression-Ignition Engines 9-7 Stirling and Ericsson Cycles 9-8 Brayton Cycle: The Ideal Cycle for Gas-Turbine Engines Development of Gas Turbines Deviation of Actual Gas-Turbine Cycles from Idealized Ones 9-9 The Brayton Cycle with Regeneration 9-10 The Brayton Cycle with Intercooling, Reheating, and Regeneration 9-11 Ideal Jet-Propulsion Cycles Modifications to Turbojet Engines 9-12 Second-Law 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 10-1 The Carnot Vapor Cycle 10-2 Rankine Cycle: The Ideal Cycle for Vapor Power Cycles Energy Analysis of the Ideal Rankine Cycle 10-3 Deviation of Actual Vapor Power Cycles from Idealized Ones 10-4 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) 10-5 The Ideal Reheat Rankine Cycle 10-6 The Ideal Regenerative Rankine Cycle Open Feedwater Heaters Closed Feedwater Heaters 10-7 Second-Law Analysis of Vapor Power Cycles 10-8 Cogeneration 10-9 Combined GasVapor Power Cycles Topics of Special Interest: Binary Vapor Cycles Summary References and Suggested Reading Problems Chapter 11 Refrigeration Cycles 11-1 Refrigerators and Heat Pumps 11-2 The Reversed Carnot Cycle 11-3 The Ideal Vapor-Compression Refrigeration Cycle 11-4 Actual Vapor-Compression Refrigeration Cycle 11-5 Selecting the Right Refrigerant 11-6 Heat Pump Systems 11-7 Innovative Vapor-Compression Refrigeration Systems Cascade Refrigeration Systems Multistage Compression Refrigeration Systems Multipurpose Refrigeration Systems with a Single Compressor Liquefaction of Gases 11-8 Gas Refrigeration Cycles 11-9 Absorption Refrigeration Systems Topics of Special Interest: Thermoelectric Power Generation and Refrigeration Systems Summary References and Suggested Reading Problems Chapter 12 Thermodynamic Property Relations 12-1 A Little MathPartial Derivatives and Associated Relations Partial Differentials Partial Differential Relations 12-2 The Maxwell Relations 12-3 The Clapeyron Equation 12-4 General Relations for du, dh, ds, Cv, and Cp Internal Energy Changes Enthalpy Changes Entropy Changes Specific Heats Cv and Cp 12-5 The Joule-Thomson Coefficient 12-6 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 13-1 Composition of a Gas Mixture: Mass and Mole Fractions 13-2 P-v-T Behavior of Gas Mixtures: Ideal and Real Gases Ideal-Gas Mixtures Real-Gas Mixtures 13-3 Properties of Gas Mixtures: Ideal and Real Gases Ideal-Gas Mixtures Real-Gas 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 Second-Law Efficiency Special-Case: Separation of a Two-Component Mixture An Application: Desalination Processes Chapter 14 GasVapor Mixtures and Air-Conditioning 14-1 Dry and Atmospheric Air 14-2 Specific and Relative Humidity of Air 14-3 Dew-Point Temperature 14-4 Adiabatic Saturation and Wet-Bulb Temperatures 14-5 The Psychrometric Chart 14-6 Human Comfort and Air-Conditioning 14-7 Air-Conditioning 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 15-1 Fuels and Combustion 15-2 Theoretical and Actual Combustion Processes 15-3 Enthalpy of Formation and Enthalpy of Combustion 15-4 First-Law Analysis of Reacting Systems Steady-Flow Systems Closed Systems 15-5 Adiabatic Flame Temperature 15-6 Entropy Change of Reacting Systems 15-7 Second-Law Analysis of Reacting systems Topics of Special Interest: Fuel Cells Summary References and Suggested Reading Problems Chapter 16 Chemical and Phase Equilibrium 16-1 Criterion for Chemical Equilibrium 16-2 The Equilibrium Constant for Ideal-Gas Mixtures 16-3 Some Remarks about the KP of Ideal-Gas Mixtures 16-4 Chemical Equilibrium for Simultaneous Reactions 16-5 Variation of KP with Temperature 16-6 Phase Equilibrium Phase Equilibrium for a Single-Component System The Phase Rule Phase Equilibrium for a Multicomponent System Summary References and Suggested Reading Problems Chapter 17 Compressible Flow 17-1 Stagnation Properties 17-2 Speed of Sound and Mach Number 17-3 One-Dimensional Isentropic Flow Variation of Fluid Velocity with Flow Area Property Relations for Isentropic Flow of Ideal Gases 17-4 Isentropic Flow through Nozzles Converging Nozzles ConvergingDiverging Nozzles 17-5 Shock Waves and Expansion Normal Shocks Oblique Shocks PrandtlMeyer Expansion Waves 17-6 Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) Property Relations for Rayleigh Flow Choked Rayleigh Flow 17-7 Steam Nozzles Summary References and Suggested Reading Problems Appendix 1 Property Tables and Charts (SI Units) Table A-1 Molar mass, gas constant, and critical-point properties Table A-2 Ideal-gas specific heats of various common gases Table A-3 Properties of common liquids, solids, and foods Table A-4 Saturated waterTemperature table Table A-5 Saturated waterPressure table Table A-6 Superheated water Table A-7 Compressed liquid water Table A-8 Saturated icewater vapor Figure A-9 T-s diagram for water Figure A-10 Mollier diagram for water Table A-11 Saturated refrigerant-134aTemperature table Table A-12 Saturated refrigerant-134aPressure table Table A-13 Superheated refrigerant-134a Figure A-14 P-h diagram for refrigerant-134a Figure A-15 NelsonObert generalized compressibility chart Table A-16 Properties of the atmosphere at high altitude Table A-17 Ideal-gas properties of air Table A-18 Ideal-gas properties of nitrogen, N2 Table A-19 Ideal-gas properties of oxygen, O2 Table A-20 Ideal-gas properties of carbon dioxide, CO2 Table A-21 Ideal-gas properties of carbon monoxide, CO Table A-22 Ideal-gas properties of hydrogen, H2 Table A-23 Ideal-gas properties of water vapor, H2O Table A-24 Ideal-gas properties of monatomic oxygen, O Table A-25 Ideal-gas properties of hydroxyl, OH Table A-26 Enthalpy of formation, Gibbs function of formation, and absolute entropy at 25°C, 1 atm Table A-27 Properties of some common fuels and hydrocarbons Table A-28 Natural Logarithms of the equilibrium constant Kp Figure A-29 Generalized enthalpy departure chart Figure A-30 Generalized entropy departure chart Figure A-31 Psychrometric chart at 1 atm total pressure Table A-32 One-dimensional isentropic compressible-flow functions for an ideal gas with k = 1.4 Table A-33 One-dimensional normal-shock functions for an ideal gas with k =1.4 Table A-34 Rayleigh flow functions for an ideal gas with k = 1.4 Appendix 2 Property Tables and Charts (English Units) Table A-1E Molar mass, gas constant, and critical-point properties Table A-2E Ideal-gas specific heats of various common gases Table A-3E Properties of common liquids, solids, and foods Table A-4E Saturated waterTemperature table Table A-5E Saturated waterPressure table Table A-6E Superheated water Table A-7E Compressed liquid water Table A-8E Saturated icewater vapor Figure A-9E T-s diagram for water Figure A-10E Mollier diagram for water Table A-11E Saturated refrigerant-134aTemperature table Table A-12E Saturated refrigerant-134aPressure table Table A-13E Superheated refrigerant-134a Figure A-14E P-h diagram for refrigerant-134a Table A-15E Table A-16E Properties of the atomosphere at high altitude Table A-17E Ideal-gas properties of air Table A-18E Ideal-gas properties of nitrogen, N2 Table A-19E Ideal-gas properties of oxygen, O2 Table A-20E Ideal-gas properties of carbon dioxide, CO2 Table A-21E Ideal-gas properties of carbon monoxide, CO Table A-22E Ideal-gas properties of hydrogen, H2 Table A-23E Ideal-gas properties of water vapor, H2O Table A-24E Table A-25E Table A-26E Enthalpy of formation, Gibbs function of formation, and absolute entropy at 77°C, 1 atm Table A-27E Properties of some common fuels and hydrocarbons Figure A-31E Psycrometric chart at 1 atm total pressure cen84959_fm.qxd 4/28/05 2:45 PM Page xvii 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 real-world 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 tomorrows engineers in a simple yet precise manner. Leads students toward a clear understanding and firm grasp of the basic principles of thermodynamics. Encourages creative thinking and development of a deeper understanding and intuitive feel for thermodynamics. Is read by students with interest and enthusiasm rather than being used as an aid to solve problems. | xvii cen84959_fm.qxd 4/28/05 2:45 PM Page xviii xviii | 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 editionsfrom 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. Yesterdays engineer spent a major portion of his or her time substituting values into the formulas and obtaining numerical results. However, formula manipulations and number crunching are now being left mainly to computers. Tomorrows engineer will need a clear understanding and a firm grasp of the basic principles so that he or she can understand even the most complex problems, formulate them, and interpret the results. A conscious effort is made to emphasize these basic principles while also providing students with a perspective of how computational tools are used in engineering practice. The traditional classical, or macroscopic, approach is used throughout the text, with microscopic arguments serving in a supporting role as appropriate. This approach is more in line with students intuition and makes learning the subject matter much easier. NEW IN THIS 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, thermo-economics, energy conversion, and conversion efficiency using familiar settings that involve mostly electrical and mechanical forms of energy. It also exposes students to some exciting real-world 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 cen84959_fm.qxd 4/28/05 2:45 PM Page xix Preface | coverage of oblique shocks and flow with heat transfer (Rayleigh flow) with some exciting photographs and extended discussions of shock waves. UPDATED STEAM AND REFRIGERANT-134A TABLES The steam and refrigerant-134a tables are updated using the most current property data from EES. Tables A-4 through A-8 and A-11 through A-13, 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 computer-EES 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 Unsteady-Flow Systems from Chapter 4. The xix cen84959_fm.qxd 4/28/05 2:45 PM Page xx xx | Preface 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 refrigerant-134a 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 A-32 and A-33) are updated and plots of functions are now included. Also, Rayleigh flow functions are added as Table A-34. 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. SELF-INSTRUCTING 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 self-instructive. The order of coverage is from simple to general. That is, it cen84959_fm.qxd 4/28/05 2:45 PM Page xxi 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 cant be fun? LEARNING OBJECTIVES AND SUMMARIES Each chapter begins with an overview of the material to be covered and chapter-specific 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 WORKED-OUT EXAMPLES WITH A SYSTEMATIC SOLUTIONS PROCEDURE Each chapter contains several worked-out 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 instructors solutions manual. A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMS The end-of-chapter 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 chapterin some cases they require review of material learned in previous chapters. Problems designated as Design and Essay are intended to encourage students to make engineering judgments, to conduct independent exploration of topics of interest, and to communicate | xxi cen84959_fm.qxd 4/28/05 2:45 PM Page xxii xxii | 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 safety-related 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 outcome-based ABET 2000 criteria) and to facilitate multiple-choice tests, over 200 multiple-choice problems are included in the end-of-chapter 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 Mass balance: E in Energy balance: mout E out ¢ S system Entropy generation Change in entropy X out X destroyed ¢ X system Exergy balance: X in S gen Net entropy transfer by heat and mass Entropy balance: S out Change in internal, kinetic, potential, etc., energies S in ¢ E system Net energy transfer by heat, work, and mass ¢ msystem Net exergy transfer by heat, work, and mass Exergy destruction 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. cen84959_fm.qxd 4/28/05 2:45 PM Page xxiii Preface 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, Second-Law 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 | xxiii cen84959_fm.qxd 4/28/05 2:45 PM Page xxiv xxiv | Preface refer the reader to end-of-chapter 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 write-up 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 write-up, 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 WisconsinMadison, this software combines equation-solving capability and engineering property data. EES can do optimization, parametric analysis, and linear and nonlinear regression, and provides publication-quality 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. INSTRUCTORS RESOURCE CD-ROM (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. COSMOS CD-ROM (COMPLETE ONLINE SOLUTIONS MANUAL ORGANIZATION SYSTEM) (AVAILABLE TO INSTRUCTORS ONLY) This CD, available to instructors only, provides electronic solutions delivered via our database management tool. McGraw-Hills COSMOS allows instructors to streamline the creation of assignments, quizzes, and tests by using problems and solutions from the textbookas well as their own custom material. Errata Sheet Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 5th ed, McGraw-Hill, 2006. (Last update: Dec. 29, 2005) Page Location Error Correction 42 45 50 177 327 461 462 462 462 462 619 916 Prob. 1-50E, 2nd line Prob. 1-76, Answer Prob. 1-128 Example 4-6, Assumptions Prob. 6-138, 2nd line Last line 1st calculation 1st line after 1st calculation 2nd calculation 1st line after 2nd calculation Fig. 11-9 Table A-20, the line T=760 K 29.1 mm Hg 5.0 It is a repeat of Prob. 1-127 Assumption #4 Bobbler (spelling error) 4665 4665, 0.923, 92.3% 7.7 4665-4306=359 359 Light-gray and light-color lines 20,135 29.1 in Hg 1.34 Delete it. Delete it (Temp doesnt have to be const) bubbler 5072 5072, 0.849, 84.9% 15.1 5072-4306=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. 1 cen84959_ch01.qxd 3/31/05 3:38 PM Page 1 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 problem-solving 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 problem-solving technique. | 1 cen84959_ch01.qxd 2 | 4/25/05 2:29 PM Page 2 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 11 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 12 Conservation of energy principle for the human body. 11 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. 11). The conservation of energy principle also forms the backbone of the diet industry: A person who has a greater energy input (food) than energy output (exercise) will gain weight (store energy in the form of fat), and a person who has a smaller energy input than output will lose weight (Fig. 12). The change in the energy content of a body or any other system is equal to the difference between the energy input and the energy E. output, and the energy balance is expressed as Ein Eout 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. 13). The high-temperature 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 well-known that a substance consists of a large number of particles called molecules. The properties of the substance naturally depend on the behavior of these particles. For example, the pressure of a gas in a container is the result of momentum transfer between the molecules and the walls of the container. However, one does not need to know the behavior of the gas cen84959_ch01.qxd 4/25/05 2:29 PM Page 3 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. 12 IMPORTANCE OF DIMENSIONS AND UNITS Any physical quantity can be characterized by dimensions. The magnitudes assigned to the dimensions are called units. Some basic dimensions such as mass m, length L, time t, and temperature T are selected as primary or fundamental dimensions, while others such as velocity V, energy E, and volume V are expressed in terms of the primary dimensions and are called secondary dimensions, or derived dimensions. 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. 14). 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 air-conditioning systems, the refrigerator, the humidifier, the pressure cooker, the water heater, the shower, the iron, and even the computer and the TV. On a larger scale, thermodynamics plays a major part in the design and 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. 15). The energy-efficient 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 13 Heat flows in the direction of decreasing temperature. Solar collectors Shower Hot water Hot water tank Cold water Heat exchanger Pump FIGURE 14 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. cen84959_ch01.qxd 4 | 3/31/05 3:38 PM Page 4 Thermodynamics The human body Air conditioning systems Airplanes Car radiators Power plants Refrigeration systems FIGURE 15 Some application areas of thermodynamics. A/C unit, fridge, radiator: © The McGraw-Hill 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 cen84959_ch01.qxd 3/31/05 3:38 PM Page 5 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 11). However, the abbreviation of a unit was to be capitalized if the unit was derived from a proper name. For example, the SI unit of force, which is named after Sir Isaac Newton (16471723), is newton (not Newton), and it is abbreviated as N. Also, the full name of a unit may be pluralized, but its abbreviation cannot. For example, the length of an object can be 5 m or 5 meters, not 5 ms or 5 meter. Finally, no period is to be used in unit 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 dual-system society, and it will stay that way until the transition to the metric system is completed. This puts an extra burden on todays engineering students, since they are expected to retain their understanding of the English system while learning, thinking, and working in terms of the SI. Given the position of the engineers in the transition period, both unit systems are used in this text, with particular emphasis on SI units. As pointed out, the SI is based on a decimal relationship between units. The prefixes used to express the multiples of the various units are listed in Table 12. They are standard for all units, and the student is encouraged to memorize them because of their widespread use (Fig. 16). | TABLE 11 The seven fundamental (or primary) dimensions and their units in SI Dimension Unit Length 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 12 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 10 1 10 2 10 3 10 6 10 9 10 12 5 cen84959_ch01.qxd 6 | 3/31/05 3:38 PM Page 6 Thermodynamics FIGURE 16 The SI unit prefixes are used in all branches of engineering. 1M 1 kg (103 g) 200 mL (0.2 L) (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 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 pound-mass (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 FIGURE 17 The definition of the force units. 1 kgf 1 ft 1 Mass 2 1 Acceleration 2 Force F 10 apples m = 1 kg 1N 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 Newtons second law, that is, or 1 apple m = 102 g 0.45359 kg 4 apples m = 1 lbm 1 lbf (11) In SI, the force unit is the newton (N), and it is defined as the force required to accelerate a mass of 1 kg at a rate of 1 m/s2. In the English system, the force unit is the pound-force (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. 17). That is, 1N 1 kg # m> s2 32.174 lbm # ft> s2 1 lbf FIGURE 18 The relative magnitudes of the force units newton (N), kilogram-force (kgf), and pound-force (lbf). ma A force of 1 N is roughly equivalent to the weight of a small apple (m 102 g), whereas a force of 1 lbf is roughly equivalent to the weight of four medium apples (mtotal 454 g), as shown in Fig. 18. Another force unit in common use in many European countries is the kilogram-force (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 Newtons second law, W mg 1N 2 (12) cen84959_ch01.qxd 3/31/05 3:38 PM Page 7 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 one-sixth of what she or he normally weighs on earth (Fig. 19). At sea level a mass of 1 kg weighs 9.807 N, as illustrated in Fig. 110. A mass of 1 lbm, however, weighs 1 lbf, which misleads people to believe that pound-mass and pound-force 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 earths 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 sea-level 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 newton-meter (N · m), which is called a joule (J). That is, 1J 1N#m (13) 103 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). | 7 FIGURE 19 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 110 The weight of a unit mass at sea level. cen84959_ch01.qxd 8 | 3/31/05 3:38 PM Page 8 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. 111). 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 11 Spotting Errors from Unit Inconsistencies While solving a problem, a person ended up with the following equation at some stage: FIGURE 111 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 right-hand 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 12 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 112 Schematic for Example 12. 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. 112. 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 cen84959_ch01.qxd 3/31/05 3:38 PM Page 9 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, 1 850 kg> m3 2 1 2 m3 2 m Discussion formulas. 1700 kg 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 N kg m s2 and lbf 32.174 lbm ft s2 They can also be expressed more conveniently as unity conversion ratios as N kg # m> s2 1 and lbf 32.174 lbm # ft> s2 1 Unity conversion ratios are identically equal to 1 and are unitless, and thus such ratios (or their inverses) can be inserted conveniently into any calculation to properly convert units. 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 13 The Weight of One Pound-Mass lbm Using unity conversion ratios, show that 1.00 lbm weighs 1.00 lbf on earth (Fig. 113). Solution A mass of 1.00 lbm is subjected to standard earth gravity. Its weight in lbf is to be determined. Assumptions Standard sea-level conditions are assumed. Properties The gravitational constant is g 32.174 ft/s2. FIGURE 113 A mass of 1 lbm weighs 1 lbf on earth. cen84959_ch01.qxd 10 | 4/25/05 2:29 PM Page 10 Thermodynamics Analysis We apply Newtons second law to calculate the weight (force) that corresponds to the known mass and acceleration. The weight of any object is equal to its mass times the local value of gravitational acceleration. Thus, Net weight: One pound (454 grams) W mg 1 1.00 lbm 2 1 32.174 ft> s2 2 a 1 lbf b 32.174 lbm # ft> s2 1.00 lbf 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. 114.) 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 Newtons second law, the actual weight of the cereal in the metric system is W FIGURE 114 A quirk in the metric system of units. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 3 ON THE DVD. SURROUNDINGS SYSTEM BOUNDARY FIGURE 115 System, surroundings, and boundary. CLOSED SYSTEM Mass NO m = constant Energy YES FIGURE 116 Mass cannot cross the boundaries of a closed system, but energy can. 13 mg 1 453.6 g 2 1 9.81 m> s2 2 a 1 kg 1N ba b 2 1000 g 1 kg # m> s 4.45 N 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. 115. 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. 116. 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 piston-cylinder device shown in Fig. 117. 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 cen84959_ch01.qxd 3/31/05 3:38 PM Page 11 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. 118a). 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. 118b. 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. 119. 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 117 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 118 A control volume can involve fixed, moving, real, and imaginary boundaries. FIGURE 119 An open system (a control volume) with one inlet and one exit. cen84959_ch01.qxd 12 | 4/25/05 2:29 PM Page 12 Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 4 ON THE DVD. m V T P ρ 1 m 2 1 V 2 T P ρ 1 m 2 1 V 2 T P ρ Extensive Extensive properties properties Intensive Intensive properties properties FIGURE 120 Criterion to differentiate intensive and extensive properties. O2 1 atm, 20°C 3 × 1016 molecules/mm3 VOID FIGURE 121 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. 14 PROPERTIES OF A SYSTEM Any characteristic of a system is called a property. Some familiar properties are pressure P, temperature T, volume V, and mass m. The list can be extended to include less familiar ones such as viscosity, thermal conductivity, 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 sizeor extentof the system. Total mass, total volume, and total momentum are some examples of extensive properties. An easy way to determine whether a property is intensive or extensive is to divide the system into two equal parts with an imaginary partition, as shown in Fig. 120. 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 10 10 m and its mass is 5.3 10 26 kg. Also, the mean free path of oxygen at 1 atm pressure and 20°C is 6.3 10 8 m. That is, an oxygen molecule travels, on average, a distance of 6.3 10 8 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. 121). The continuum model is applicable as long as the characteristic length of the system (such as its cen84959_ch01.qxd 4/25/05 3:34 PM Page 13 Chapter 1 | 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. 15 INTERACTIVE TUTORIAL DENSITY AND SPECIFIC GRAVITY SEE TUTORIAL CH. 1, SEC. 5 ON THE DVD. Density is defined as mass per unit volume (Fig. 122). Density: r 1 kg> m3 2 m V (14) EXPERIMENT The reciprocal of density is the specific volume v, which is defined as volume per unit mass. That is, v 1 r V m (15) 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 well-known 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 end-of-chapter problem 1129. © Ronald Mullisen V = 12 m 3 m = 3 kg (16) 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 1 kg/L density in kg/m3) since the density of water at 4°C is 1 g/cm3 1000 kg/m3. The specific gravity of mercury at 0°C, for example, is 13.6. 13.6 kg/L 13,600 kg/m3. Therefore, its density at 0°C is 13.6 g/cm3 The specific gravities of some substances at 0°C are given in Table 13. Note that substances with specific gravities less than 1 are lighter than water, and thus they would float on water. ρ = 0.25 kg/m 3 .25 kg/m 1 = 4 m 3/kg v = /kg ρ FIGURE 122 Density is mass per unit volume; specific volume is volume per unit mass. cen84959_ch01.qxd 14 | 4/25/05 2:29 PM Page 14 Thermodynamics TABLE 13 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 0.30.9 19.2 1.72.0 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 123 A system at two different states. 20° C 23°C °C 30 35° C 40° C 42° C (a) Before 32° C 32° C °C 32 32° C 32° C 32° C (b ) After FIGURE 124 A closed system reaching thermal equilibrium. gs rg 1 N> m3 2 (17) where g is the gravitational acceleration. The densities of liquids are essentially constant, and thus they can often be approximated as being incompressible substances during most processes without sacrificing much in accuracy. 16 STATE AND EQUILIBRIUM Consider a system not undergoing any change. At this point, all the properties can be measured or calculated throughout the entire system, which gives us a set of properties that completely describes the condition, or the state, of the system. At a given state, all the properties of a system have fixed values. If the value of even one property changes, the state will change to a different one. In Fig. 123 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. 124. 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. cen84959_ch01.qxd 4/25/05 2:29 PM Page 15 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. 125). Temperature and pressure, however, are independent properties for single-phase 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 phase-change process; thus, temperature and pressure are not sufficient to fix the state of a two-phase system. Phase-change processes are discussed in detail in Chap. 3. | 15 Nitrogen T = 25 °C v = 0.9 m 3/kg FIGURE 125 The state of nitrogen is fixed by two independent, intensive properties. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 7 ON THE DVD. Property A 17 PROCESSES AND CYCLES Any change that a system undergoes from one equilibrium state to another is called a process, and the series of states through which a system passes during a process is called the path of the process (Fig. 126). 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 quasi-equilibrium, process. A quasi-equilibrium 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. 127. When a gas in a piston-cylinder 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 high-pressure region there. Because of this pressure difference, the system can no longer be said to be in equilibrium, and this makes the entire process nonquasi-equilibrium. 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 quasi-equilibrium process. It should be pointed out that a quasi-equilibrium 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 126 A process between states 1 and 2 and the process path. (a) Slow compression (quasi-equilibrium) (b ) Very fast compression (nonquasi-equilibrium) FIGURE 127 Quasi-equilibrium and nonquasiequilibrium compression processes. cen84959_ch01.qxd 16 | 3/31/05 3:38 PM Page 16 Thermodynamics P Final state 2 Process path Initial state 1 V2 V1 V System (2) (1) FIGURE 128 The P-V diagram of a compression process. Mass in 300°C 250°C Control volume work-producing devices deliver the most work when they operate on quasiequilibrium processes. Therefore, quasi-equilibrium 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 128 shows the P-V 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 nonquasi-equilibrium 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 nonquasi-equilibrium 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 129 During a steady-flow process, fluid properties within the control volume may change with position but not with time. The Steady-Flow 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 steady-flow devices. Processes involving such devices can be represented reasonably well by a somewhat idealized process, called the steady-flow process, which can be defined as a process during which a fluid flows through a control volume steadily (Fig. 129). 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. 130). Steady-flow 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 cen84959_ch01.qxd 4/25/05 2:29 PM Page 17 Chapter 1 time in a periodic manner, and the flow through these devices can still be analyzed as a steady-flow process by using time-averaged values for the properties. Mass in | 17 Control volume mcV = const. 18 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 red-hot. 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 mercury-in-glass thermometer, for example, is based on the expansion of mercury with temperature. Temperature is also measured by using several other temperature-dependent 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. 131). 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 130 Under steady-flow 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 131 Two bodies reaching thermal equilibrium after being brought into contact in an isolated enclosure. cen84959_ch01.qxd 18 | 3/31/05 3:38 PM Page 18 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, 17021744, who devised it) and the Fahrenheit scale (named after the German instrument maker G. Fahrenheit, 16861736), respectively. On the Celsius scale, the ice and steam points were originally assigned the values of 0 and 100°C, respectively. The corresponding values on the Fahrenheit scale are 32 and 212°F. These are often referred to as two-point 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 (18241907). 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 (18201872). The temperature unit on this scale is the rankine, which is designated by R. A temperature scale that turns out to be nearly identical to the Kelvin scale is the ideal-gas temperature scale. The temperatures on this scale are measured using a constant-volume 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 (18) 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 ideal-gas 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 cen84959_ch01.qxd 3/31/05 3:38 PM Page 19 Chapter 1 points on a plane, these two measurements are sufficient to determine the constants a and b in Eq. 18. Then the unknown temperature T of a medium 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 P-T 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. 132. 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. 18. In that case Eq. 18 reduces to T bP, and thus we need to specify the temperature at only one point to define an absolute gas temperature scale. It should be noted that the absolute gas temperature scale is not a 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 low-pressure 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. 133). The Kelvin scale is related to the Celsius scale by T 1K 2 T 1 °C 2 273.15 (19) The Rankine scale is related to the Fahrenheit scale by T 1R 2 T 1 °F 2 459.67 (110) It is common practice to round the constant in Eq. 19 to 273 and that in Eq. 110 to 460. The temperature scales in the two unit systems are related by T 1R 2 T 1 °F 2 1.8T 1 K 2 1.8T 1 °C 2 (111) 32 (112) A comparison of various temperature scales is given in Fig. 134. 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 solidliquid 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) 0 FIGURE 132 P versus T plots of the experimental data obtained from a constant-volume 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 0 P (kPa) 120 80 40 0 0 Absolute vacuum V = constant FIGURE 133 A constant-volume gas thermometer would read 273.15°C at absolute zero pressure. cen84959_ch01.qxd 20 | Thermodynamics °C 0.01 273.15 3/31/05 3:38 PM Page 20 K 273.16 0 °F 32.02 459.67 R Triple 491.69 point of water 0 Absolute zero FIGURE 134 Comparison of temperature scales. assigned the value 273.16 K. The Celsius scale was also redefined at this conference in terms of the ideal-gas temperature scale and a single fixed point, which is again the triple point of water with an assigned value of 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 (ITS-90) The International Temperature Scale of 1990, which supersedes the International Practical Temperature Scale of 1968 (IPTS-68), 1948 (ITPS-48), and 1927 (ITS-27), 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 ITS-90 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 ITS-90 and the Kelvin scale and is the most important thermometric fixed point used in the calibration of thermometers to ITS-90. 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 ITS-90 and ITPS-68, but the steam point is 99.975°C in ITS-90 (with an uncertainly of 0.005°C) whereas it was 100.000°C in IPTS-68. The change is due to precise measurements made by gas thermometry by paying particular attention to the effect of sorption (the impurities in a gas absorbed by the walls of the bulb at the reference temperature being desorbed at higher temperatures, causing the measured gas pressure to increase). The ITS-90 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 ITS-90, 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 pressuretemperature 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). cen84959_ch01.qxd 4/25/05 2:29 PM Page 21 Chapter 1 We emphasize that the magnitudes of each division of 1 K and 1°C are identical (Fig. 135). 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 1K 2 ¢ T 1 °C 2 ¢T 1R 2 1°C 1.8 R 21 1.8°F (113) ¢ T 1 °F 2 (114) Some thermodynamic relations involve the temperature T and often the question arises of whether it is in K or °C. If the relation involves temperature differences (such as a b T), it makes no difference and either can be used. However, if the relation involves temperatures only instead of temperature differences (such as a bT) then K must be used. When in doubt, it is always safe to use K because there are virtually no situations in which the use of K is incorrect, but there are many thermodynamic relations that will yield an erroneous result if °C is used. EXAMPLE 14 1K | FIGURE 135 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 1K 2 ¢ T 1 °C 2 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. 111 and 114: ¢T 1R 2 and 1.8 ¢ T 1 K 2 ¢ T 1 °F 2 1 1.8 2 1 10 2 ¢T 1R 2 18 R 18°F Discussion Note that the units °C and K are interchangeable when dealing with temperature differences. 19 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. cen84959_ch01.qxd 22 | 3/31/05 3:38 PM Page 22 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 kilogram-force per square centimeter: 1 bar 1 atm 1 kgf> cm Afeet = 50 in2 2 105 Pa 101,325 Pa 9.807 N> cm 2 0.1 MPa 100 kPa 101.325 kPa 9.807 1.01325 bars 10 N> m2 4 9.807 104 Pa 0.9807 bar 0.9679 atm P = 3 psi P = 6 psi W 150 lbf P = sn = = = 3 psi Afeet 50 in2 FIGURE 136 The normal stress (or pressure) on the feet of a chubby person is much greater than on the feet of a slim person. FIGURE 137 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 pound-force per square 14.696 psi. The pressure units kgf/cm2 inch (lbf/in2, or psi), and 1 atm 2 are also denoted by kg/cm2 and lb/in2, respectively, and they are and lbf/in 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. 136). 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 pressure-measuring devices, however, are calibrated to read zero in the atmosphere (Fig. 137), 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 (115) Pvac Patm Pabs (116) This is illustrated in Fig. 138. 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. cen84959_ch01.qxd 3/31/05 3:38 PM Page 23 Chapter 1 | P gage Patm Pvac P abs P atm Patm Pabs Absolute Absolute P abs = 0 vacuum vacuum FIGURE 138 Absolute, gage, and vacuum pressures. EXAMPLE 15 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. 116 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. 139). Pgage FIGURE 139 The pressure of a fluid at rest increases with depth (as a result of added weight). 23 cen84959_ch01.qxd 24 | 3/31/05 3:38 PM Page 24 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. 140. Assuming the density of the fluid r to be constant, a force balance in the vertical z-direction gives z P1 a Fz x maz 0: P2 ¢ x P1 ¢ x rg ¢ x ¢ z 0 (117) where W mg rg x z is the weight of the fluid element. Dividing by x and rearranging gives z W ¢P P2 x 0 FIGURE 140 Free-body diagram of a rectangular fluid element in equilibrium. Ptop = 1 atm AIR (A 5-m-high room) P bottom = 1.006 atm FIGURE 141 In a room filled with a gas, the variation of pressure with height is negligible. P2 P1 rg ¢ z gs ¢ z (118) 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. 118 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. 141). If we take point 1 to be at the free surface of a liquid open to the atmosphere (Fig. 142), where the pressure is the atmospheric pressure Patm, then the pressure at a depth h from the free surface becomes P Patm rgh or Pgage rgh (119) Liquids are essentially incompressible substances, and thus the variation of density with depth is negligible. This is also the case for gases when the 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. 117 by x z, and taking the limit as z 0. It gives dP dz rg (120) 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, cen84959_ch01.qxd 3/31/05 3:38 PM Page 25 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 (121) P1 = Patm 1 For constant density and constant gravitational acceleration, this relation reduces to Eq. 118, 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 (15481620) published in 1586 the principle illustrated in Fig. 143. 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 Pascals law, after Blaise Pascal (16231662). Pascal also knew that the force applied by a fluid is proportional to the surface area. He realized that two hydraulic cylinders of different areas could be connected, and the larger could be used h 2 P2 = Patm + rgh FIGURE 142 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 H G Mercury PH PI F I FIGURE 143 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. cen84959_ch01.qxd 26 | 4/25/05 2:29 PM Page 26 Thermodynamics to exert a proportionally greater force than that applied to the smaller. Pascals machine has been the source of many inventions that are a part of our daily lives such as hydraulic brakes and lifts. This is what enables us to lift P2 since a car easily by one arm, as shown in Fig. 144. Noting that P1 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 2 P1 FIGURE 144 Lifting of a large weight by a small force by the application of Pascals law. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 10 ON THE DVD. Gas h 1 2 P2 F1 A1 S S F2 F1 A2 A1 (122) 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 1000-kg car by applying a force of just 100 kgf ( 981 N). 110 THE MANOMETER We notice from Eq. 118 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 U-tube 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. 145 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. 119 to be P2 FIGURE 145 The basic manometer. F2 A2 Patm rgh (123) where r is the density of the fluid in the tube. Note that the cross-sectional 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 16 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. 146. If the local atmospheric pressure is 96 kPa, determine the absolute pressure within the tank. cen84959_ch01.qxd 3/31/05 3:38 PM Page 27 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: SG 1 r H2O 2 1 0.85 2 1 1000 kg> m3 2 P Patm h = 55 cm SG = 0.85 FIGURE 146 Schematic for Example 16. rgh 1 850 kg> m3 2 1 9.81 m> s2 2 1 0.55 m 2 a 96 kPa P=? 850 kg> m3 Then from Eq. 123, 1N 1 kPa # m> s2 b a 1000 N> m2 b 1 kg Patm Fluid 1 100.6 kPa Discussion 27 Patm = 96 kPa Solution The reading of a manometer attached to a tank and the atmo- r | 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 Ptop), and (3) two points at given fluid and decreases upward (i.e., Pbottom the same elevation in a continuous fluid at rest are at the same pressure. The last principle, which is a result of Pascals law, allows us to jump from one fluid column to the next in manometers without worrying about pressure change as long as we dont 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. 147 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 Patm r1gh1 r2gh2 r3gh3 Fluid 3 h3 1 FIGURE 147 In stacked-up 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 P1 In the special case of all fluids having the same density, this relation reduces to Eq. 123, as expected. Manometers are particularly well-suited 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. 148. 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 148 Measuring the pressure drop across a flow section or a flow device by a differential manometer. cen84959_ch01.qxd 28 | 3/31/05 3:38 PM Page 28 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: r1g 1 a P1 h2 r2gh r1ga P2 (124) 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 1 r2 r1 2 gh (125) Note that the distance a has no effect on the result, but must be included in r2 the analysis. Also, when the fluid flowing in the pipe is a gas, then r1 and the relation in Eq. 125 simplifies to P1 P2 r2gh. EXAMPLE 17 Oil AIR 1 WATER h1 2 h2 h3 Mercury FIGURE 149 Schematic for Example 17. (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. 149. 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 airwater 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 airwater 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 Patm rwatergh1 g 1 rmercuryh3 85.6 kPa roilgh2 rmercurygh3 rwaterh1 roilh2 2 1 9.81 m> s 2 3 1 13,600 kg> m3 2 1 0.35m 2 2 1 850 kg> m3 2 1 0.2 m 2 4 a 1000 kg> m3 2 1 0.1 m 2 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. cen84959_ch01.qxd 4/25/05 2:29 PM Page 29 Chapter 1 | 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 (18081884), which consists of a hollow metal tube bent like a hook whose end is closed and connected to a dial indicator needle (Fig. 150). 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 pressure-sensing 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. Strain-gage 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 solid-state 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 press-electric) effect. Piezoelectric pressure transducers have a much faster frequency response compared to the diaphragm units and are very suitable for high-pressure applications, but they are generally not as sensitive as the diaphragm-type transducers. 111 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 (16081647) was the first to conclusively prove that the atmospheric pressure can be measured by inverting a mercury-filled tube into a mercury container that is open to the atmosphere, C-type Spiral Twisted tube Helical Tube cross section FIGURE 150 Various types of Bourdon tubes used to measure pressure. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 1, SEC. 11 ON THE DVD. cen84959_ch01.qxd 30 | 3/31/05 3:38 PM Page 30 Thermodynamics as shown in Fig. 151. 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 151 The basic barometer. A1 A2 A3 FIGURE 152 The length or the cross-sectional 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 (126) 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 cross-sectional area of the tube have no effect on the height of the fluid column of a barometer (Fig. 152). 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 far-reaching 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 2.0-L car engine will act like a 1.7-L 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. 153). 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. cen84959_ch01.qxd 3/31/05 3:38 PM Page 31 Chapter 1 EXAMPLE 18 Measuring Atmospheric Pressure w ith 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 10 C, 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 10 C. Properties The density of mercury is given to be 13,570 kg/m3. Analysis From Eq. 126, the atmospheric pressure is determined to be Patm rgh 1 13,570 kg> m3 2 1 9.81 m> s2 2 1 0.74 m 2 a FIGURE 153 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 # m> s2 b a 1000 N> m2 b 1 kg 98.5 kPa Discussion Note that density changes with temperature, and thus this effect should be considered in calculations. EXAMPLE 19 Effect of Piston Weight on Pressure in a Cylinder The piston of a vertical pistoncylinder device containing a gas has a mass of 60 kg and a cross-sectional area of 0.04 m2, as shown in Fig. 154. 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 pistoncylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the free-body diagram of the piston as shown in Fig. 154 and balancing the vertical forces yield PA Patm A W Solving for P and substituting, P Patm mg A 0.97 bar Patm = 0.97 bar m = 60 kg Patm A = 0.04 m 2 1 60 kg 2 1 9.81 m> s2 2 1 0.04 m2 2 a 1N 1 bar # m> s2 b a 10 5 N> m2 b 1 kg 1.12 bar (b) The volume change will have no effect on the free-body 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 154 Schematic for Example 19, and the free-body diagram of the piston. cen84959_ch01.qxd 32 | 3/31/05 3:38 PM Page 32 Thermodynamics EXAMPLE 110 Hydrostatic Pressure in a Solar Pond w ith 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. 155, and the density can be expressed as r r0 B 1 tan2 a pz b 4H 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 1 1040 kg> m3 2 1 9.81 m> s2 2 1 0.8 m 2 a 1 kN b 1000 kg # m> s2 8.16 kPa since 1 kN/m2 1 kPa. The differential change in hydrostatic pressure across a vertical distance of dz is given by dP rg dz r0 1 B 0 Integrating from the top of the gradient zone (point 1 where z location z in the gradient zone (no subscript) gives z P P1 z rg dz S P 0 P1 tan2 a Sun Increasing salinity and density r0 = 1040 kg/m3 Surface zone 1 z H=4m Gradient zone Storage zone FIGURE 155 Schematic for Example 110. 2 0) to any pz b g dz 4H 4/25/05 2:29 PM Page 33 Chapter 1 r0g 4H sinh p 1 a tan 3 pz b 4H 2.5 Then the pressure at the bottom of the gradient zone (z becomes P2 8.16 kPa sinh 1 1 1040 kg> m3 2 1 9.81 m> s2 2 a tan H 4 m) 4 14 m 2 p p4 1 kN ba b 44 1000 kg # m> s2 54.0 kPa 1 gage 2 Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig. 156. 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. 112 3.5 z, m P1 33 4 Performing the integration gives the variation of gage pressure in the gradient zone to be P | 2 1.5 1 0.5 0 0 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 156 The variation of gage pressure with depth in the gradient zone of the solar pond. PROBLEM-SOLVING 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 real-world problems. Solving such problems, especially complicated ones, require a systematic approach. By using a step-by-step approach, an engineer can reduce the solution of a complicated problem into the solution of a series of simple problems (Fig. 157). 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 cen84959_ch01.qxd FIGURE 157 A step-by-step approach can greatly simplify problem solving. cen84959_ch01.qxd 34 | 3/31/05 3:39 PM Page 34 Thermodynamics 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 158 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. 158). 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, dont give a false implication of high precision by copying all the digits from the screen of the calculatorround the results to an appropriate number of significant digits (see p. 38). Step 7: Reasoning, Verification, and Discussion Energy use: Energy use: $80/yr $80/yr Energy saved Energy saved by insulation: by insulation: $200/yr $200/yr IMPOSSIBLE! IMPOSSIBLE! FIGURE 159 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. 159). 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. cen84959_ch01.qxd 3/31/05 3:39 PM Page 35 Chapter 1 | 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 in-depth 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 high-salaried engineers since any person who knows how to use a word-processing 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 word-processing 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. 160). 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 160 An excellent word-processing program does not make a person a good writer; it simply makes a good writer a more efficient writer. © Vol. 80/PhotoDisc 35 cen84959_ch01.qxd 36 | 3/31/05 3:39 PM Page 36 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 todays engineers. They must still have a thorough understanding of the fundamentals, develop a feel of the physical phenomena, be able to put the data into proper perspective, and make sound engineering judgments, just like their predecessors. However, they must do it much better, and much faster, using more realistic models because of the powerful tools available today. The engineers in the past had to rely on hand calculations, slide rules, and later hand calculators and computers. Today they rely on software packages. The easy access to such power and the possibility of a simple misunderstanding or misinterpretation causing great damage make it more important today than ever to have solid training in the fundamentals of engineering. In this text we make an extra effort to put the emphasis on developing an intuitive and physical understanding of natural phenomena instead of on the mathematical details of solution procedures. 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 built-in 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 111 and 112. The use and capabilities of EES are explained in Appendix 3 on the enclosed DVD. EXAMPLE 111 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 double-clicking on its icon, open a new file, and type the following on the blank screen that appears: xy x^2 y^2 4 x y 20 cen84959_ch01.qxd 3/31/05 3:39 PM Page 37 Chapter 1 | 37 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 112 Oil Analyzing a Multifluid Manometer with EES Reconsider the multifluid manometer discussed in Example 17 and replotted in Fig. 161. 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 double-clicking 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*h2-rm*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 h3 0.35 by P1 129647 and r m 13600 by r m 1030, 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 161 Schematic for Example 112. h3 cen84959_ch01.qxd 38 | 3/31/05 3:39 PM Page 38 Thermodynamics within brackets [ ] after the specified value. When this feature is utilized, the previous equations would take the following form: Given: Given: Volume: Volume: Density: Density: V = 3.75 L .75 r = 0.845 kg / L .845 kg (3 significant digits) (3 significant digits) Also, Also, 3.75 .75 0.845 = 3.16875 .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*h2-rm*g*h3=Patm Find: Find: Mass: Mass: m = rV = 3.16875 kg .16875 kg Rounding to significant digits: Rounding to 3 significant digits: m = 3.17 kg .17 kg FIGURE 162 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 3.75-L 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. 162). 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. cen84959_ch01.qxd 3/31/05 3:39 PM Page 39 Chapter 1 | 39 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 mass-dependent 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 quasi-static or quasi-equilibrium 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 1K 2 T 1R 2 T 1 °C 2 T 1 °F 2 273.15 459.67 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 °C 2 and ¢T 1R 2 ¢ T 1 °F 2 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 Pvac Patm 1 for pressures above Patm 2 Patm Pabs 1 for 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 dz rg 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. Pascals principle states that the pressure applied to a confined fluid increases the pressure throughout by the same amount. The atmospheric pressure is measured by a barometer and is given by 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 380-79, 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. cen84959_ch01.qxd 40 | 3/31/05 3:39 PM Page 40 Thermodynamics PROBLEMS* Thermodynamics 11C What is the difference between the classical and the statistical approaches to thermodynamics? 12C 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? 13C 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 14C What is the difference between pound-mass and pound-force? 15C What is the difference between kg-mass and kgforce? 16C 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? 112 A 5-kg 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. 113 Solve Prob. 112 using EES (or other) software. Print out the entire solution, including the numerical results with proper units. 114 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 115C 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. 17 A 3-kg 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. 18 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 19 At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g a bz, 3.32 10 6 s 2. Determine where a 9.807 m/s2 and b the height above sea level where the weight of an object will decrease by 1 percent. Answer: 29,539 m 110E A 150-lbm 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 111 The acceleration of high-speed aircraft is sometimes expressed in gs (in multiples of the standard acceleration of gravity). Determine the upward force, in N, that a 90-kg man would experience in an aircraft whose acceleration is 6 gs. *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 P115C © The McGraw-Hill Companies, Inc./Jill Braaten, photographer 116C 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. 117C What is the difference between intensive and extensive properties? 118C For a system to be in thermodynamic equilibrium, do the temperature and the pressure have to be the same everywhere? 119C What is a quasi-equilibrium process? What is its importance in engineering? 120C Define the isothermal, isobaric, and isochoric processes. 121C What is the state postulate? 122C Is the state of the air in an isolated room completely specified by the temperature and the pressure? Explain. cen84959_ch01.qxd 3/31/05 3:39 PM Page 41 Chapter 1 | 41 123C What is a steady-flow process? Pressure, Manometer, and Barometer 124C What is specific gravity? How is it related to density? 134C What is the difference between gage pressure and absolute pressure? 125 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 126C What is the zeroth law of thermodynamics? 127C What are the ordinary and absolute temperature scales in the SI and the English system? 128C 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. 129 The deep body temperature of a healthy person is 37°C. What is it in kelvins? 130E Consider a system whose temperature is 18°C. Express this temperature in R, K, and °F. 135C Explain why some people experience nose bleeding and some others experience shortness of breath at high elevations. 136C Someone claims that the absolute pressure in a liquid of constant density doubles when the depth is doubled. Do you agree? Explain. 137C 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? 138C Express Pascals law, and give a real-world example of it. 139C 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? 140 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. 141E 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. 142 The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig. P142. Determine the gage pressure of air in the tank if Oil AIR 1 WATER h1 2 131 The temperature of a system rises by 15°C during a heating process. Express this rise in temperature in kelvins. 132E The temperature of a system drops by 45°F during a cooling process. Express this drop in temperature in K, R, and °C. 133 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 P142 h3 cen84959_ch01.qxd 42 | 3/31/05 3:39 PM Page 42 Thermodynamics 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. 143 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. 144 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. 145 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. 146E Show that 1 kgf/cm2 14.223 psi. 147E A 200-pound 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. 148 Consider a 70-kg 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 P153 149 A vacuum gage connected to a tank reads 15 kPa at a location where the barometric reading is 750 mm Hg. Deter13,590 mine the absolute pressure in the tank. Take rHg kg/m3. Answer: 85.0 kPa 150E 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 154 Solve Prob. 153 using EES (or other) software. Print out the entire solution, including the numerical results with proper units. 151 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. 155 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. 152 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 156E 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. 153 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. 157 A gas is contained in a vertical, frictionless pistoncylinder device. The piston has a mass of 4 kg and a cross-sectional 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 cen84959_ch01.qxd 3/31/05 3:39 PM Page 43 Chapter 1 60 N Patm = 95 kPa | 43 ence in the manometer levels is 15 mm, and the atmospheric pressure is 100 kPa. (a) Judging from Fig. P162, 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 P157 158 Reconsider Prob. 157. 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. 159 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 P162 163 Repeat Prob. 162 for a differential mercury height of 45 mm. 164 Blood pressure is usually measured by wrapping a closed air-filled 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. 165 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 P159 160 Reconsider Prob. 159. 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 161 A manometer containing oil (r 850 kg/m3) is attached to a tank filled with air. If the oil-level 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 162 A mercury manometer (r 13,600 kg/m3) is connected to an air duct to measure the pressure inside. The differ- FIGURE P165 cen84959_ch01.qxd 44 | 3/31/05 3:39 PM Page 44 Thermodynamics 166 Consider a 1.8-m-tall 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 167 Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (r 790 kg/m3) from the other. One arm contains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 4. Determine the height of each fluid in that arm. 10 in 25 in 6 in Mercury SG = 13.6 172E Repeat Prob. 171E by replacing air by oil with a specific gravity of 0.69. Water 173 The gage pressure of the air in the tank shown in Fig. P173 is measured to be 80 kPa. Determine the differential height h of the mercury column. FIGURE P167 168 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 P173 174 40 cm 70 cm Water 30 cm Air 60 cm Oil SG = 0.72 80 kPa 169 Freshwater and seawater flowing in parallel horizontal pipelines are connected to each other by a double U-tube manometer, as shown in Fig. P169. 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 P171E Oil 70 cm 2 in Sea water Repeat Prob. 173 for a gage pressure of 40 kPa. 175 The top part of a water tank is divided into two compartments, as shown in Fig. P175. Now a fluid with an Unknown liquid 10 cm Mercury 80 cm FIGURE P169 170 Repeat Prob. 169 by replacing the air with oil whose specific gravity is 0.72. 171E The pressure in a natural gas pipeline is measured by the manometer shown in Fig. P171E 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 P175 95 cm cen84959_ch01.qxd 4/19/05 11:04 AM Page 45 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. | 45 178 A multifluid container is connected to a U-tube, as shown in Fig. P178. 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 176 Consider a double-fluid manometer attached to an air pipe shown in Fig. P176. 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 Oil SG = 0.90 30 cm Water 90 cm P = 76 kPa 40 cm Glycerin SG = 1.26 20 cm SG2 15 cm 22 cm FIGURE P178 SG1 = 13.55 Solving Engineering Problems and EES 179C What is the value of the engineering software packages in (a) engineering education and (b) engineering practice? FIGURE P176 177 Consider the system shown in Fig. P177. If a change of 0.7 kPa in the pressure of air causes the brinemercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant, determine the ratio of A2/A1. 180 Determine a positive real root of this equation using EES: 2x3 181 10x0.5 Air y2 3xy Water 182 2x 3x Area, A1 y y 183 z 5 z 2z 2 8 Solve this system of three equations with three unknowns using EES: x2y z 3y0.5 x FIGURE P177 3.5 2y xy Mercury SG = 13.56 7.75 Solve this system of three equations with three unknowns using EES: 2 Area, A2 3 Solve this system of two equations with two unknowns using EES: x3 Brine pipe SG = 1.1 3x x y 1 xz z 2 2 cen84959_ch01.qxd 46 | 3/31/05 3:39 PM Page 46 Thermodynamics 184E 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 187 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 185 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 P187 188 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. 19, determine the weight of an 80-kg person at sea level (z 0), in Denver (z 1610 m), and on the top of Mount Everest (z 8848 m). 189 A man goes to a traditional market to buy a steak for dinner. He finds a 12-oz steak (1 lbm 16 oz) for $3.15. He then goes to the adjacent international market and finds a 320-g steak of identical quality for $2.80. Which steak is the better buy? 190 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 P185 186 A vertical pistoncylinder 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 191E 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? 192E The boiling temperature of water decreases by about 3°C for each 1000-m rise in altitude. What is the decrease in the boiling temperature in (a) K, (b) °F, and (c) R for each 1000-m rise in altitude? 193E 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? 194E 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 P186 195E 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. cen84959_ch01.qxd 3/31/05 3:39 PM Page 47 Chapter 1 196 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. 197 Joe Smith, an old-fashioned 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. 1101 Balloons are often filled with helium gas because it weighs only about one-seventh 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 = 1 rair 7 198E It is well-known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the chilling effect of the wind. This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities. The equivalent wind chill temperature in °F is given by [ASHRAE, Handbook of Fundamentals (Atlanta, GA, 1993), p. 8.15] Tequiv 91.4 1 91.4 1 0.475 Tambient 2 0.0203V 0.304 2V 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 (33.0 Tambient ) (0.475 0.0126V 0.240 V ) 199E Reconsider Problem 198E. 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. 1100 An air-conditioning system requires a 20-m-long section of 15-cm diameter duct work to be laid underwater. m = 140 kg FIGURE P1101 1102 Reconsider Prob. 1101. 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. 1103 Determine the maximum amount of load, in kg, the balloon described in Prob. 1101 can carry. Answer: 520.5 kg 1104E The pressure in a steam boiler is given to be 92 kgf/cm2. Express this pressure in psi, kPa, atm, and bars. 1105 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 pilots 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 cen84959_ch01.qxd 48 | 3/31/05 3:39 PM Page 48 Thermodynamics 1106 The lower half of a 10-m-high 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 1109 A glass tube is attached to a water pipe, as shown in Fig. P1109. 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 P1106 Water 1107 A vertical, frictionless pistoncylinder 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. 1108 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 cross-sectional area of 4 mm2. Assume an atmospheric pressure of 101 kPa, and draw the free-body diagram of the petcock. Answer: 40.8 g Patm = 101 kPa Petcock A = 4 mm2 FIGURE P1109 1110 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). 1111 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. P1111. 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 P1111 FIGURE P1108 cen84959_ch01.qxd 3/31/05 3:39 PM Page 49 Chapter 1 1112E Consider a U-tube 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 U-tube until the contact surface of the two fluids moves to the bottom of the U-tube, 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 P1114 Air Oil | Water 30 in FIGURE P1112E 1113 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 P1113 1114 A gasoline line is connected to a pressure gage through a double-U manometer, as shown in Fig. P1114. If the reading of the pressure gage is 370 kPa, determine the gage pressure of the gasoline line. 1115 Repeat Prob. 1114 for a pressure gage reading of 180 kPa. 1116E A water pipe is connected to a double-U manometer as shown in Fig. P1116E 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 P1116E 1117 It is well-known 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 temperbz, where T0 is the ature can be approximated by T T0 temperature at sea level, which can be taken to be 288.15 K, and b 0.0065 K/m. The gravitational acceleration also z /6,370,320)2 where changes with altitude as g(z) g0 /(1 9.807 m/s2 and z is the elevation from sea level in m. g0 Obtain a relation for the variation of pressure in the troposphere (a) by ignoring and (b) by considering the variation of g with altitude. 1118 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 z-direction is given as dP rg dz, obtain a relation for pressure as a function of cen84959_ch01.qxd 50 | 3/31/05 3:39 PM Page 50 Thermodynamics elevation z. Take the pressure and density at z and r0, respectively. 0 to be P0 1119 Pressure transducers are commonly used to measure pressure by generating analog signals usually in the range of 4 mA to 20 mA or 0 V-dc to 10 V-dc in response to applied pressure. The system whose schematic is shown in Fig. P1119 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. 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 kPa. If the density of air is 1.0 kg/m3, the height of the building is (a) 17 m (d) 204 m (c) 170 m 1122 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) 1.25 kJ (d) 8.1 kJ (b) 2.50 kJ (e) 4.1 kJ (c) 5.0 kJ 1123 Consider a 2-m 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 1124 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 1125 During a heating process, the temperature of an object rises by 20°C. This temperature rise is equivalent to a temperature rise of (a) 20°F (d) 36 R Pressure transducer (b) 20 m (e) 252 m (b) 52°F (e) 293 K (c) 36 K Design, Essay, and Experiment Problems 1126 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 doctors 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 Manometer 1127 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 Rigid container 1128 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 P1119 Fundamentals of Engineering (FE) Exam Problems 1120 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 (b) 9800 Pa (e) 441,000 Pa (c) 50,000 Pa 1121 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 1129 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 write-up, and the data provided on the DVD accompanying this book. cen84959_ch02.qxd 3/31/05 5:02 PM Page 51 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. | 51 cen84959_ch02.qxd 4/25/05 2:32 PM Page 52 52 | Thermodynamics 21 Room FIGURE 21 A refrigerator operating with its door open in a well-sealed and well-insulated room. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 1 ON THE DVD. Well-sealed and well-insulated room Fan FIGURE 22 A fan running in a well-sealed and well-insulated 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 lets test ourselves to see how well we understand and truly believe in this principle. Consider a room whose door and windows are tightly closed, and whose walls are well-insulated so that heat loss or gain through the walls is negligible. Now lets place a refrigerator in the middle of the room with its door open, and plug it into a wall outlet (Fig. 21). You may even use a small fan to circulate the air in order to maintain temperature uniformity in the room. Now, what do you think will happen to the average temperature of air in the room? Will it be increasing or decreasing? Or will it remain constant? Probably the first thought that comes to mind is that the average air 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 roomincluding the air and the refrigeratoras the system, which is an adiabatic closed system since the room is well-sealed and well-insulated, 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. 22)? Note that energy is conserved during the process of operating the refrigerator placed in a roomthe 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. cen84959_ch02.qxd 4/25/05 2:32 PM Page 53 Chapter 2 | 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. 22 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 1 kJ> kg 2 E m m V2 2 1 kJ 2 SEE TUTORIAL CH. 2, SEC. 2 ON THE DVD. (21) Thermodynamics provides no information about the absolute value of the total energy. It deals only with the change of the total energy, which is what matters in engineering problems. Thus the total energy of a system can be assigned a value of zero (E 0) at some convenient reference point. The change in total energy of a system is independent of the reference point selected. The decrease in the potential energy of a falling rock, for example, depends on only the elevation difference and not the reference level selected. In thermodynamic analysis, it is often helpful to consider the various forms of energy that make up the total energy of a system in two groups: 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. 23). 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 INTERACTIVE TUTORIAL (22) FIGURE 23 The macroscopic energy of an object changes with velocity and elevation. cen84959_ch02.qxd 3/31/05 5:02 PM Page 54 54 | Thermodynamics or, on a unit mass basis, 1 kJ> kg 2 V2 2 ke (23) where V denotes the velocity of the system relative to some fixed reference frame. The kinetic energy of a rotating solid body is given by 1 Iv2 where I 2 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 (24) 1 kJ> kg 2 or, on a unit mass basis, pe 1 kJ 2 (25) mgz gz 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 pe u m V2 2 mgz 1 kJ 2 (26) or, on a unit mass basis, e Vavg Steam ke V2 2 gz 1 kJ> kg 2 (27) 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 u m = r AcVavg Mass flow rate: # rV # m E = me FIGURE 24 Mass and energy flow rates associated with the flow of steam in a pipe of inner diameter D with an average velocity of Vavg. rAcVavg 1 kg> s 2 (28) 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. 24) Energy flow rate: which is analogous to E # E # me me. 1 kJ> s or kW 2 (29) cen84959_ch02.qxd 3/31/05 5:02 PM Page 55 Chapter 2 | 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 back-and-forth 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. 25). 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 phase-change 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 phase-change 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. 26). 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 25 The various forms of microscopic energies that make up sensible energy. Sensible and latent energy Chemical energy Nuclear energy FIGURE 26 The internal energy of a system is the sum of all forms of the microscopic energies. cen84959_ch02.qxd 3/31/05 5:02 PM Page 56 56 | Thermodynamics Microscopic kinetic energy of molecules (does not turn the wheel) Water Dam Macroscopic kinetic energy (turns the wheel) FIGURE 27 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. 27). 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 U-235 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 large-scale nuclear reactors were built in 1944 for the purpose of producing material for nuclear weapons. When a cen84959_ch02.qxd 4/28/05 5:09 PM Page 57 Chapter 2 uranium-235 atom absorbs a neutron and splits during a fission process, it produces a cesium-140 atom, a rubidium-93 atom, 3 neutrons, and 3.2 10 11 J of energy. In practical terms, the complete fission of 1 kg of uranium-235 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 helium-3 atom, a free neutron, and 5.1 10 13 J of energy (Fig. 28). 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 A-bomb). In fact, the uncontrolled fusion reaction in a hydrogen bomb (the H-bomb) 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 21 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 1 rV 2 gasoline 1 0.75 kg> L 2 1 5 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 U-235 1 mgasoline 2 1 Heating value 2 1 3.75 kg> day 2 1 44,000 kJ> kg 2 165,000 kJ> day 57 3.2 × 10 11 J Ce-140 n n 3 neutrons n n neutron Rb-93 (a) Fission of uranium H-2 He-3 n neutron H-2 5.1 × 10 13 J (b) Fusion of hydrogen FIGURE 28 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 U-235. If a new car comes equipped with 0.1-kg of the nuclear fuel U-235, determine if this car will ever need refueling under average driving conditions (Fig. 29). mgasoline Uranium | Nuclear fuel FIGURE 29 Schematic for Example 21. cen84959_ch02.qxd 3/31/05 5:02 PM Page 58 58 | Thermodynamics The complete fission of 0.1 kg of uranium-235 releases 1 6.73 1010 kJ> kg 2 1 0.1 kg 2 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 Daily energy use 6.73 109 kJ 165,000 kJ> day 40,790 days 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 mechanN· ical energy. In fact, the pressure unit Pa is equivalent to Pa N/m2 3 3, which is energy per unit volume, and the product Pv or its J/m m/m 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 P r V2 2 gz (210) cen84959_ch02.qxd 3/31/05 5:02 PM Page 59 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 # Emech #P ma r # memech gz b V2 2 (211) . where m is the mass flow rate of the fluid. Then the mechanical energy change of a fluid during incompressible (r constant) flow becomes V2 2 P1 P2 ¢ emech 2 V1 r g 1 z2 2 z1 2 1 kJ> kg 2 (212) and # ¢ E mech # m ¢ emech # P2 ma V2 2 P1 r 2 V1 2 z1 2 b g 1z2 1 kW 2 (213) Therefore, the mechanical energy of a fluid does not change during flow if its pressure, density, velocity, and elevation remain constant. In the absence of any 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 22 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. 210). 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 2 e ke V 2 1 8.5 m> s 2 2 2 a 1 J> kg 1 m2> s2 b 36.1 J> kg © Vol. 36/PhotoDisc (b) Wind energy for an air mass of 10 kg is E me 1 10 kg 2 1 36.1 J> kg 2 361 J (c) Wind energy for a mass flow rate of 1154 kg/s is # E # me 1 1154 kg> s 2 1 36.1 J> kg 2 a FIGURE 210 Potential site for a wind farm as discussed in Example 22. 1 kW b 1000 J> s 41.7 kW Discussion It can be shown that the specified mass flow rate corresponds to a 12-m 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 one-third of this potential to electric power. 59 cen84959_ch02.qxd 4/25/05 2:32 PM Page 60 60 | Thermodynamics 23 INTERACTIVE TUTORIAL SEE TUTORIAL CH. 2, SEC. 3 ON THE DVD. System boundary Heat CLOSED SYSTEM Work (m = constant) FIGURE 211 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 212 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. 211). 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. 212). That is, an energy interaction is heat only if it takes place because of a temperature difference. Then it follows that there cannot be any heat transfer between two systems that are at the same temperature. Several phrases in common use todaysuch as heat flow, heat addition, heat rejection, heat absorption, heat removal, heat gain, heat loss, heat storage, heat generation, electrical heating, resistance heating, frictional heating, gas heating, heat of reaction, liberation of heat, specific heat, sensible heat, latent heat, waste heat, body heat, process heat, heat sink, and heat sourceare not consistent with the strict thermodynamic meaning of the term heat, which limits its use to the transfer of thermal energy during a process. However, these phrases are deeply rooted in our vocabulary, and they are used by both ordinary people and scientists without causing any misunderstanding since they are usually interpreted properly instead of being taken literally. (Besides, no acceptable alternatives exist for some of these phrases.) For example, the phrase body heat is understood to mean the thermal energy content of a body. Likewise, heat flow is understood to mean the transfer of thermal energy, not the flow of a fluidlike 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. 213. 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. cen84959_ch02.qxd 3/31/05 5:02 PM Page 61 Chapter 2 A process during which there is no heat transfer is called an adiabatic process (Fig. 214). 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 1 kJ> kg 2 Q m | 61 2 kJ thermal energy HEAT SURROUNDING AIR 2 kJ heat BAKED POTATO System boundary 2 kJ thermal energy FIGURE 213 Energy is recognized as heat transfer only as it crosses the system boundary. (214) 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. 215). 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 1 kJ 2 # Q dt Q t1 where t t2 # Q ¢t 1 kJ 2 ADIABATIC SYSTEM (215) . When Q remains constant during a process, this relation reduces to Q Q=0 (216) FIGURE 214 During an adiabatic process, a system exchanges no heat with its surroundings. 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 (17441794) in 1789. The caloric theory asserts that heat is a fluidlike substance called the caloric that is a massless, colorless, odorless, and tasteless substance that can be poured from one body into another (Fig. 216). When caloric was added to a body, its Q = 30 kJ m = 2 kg t = 5 s 30 30 kJ heat heat Q = 6 kW q = 15 kJ/k g J/ FIGURE 215 . The relationships among q, Q, and Q. cen84959_ch02.qxd 4/25/05 2:32 PM Page 62 62 | Thermodynamics Contact surface Hot body Cold body Caloric FIGURE 216 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 m = 2 kg t = 5 s 30 kJ 30 kJ work work W = 6 kW w = 15 k J/k g J/k 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 ones hands together or rubbing two pieces of wood together. In 1798, the American Benjamin Thompson (Count Rumford) (17541814) showed in his papers that heat can be generated continuously through friction. The validity of the caloric theory was also challenged by several others. But it was the careful experiments of the Englishman James P. Joule (18181889) published in 1843 that finally convinced the skeptics that heat was not a substance after all, and thus put the caloric theory to rest. Although the caloric theory was totally abandoned in the middle of the 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. 24 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 217 # The relationships among w, W, and W. W m 1 kJ> kg 2 (217) . The work done per unit time is called power and is denoted W (Fig. 217). The unit of power is kJ/s, or kW. cen84959_ch02.qxd 3/31/05 5:02 PM Page 63 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. 218). For example, a work input of 5 kJ can be expressed as Win 5 3 kJ. When the kJ, while a heat loss of 3 kJ can be expressed as Qout 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 218 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). P VA = 3 m3; WA = 8 kJ 1 A V1 ¢V 1 2 m3 2 dW B ss ce V2 That is, the volume change during process 12 is always the volume at state 2 minus the volume at state 1, regardless of the path followed (Fig. 219). The total work done during process 12, however, is 1 s es oc Pr 2 2 dV VB = 3 m3; WB = 12 kJ 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 W12 1 not ¢ W 2 5 m3 V FIGURE 219 Properties are point functions; but heat and work are path functions (their magnitudes depend on the path followed). cen84959_ch02.qxd 3/31/05 5:02 PM Page 64 64 | 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 23 Burning of a Candle in an Insulated Room A candle is burning in a well-insulated 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 well-insulated room is considered. It is to FIGURE 220 Schematic for Example 23. 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. 220. 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 24 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. 221. 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 221 Schematic for Example 24. whether there is any heat transfer during this process. Analysis This is not a well-defined 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 25 Heating of an Oven by Work Transfer A well-insulated 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. cen84959_ch02.qxd 3/31/05 5:02 PM Page 65 Chapter 2 Solution A well-insulated 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. 222. 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 26 | 65 System boundary Electric oven Heating element FIGURE 222 Schematic for Example 25. Heating of an Oven by Heat Transfer Answer the question in Example 25 if the system is taken as only the air in the oven without the heating element. System boundary Solution The question in Example 25 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. 223. 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 223 Schematic for Example 26. Electrical Work It was pointed out in Example 25 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 1W 2 2 VI dt 1 1 kJ 2 (219) When both V and I remain constant during the time interval t, it reduces to We VI ¢ t 1 kJ 2 We = VI = I 2R = V2/R R V (218) . where We is the electrical power and I is the number of electrical charges flowing per unit time, that is, the current (Fig. 224). In general, both V and I vary with time, and the electrical work done during a time interval t is expressed as We I (220) FIGURE 224 Electrical power in terms of resistance R, current I, and potential difference V. cen84959_ch02.qxd 4/25/05 2:32 PM Page 66 66 | Thermodynamics 25 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. 225). 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 1 kJ 2 Fs (221) If the force F is not constant, the work done is obtained by adding (i.e., integrating) the differential amounts of work, s FIGURE 225 The work done is proportional to the force applied (F ) and the distance traveled (s). FIGURE 226 If there is no movement, no work is done. © Reprinted with special permission of King Features Syndicate. 2 W 1 kJ 2 F ds 1 (222) Obviously one needs to know how the force varies with displacement to perform this integration. Equations 221 and 222 give only the magnitude of the work. The sign is easily determined from physical considerations: The work done on a system by an external force acting in the direction of motion is negative, and work done by a system against an external force acting in the opposite direction to motion is positive. There are two requirements for a work interaction between a system and its surroundings to exist: (1) there must be a force acting on the boundary, and (2) the boundary must move. Therefore, the presence of forces on the 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. 226). Some common forms of mechanical work are discussed next. Shaft Work Energy transmission with a rotating shaft is very common in engineering practice (Fig. 227). 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. 228) T Boat Fr S F T r (223) This force acts through a distance s, which is related to the radius r by s 1 2pr 2 n (224) Then the shaft work is determined from Engine FIGURE 227 Energy transmission through rotating shafts is commonly encountered in practice. Wsh Fs a T b 1 2prn 2 r 2pnT 1 kJ 2 (225) The power transmitted through the shaft is the shaft work done per unit time, which can be expressed as # Wsh # 2pnT 1 kW 2 . where n is the number of revolutions per unit time. (226) cen84959_ch02.qxd 3/31/05 5:02 PM Page 67 Chapter 2 | · EXAMPLE 27 · Wsh = 2 π nT 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. 229. The shaft power is determined directly from # Wsh # 2pnT 83.8 kW 1 2p 2 a 4000 1 1 min 1 kJ b 1 200 N # m 2 a ba b min 60 s 1000 N # m FIGURE 228 Shaft work is proportional to the torque applied and the number of revolutions of the shaft. 1 or 112 hp 2 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. 230). 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 229 Schematic for Example 27. (227) To determine the total spring work, we need to know a functional relationship between F and x. For linear elastic springs, the displacement x is proportional to the force applied (Fig. 231). That is, F kx 1 kN 2 (228) 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. 228 into Eq. 227 and integrating yield Wspring 1 2 2 k 1 x2 x2 2 1 1 kJ 2 (229) where x1 and x2 are the initial and the final displacements of the spring, respectively, measured from the undisturbed position of the spring. There are many other forms of mechanical work. Next we introduce some of them briefly. Work Done on Elastic Solid Bars Solids are often modeled as linear springs because under the action of a force they contract or elongate, as shown in Fig. 232, 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 230 Elongation of a spring under the influence of a force. 67 cen84959_ch02.qxd 3/31/05 5:02 PM Page 68 68 | 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: 2 Rest position x1 = 1 mm x2 = 2 mm 2 F dx Welastic sn A dx 1 1 1 kJ 2 (230) where A is the cross-sectional 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 231 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. 233). 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 liquidair interfaces. These microscopic forces are perpendicular to any line in the surface, and the force generated by these forces per unit length is called the surface tension 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 ss dA 1 1 kJ 2 (231) where dA 2b dx is the change in the surface area of the film. The factor 2 is due to the fact that the film has two surfaces in contact with air. The force acting on the movable wire as a result of surface tension effects is F 2bss where ss is the surface tension force per unit length. x Work Done to Raise or to Accelerate a Body F FIGURE 232 Solid bars behave as springs under the influence of a force. Rigid wire frame Surface of film Movable wire F b dx x FIGURE 233 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. 234). 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. cen84959_ch02.qxd 3/31/05 5:02 PM Page 69 Chapter 2 EXAMPLE 28 Power Needs of a Car to Climb a Hill | 69 Motor Consider a 1200-kg 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. 235). 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 1 1200 kg 2 1 9.81 m> s2 2 1 90 km> h 2 1 sin 30° 2 a 147 kJ> s 147 kW 1 or 197 hp 2 1 m> s 3.6 km> h ba 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 29 FIGURE 234 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 900-kg car shown in Fig. 236 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 1 2 2m 1 V 2 V22 1 1 2 1 900 kg 2 c a 80,000 m 2 b 3600 s 02 d a 1 kJ> kg 1000 m2> s2 b 30° 222 kJ The average power is determined from # Wa Wa ¢t 222 kJ 20 s 11.1 kW 1 or 14.9 hp 2 FIGURE 235 Schematic for Example 28. 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 25 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 0 80 km/h m = 900 kg FIGURE 236 Schematic for Example 29. cen84959_ch02.qxd 4/25/05 2:32 PM Page 70 70 | 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 237 Energy cannot be created or destroyed; it can only change forms. 26 THE FIRST LAW OF THERMODYNAMICS So far, we have considered various forms of energy such as heat Q, work W, and total energy E individually, and no attempt is made to relate them to each other during a process. The first law of thermodynamics, also known as the conservation of energy principle, provides a sound basis for studying the relationships among the various forms of energy and energy interactions. Based on experimental observations, the first law of thermodynamics states that energy can be neither created nor destroyed during a process; it can only change forms. Therefore, every bit of energy should be accounted for during a process. We all know that a rock at some elevation possesses some potential energy, and part of this potential energy is converted to kinetic energy as the rock falls (Fig. 237). Experimental data show that the decrease in potential energy (mg z) exactly equals the increase in kinetic energy 3 m 1 V 2 V 2 2> 2 4 when 2 1 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 far-reaching 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- cen84959_ch02.qxd 4/28/05 4:21 PM Page 71 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. 238). 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. 239). 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 well-insulated (i.e., adiabatic) room heated by an electric heater as our system (Fig. 240). 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. 241). 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. 242). 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. 243). 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 238 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 239 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 240 The work (electrical) done on an adiabatic system is equal to the increase in the energy of the system. cen84959_ch02.qxd 3/31/05 5:02 PM Page 72 72 | 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 b entering the system a Total energy b leaving the system Wb,in = 10 kJ Change in the total b energy of the system or Ein FIGURE 241 The work (shaft) done on an adiabatic system is equal to the increase in the energy of the system. a Eout ¢ Esystem 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 E = 10 kJ (Adiabatic) FIGURE 242 The work (boundary) done on an adiabatic system is equal to the increase in the energy of the system. Efinal ¢E ¢U where ¢U ¢ KE Wsh, in = 6 kJ Qin = 15 kJ FIGURE 243 The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings. E2 E1 (232) Note that energy is a property, and the value of a property does not change unless the state of the system changes. Therefore, the energy change of a system is zero if the state of the system does not change during the process. Also, energy can exist in numerous forms such as internal (sensible, latent, chemical, and nuclear), kinetic, potential, electric, and magnetic, and their sum constitutes the total energy E of a system. In the absence of electric, magnetic, and surface tension effects (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 Qout = 3 kJ E = (15 3) + 6 = 18 kJ Einitial ¢ PE ¢ KE m 1 u2 1 2 2 m 1V 2 mg 1 z2 ¢ PE (233) u1 2 2 V1 2 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. 244). 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. 233 reduces to E U for such systems. Also, the energy cen84959_ch02.qxd 3/31/05 5:02 PM Page 73 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 hot-water tank (the control volume) decreases as a result of this mass interaction (Fig. 245). 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 1 Q in Q out 2 1 Win Wout 2 1 E mass,in E mass,out 2 ¢ E system (234) 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 = V KE = 0 2 E = U FIGURE 244 For stationary systems, KE 0; thus E U. PE cen84959_ch02.qxd 3/31/05 5:02 PM Page 74 74 | Thermodynamics Mass in W Energy balance for any system undergoing any kind of process can be expressed more compactly as E out E in 1 kJ 2 1 kW 2 Mass out Change in internal, kinetic, potential, etc., energies or, in the rate form, as dE system> dt . E out . E in Rate of net energy transfer by heat, work, and mass FIGURE 245 The energy content of a control volume can be changed by mass flow as well as heat and work interactions. (236) Net energy transfer by heat, work, and mass Q (235) ¢ E system Control volume 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 1 dE> dt 2 ¢ t ¢E 1 kJ 2 (237) The energy balance can be expressed on a per unit mass basis as ein eout 1 kJ> kg 2 ¢ esystem (238) which is obtained by dividing all the quantities in Eq. 235 by the mass m of the system. Energy balance can also be expressed in the differential form as P dEin V 0, thus Q dEsystem or dein deout desystem (239) 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 FIGURE 246 For a cycle E dEout W. Wnet,out or # Wnet,out # Q net,in 1 for a cycle 2 (240) That is, the net work output during a cycle is equal to net heat input (Fig. 246). EXAMPLE 210 Qout = 500 kJ Q net,in 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 247 Schematic for Example 210. 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 systems energy that may change during this process. 2 Energy stored in the paddle wheel is negligible. Analysis Take the contents of the tank as the system (Fig. 247). 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 75 Chapter 2 E in E out | 75 ¢ E system Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies Q out Wsh,in 100 kJ 500 kJ U2 ¢U U2 U2 U1 800 kJ 400 kJ Therefore, the final internal energy of the system is 400 kJ. EXAMPLE 211 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. 248). Determine if this claim is reasonable. Solution A fan is claimed to increase the velocity of air to a specified value while consuming electric power at a specified rate. The validity of this claim is to be investigated. Assumptions The ventilating room is relatively calm, and air velocity in it is negligible. Analysis First, lets examine the energy conversions involved: The motor of the fan converts part of the electrical power it consumes to mechanical (shaft) power, which is used to rotate the fan blades in air. The blades are shaped such that they impart a large fraction of the mechanical power of the shaft to air by mobilizing it. In the limiting ideal case of no losses (no conversion of electrical and mechanical energy to thermal energy) in steady operation, the electric power input will be equal to the rate of increase of the kinetic energy of air. Therefore, for a control volume that encloses the fanmotor unit, the energy balance can be written as # E in dE system > dt 0 1steady2 # E out 0 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 # Welect,in # B 2mair # mair keout B 2 1 0.25 kg> s 2 2 # Vout mair 2 Solving for Vout and substituting gives the maximum air outlet velocity to be Vout 20 J> s a 1 m2> s2 1 J> kg b 6.3 m> s 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/sbut 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 248 Schematic for Example 211. © Vol. 0557/PhotoDisc Fan cen84959_ch02.qxd 3/31/05 5:02 PM Page 76 76 | Thermodynamics Qout Room Welect. in Fan EXAMPLE 212 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. 249). 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 249 Schematic for Example 212. 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 # E in dE system > dt 0 1steady2 # E out 0 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 Substituting, 200 W # Q out UA 1 Ti To 2 1 6 W> m2 # °C 2 1 30 m2 2 1 Ti 25°C 2 It gives Ti 26.1°C Therefore, the room air temperature will remain constant after it reaches 26.1°C. Discussion Note that a 200-W fan heats a room just like a 200-W 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 213 FIGURE 250 Fluorescent lamps lighting a classroom as discussed in Example 213. © 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. 250). The lights in the classroom are kept on for 12 hours a day and 250 days a year. For a unit electricity cost of cen84959_ch02.qxd 4/19/05 11:24 AM Page 77 Chapter 2 | 77 7 cents per kWh, determine annual energy cost of lighting for this classroom. Also, discuss the effect of lighting on the heating and air-conditioning 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 lightings effect on the heating and air-conditioning 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 1 Power consumed per lamp 2 1 No. of lamps 2 1 80 W> lamp 2 1 30 lamps 2 2400 W Operating hours 2.4 kW 1 12 h> day 2 1 250 days> year 2 3000 h> year Then the amount and cost of electricity used per year become Lighting energy 1 Lighting power 2 1 Operating hours 2 1 2.4 kW 2 1 3000 h> year 2 Lighting cost 7200 kWh> year 1 Lighting energy 2 1 Unit cost 2 1 7200 kWh> year 2 1 $0.07> kWh 2 $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 air-conditioning 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 214 Conservation of Energy for an Oscillating S teel Ball The motion of a steel ball in a hemispherical bowl of radius h shown in Fig. 251 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 0 B 2 FIGURE 251 Schematic for Example 214. cen84959_ch02.qxd 4/25/05 2:32 PM Page 78 78 | 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 1 ke2 pe2 2 1 ke1 pe1 2 or 2 V1 2 gz1 V2 2 2 gz2 wfriction 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 V2 1 2 gz1 V2 2 2 gz2 or V2 2 gz C constant 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. 27 FIGURE 252 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. 252) Performance Desired output Required output (241) If you are shopping for a water heater, a knowledgeable salesperson will tell you that the efficiency of a conventional electric water heater is about 90 percent (Fig. 253). You may find this confusing, since the heating elements of electric water heaters are resistance heaters, and the efficiency of cen84959_ch02.qxd 3/31/05 5:02 PM Page 79 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 hot-water 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. 254). Then the performance of combustion equipment can be characterized by combustion efficiency, defined as hcombustion Q HV Amount of heat released during combustion Heating value of the fuel burned 79 Water heater Type (242) 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 start-up 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, high-efficiency Electric, conventional Electric, high-efficiency 55% 62% 90% 94% FIGURE 253 Typical efficiencies of conventional and high-efficiency electric and natural gas water heaters. © The McGraw-Hill 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 254 The definition of the heating value of gasoline. cen84959_ch02.qxd 3/31/05 5:02 PM Page 80 80 | Thermodynamics TABLE 21 The efficacy of different lighting systems Type of lighting Combustion Candle Efficacy, lumens/W 0.2 Incandescent Ordinary Halogen 620 1625 Fluorescent Ordinary High output Compact 4060 7090 5080 High-intensity discharge Mercury vapor 5060 Metal halide 56125 High-pressure sodium 100150 Low-pressure sodium up to 200 15 W 60 W FIGURE 255 A 15-W compact fluorescent lamp provides as much light as a 60-W incandescent lamp. AFUE of some new high-efficiency 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, h overall h combustion h thermal h generator # Wnet,electric # HHV mnet (243) The overall efficiencies are about 2630 percent for gasoline automotive engines, 3440 percent for diesel engines, and 4060 percent for large power plants. We are all familiar with the conversion of electrical energy to light by incandescent lightbulbs, fluorescent tubes, and high-intensity 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 21. Note that a compact fluorescent lightbulb produces about four times as much light as an incandescent lightbulb per W, and thus a 15-W fluorescent bulb can replace a 60-W incandescent lightbulb (Fig. 255). 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. Sodium-filled high-intensity 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 81 Chapter 2 the food to the energy consumed by the appliance (Fig. 256). Electric ranges are more efficient than gas ranges, but it is much cheaper to cook with natural gas than with electricity because of the lower unit cost of natural gas (Table 22). The cooking efficiency depends on user habits as well as the individual appliances. Convection and microwave ovens are inherently more efficient than conventional ovens. On average, convection ovens save about one-third and microwave ovens save about two-thirds 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 flat-bottomed 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 self-cleaning feature of ovens right after cooking, and keeping inside surfaces of microwave ovens clean. Using energy-efficient 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 22 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 Energy-Efficient Economy, 1996, p. 192.] Cooking appliance Electric oven Convection oven (elect.) Gas oven Frying pan Toaster oven Electric slow cooker Microwave oven Cooking temperature 350 F (177 C) 325 F (163 C) 350 F (177 C) 420 F (216 C) 425 F (218 C) 200 F (93 C) 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 256 The efficiency of a cooking appliance represents the fraction of the energy supplied to the appliance that is transferred to the food. cen84959_ch02.qxd 3/31/05 5:02 PM Page 82 82 | Thermodynamics 38% Gas Range 73% Electric Range EXAMPLE 215 Cost of Cooking with Electric and Gas Ranges The efficiency of cooking appliances affects the internal heat gain from them since an inefficient appliance consumes a greater amount of energy for the same task, and the excess energy consumed shows up as heat in the living space. The efficiency of open burners is determined to be 73 percent for electric units and 38 percent for gas units (Fig. 257). Consider a 2-kW 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 # Qutilized FIGURE 257 Schematic of the 73 percent efficient electric heating unit and 38 percent efficient gas burner discussed in Example 215. 1 Energy input 2 1 Efficiency 2 1 2 kW 2 1 0.73 2 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 Cost of energy input Efficiency $0.09> kWh 0.73 $0.123> kWh Noting that the efficiency of a gas burner is 38 percent, the energy input to a gas burner that supplies utilized energy at the same rate (1.46 kW) is # Qinput, gas # Qutilized Efficiency 1.46 kW 0.38 3.84 kW 1 13,100 Btu> h 2 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 Cost of energy input Efficiency $0.049> kWh $0.55> 29.3 kWh 0.38 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 cost-conscious 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 83 Chapter 2 converted entirely from one mechanical form to another, and the mechanical efficiency of a device or process can be defined as (Fig. 258) h mech Mechanical energy output Mechanical energy input E mech,out E mech,in 1 E mech,loss 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 Mechanical energy increase of the fluid Mechanical energy input # ¢ E mech,fluid # Wshaft,in # Wpump,u # Wpump (245) # # # 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 h turbine Mechanical energy output Mechanical energy decrease of the fluid # Wshaft,out # 0 ¢ E mech,fluid 0 # Wturbine (246) # Wturbine,e # Wshaft,out # Welect,in (247) # # # E mech,in E mech,out is the rate of decrease in the where 0 ¢ E mech,fluid 0 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 full-load motor efficiencies range from about 35 percent for small motors to over 97 percent for large high-efficiency 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 Mechanical power output Electric power input 83 Fan 50 W (244) | 1 · m = 0.50 kg/s 2 V1 = 0, V2 = 12 m/s z1 = z2 P1 = P2 · ·2 Emech,fluid mV 2/2 hmech, fan = = · · Wshaft,in Wshaft,in (0.50 kg/s)(12 m/s)2/2 = 50 W = 0.72 FIGURE 258 The mechanical efficiency of a fan is the ratio of the kinetic energy of air at the fan exit to the mechanical power input. cen84959_ch02.qxd 3/31/05 5:02 PM Page 84 84 | Thermodynamics and hturbine = 0.75 hgenerator = 0.97 Generator: Turbine Electric power output Mechanical power input h generator Generator h turbinegen = hturbineh generator = 0.75 × 0.97 = 0.73 FIGURE 259 The overall efficiency of a turbinegenerator 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. Lake h = 50 m Turbine Generator · m = 5000 kg/s FIGURE 260 Schematic for Example 216. (248) A pump is usually packaged together with its motor, and a turbine with its generator. Therefore, we are usually interested in the combined or overall efficiency of pumpmotor and turbinegenerator combinations (Fig. 259), which are defined as h pump h pumph motor motor # Wpump,u # Welect,in # ¢ E mech,fluid # Welect,in (249) and h turbine gen h turbineh generator # Welect,out # Wturbine,e # Welect,out # 0 ¢ E mech,fluid (250) All the efficiencies just defined range between 0 and 100 percent. The lower limit of 0 percent corresponds to the conversion of the entire mechanical or electric energy input to thermal energy, and the device in this case functions like a resistance heater. The upper limit of 100 percent corresponds to the case of perfect conversion with no friction or other irreversibilities, and thus no conversion of mechanical or electric energy to thermal energy. EXAMPLE 216 hgenerator = 0.95 1862 kW # Welect,out # Wshaft,in Performance of a Hydraulic TurbineGenerator The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbinegenerator at a location where the depth of the water is 50 m (Fig. 260). 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 turbinegenerator 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 P r 0 gh 0.491 kJ> kg 1 9.81 m> s2 2 1 50 m 2 a 1 kJ> kg 1000 m2> s2 b cen84959_ch02.qxd 3/31/05 5:02 PM Page 85 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 1 emech,in # h turbine h overall gen emech,out 2 1 5000 kg> s 2 1 0.491 kJ> kg 2 # Welect,out 1862 kW 0.76 # 2455 kW 0 ¢ E mech,fluid 0 2455 kW (b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from hturbine gen hturbinehgenerator S hturbine hturbine gen hgenerator 0.76 0.95 0.80 (c) The shaft power output is determined from the definition of mechanical efficiency, # Wshaft,out # h turbine 0 ¢ E mech,fluid 0 1 0.80 2 1 2455 kW 2 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 217 Cost Savings Associated with High-Efficiency Motors 60 hp h = 89.0% Standard Motor A 60-hp 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 high-efficiency motor (Fig. 261). 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 high-efficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $4520 and $5160, respectively. Solution A worn-out standard motor is to be replaced by a high-efficiency 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 # Welectric in,efficient Power savings # Wshaft> h st 1 Rated power 2 1 Load factor 2 > h st # Wshaft> h eff 1 Rated power 2 1 Load factor 2 > h eff # # Welectric in,standard Welectric in,efficient 1 Rated power 2 1 Load factor 2 1 1> h st 1> h eff 2 60 hp h = 93.2% High-Efficiency Motor FIGURE 261 Schematic for Example 217. | 85 cen84959_ch02.qxd 4/25/05 2:32 PM Page 86 86 | Thermodynamics where hst is the efficiency of the standard motor, and heff is the efficiency of the comparable high-efficiency motor. Then the annual energy and cost savings associated with the installation of the high-efficiency motor become Energy savings 1 Power savings 2 1 Operating hours 2 1 Rated power 2 1 Operating hours 2 1 Load factor 2 1 1> h st 1 60 hp 2 1 0.7457 kW> hp 2 1 3500 h> year 2 1 1 2 1 1> 0.89 7929 kWh> year Cost savings 1> h eff 2 1> 0.93.2 2 1 Energy savings 2 1 Unit cost of energy 2 1 7929 kWh> year 2 1 $0.08> kWh 2 $634> year Also, Excess initial cost Purchase price differential $5160 $4520 $640 This gives a simple payback period of Simple payback period Excess initial cost Annual cost savings $640 $634> year 1.01 year Discussion Note that the high-efficiency 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. 28 FIGURE 262 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. 262). 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 87 Chapter 2 usually grouped as hydrocarbons (HC), nitrogen oxides (NOx), and carbon monoxide (CO) (Fig. 263). 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 14-day 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, todays industrial facilities and vehicles emit a fraction of the pollutants they used to emit a few decades ago. The HC emissions of cars, for example, decreased from about 8 gpm (grams per mile) in 1970 to 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 smogthe dark yellow or brown haze that builds up in a large stagnant air mass and hangs over populated areas on calm hot summer days. Smog is made up mostly of ground-level 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 ground-level ozone should not be confused with the useful ozone layer high in the stratosphere that protects the earth from the suns harmful ultraviolet rays. Ozone at ground level is a pollutant with several adverse health effects. The primary source of both nitrogen oxides and hydrocarbons is the motor vehicles. Hydrocarbons and nitrogen oxides react in the presence of sunlight on hot calm days to form ground-level ozone, which is the primary | 87 NOx CO HC FIGURE 263 Motor vehicles are the largest source of air pollution. cen84959_ch02.qxd 3/31/05 5:02 PM Page 88 88 | Thermodynamics SUN O3 NOx HC SMOG FIGURE 264 Ground-level 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. 264). The smog formation usually peaks in late afternoons when the temperatures are highest and there is plenty of sunlight. Although ground-level 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 individuals 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 smog-prone 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 bodys organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars. 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 265 Sulfuric acid and nitric acid are formed when sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight. 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 high-sulfur coal. The Clean Air Act of 1970 has limited the SO2 emissions severely, which forced the plants to install SO2 scrubbers, to switch to low-sulfur 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. 265). The acids formed usually dissolve in the suspended water droplets in clouds or fog. These acid-laden 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 89 Chapter 2 | 89 a certain amount of acid, but the amounts produced by the power plants using inexpensive high-sulfur 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. 266). 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 peoples activities do make a difference in CO2 concentration in the atmosphere. The greenhouse effect makes life on earth possible by keeping the earth warm (about 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 earths surface FIGURE 266 The greenhouse effect on earth. cen84959_ch02.qxd 3/31/05 5:02 PM Page 90 90 | Thermodynamics FIGURE 267 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 worlds leading climate scientists concluded that the earth has already warmed about 0.5°C during the last century, and they estimate that the earths 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 worlds industrialized countries adopted the Kyoto protocol and committed to reduce their CO2 and other greenhouse gas emissions by 5 percent below the 1990 levels by 2008 to 2012. 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. 267). This and other emissions can be reduced significantly by buying an energyefficient car that burns less fuel over the same distance, and by driving sensibly. Saving fuel also saves money and the environment. For example, choosing a vehicle that gets 30 rather than 20 miles per gallon will prevent 2 tons of CO2 from being released to the 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 91 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 emission-free renewable resources such as hydroelectric, solar, wind, and geothermal energy (Fig. 268). Therefore, the use of renewable energy should be encouraged worldwide, with incentives, as necessary, to make the earth a better place to live in. The advancements in thermodynamics have contributed greatly in recent decades to improve conversion efficiencies (in some cases doubling them) and thus to reduce pollution. As individuals, we can also help by practicing energy conservation measures and by making energy efficiency a high priority in our purchases. EXAMPLE 218 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 1 NOx emission per therm 2 1 No. of therms per year 2 1 0.0047 kg> therm 2 1 18 8.5 CO2 savings 104 kg> year 106 therm> year 2 1 CO2 emission per therm 2 1 No. of therms per year 2 1 6.4 kg> therm 2 1 18 1.2 108 kg> year 106 therm> year 2 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. | 91 FIGURE 268 Renewable energies such as wind are called green energy since they emit no pollutants or greenhouse gases. © Corbis Royalty Free cen84959_ch02.qxd 4/25/05 2:32 PM Page 92 92 | 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 high-temperature 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. 269). # 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 269 Heat conduction from warm air to a cold canned drink through the wall of the aluminum can. # Qcond SEE TUTORIAL CH. 2, SEC. 9 ON THE DVD. ¢T ¢x 1W 2 (251) 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 23). 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 # Qcond INTERACTIVE TUTORIAL kt A kt A dT dx 1W 2 (252) which is known as Fouriers law of heat conduction. It indicates that the rate of heat conduction in a direction is proportional to the temperature 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. 252 to make heat transfer in the positive x direction a positive quantity. Temperature is a measure of the kinetic energies of the molecules. In a liquid or gas, the kinetic energy of the molecules is due to the random motion of the molecules as well as the vibrational and rotational motions. When two 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. cen84959_ch02.qxd 3/31/05 5:02 PM Page 93 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 23. Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas that is in motion, and it involves the combined effects of conduction and fluid motion. The faster the fluid motion, the greater the convection heat transfer. In the absence of any bulk fluid motion, heat transfer between a solid surface and the adjacent fluid is by pure conduction. The 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. 270). Energy is first transferred to the air layer adjacent to the surface of the block by conduction. This energy is then carried away from the surface by 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. 271). For example, in the absence of a fan, heat transfer from the surface of the hot block in Fig. 270 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 Newtons law of cooling, expressed as # Qconv hA 1 Ts Tf 2 1W 2 (253) where h is the convection heat transfer coefficient, A is the surface area through which heat transfer takes place, Ts is the surface temperature, and Tf is bulk fluid temperature away from the surface. (At the surface, the fluid temperature equals the surface temperature of the solid.) | 93 TABLE 23 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 270 Heat transfer from a hot surface to air by convection. cen84959_ch02.qxd 3/31/05 5:02 PM Page 94 94 | Thermodynamics Forced convection Natural convection AIR AIR hot egg hot egg FIGURE 271 The cooling of a boiled egg by forced and natural convection. Person 30°C Air 5°C Fire 900°C Radiation FIGURE 272 Unlike conduction and convection, heat transfer by radiation can occur between two bodies, even when they are separated by a medium colder than both of them. 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 225 for the free convection of gases, 501000 for the free convection of liquids, 25250 for the forced convection of gases, 5020,000 for the forced convection of liquids, and 2500100,000 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. 272). In fact, energy transfer by radiation is fastest (at the speed of light) and it suffers no attenuation in a vacuum. This is exactly how the energy of the sun reaches the earth. In heat transfer studies, we are interested in thermal radiation, which is the form of radiation emitted by bodies because of their temperature. It differs from other forms of electromagnetic radiation such as X-rays, 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 StefanBoltzmann law as # Q emit,max sAT s4 1W 2 (254) 8 W/m2 K4 where A is the surface area and s 5.67 10 · is the StefanBoltzmann 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 1W 2 (255) 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 24. 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. cen84959_ch02.qxd 4/26/05 4:41 PM Page 95 Chapter 2 In general, both e and a of a surface depend on the temperature and the wavelength of the radiation. Kirchhoffs 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. 273) # Q abs # aQ incident 1W 2 (256) # 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. 274) # Qrad esA 1 T 4 s T4 2 surr 1W 2 | 95 TABLE 24 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 0.920.97 0.850.93 0.930.96 0.95 0.820.92 0.930.96 0.96 0.920.96 · Qincident · · Qref = (1 α ) Qincident (257) 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 273 The absorption of radiation incident on an opaque surface of absorptivity a. EXAMPLE 219 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. 275). Solution A person is standing in a breezy room. The total rate of heat loss from the person is to be determined. Assumptions 1 The emissivity and heat transfer coefficient are constant and 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 274 Radiation heat transfer between a body and the inner surfaces of a much larger enclosure that completely surrounds it. cen84959_ch02.qxd 3/31/05 5:02 PM Page 96 96 | Thermodynamics Room air 20°C Qconv that the experimentally determined value for the rate of convection heat transfer in this case is 6 W per unit surface area (m2) per unit temperature difference (in K or °C) between the person and the air away from the person. Thus, the rate of convection heat transfer from the person to the air in the room is, from Eq. 253, hA 1 Ts # Qconv 29°C Tf 2 1 6 W> m # °C 2 1 1.6 m2 2 1 29 Qrad 2 20 2 °C 86.4 W Qcond FIGURE 275 Heat transfer from the person described in Example 219. 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. 257, # Qrad esA 1 T 4 s 1 0.95 2 1 5.67 T4 2 surr 10 8 W> m2 # K4 2 1 1.6 m2 2 3 1 29 273 2 4 1 20 273 2 4 4 K4 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 97 Chapter 2 V 2 gz Change in internal, kinetic, potential, etc., energies It can also be expressed in the rate form as . . E in Eout dE system> dt 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 # E mech # memech #P ma r V2 2 gz b 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: Electrical work: We Shaft work: Wsh Spring work: Wspring h turbine h motor h generator h pump # Wshaft,out # 0 ¢ E mech,fluid 0 1 k 1x 2 2 2 # Wshaft,out # Welect,in Electric power output Mechanical power input # Welect,out # Wshaft,in motor h turbinegen x 22 1 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 # Wturbine # Wturbine,e Mechanical power output Electric power input h pumph motor VI ¢ t 2pnT 1 kW 2 P r emech 97 1 kJ 2 ¢ E system 2 E out 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 The energy flow rate associated with a fluid flowing at a rate . of m is # # E me | h turbineh generator # ¢ E mech,fluid # Welect,in # Welect,out # 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 Air-Conditioning Engineers, Inc., 1993. 2. Y. A. Çengel. An Intuitive and Unified Approach to Teaching Thermodynamics. ASME International Mechanical Engineering Congress and Exposition, Atlanta, Georgia, AES-Vol. 36, pp. 251260, November 1722, 1996. cen84959_ch02.qxd 3/31/05 5:02 PM Page 98 98 | Thermodynamics PROBLEMS* Forms of Energy 21C Portable electric heaters are commonly used to heat small rooms. Explain the energy transformation involved during this heating process. 22C 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? 23C What is the difference between the macroscopic and microscopic forms of energy? 24C What is total energy? Identify the different forms of energy that constitute the total energy. 25C List the forms of energy that contribute to the internal energy of a system. 26C How are heat, internal energy, and thermal energy related to each other? 27C What is mechanical energy? How does it differ from thermal energy? What are the forms of mechanical energy of a fluid stream? 28 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 210 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 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. 211 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. 212 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. 213 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. 214 A person gets into an elevator at the lobby level of a hotel together with his 30-kg 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 215C In what forms can energy cross the boundaries of a closed system? 216C When is the energy crossing the boundaries of a closed system heat and when is it work? FIGURE P28 217C What is an adiabatic process? What is an adiabatic system? 29 Electric power is to be generated by installing a hydraulic turbinegenerator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. Determine the power generation potential. 218C A gas in a pistoncylinder 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 CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 219C 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. 220C 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? 221C 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. 222C What are point and path functions? Give some examples. cen84959_ch02.qxd 3/31/05 5:02 PM Page 99 Chapter 2 223C What is the caloric theory? When and why was it abandoned? Mechanical Forms of Work 224C 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? 225C 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? 226 Determine the energy required to accelerate an 800kg car from rest to 100 km/h on a level road. Answer: 309 kJ 227 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. 228E Determine the torque applied to the shaft of a car that transmits 450 hp and rotates at a rate of 3000 rpm. 229 Determine the work required to deflect a linear spring with a spring constant of 70 kN/m by 20 cm from its rest position. 230 The engine of a 1500-kg 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? 231 A ski lift has a one-way 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. | 99 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 233 A damaged 1200-kg 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 234C For a cycle, is the net work necessarily zero? For what kind of systems will this be the case? 235C 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. 236C What are the different mechanisms for transferring energy to or from a control volume? 237 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 paddle-wheel 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 232 Determine the power required for a 2000-kg car to climb a 100-m-long 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 P237 30° FIGURE P232 238E A vertical pistoncylinder device contains water and is being heated on top of a range. During the process, 65 Btu cen84959_ch02.qxd 4/11/05 2:21 PM Page 100 100 | Thermodynamics 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 239 A classroom that normally contains 40 people is to be air-conditioned with window air-conditioning 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 air-conditioning units required. Answer: 2 units 240 The lighting requirements of an industrial facility are being met by 700 40-W standard fluorescent lamps. The lamps are close to completing their service life and are to be replaced by their 34-W high-efficiency 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 high-efficiency fluorescent lamps. Also, determine the simple payback period. 241 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. 242 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. 243 Consider a room that is initially at the outdoor temperature of 20°C. The room contains a 100-W lightbulb, a 110-W TV set, a 200-W refrigerator, and a 1000-W 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. 244 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 245E 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. 246 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. 247 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 P247 248 The 60-W 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. 249E 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. 250 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 cen84959_ch02.qxd 4/20/05 5:06 PM Page 101 Chapter 2 this escalator. What would your answer be if the escalator velocity were to be doubled? 251 Consider a 1400-kg 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 252C What is mechanical efficiency? What does a mechanical efficiency of 100 percent mean for a hydraulic turbine? 253C How is the combined pumpmotor efficiency of a pump and motor system defined? Can the combined pumpmotor efficiency be greater than either the pump or the motor efficiency? 254C Define turbine efficiency, generator efficiency, and combined turbinegenerator efficiency. 255C Can the combined turbine-generator efficiency be greater than either the turbine efficiency or the generator efficiency? Explain. 256 Consider a 3-kW 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. 257 A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor that has an efficiency of 95.4 percent. Determine the reduction in the heat gain of the room due to higher efficiency under full-load conditions. 258 A 90-hp (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. 259 A 75-hp (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 high-efficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5449 and $5520, respectively. | 101 260E 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 hand-held 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. 261E Reconsider Prob. 260E. 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. 262 An exercise room has eight weight-lifting machines that have no motors and four treadmills each equipped with a 2.5-hp (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. 263 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. 264 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 0.25-hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25-hp equipment is 54 percent. Determine the rate of heat supply by the fanmotor assembly to the room. 265 Electric power is to be generated by installing a hydraulic turbinegenerator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 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 turbinegenerator efficiency of this plant. Neglect losses in the pipes. 266 At a certain location, wind is blowing steadily at 12 m/s. Determine the mechanical energy of air per unit mass cen84959_ch02.qxd 3/31/05 5:02 PM Page 102 102 | Thermodynamics and the power generation potential of a wind turbine with 50m-diameter 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. 267 Reconsider Prob. 266. 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. 268 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. 269 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 pumpmotor unit and (b) the pressure difference between the inlet and the exit of the pump. 272 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 24-hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.06/kWh for electricity. 273E 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. 274 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 P269 FIGURE P274 270 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. 271 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 2-kW resistance heater. Determine if it is necessary to turn the heater on when the motor runs at full load. 275 A 7-hp (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. 276 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. 277 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, cen84959_ch02.qxd 3/31/05 5:02 PM Page 103 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 P277 278E A 73-percent 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 constant-diameter 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 279C How does energy conversion affect the environment? What are the primary chemicals that pollute the air? What is the primary source of these pollutants? 286 Repeat Prob. 285 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. 287E 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. 288 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 281C 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? 283C Why is carbon monoxide a dangerous air pollutant? How does it affect human health at low and at high levels? 284E 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 5-year period if he trades his Taurus for an Explorer. 285 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. 280C What is smog? What does it consist of? How does ground-level ozone form? What are the adverse effects of ozone on human health? 282C What is the greenhouse effect? How does the excess CO2 gas in the atmosphere cause the greenhouse effect? What are the potential long-term consequences of greenhouse effect? How can we combat this problem? | FIGURE P288 Special Topic: Mechanisms of Heat Transfer 289C What are the mechanisms of heat transfer? 290C Does any of the energy of the sun reach the earth by conduction or convection? 291C Which is a better heat conductor, diamond or silver? 292C How does forced convection differ from natural convection? 293C Define emissivity and absorptivity. What is Kirchhoffs law of radiation? 294C What is blackbody? How do real bodies differ from a blackbody? cen84959_ch02.qxd 3/31/05 5:02 PM Page 104 104 | Thermodynamics 295 The inner and outer surfaces of a 5-m 6-m 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. 2102 Hot air at 80°C is blown over a 2-m 4-m 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 30 cm 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 FIGURE P295 296 The inner and outer surfaces of a 0.5-cm-thick 2-m 2-m 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 1-cm thick? Reconsider Problem 296. Using EES (or other) software, investigate the effect of glass thickness on heat loss for the specified glass surface temperatures. Let the glass thickness vary from 0.2 to 2 cm. Plot the heat loss versus the glass thickness, and discuss the results. 2103 A 1000-W 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. 1000-W iron Air 20°C 297 298 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. 299 For heat transfer purposes, a standing man can be modeled as a 30-cm diameter, 170-cm 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 P2103 2104 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 2100 A 5-cm-diameter 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. 2101 Reconsider Problem 2100. Using EES (or other) software, investigate the effect of the FIGURE P2104 cen84959_ch02.qxd 4/20/05 5:06 PM Page 105 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. 2105 Reconsider Problem 2104. 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. 2106 A 5-cm-external-diameter, 10-m-long hot-water 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. 2107 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. 2108 Reconsider Problem 2107. 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. 2109 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 P2109 2110 The inner and outer glasses of a 2-m 2-m double pane window are at 18°C and 6°C, respectively. If the 1-cm space between the two glasses is filled with still air, determine the rate of heat transfer through the window, in kW. | 105 2111 Two surfaces of a 2-cm-thick 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 2112 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 400-kg 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? 2113 Consider a homeowner who is replacing his 25-yearold 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 high-efficiency model. 2114 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 mid-1990s, which is about the price of electricity generated at coal-fired 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 worlds 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. cen84959_ch02.qxd 4/20/05 5:06 PM Page 106 106 | 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. 2118 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 P2114 © Vol. 57/PhotoDisc In 2003, 8133 MW of new wind energy generating capacity were installed worldwide, bringing the worlds 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 two-bladed 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 80-m-diameter 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. 2115 Repeat Prob. 2114 for an average wind velocity of 25 km/h. 2116E 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 lowest-cost energy source for water heaters. 2117 A homeowner is considering these heating systems for heating his house: Electric resistance heating with $0.09/kWh 2119 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 gas-heated 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. 2120 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. 2121 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). 2122 The pump of a water distribution system is powered by a 15-kW 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 P2122 cen84959_ch02.qxd 3/31/05 5:02 PM Page 107 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 2123 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 turbinegenerator is 80 percent. Disregarding frictional losses in piping, estimate the electric power output of this plant. Answer: 94.2 MW | 107 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 pumpturbine system can generate per year. Reservoir 40 m 1 Pump turbine 100 m3/s 120 m Lake 2 Generator Turbine FIGURE P2124 h turbinegen = 80% FIGURE P2123 2124 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 pumpmotor operates as a turbinegenerator during reverse flow. Preliminary analysis shows that a water flow rate of 2 m3/s can be used in either direction. The combined pumpmotor and turbinegenerator 2125 A diesel engine with an engine volume of 4.0 L and an engine speed of 2500 rpm operates on an airfuel 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. 2126 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 108 108 | Thermodynamics Fundamentals of Engineering (FE) Exam Problems 2127 A 2-kW 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 2128 On a hot summer day, the air in a well-sealed room is circulated by a 0.50-hp 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 2129 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 2130 A 900-kg 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 2131 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 2132 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 turbinegenerator 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 2133 A 75-hp 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 2134 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 2135 A 2-kW 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 2136 A glycerin pump is powered by a 5-kW 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 2137 A 10-cm high and 20-cm 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 2138 A 50-cm-long, 0.2-cm-diameter 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 2139 A 3-m2 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 2140 Heat is transferred steadily through a 0.2-m 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 cen84959_ch02.qxd 3/31/05 5:02 PM Page 109 Chapter 2 2141 The roof of an electrically heated house is 7-m long, 10-m wide, and 0.25-m 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 2142 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. 2143 Solar energy reaching the earth is about 1350 W/m2 outside the earths atmosphere, and 950 W/m2 on earths 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. 2144 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. 2145 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. 2146 The performance of a device is defined as the ratio of the desired output to the required input, and this definition can be extended to nontechnical fields. For example, your 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. 2147 Your neighbor lives in a 2500-square-foot (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. 2148 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 30-year service life, a 4-kW 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. cen84959_ch02.qxd 3/31/05 5:02 PM Page 110 cen84959_ch03.qxd 4/1/05 12:31 PM Page 111 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 phase-change processes. We then illustrate the various property diagrams and P-v-T surfaces of pure substances. After demonstrating the use of the property tables, the hypothetical substance ideal gas and the ideal-gas equation of state are discussed. The compressibility factor, which accounts for the deviation of real gases from ideal-gas behavior, is introduced, and some of the best-known equations of state such as the van der Waals, Beattie-Bridgeman, and Benedict-Webb-Rubin equations are presented. Objectives The objectives of Chapter 3 are to: Introduce the concept of a pure substance. Discuss the physics of phase-change processes. Illustrate the P-v, T-v, and P-T property diagrams and P-v-T 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 ideal-gas equation of state. Apply the ideal-gas equation of state in the solution of typical problems. Introduce the compressibility factor, which accounts for the deviation of real gases from ideal-gas behavior. Present some of the best-known equations of state. | 111 cen84959_ch03.qxd 4/25/05 2:47 PM Page 112 112 | Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 1 ON THE DVD. N2 AIR FIGURE 31 Nitrogen and gaseous air are pure substances. VAPOR LIQUID ( a ) H 2O VAPOR LIQUID (b) AIR FIGURE 32 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 33 The molecules in a solid are kept at their positions by the large springlike intermolecular forces. 31 PURE SUBSTANCE A substance that has a fixed chemical composition throughout is called a pure substance. Water, nitrogen, helium, and carbon dioxide, for example, are all pure substances. A pure substance does not have to be of a single chemical element or compound, however. A mixture of various chemical elements or compounds also qualifies as a pure substance as long as the mixture is homogeneous. Air, for example, is a mixture of several gases, but it is often considered to be a pure substance because it has a uniform chemical composition (Fig. 31). However, a mixture of oil and water is not a pure substance. Since oil is not soluble in water, it will collect on top of the water, forming two chemically dissimilar regions. A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same (Fig. 32). A mixture of ice and liquid water, for example, is a pure substance because both phases have the same chemical composition. A mixture of liquid air and 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. 32 PHASES OF A PURE SUBSTANCE We all know from experience that substances exist in different phases. At room temperature and pressure, copper is a solid, mercury is a liquid, and nitrogen is a gas. Under different conditions, each may appear in a different phase. Even though there are three principal phasessolid, liquid, and gasa substance may have several phases within a principal phase, each with a different molecular structure. Carbon, for example, may exist as graphite or diamond in the solid phase. Helium has two liquid phases; iron has three solid phases. Ice may exist at seven different phases at high 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 three-dimensional pattern (lattice) that is repeated throughout (Fig. 33). 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. 34). Note that the attractive forces between molecules turn to repulsive forces as the cen84959_ch03.qxd 4/26/05 4:40 PM Page 113 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. 35). 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. 33 113 FIGURE 34 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. PHASE-CHANGE 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 35 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. cen84959_ch03.qxd 4/1/05 12:31 PM Page 114 114 | Thermodynamics important phase-change 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 36 At 1 atm and 20°C, water exists in the liquid phase (compressed liquid). STATE 2 P = 1 atm T = 100°C Consider a pistoncylinder device containing liquid water at 20°C and 1 atm pressure (state 1, Fig. 36). 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. 37). At this point water is still a liquid, but any heat addition will cause some of the liquid to vaporize. That is, a phase-change 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 37 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 38 As more heat is transferred, part of the saturated liquid vaporizes (saturated liquidvapor mixture). Once boiling starts, the temperature stops rising until the liquid is completely vaporized. That is, the temperature will remain constant during the entire phase-change 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. 38), 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. 39). At this point, the entire cylinder is filled with vapor that is on the borderline of the liquid phase. Any heat loss from this vapor will cause some of the vapor to condense (phase change from vapor to liquid). A vapor that is about to condense is called a saturated vapor. Therefore, state 4 is a saturated vapor state. A substance at states between 2 and 4 is referred to as a saturated liquidvapor mixture since the liquid and vapor phases coexist in equilibrium at these states. Once the phase-change 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. 310). 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 cen84959_ch03.qxd 4/1/05 12:31 PM Page 115 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 constant-pressure phase-change process is illustrated on a T-v diagram in Fig. 311. 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 ITS-90 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 311 T-v diagram for the heating process of water at constant pressure. | 115 STATE 4 P = 1 atm T = 100°C Heat FIGURE 39 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 310 As more heat is transferred, the temperature of the vapor starts to rise (superheated vapor). cen84959_ch03.qxd 4/25/05 2:47 PM Page 116 116 | Thermodynamics EXPERIMENT Use actual data from the experiment shown here to obtain the latent heat of fusion of water. See end-of-chapter problem 3146. © Ronald Mullisen TABLE 31 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 312 The liquidvapor 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 31 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 phase-change 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 phase-change process, pressure and temperature are obviously dependent properties, and there is a definite relation between them, that is, f (Psat). A plot of Tsat versus Psat, such as the one given for water in Tsat Fig. 312, is called a liquidvapor saturation curve. A curve of this kind is characteristic of all pure substances. It is clear from Fig. 312 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 32. For each 1000 m increase in elevation, the boiling temperature Psat, kPa 600 400 200 0 0 50 100 150 200 Tsat,°C cen84959_ch03.qxd 4/1/05 12:31 PM Page 117 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 refrigerant-134a 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 refrigerant-134a 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 refrigerant-134a 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 A3a). 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 low-temperature 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. 313). 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 32 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 313 The temperature of liquid nitrogen exposed to the atmosphere remains constant at 196°C, and thus it maintains the test chamber at 196°C. cen84959_ch03.qxd 4/25/05 2:47 PM Page 118 118 | Thermodynamics Temperature °C Start of cooling (25°C, 100 kPa) 25 End of cooling (0°C, 0.61 kPa) 0 0 0.61 1 3.17 10 100 Pressure (kPa) FIGURE 314 The variation of the temperature of fruits and vegetables with pressure during vacuum cooling from 25 C to 0 C. Vacuum pump Insulation Air + Vapor Low vapor pressure Evaporation High vapor pressure Ice Water FIGURE 315 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. 314). 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 well-suited 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. 315). Package icing is commonly used in small-scale cooling applications to remove heat and keep the products cool during transit by taking advantage of the large latent heat of fusion of water, but its use is limited to products that are not harmed by contact with ice. Also, ice provides moisture as well as refrigeration. 34 PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES The variations of properties during phase-change processes are best studied and understood with the help of property diagrams. Next, we develop and discuss the T-v, P-v, and P-T diagrams for pure substances. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 4 ON THE DVD. 1 The T-v Diagram The phase-change process of water at 1 atm pressure was described in detail in the last section and plotted on a T-v diagram in Fig. 311. Now we repeat this process at different pressures to develop the T-v 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. 316, but there are some noticeable differences. First, water starts boiling at a much higher tempera- cen84959_ch03.qxd 4/1/05 12:31 PM Page 119 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 316 T-v diagram of constant-pressure phase-change 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< ture (179.9°C) at this pressure. Second, the specific volume of the saturated liquid is larger and the specific volume of the saturated vapor is smaller than the corresponding values at 1 atm pressure. That is, the horizontal line that connects the saturated liquid and saturated vapor states is much shorter. As the pressure is increased further, this saturation line continues to shrink, as shown in Fig. 316, and it becomes a point when the pressure reaches 22.06 MPa for the case of water. This point is called the critical point, and it is defined as the point at which the saturated liquid and saturated vapor states are identical. The temperature, pressure, and specific volume of a substance at the critical point are called, respectively, the critical temperature Tcr, critical pressure Pcr, and critical specific volume vcr. The critical-point properties of 22.06 MPa, Tcr 373.95°C, and vcr 0.003106 m3/kg. water are Pcr For helium, they are 0.23 MPa, 267.85°C, and 0.01444 m3/kg. The critical properties for various substances are given in Table A1 in the appendix. At pressures above the critical pressure, there is not a distinct phasechange process (Fig. 317). Instead, the specific volume of the substance continually increases, and at all times there is only one phase present. Eventually, it resembles a vapor, but we can never tell when the change Phase change LI QU ID vcr v FIGURE 317 At supercritical pressures (P Pcr), there is no distinct phase-change (boiling) process. cen84959_ch03.qxd 4/1/05 12:31 PM Page 120 120 | Thermodynamics has occurred. Above the critical state, there is no line that separates the compressed liquid region and the superheated vapor region. However, it is customary to refer to the substance as superheated vapor at temperatures above the critical temperature and as compressed liquid at temperatures below the critical temperature. The saturated liquid states in Fig. 316 can be connected by a line called the saturated liquid line, and saturated vapor states in the same figure can be connected by another line, called the saturated vapor line. These two lines meet at the critical point, forming a dome as shown in Fig. 318. All the compressed liquid states are located in the region to the left of the saturated liquid line, called the compressed liquid region. All the superheated vapor states are located to the right of the saturated vapor line, called the superheated vapor region. In these two regions, the substance exists in a single phase, a liquid or a vapor. All the states that involve both phases in equilibrium are located under the dome, called the saturated liquidvapor mixture region, or the wet region. 2 The P-v Diagram The general shape of the P-v diagram of a pure substance is very much like the T-v diagram, but the T constant lines on this diagram have a downward trend, as shown in Fig. 319. Consider again a pistoncylinder device that contains liquid water at 1 MPa and 150°C. Water at this state exists as a compressed liquid. Now the weights on top of the piston are removed one by one so that the pressure inside the cylinder decreases gradually (Fig. 320). The water is allowed to exchange heat with the surroundings so its temperature remains constant. As T st. on =c r e SATURATED LIQUID VAPOR REGION lin ted po Satura SUPERHEATED VAPOR REGION va liquid P 1 ted FIGURE 318 T-v diagram of a pure substance. ra COMPRESSED LIQUID REGION line tu P 2 Sa = co ns t. > P 1 Critical point v cen84959_ch03.qxd 4/1/05 12:31 PM Page 121 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 saturation-pressure 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 phase-change 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 phase-change processes. Connecting the saturated liquid and the saturated vapor states by a curve, we obtain the P-v diagram of a pure substance, as shown in Fig. 319. Extending the Diagrams to Include the Solid Phase The two equilibrium diagrams developed so far represent the equilibrium states involving the liquid and the vapor phases only. However, these diagrams can easily be extended to include the solid phase as well as the solidliquid and the solidvapor saturation regions. The basic principles discussed in conjunction with the liquidvapor phase-change process apply equally to the solidliquid and solidvapor phase-change processes. Most substances contract during a solidification (i.e., freezing) process. Others, like water, expand as they freeze. The P-v diagrams for both groups of substances are given in Figs. 321 and 322. These two diagrams differ only in FIGURE 319 P-v diagram of a pure substance. P = 1 MPa T = 150°C Heat FIGURE 320 The pressure in a pistoncylinder device can be reduced by reducing the weight of the piston. cen84959_ch03.qxd 4/1/05 12:31 PM Page 122 122 | Thermodynamics P VAPOR LIQUID SOLID SOLID + LIQUID Critical point LIQUID + VAPOR Triple line FIGURE 321 P-v diagram of a substance that contracts on freezing. SOLID + VAPOR v P LIQUID SOLID + LIQUID Critical point VAPOR SOLID LIQUID + VAPOR FIGURE 322 P-v diagram of a substance that expands on freezing (such as water). Triple line SOLID + VAPOR v the solidliquid saturation region. The T-v diagrams look very much like the P-v diagrams, especially for substances that contract on freezing. The fact that water expands upon freezing has vital consequences in nature. If water contracted on freezing as most other substances do, the ice formed would be heavier than the liquid water, and it would settle to the bottom of rivers, lakes, and oceans instead of floating at the top. The suns cen84959_ch03.qxd 4/1/05 12:31 PM Page 123 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. 323). On P-v or T-v diagrams, these triple-phase 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 P-T diagrams and, therefore, is often called the triple point. The triple-point temperatures and pressures of various substances are given in Table 33. For water, the triple-point 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 triple-point pressure. The same can be said for temperature for substances that contract on freezing. However, TABLE 33 Triple-point 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 10 4 0.12 5.1 7.04 13.9 1.65 10 7 11.7 43.2 21.92 12.6 87.85 0.152 3.5 10 3 2.0 10 4 1.67 5.3 10 3 151.7 0.61 81.5 0.065 | VAPOR LIQUID SOLID FIGURE 323 At triple-point pressure and temperature, a substance exists in three phases in equilibrium. 123 cen84959_ch03.qxd 4/1/05 12:31 PM Page 124 124 | Thermodynamics VAPOR SOLID FIGURE 324 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 triple-point 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. 324). Passing from the solid phase directly into the vapor phase is called sublimation. For substances that have a triple-point 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 P-T Diagram Figure 325 shows the P-T 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 P-T 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 325 P-T diagram of pure substances. li ub ma tio n T cen84959_ch03.qxd 4/1/05 12:31 PM Page 125 Chapter 3 | The P-v-T Surface Liquid SolidLiquid 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 P-v-T behavior of a substance as a surface in space, as shown in Figs. 326 and 327. 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 326 P-v-T 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 327 P-v-T surface of a substance that expands on freezing (like water). 125 cen84959_ch03.qxd 4/25/05 2:47 PM Page 126 126 | 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 quasi-equilibrium process lie on the P-v-T surface since such a process must pass through equilibrium states. The single-phase regions appear as curved surfaces on the P-v-T surface, and the two-phase regions as surfaces perpendicular to the P-T plane. This is expected since the projections of two-phase regions on the P-T plane are lines. All the two-dimensional diagrams we have discussed so far are merely projections of this three-dimensional surface onto the appropriate planes. A P-v diagram is just a projection of the P-v-T surface on the P-v plane, and a T-v diagram is nothing more than the birds-eye view of this surface. The P-v-T surfaces present a great deal of information at once, but in a thermodynamic analysis it is more convenient to work with two-dimensional diagrams, such as the P-v and T-v diagrams. 35 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 A6 and A6E, 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. EnthalpyA Combination Property P1v1 Control volume u2 P2v2 FIGURE 328 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. 328), we frequently encounter the combination of properties u P v. For the sake of simplicity and convenience, this combination is defined as a new property, enthalpy, and given the symbol h: h u Pv 1 kJ> kg 2 (31) cen84959_ch03.qxd 4/1/05 12:31 PM Page 127 Chapter 3 or, H U PV 1 kJ 2 | 127 (32) Both the total enthalpy H and specific enthalpy h are simply referred to as enthalpy since the context clarifies which one is meant. Notice that the equations given above are dimensionally homogeneous. That is, the unit of the pressurevolume product may differ from the unit of the internal energy by only a factor (Fig. 329). 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 329 The product pressure energy units. volume has 1a Saturated Liquid and Saturated Vapor States The properties of saturated liquid and saturated vapor for water are listed in Tables A4 and A5. Both tables give the same information. The only difference is that in Table A4 properties are listed under temperature and in Table A5 under pressure. Therefore, it is more convenient to use Table A4 when temperature is given and Table A5 when pressure is given. The use of Table A4 is illustrated in Fig. 330. 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 1 that 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 330 A partial list of Table A4. Specific volume of saturated vapor cen84959_ch03.qxd 4/1/05 12:31 PM Page 128 128 | Thermodynamics T,°C EXAMPLE 31 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 T-v diagram in Fig. 331. Since saturation conditions exist in the tank, the pressure must be the saturation pressure at 90°C: 7 P= 90 1 0. P vf Psat @ 90°C 70.183 kPa The specific volume of the saturated liquid at 90°C is v v FIGURE 331 Schematic and T-v diagram for Example 31. vf @ 90°C 0.001036 m3> kg 1 Table A4 2 Then the total volume of the tank becomes mv V EXAMPLE 32 P, psia 1 50 kg 2 1 0.001036 m3> kg 2 0.0518 m3 Temperature of Saturated Vapor in a Cylinder A pistoncylinder device contains 2 ft3 of saturated water vapor at 50-psia 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 P-v diagram in Fig. 332. Since the cylinder contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this pressure: T = 280.99 F 50 1 Table A4 2 T Tsat @ 50 psia 280.99°F 1 Table A5E 2 The specific volume of the saturated vapor at 50 psia is vg FIGURE 332 Schematic and P-V diagram for Example 32. v v vg @ 50 psia 8.5175 ft3> lbm 1 Table A5E 2 Then the mass of water vapor inside the cylinder becomes m EXAMPLE 33 V v 2 ft3 8.5175 ft3> lbm 0.235 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 P-v diagram in Fig. 333. The volume change per unit mass during a vaporization process is vfg, which cen84959_ch03.qxd 4/1/05 12:31 PM Page 129 Chapter 3 is the difference between vg and vf. Reading these values from Table A5 at 100 kPa and substituting yield vfg vg vf 1.6941 0.001043 ¢V 1 0.2 kg 2 1 1.6931 m > kg 2 mvfg 129 P, kPa 1.6931 m3> kg Thus, 3 | Sat. liquid P = 100 kPa 0.3386 m Sat. vapor P = 100 kPa vf 3 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 1 0.2 kg 2 1 2257.5 kJ> kg 2 451.5 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 333 Schematic and P-v diagram for Example 33. 1b Saturated LiquidVapor Mixture mg 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 t a tes uid s ted liq Satura mf s te mvapor sta mliquid or mtotal ap where dv (33) 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. 334). 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 334 The relative amounts of liquid and vapor phases in a saturated mixture are specified by the quality x. cen84959_ch03.qxd 4/14/05 4:09 PM Page 130 130 | Thermodynamics Saturated vapor vg vf Saturated liquid vavg Saturated liquidvapor mixture to imagine that the two phases are mixed well, forming a homogeneous mixture (Fig. 335). Then the properties of this mixture will simply be the average properties of the saturated liquidvapor mixture under consideration. Here is how it is done. Consider a tank that contains a saturated liquidvapor mixture. The volume occupied by saturated liquid is Vf, and the volume occupied by saturated vapor is Vg. The total volume V is the sum of the two: V V mf mv ¡ mtvavg mt m g ¡ mtvavg Dividing by mt yields 11 vavg since x FIGURE 335 A two-phase system can be treated as a homogeneous mixture for convenience. AB AC v avg v f A 1 mt x 2 vf mgvg m g 2 vf mgvg xvg vg vf 1 m3> kg 2 xvfg (34) vf. Solving for quality, we obtain x vavg vf (35) vfg u avg B vavg uf xu fg 1 kJ> kg 2 (36) havg hf xh fg 1 kJ> kg 2 (37) C vfg vf mf vf Based on this equation, quality can be related to the horizontal distances on a P-v or T-v diagram (Fig. 336). At a given temperature or pressure, the numerator of Eq. 35 is the distance between the actual state and the saturated liquid state, and the denominator is the length of the entire horizontal line that connects the saturated liquid and saturated vapor states. A state of 50 percent quality lies in the middle of this horizontal line. The analysis given above can be repeated for internal energy and enthalpy with the following results: P or T x= Vg mg /mt. This relation can also be expressed as vavg where vfg Vf vg FIGURE 336 Quality is related to the horizontal distances on P-v and T-v diagrams. v All the results are of the same format, and they can be summarized in a single equation as y avg yf 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. 337). That is, yf y avg yg Finally, all the saturated-mixture states are located under the saturation curve, and to analyze saturated mixtures, all we need are saturated liquid and saturated vapor data (Tables A4 and A5 in the case of water). cen84959_ch03.qxd 4/1/05 12:31 PM Page 131 Chapter 3 EXAMPLE 34 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 liquidvapor mixture is shown in Fig. 338. 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 1 Table A4 2 70.183 kPa (b) At 90°C, we have vf 0.001036 m3/kg and vg 2.3593 m3/kg (Table A4). One way of finding the volume of the tank is to determine the volume occupied by each phase and then add them: V Vf mfvf Vg mgvg 1 8 kg 2 1 0.001036 m3> kg 2 1 2 kg 2 1 2.3593 m3> kg 2 4.73 m3 vf vf < v < vg vg v FIGURE 337 The v value of a saturated liquidvapor 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 v mg 2 kg 10 kg mt T,°C 0.2 xvfg vf 0.001036 m > kg 3 0.473 m > kg 1 0.2 2 3 1 2.3593 T = 90°C mg = 2 kg 0.001036 2 m > kg 4 3 3 m f = 8 kg and V mv 1 10 kg 2 1 0.473 m > kg 2 3 3 4.73 m 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 35 Properties of Saturated LiquidVapor Mixture An 80-L vessel contains 4 kg of refrigerant-134a 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 refrigerant-134a. Some properties of the refrigerant are to be determined. Analysis (a) The state of the saturated liquidvapor mixture is shown in Fig. 339. At this point we do not know whether the refrigerant is in the compressed liquid, superheated vapor, or saturated mixture region. This can FIGURE 338 Schematic and T-v diagram for Example 34. cen84959_ch03.qxd 4/11/05 12:23 PM Page 132 132 | 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: R-134a P = 160 kPa m = 4 kg 0.080 m3 4 kg V m v 0.02 m3> kg At 160 kPa, we read 0.0007437 m3> kg vf T= 160 15.60 C m3/kg vf = 0.0007437 vg = 0.12348 v, hg = 241.11 h, kJ/kg h f = 31.21 FIGURE 339 Schematic and P-v diagram for Example 35. 0.12348 m3> kg vg (Table A12) Obviously, vf v vg, and, the refrigerant is in the saturated mixture region. Thus, the temperature must be the saturation temperature at the specified pressure: T Tsat @ 160 kPa 15.60°C (b) Quality can be determined from v x vf 0.02 0.0007437 0.12348 0.0007437 vfg (c) At 160 kPa, we also read from Table A12 that hf 209.90 kJ/kg. Then, h hf xhfg 31.21 kJ> kg 64.2 kJ> kg (d) The mass of the vapor is mg xmt 0.157 31.21 kJ/kg and hfg 1 0.157 2 1 209.90 kJ> kg 2 1 0.157 2 1 4 kg 2 0.628 kg and the volume occupied by the vapor phase is Vg mgvg 1 0.628 kg 2 1 0.12348 m3> kg 2 0.0775 m3 1 or 77.5 L 2 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 340 A partial listing of Table A6. 2748.1 2855.8 2961.0 Property tables are also available for saturated solidvapor mixtures. Properties of saturated icewater vapor mixtures, for example, are listed in Table A8. Saturated solidvapor mixtures can be handled just as saturated liquidvapor mixtures. 2 Superheated Vapor In the region to the right of the saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor. Since the superheated region is a single-phase 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. 340. 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. cen84959_ch03.qxd 4/1/05 12:31 PM Page 133 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 36 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 A6E) to be u Temperature of Superheated Vapor T 0.5 MPa and h 0.5 Determine the temperature of water at a state of P 2890 kJ/kg. MP a EXAMPLE 37 1145.1 Btu> lbm 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. 341, we again have superheated vapor. Under 0.5 MPa in Table A6 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 A7 is the only compressed liquid table in this text. The format of Table A7 is very much like the format of the superheated vapor tables. One reason for the lack of compressed liquid data is the relative independence of compressed liquid properties from pressure. Variation of properties of compressed liquid with pressure is very mild. Increasing the pressure 100 times often causes properties to change less than 1 percent. hg h > hg h FIGURE 341 At a specified P, superheated vapor exists at a higher h than the saturated vapor (Example 37). cen84959_ch03.qxd 4/1/05 12:31 PM Page 134 134 | Thermodynamics In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature (Fig. 342). This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure. Thus, Given: Given: P and T nd = v ~ vf @T u ~ uf @T = ~ h = hf @T y yf @ T FIGURE 342 A compressed liquid may be approximated as a saturated liquid at the given temperature. (38) for compressed liquids, where y is v, u, or h. Of these three properties, the property whose value is most sensitive to variations in the pressure is the enthalpy h. Although the 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) (39) instead of taking it to be just hf . Note, however, that the approximation in Eq. 39 does not yield any significant improvement at moderate to high 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 38 Approximating Compressed Liquid a s 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 343 Schematic and T-u diagram for Example 38. 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. 343. (a) From the compressed liquid table (Table A7) 5 MPa fu 80°C P T 333.82 kJ> kg (b) From the saturation table (Table A4), we read u uf @ 80°C 334.97 kJ> kg The error involved is 334.97 333.82 333.82 which is less than 1 percent. 100 0.34% cen84959_ch03.qxd 4/1/05 12:31 PM Page 135 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 refrigerant-134a, 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 39 The Use of Steam Tables to Determine Properties Determine the missing properties and the phase descriptions in the following table for water: T, °C (a) (b) (c) (d) (e) P, kPa u, kJ/kg 200 Phase description 0.6 125 1600 2950 1000 500 850 75 x 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 liquidvapor 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 1 Table A5 2 At 200 kPa, we also read from Table A5 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 1719.26 kJ> kg 1 0.6 2 1 2024.6 kJ> kg 2 (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 cen84959_ch03.qxd 4/1/05 12:31 PM Page 136 136 | Thermodynamics saturation table (Table A4) 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 if uf if we have compressed liquid u 7 ug u ug we have saturated mixture 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 liquidvapor mixture. Then the pressure must be the saturation pressure at the given temperature: P Psat @ 125°C 232.23 kPa 1 Table A4 2 The quality is determined from x u uf 1600 524.83 2009.5 ufg 0.535 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 1 Table A6 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 A5) 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 0 kP a T, °C T 6 Tsat @ given P we have compressed liquid 151.83 if T we have saturated mixture 75 if T 7 Tsat @ given P P = if ~ u = u f @ 75°C FIGURE 344 At a given P and T, a pure substance will exist as a compressed liquid if T Tsat @ P. u Tsat @ given P we have superheated vapor 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. 344), 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 1 Table A4 2 cen84959_ch03.qxd 4/25/05 2:47 PM Page 137 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 u 36 Tsat @ 850 kPa u f @ 850 kPa 172.94°C 731.00 kJ> kg 1 Table A5 2 THE IDEAL-GAS 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 best-known equation of state for substances in the gas phase is the ideal-gas equation of state. This equation predicts the P-v-T 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. Gay-Lussac, Frenchmen, experimentally determined that at low pressures the volume of a gas is proportional to its temperature. That is, P INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 6 ON THE DVD. T Ra b v or Pv RT (310) where the constant of proportionality R is called the gas constant. Equation 310 is called the ideal-gas equation of state, or simply the ideal-gas 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. 345) and is determined from R Ru M 1 kJ> kg # K or kPa # m3> kg # K 2 where Ru is the universal gas constant and M is the molar mass (also Substance ubstance Air Air Helium Helium Argon Argon Nitrogen Nitrogen R, kJ/kg·K kJ/kg 0.2870 0.2870 2.0769 2.0769 0.2081 0.2081 0.2968 0.2968 FIGURE 345 Different substances have different gas constants. cen84959_ch03.qxd 4/26/05 4:40 PM Page 138 138 | 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 (311) The molar mass M can simply be defined as the mass of one mole (also called a gram-mole, abbreviated gmol) of a substance in grams, or the mass of one kmol (also called a kilogram-mole, 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 1 kg 2 (312) The values of R and M for several substances are given in Table A1. The ideal-gas equation of state can be written in several different forms: FIGURE 346 Properties per unit mole are denoted with a bar on the top. mv ¡ PV mRT (313) NR u ¡ PV NR uT (314) R uT (315) V mR 1 MN 2 R V N v ¡ Pv 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 unit-mole basis throughout this text (Fig. 346). By writing Eq. 313 twice for a fixed mass and simplifying, the properties of an ideal gas at two different states are related to each other by P1V1 T1 FIGURE 347 The ideal-gas relation often is not applicable to real gases; thus, care should be exercised when using it. © Reprinted with special permission of King Features Syndicate. P2V2 T2 (316) An ideal gas is an imaginary substance that obeys the relation Pv RT (Fig. 347). It has been experimentally observed that the ideal-gas relation given closely approximates the P-v-T 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. cen84959_ch03.qxd 4/25/05 2:47 PM Page 139 Chapter 3 EXAMPLE 310 Mass of Air in a Room 5m Solution The mass of air in a room is to be determined. Analysis A sketch of the room is given in Fig. 348. Air at specified conditions can be treated as an ideal gas. From Table A1, 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 14 m 2 15 m 2 16 m 2 PV RT 4m AIR P = 100 kPa T = 25°C m=? 5m 120 m3 The mass of air in the room is determined from the ideal-gas relation to be m 139 6m Determine the mass of the air in a room whose dimensions are 4 m 6 m at 100 kPa and 25°C. V | 1 100 kPa 2 1 120 m 2 3 1 0.287 kPa # m3> kg # K 2 1 298 K 2 140.3 kg FIGURE 348 Schematic for Example 310. 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. 349. 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 air-conditioning 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, ideal-gas relations should not be used. 37 COMPRESSIBILITY FACTORA MEASURE OF DEVIATION FROM IDEAL-GAS BEHAVIOR The ideal-gas equation is very simple and thus very convenient to use. However, as illustrated in Fig. 349, gases deviate from ideal-gas behavior significantly at states near the saturation region and the critical point. This deviation from ideal-gas behavior at a given temperature and pressure can accurately be accounted for by the introduction of a correction factor called the compressibility factor Z defined as Z Pv RT (317) Pv ZRT (318) or INTERACTIVE TUTORIAL SEE TUTORIAL CH. 3, SEC. 7 ON THE DVD. cen84959_ch03.qxd 4/1/05 12:31 PM Page 140 140 | 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 349 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 <1 FIGURE 350 The compressibility factor is unity for ideal gases. vactual videal (319) where videal RT/P. Obviously, Z 1 for ideal gases. For real gases Z can be greater than or less than unity (Fig. 350). The farther away Z is from unity, the more the gas deviates from ideal-gas behavior. We have said that gases follow the ideal-gas equation closely at low pressures and high temperatures. But what exactly constitutes low pressure or high temperature? Is 100°C a low temperature? It definitely is for most substances but not for air. Air (or nitrogen) can be treated as an ideal gas at this temperature and atmospheric pressure with an error under 1 percent. This is because nitrogen is well over its critical temperature ( 147°C) and away from the saturation region. At this temperature and pressure, however, most substances would exist in the solid phase. Therefore, the pressure or temperature of a substance is high or low relative to its critical temperature or pressure. cen84959_ch03.qxd 4/1/05 12:31 PM Page 141 Chapter 3 | 141 Gases behave differently at a given temperature and pressure, but they behave very much the same at temperatures and pressures normalized with respect to their critical temperatures and pressures. The normalization is done as PR P Pcr and TR T Tcr (320) Here PR is called the reduced pressure and TR the reduced temperature. The Z factor for all gases is approximately the same at the same reduced pressure and temperature. This is called the principle of corresponding states. In Fig. 351, the experimentally determined Z values are plotted against PR and TR for several gases. The gases seem to obey the principle of corresponding states reasonably well. By curve-fitting all the data, we 1.1 1.0 TR = 2.00 0.9 TR = 1.50 0.8 TR = 1.30 Z= Pv RT 0.7 0.6 TR = 1.20 0.5 TR = 1.10 Legend: 0.4 Methane Iso-pentane Ethylene n-Heptane Ethane Nitrogen Propane Carbon dioxide n-Butane Water Average curve based on data on hydrocarbons TR = 1.00 0.3 0.2 0.1 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 Reduced pressure PR FIGURE 351 Comparison of Z factors for various gases. Source: Gour-Jen Su, Modified Law of Corresponding States, Ind. Eng. Chem. (international ed.) 38 (1946), p. 803. 5.5 6.0 6.5 7.0 cen84959_ch03.qxd 4/1/05 12:31 PM Page 142 142 | Thermodynamics obtain the generalized compressibility chart that can be used for all gases (Fig. A15). The following observations can be made from the generalized compressibility chart: as REAL GAS P 0 IDEAL GAS 1. At very low pressures (PR 1), gases behave as an ideal gas regardless of temperature (Fig. 352), 2. At high temperatures (TR 2), ideal-gas behavior can be assumed with 1). good accuracy regardless of pressure (except when PR 3. The deviation of a gas from ideal-gas behavior is greatest in the vicinity of the critical point (Fig. 353). FIGURE 352 At very low pressures, all gases approach ideal-gas behavior (regardless of their temperature). T Nonideal-gas behavior Ideal-gas behavior EXAMPLE 311 The Use of Generalized Charts Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the generalized compressibility chart. Compare the values obtained to the actual value of 0.021796 m3/kg and determine the error involved in each case. Solution The specific volume of refrigerant-134a is to be determined Ideal-gas behavior assuming ideal- and nonideal-gas behavior. Analysis The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are determined from Table A1 to be v FIGURE 353 Gases deviate from the ideal-gas behavior the most in the neighborhood of the critical point. R 0.0815 kPa # m3> kg # K Pcr 4.059 MPa Tcr 374.2 K (a) The specific volume of refrigerant-134a under the ideal-gas assumption is 1 0.0815 kPa # m3> kg # K 2 1 323 K 2 RT P v 1000 kPa 0.026325 m3> kg Therefore, treating the refrigerant-134a 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: PR TR P Pcr T Tcr 1 MPa 0.246 4.059 MPa 323 K 0.863 374.2 K Z 0.84 Thus v Zvideal 1 0.84 2 1 0.026325 m3> kg 2 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. cen84959_ch03.qxd 4/1/05 12:31 PM Page 143 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 pseudo-reduced specific volume vR as vR vactual RTcr> Pcr (321) vR = 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 time-consuming iterations (Fig. 354). EXAMPLE 312 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 ideal-gas equation, and (c) the generalized compressibility chart. v RTcr / P cr | Z = (Fig. A15) FIGURE 354 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. 355. The gas constant, the critical pressure, and the critical temperature of steam are determined from Table A1E to be R 0.5956 psia # ft3> lbm # R Pcr 3200 psia Tcr 1164.8 R H 2O (a) The pressure at the specified state is determined from Table A6E to be v T 0.51431 ft > lbm f 600°F 3 P 1000 psia This is the experimentally determined value, and thus it is the most accurate. (b) The pressure of steam under the ideal-gas assumption is determined from the ideal-gas relation to be P RT v 1 0.5956 psia # ft3> lbm # R 2 1 1060 R 2 0.51431 ft3> lbm 1228 psia 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. A15), we first need to calculate the pseudo-reduced specific volume and the reduced temperature: vR TR 1 0.51431 ft3> lbm 2 1 3200 psia 2 vactual RTcr> Pcr 1 0.5956 psia # ft3> lbm # R 2 1 1164.8 R 2 T 1060 R 0.91 Tcr 1164.8 R 2.372 PR 0.33 143 T = 600°F v = 0.51431 ft3/lbm P=? FIGURE 355 Schematic for Example 312. cen84959_ch03.qxd 4/25/05 2:47 PM Page 144 144 | Thermodynamics Thus, P P, psia Exact Z chart Ideal gas 1000 1056 1228 PRPcr 1 0.33 2 1 3200 psia 2 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. 356). 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 3-12) FIGURE 356 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. 38 OTHER EQUATIONS OF STATE The ideal-gas equation of state is very simple, but its range of applicability is limited. It is desirable to have equations of state that represent the P-v-T 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. 357), but we shall discuss only three: the van der Waals equation because it is one of the earliest, the Beattie-Bridgeman equation of state because it is one of the best known and is reasonably accurate, and the Benedict-Webb-Rubin 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 Redlich-Kwang Beattie-Bridgeman Benedict-Webb-Rubin Strobridge Virial FIGURE 357 Several equations of state have been proposed throughout history. a b 1v v2 b2 RT (322) Van der Waals intended to improve the ideal-gas equation of state by including two of the effects not considered in the ideal-gas 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 one-thousandth of the volume of the room. As the pressure increases, the volume occupied by the molecules becomes an increasingly significant part of the total volume. Van der Waals proposed to correct this by replacing v in the ideal-gas 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 P-v diagram has a horizontal inflection point at the critical point (Fig. 358). Thus, the first and the second derivatives of P with respect to v at the critical point must be zero. That is, a 0P b 0v T 0 Tcr const and a 0 2P b 0 v2 T 0 Tcr const cen84959_ch03.qxd 4/1/05 12:31 PM Page 145 Chapter 3 By performing the differentiations and eliminating vcr, the constants a and b are determined to be 27R 2 T 2 cr 64Pcr a and b RTcr 8Pcr Beattie-Bridgeman Equation of State The Beattie-Bridgeman equation, proposed in 1928, is an equation of state based on five experimentally determined constants. It is expressed as R uT v2 a1 c b 1v v T3 B2 A v2 (324) where A A0 a 1 a b v and B0 a 1 B b b v (325) The constants appearing in the above equation are given in Table 34 for various substances. The Beattie-Bridgeman 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. Benedict-Webb-Rubin Equation of State Benedict, Webb, and Rubin extended the Beattie-Bridgeman equation in 1940 by raising the number of constants to eight. It is expressed as P R uT v a B0R uT A0 C0 T 2 b bR uT 1 v2 v a 3 aa v6 c a1 v T2 3 g v 2 be g >v 2 (326) The values of the constants appearing in this equation are given in Table 34. This equation can handle substances at densities up to about 2.5rcr. In 1962, Strobridge further extended this equation by raising the number of constants to 16 (Fig. 359). Virial Equation of State The equation of state of a substance can also be expressed in a series form as P RT v a 1T 2 v 2 b 1T 2 v 3 c 1T 2 v 4 145 P (323) The constants a and b can be determined for any substance from the criticalpoint data alone (Table A1). The accuracy of the van der Waals equation of state is often inadequate, but it can be improved by using values of a and b that are based on the actual behavior of the gas over a wider range instead of a single point. Despite its limitations, the van der Waals equation of state has a historical value in that it was one of the first attempts to model the behavior of real gases. The van der Waals equation of state can also be expressed on a unitmole basis by replacing the v in Eq. 322 by v and the R in Eqs. 322 and 323 by Ru. P | d 1T 2 v5 ... (327) Critical point T cr =c on sta nt v FIGURE 358 Critical isotherm of a pure substance has an inflection point at the critical state. cen84959_ch03.qxd 4/19/05 9:40 AM Page 146 146 | Thermodynamics TABLE 34 Constants that appear in the Beattie-Bridgeman and the Benedict-Webb-Rubin 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 b 0.04611 0.03931 0.10476 0.01400 0.02096 0.05046 0.04624 c 104 104 105 40 504 4.20 104 4.80 104 0.001101 0.0 0.07235 0.0 0.04359 0.00691 0.004208 4.34 5.99 6.60 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 Webb-Rubin equation are as follows: Gas a n-Butane, 190.68 C4H10 Carbon dioxide, CO2 13.86 Carbon monoxide, CO 3.71 Methane, CH4 5.00 Nitrogen, N2 2.54 A0 8.314 kPa · m3/kmol · K, the eight constants in the Benedict- b B0 0.039998 0.12436 3.205 107 1.006 108 1.101 10 3 0.0340 277.30 0.007210 0.04991 1.511 106 1.404 107 8.470 10 5 0.00539 135.87 187.91 106.73 0.002632 0.003380 0.002328 0.05454 0.04260 0.04074 1.054 2.578 7.379 105 105 104 8.673 2.286 8.164 105 1.350 106 1.244 105 1.272 10 10 10 4 0.0060 0.0060 0.0053 1021.6 c C0 a g 4 4 Source: Kenneth Wark, Thermodynamics, 4th ed. (New York: McGraw-Hill, 1983), p. 815, table A-21M. 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. Beattie-Bridgeman: 5 constants. Accurate for ρ < 0.8rcr. Benedict-Webb-Rubin: 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 359 Complex equations of state represent the P-v-T 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 ideal-gas equation of state. The P-v-T behavior of a substance can be represented accurately with the virial equation of state over a wider range by including a sufficient number of terms. The equations of state discussed here are applicable to the gas phase of the substances only, and thus should not be used for liquids or liquidvapor mixtures. Complex equations represent the P-v-T 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 specific-volume calculations since all the earlier equations are implicit in v and require a trial-and-error approach. The accuracy of the van der Waals, cen84959_ch03.qxd 4/1/05 12:31 PM Page 147 Chapter 3 | 147 3.7% 0.1% 0.4% 2.9% 0.3% 0.7% 200 5.3% 0.1% 0.1% 6.7% 0.7% 0.1% 11.6% 6.3% 1.2% a 4.2% 0.1% 0.2% 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 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) Beattie-Bridgeman (middle) Benedict-Webb-Rubin (bottom) 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% 0.0% MP a 20.7% 14.1% 2.1% 4 MPa 10 MPa 20 MPa 4.7% 0.2% 0.2% 300 0.2 MP T, K >100% >100% >100% 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 Beattie-Bridgeman, and Benedict-Webb-Rubin equations of state is illustrated in Fig. 360. It is apparent from this figure that the Benedict-WebbRubin equation of state is usually the most accurate. 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 ideal-gas equation of state, (b) the van der Waals equation of state, (c) the Beattie-Bridgeman equation of state, and (d) the Benedict-Webb-Rubin 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. 0.8% 0.4% 0.1% 0.1% 0.8% 0.3% 10 FIGURE 360 Percentage of error involved in various equations of state for nitrogen (% error EXAMPLE 313 125 5.7% 59.3% 18.7% 0.0 100 100 [(|vtable vequation|)/vtable] v, m3/kmol 100). cen84959_ch03.qxd 4/1/05 12:31 PM Page 148 148 | Thermodynamics Properties The gas constant of nitrogen gas is 0.2968 kPa m3/kg K (Table A1). Analysis (a) Using the ideal-gas equation of state, the pressure is found to be P 1 0.2968 kPa # m3> kg # K 2 1 175 K 2 RT v 0.00375 m3> kg 13,851 kPa which is in error by 38.5 percent. (b) The van der Waals constants for nitrogen are determined from Eq. 323 to be 0.175 m6 # kPa> kg2 a 0.00138 m3> kg b From Eq. 322, RT P a v2 b v 9471 kPa which is in error by 5.3 percent. (c) The constants in the Beattie-Bridgeman equation are determined from Table 34 to be A 102.29 B 0.05378 c 4.2 104 Also, v Mv (28.013 kg/mol)(0.00375 m3/kg) stituting these values into Eq. 324, we obtain P RuT v2 a1 c b 1v vT3 B2 A v2 0.10505 m3/kmol. Sub- 10,110 kPa which is in error by 1.1 percent. (d) The constants in the Benedict-Webb-Rubin equation are determined from Table 34 to be A0 a b c 7.379 10 a 1.272 10 4 8.164 g 4 0.04074 C0 0.002328 106.73 B0 2.54 0.0053 105 Substituting these values into Eq. 326 gives P R uT v aa v6 a B0R uT c a1 v T2 3 A0 g v2 C0 T be 2 b 1 v2 bR uT v a 3 g>v 2 10,009 kPa which is in error by only 0.09 percent. Thus, the accuracy of the BenedictWebb-Rubin equation of state is rather impressive in this case. cen84959_ch03.qxd 4/11/05 12:23 PM Page 149 Chapter 3 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. 361). That is, Patm Pa Pv Pv fPsat @ T (329) 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 0.6 1 3.17 kPa 2 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. Patm = Pa + Pv Air Water vapor (328) (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 fPsat @ 25°C 149 Vapor Pressure and Phase Equilibrium TOPIC OF SPECIAL INTEREST* Pv | FIGURE 361 Atmospheric pressure is the sum of the dry air pressure Pa and the vapor pressure Pv. cen84959_ch03.qxd 4/1/05 12:31 PM Page 150 150 | Thermodynamics Water Salty water Salt (a) Before (b) After FIGURE 362 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 363 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. 362). The commodity simply creeps away during redistribution, and thus the flow is a diffusion process. We know from experience that a wet T-shirt 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 T-shirt 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. 363), Phase equilibrium criterion for water exposed to air: Pv Psat @ T (330) Therefore, if the vapor pressure in the air is less than the saturation pressure of water at the water temperature, some liquid will evaporate. The larger the difference between the vapor and saturation pressures, the higher the rate of evaporation. The evaporation has a cooling effect on water, and thus reduces its temperature. This, in turn, reduces the saturation pressure of water and thus the rate of evaporation until some kind of quasi-steady 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 cen84959_ch03.qxd 4/1/05 12:31 PM Page 151 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 liquidvapor 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. 364). Boiling, on the other hand, occurs at the solidliquid interface when a liquid is brought into contact with a surface maintained at a temperature Ts sufficiently above the saturation temperature Tsat of the liquid. At 1 atm, for example, liquid water in contact with a solid surface at 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 364 A liquid-to-vapor phase change process is called evaporation if it occurs at a liquidvapor interface, and boiling if it occurs at a solidliquid interface. 151 cen84959_ch03.qxd 4/11/05 12:23 PM Page 152 152 | Thermodynamics Air 25°C φ = 10% Pv EXAMPLE 314 Pv = Psat @ T T Lake FIGURE 365 Schematic for Example 314. 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. 365). 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 31, is 3.17 kPa. Then the vapor pressures at relative humidities of 10, 80, and 100 percent are determined from Eq. 329 to be Relative humidity Pv1 10%: f1Psat @ 25°C 0.1 1 3.17 kPa 2 0.8 1 3.17 kPa 2 1.0 1 3.17 kPa 2 0.317 kPa Relative humidity Pv 2 80%: f2Psat @ 25°C 2.536 kPa Relative humidity Pv 3 100%: f3Psat @25°C 3.17 kPa The saturation temperatures corresponding to these pressures are determined from Table 31 (or Table A5) 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 well-insulated pan, you can actually freeze the water if the air is very dry and relatively cool. cen84959_ch03.qxd 4/1/05 12:31 PM Page 153 Chapter 3 | 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 phase-change 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 liquidvapor mixture, the mass fraction of vapor is called the quality and is expressed as mvapor x 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 y yf Pv RT where R is the gas constant. Caution should be exercised in using this relation since an ideal gas is a fictitious substance. or vactual videal Z The Z factor is approximately the same for all gases at the same reduced temperature and reduced pressure, which are defined as T Tcr TR and P Pcr PR 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 vactual RTcr> Pcr vR The P-v-T 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 v2 b2 RT and b RTcr 8Pcr a1 c b 1v vT3 B2 B0 a 1 b b v bRuT a where a 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 triple-line 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 best-known equation of state is the ideal-gas equation of state, given as Pv RT Z 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, y Real gases exhibit ideal-gas behavior at relatively low pressures and high temperatures. The deviation from ideal-gas behavior can be properly accounted for by using the compressibility factor Z, defined as 27R 2 T 2 cr 64Pcr Beattie-Bridgeman: P RuT v2 A v2 where A A0 a 1 a b v Benedict-Webb-Rubin: RuT P a B0RuT A0 v c a1 v T2 3 g v 2 be and B C0 1 b T2 v2 v3 aa v6 g>v 2 where Ru is the universal gas constant and v is the molar specific volume. cen84959_ch03.qxd 4/1/05 12:31 PM Page 154 154 | Thermodynamics REFERENCES AND SUGGESTED READINGS 1. ASHRAE Handbook of Fundamentals. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and Air-Conditioning 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, Phase-Change Processes, Property Diagrams 312C How does the boiling process at supercritical pressures differ from the boiling process at subcritical pressures? 31C Is iced water a pure substance? Why? Property Tables 32C What is the difference between saturated liquid and compressed liquid? 313C 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. 33C What is the difference between saturated vapor and superheated vapor? 34C 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? 35C 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? 36C Is it true that water boils at higher temperatures at higher pressures? Explain. 37C If the pressure of a substance is increased during a boiling process, will the temperature also increase or will it remain constant? Why? 38C Why are the temperature and pressure dependent properties in the saturated mixture region? 39C What is the difference between the critical point and the triple point? 310C Is it possible to have water vapor at 10°C? 311C 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 CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 314C 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. 315C 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? 316C 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. 317C 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? 318C Does the reference point selected for the properties of a substance have any effect on thermodynamic analysis? Why? 319C What is the physical significance of hfg? Can it be obtained from a knowledge of hf and hg? How? 320C 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? 321C What is quality? Does it have any meaning in the superheated vapor region? 322C 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? 323C Does hfg change with pressure? How? cen84959_ch03.qxd 4/1/05 12:31 PM Page 155 Chapter 3 324C Can quality be expressed as the ratio of the volume occupied by the vapor phase to the total volume? Explain. 325C In the absence of compressed liquid tables, how is the specific volume of a compressed liquid at a given P and T determined? 326 Complete this table for H2O: 333E P, psia 80 15 10 h, Btu/lbm P, kPa 50 250 110 v, m3/kg Phase description 4.16 334 T, °C 200 400 600 Saturated vapor 327 Reconsider Prob. 326. Using EES (or other) software, determine the missing properties of water. Repeat the solution for refrigerant-134a, refrigerant22, and ammonia. 335 T, °C 328E 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 329E Reconsider Prob. 328E. Using EES (or other) software, determine the missing properties of water. Repeat the solution for refrigerant-134a, refrigerant22, and ammonia. 330 T, °C P, kPa h, kJ/kg 200 x Phase description 0.7 950 500 800 T, °C T, °C P, kPa v, Phase description 320 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 336 A 1.8-m3 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 337 A pistoncylinder 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 refrigerant-134a 0.015 180 600 Saturated vapor Complete this table for refrigerant-134a: P, kPa 20 12 8 Phase description FIGURE P336 3162.2 m3/kg 332 v, m3/kg 0.0 Complete this table for refrigerant-134a: 80 P, kPa 1800 331 8 30 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 x 78 110 T, °C 155 Complete this table for refrigerant-134a: T, °F | u, kJ/kg Phase description 95 Q R-134a 0.85 kg 10°C Saturated liquid 400 600 300 FIGURE P337 cen84959_ch03.qxd 4/1/05 12:31 PM Page 156 156 | Thermodynamics 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 refrigerant-134a. 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water. 338E 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? 342 Repeat Prob. 341 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. 343 Water is boiled at 1 atm pressure in a 25-cm-internaldiameter 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. 344 Repeat Prob. 343 for a location at 2000-m elevation where the standard atmospheric pressure is 79.5 kPa. Pressure cooker 250°F 345 Saturated steam coming off the turbine of a steam power plant at 30°C condenses on the outside of a 3-cmouter-diameter, 35-m-long 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 P338E 339E 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. 340 A person cooks a meal in a 30-cm-diameter pot that is covered with a well-fitting 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. 346 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. 347 Water in a 5-cm-deep pan is observed to boil at 98°C. At what temperature will the water in a 40-cm-deep pan boil? Assume both pans are full of water. 348 A cooking pan whose inner diameter is 20 cm is filled with water and covered with a 4-kg 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 341 Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3kW electric burner. If H2O Vapor FIGURE P348 349 Reconsider Prob. 348. 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 P341 40% 350 Water is being heated in a vertical pistoncylinder device. The piston has a mass of 20 kg and a cross-sectional area of 100 cm2. If the local atmospheric pressure is 100 kPa, determine the temperature at which the water starts boiling. cen84959_ch03.qxd 4/1/05 12:31 PM Page 157 Chapter 3 351 A rigid tank with a volume of 2.5 m3 contains 15 kg of saturated liquidvapor 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 T-v diagram with respect to saturation lines. Answer: 187.0°C 352 A rigid vessel contains 2 kg of refrigerant-134a at 800 kPa and 120°C. Determine the volume of the vessel and the total internal energy. Answers: 0.0753 m3, 655.7 kJ 353E A 5-ft3 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. 354 A 0.5-m3 vessel contains 10 kg of refrigerant-134a 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 pistoncylinder 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. 355 (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 P-v diagram with respect to saturation lines. | 157 358E Reconsider Prob. 357E. 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. 357E on a T-v diagram using the property plot feature of EES. 359 A pistoncylinder 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 T-v diagram with respect to saturation lines. Answers: (a) 49.61 kg, (b) 120.21°C, (c) 125,943 kJ 360 A 0.3-m3 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 361 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. 362 H 2O P = 800 kPa Reconsider Prob. 361. Using EES (or other) software, determine the indicated properties of compressed liquid, and compare them to those obtained using the saturated liquid approximation. 363E A 15-ft3 rigid tank contains a saturated mixture of refrigerant-134a 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. 364 A pistoncylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled at constant pressure until one-half of the mass condenses. FIGURE P355 356 Reconsider Prob. 355. 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. 355 on a P-v diagram using the property plot feature of EES. 357E 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 T-v 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 T-v diagram. (b) Find the final temperature. (c) Determine the volume change. 365 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 366 Water is boiled in a pan covered with a poorly fitting lid at a specified location. Heat is supplied to the pan by a 2-kW 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 cen84959_ch03.qxd 4/1/05 12:31 PM Page 158 158 | Thermodynamics 367 A rigid tank initially contains 1.4-kg 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 374 Reconsider Prob. 373. 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. 375 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 P367 368 A pistoncylinder 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 P375 Q FIGURE P368 376E 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 369C 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. 377 The pressure gage on a 2.5-m3 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. 370C Under what conditions is the ideal-gas assumption suitable for real gases? 371C What is the difference between R and Ru? How are these two related? 372C What is the difference between mass and molar mass? How are these two related? 373 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 P = 500 kPa g O2 V = 2.5 m 3 T = 28˚C FIGURE P377 cen84959_ch03.qxd 4/1/05 12:31 PM Page 159 Chapter 3 | 159 378E 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 389E Refrigerant-134a at 400 psia has a specific volume of 0.13853 ft3/lbm. Determine the temperature of the refrigerant based on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the refrigerant tables. 379 A 400-L rigid tank contains 5 kg of air at 25°C. Determine the reading on the pressure gage if the atmospheric pressure is 97 kPa. 390 A 0.016773-m3 tank contains 1 kg of refrigerant-134a at 110°C. Determine the pressure of the refrigerant, using (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the refrigerant tables. Answers: (a) 1.861 MPa, 380 A 1-m3 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 391 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 392 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 381C What is the physical significance of the compressibility factor Z? 393 What is the percentage of error involved in treating carbon dioxide at 7 MPa and 380 K as an ideal gas? 382C What is the principle of corresponding states? 394 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 ideal-gas equation and (b) the generalized compressibility chart. Also, determine (c) the error involved in each case. 383C How are the reduced pressure and reduced temperature defined? 384 Determine the specific volume of superheated water vapor at 10 MPa and 400°C, using (a) the ideal-gas 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. 384. 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 ideal-gas approximation against temperature, and discuss the results. 3 MPa 500 K 2 kg/s 385 386 Determine the specific volume of refrigerant-134a vapor at 0.9 MPa and 70°C based on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) data from tables. Also, determine the error involved in the first two cases. 387 Determine the specific volume of nitrogen gas at 10 MPa and 150 K based on (a) the ideal-gas 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 388 Determine the specific volume of superheated water vapor at 3.5 MPa and 450°C based on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the steam tables. Determine the error involved in the first two cases. CO2 450 K FIGURE P394 Other Equations of State 395C 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? 396 A 3.27-m3 tank contains 100 kg of nitrogen at 175 K. Determine the pressure in the tank, using (a) the ideal-gas equation, (b) the van der Waals equation, and (c) the BeattieBridgeman equation. Compare your results with the actual value of 1505 kPa. 397 A 1-m3 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 398 Reconsider Prob. 397. 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 cen84959_ch03.qxd 4/1/05 12:31 PM Page 160 160 | Thermodynamics 1 MPa in 0.1 MPa increments. Plot the percent error involved in the ideal-gas approximation against pressure, and discuss the results. 399E Refrigerant-134a at 100 psia has a specific volume of 0.54022 ft3/lbm. Determine the temperature of the refrigerant based on (a) the ideal-gas equation, (b) the van der Waals equation, and (c) the refrigerant tables. 3100 Nitrogen at 150 K has a specific volume of 0.041884 m3/kg. Determine the pressure of the nitrogen, using (a) the ideal-gas 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 3108 The combustion in a gasoline engine may be approximated by a constant volume heat addition process. There exists the airfuel 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 3101 Reconsider Prob. 3100. Using EES (or other) software, compare the pressure results of the ideal-gas and Beattie-Bridgeman 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 3102 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. 3103 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? 3104 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 P3108 3109 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 3105 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. 3106E A thermos bottle is half-filled 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. 3107 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 P3109 3110 Carbon-dioxide 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 P3110 450 K cen84959_ch03.qxd 4/1/05 12:31 PM Page 161 Chapter 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. 3111 A pistoncylinder 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 P3111 3112 Combustion in a diesel engine may be modeled as a constant-pressure 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 airfuel 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 T-v diagram sketch the constant specific volume process through the state T 120°C, v 0.7163 m3/kg 100 kPa to P2 300 kPa. For this data set place from P1 the temperature values at states 1 and 2 on its axis. Place the value of the specific volume on its axis. 3114 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. 3115 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 hot-air 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, hot-air 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. Hot-air 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 P3112 3113 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 P-v 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 P-v diagram. | FIGURE P3115 © Vol. 1/PhotoDisc cen84959_ch03.qxd 4/1/05 12:31 PM Page 162 162 | 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 FB H 2O V= 4L m = 2 kg T = 50°C 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 20-m-diameter hot-air 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 65-kg people. Determine the average temperature of the air in the balloon. What would your response be if the atmospheric air temperature were 30°C? 3116 Reconsider Prob. 3115. 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. 3117 Consider an 18-m-diameter hot-air balloon that, together with its cage, has a mass of 120 kg when empty. The air in the balloon, which is now carrying two 70-kg 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? 3118E Water in a pressure cooker is observed to boil at 260°F. What is the absolute pressure in the pressure cooker, in psia? 3119 A rigid tank with a volume of 0.117 m3 contains 1 kg of refrigerant-134a 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 P-v diagram with respect to saturation lines. 3120 A 4-L rigid tank contains 2 kg of saturated liquidvapor 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 P3120 3121 A 10-kg mass of superheated refrigerant-134a 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 T-v 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 3122 A 0.5-m3 rigid tank containing hydrogen at 20°C and 600 kPa is connected by a valve to another 0.5-m3 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 P3122 3123 Reconsider Prob. 3122. 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. 3124 A 20-m3 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 3125 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 3126 A tank whose volume is unknown is divided into two parts by a partition. One side of the tank contains 0.01 m3 cen84959_ch03.qxd 4/1/05 12:31 PM Page 163 Chapter 3 of refrigerant-134a 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. R-134a V = 0.01 m3 P = 0.8 MPa Evacuated Reconsider Prob. 3126. Using EES (or other) software, investigate the effect of the initial pressure of refrigerant-134a 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. 3132 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. T, °C 200 270.3 30 130 400 3127 3128 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. Leak Propane FIGURE P3128 3129 Repeat Prob. 3128 for isobutane. 3130 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 313l 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. P, kPa FIGURE P3126 | 300 500 v, m3/kg u, kJ/kg Condition description and quality (if applicable) 1.5493 0.500 3084 3133 Complete the blank cells in the following table of properties of refrigerant-134a. In the last column describe the condition of refrigerant-134a as compressed liquid, saturated mixture, superheated vapor, or insufficient information; and, if applicable, give the quality. P, kPa 320 1000 180 200 T, °C v, m3/kg u, kJ/kg Condition description and quality (if applicable) 12 39.37 40 0.17794 0.0700 249 3134 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 refrigerant-134a. Use arrows to indicate the direction of the process, and label the initial and final states: (a) On the P-v diagram sketch the constant temperature process through the state P 280 kPa, v 0.06 m3/kg as 400 kPa to P2 200 kPa. pressure changes from P1 Place the value of the temperature on the process curve on the P-v diagram. (b) On the T-v diagram sketch the constant specific volume process through the state T 20°C, v 0.02 m3/kg from 1200 kPa to P2 300 kPa. For this data set place the P1 temperature values at states 1 and 2 on its axis. Place the value of the specific volume on its axis. cen84959_ch03.qxd 4/20/05 5:07 PM Page 164 164 | Thermodynamics Fundamentals of Engineering (FE) Exam Problems 3135 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 3136 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 3137 A 300-m3 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 3138 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type 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 3139 A 1-m3 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 3140 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 3141 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 3142 A 3-m3 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 3143 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 3144 A solid normally absorbs heat as it melts, but there is a known exception at temperatures close to absolute zero. Find out which solid it is and give a physical explanation for it. 3145 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. 3146 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 phase-change is the enthalpy of fusion. Obtain the enthalpy of fusion for water using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. cen84959_ch04.qxd 4/20/05 5:10 PM Page 165 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. | 165 cen84959_ch04.qxd 4/25/05 3:38 PM Page 166 166 | Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 1 ON THE DVD. The moving boundary GAS FIGURE 41 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 pistoncylinder device. During this process, part of the boundary (the inner face of the piston) moves back and forth. Therefore, the expansion and compression work is often called moving boundary work, or simply boundary work (Fig. 41). Some call it the P dV work for reasons explained later. Moving boundary work is the primary form of work involved in automobile engines. During their 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 quasi-equilibrium 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 quasi-equilibrium processes are used in place of nonquasi-equilibrium processes. Below, the work associated with a moving boundary is evaluated for a quasi-equilibrium process. Consider the gas enclosed in the pistoncylinder device shown in Fig. 42. The initial pressure of the gas is P, the total volume is V, and the crosssectional area of the piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium manner, the differential work done during this process is dWb A ds P GAS FIGURE 42 A gas does a differential amount of work dWb as it forces the piston to move by a differential amount ds. F ds PA ds P dV (41) That is, the boundary work in the differential form is equal to the product of the absolute pressure P and the differential change in the volume dV of the system. This expression also explains why the moving boundary work is sometimes called the P dV work. Note in Eq. 41 that P is the absolute pressure, which is always positive. However, the volume change dV is positive during an expansion process (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. 41 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: 2 Wb P dV 1 1 kJ 2 (42) cen84959_ch04.qxd 4/20/05 5:10 PM Page 167 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 P-V diagram. The quasi-equilibrium expansion process described is shown on a P-V 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 12 is obtained by adding these differential areas: 2 Area A dA P dV 1 Process path 2 dA = P dV V1 V (43) A comparison of this equation with Eq. 42 reveals that the area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. (On the P-v 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 work-producing devices. The work produced by these devices during one part of the cycle would have to be consumed during another part, and there would be no net work output. The cycle shown in Fig. 45 produces a net work output because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B), and the difference between these two is the net work done during the cycle (the colored area). If the relationship between P and V during an expansion or a compression process is given in terms of experimental data instead of in a functional form, obviously we cannot perform the integration analytically. But we can always plot the P-V diagram of the process, using these data points, and calculate the area underneath graphically to determine the work done. Strictly speaking, the pressure P in Eq. 42 is the pressure at the inner surface of the piston. It becomes equal to the pressure of the gas in the cylinder only if the process is quasi-equilibrium and thus the entire gas in the cylinder is at the same pressure at any given time. Equation 42 can also be used for nonquasi-equilibrium 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 nonquasi-equilibrium process since properties are defined for equilibrium states only.) Therefore, we can generalize the boundary work relation by expressing it as 2 Pi dV V2 dV 1 Wb 167 P 2 1 | (44) 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 43 The area under the process curve on a P-V diagram represents the boundary work. P WA = 10 kJ 1 WB = 8 kJ WC = 5 kJ A B C 2 V1 V2 V FIGURE 44 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 45 The net work done during a cycle is the difference between the work done by the system and the work done on the system. cen84959_ch04.qxd 4/20/05 5:10 PM Page 168 168 | 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, 2 Wb Wfriction Watm Wcrank 1 1 Ffriction Patm A Fcrank 2 dx (45) Of course the work used to overcome friction appears as frictional heat and the energy transmitted through the crankshaft is transmitted to other 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 quasi-equilibrium processes of gases only. It can also be used for solids and liquids. EXAMPLE 41 Boundary Work for a Constant-Volume 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 P-V diagram of the process are shown in Fig. 46. The boundary work can be determined from Eq. 42 to be 2 Wb 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 constant-volume process is always zero. This is also evident from the P-V diagram of the process (the area under the process curve is zero). P, kPa AIR P1 = 500 kPa T1 = 150°C FIGURE 46 Schematic and P-V diagram for Example 41. P2 = 400 kPa T2 = 65°C Heat 500 1 400 2 V cen84959_ch04.qxd 4/20/05 5:10 PM Page 169 Chapter 4 EXAMPLE 42 | Boundary Work for a Constant-Pressure Process A frictionless pistoncylinder device contains 10 lbm of steam at 60 psia and 320 F. Heat is now transferred to the steam until the temperature reaches 400 F. 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 P-v diagram of the process are shown in Fig. 47. Assumption The expansion process is quasi-equilibrium. 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 constant-pressure process, and, from Eq. 42 2 Wb 2 P dV P0 1 dV 1 or mP0 1 v2 Wb P0 1 V2 V1 2 (46) v1 2 since V mv. From the superheated vapor table (Table A6E), the specific volumes are determined to be v1 7.4863 ft3/lbm at state 1 (60 psia, 320 F) and v2 8.3548 ft3/lbm at state 2 (60 psia, 400 F). Substituting these values yields Wb 1 10 lbm 2 1 60 psia 2 3 1 8.3548 7.4863 2 ft3> lbm 4 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 P-V 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 47 Schematic and P-v diagram for Example 42. 169 cen84959_ch04.qxd 4/20/05 5:10 PM Page 170 170 | Thermodynamics EXAMPLE 43 Isothermal Compression of an Ideal Gas A pistoncylinder 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 pistoncylinder device is compressed isothermally. The boundary work done is to be determined. Analysis A sketch of the system and the P-V diagram of the process are shown in Fig. 48. Assumptions 1 The compression process is quasi-equilibrium. 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 critical-point values. Analysis For an ideal gas at constant temperature T0, PV mRT0 C or C V P where C is a constant. Substituting this into Eq. 42, we have 2 Wb 2 P dV 1 1 C dV V 2 C 1 dV V C ln V2 V1 P1V1 ln V2 V1 (47) In Eq. 47, 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. 47 yields Wb 1 100 kPa 2 1 0.4 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 48 Schematic and P-V diagram for Example 43. 0.4 V, m3 cen84959_ch04.qxd 4/25/05 2:48 PM Page 171 Chapter 4 | 171 P n n P1V1 = P2V 2 1 P1 PV n = const. GAS P2 PV n = C = const. 2 V1 V2 FIGURE 49 Schematic and P-V diagram for a polytropic process. V Polytropic Process During actual expansion and compression processes of gases, pressure and C, where n and C are constants. A volume are often related by PV n process of this kind is called a polytropic process (Fig. 49). 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 n CV EXPERIMENT (48) Substituting this relation into Eq. 42, we obtain 2 Wb 2 P dV CV 1 n dV C V2 n 1 1 V1 n n1 since C P1V1n P2V2n. For an ideal gas (PV also be written as mR 1 T2 Wb 1 For the special case of n n T1 2 2 P dV 1 1 P2V2 1 P1V1 n (49) mRT), this equation can 1 kJ 2 (410) 1 the boundary work becomes 2 Wb n 1 CV 1 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 44 Expansion of a Gas against a Spring A pistoncylinder 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 cross-sectional 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 end-of-chapter problem 4174. © Ronald Mullisen cen84959_ch04.qxd 4/20/05 5:10 PM Page 172 172 | Thermodynamics the gas, and (c) the fraction of this work done against the spring to compress it. Solution A gas in a pistoncylinder device equipped with a linear spring expands as a result of heating. The final gas pressure, the total work done, and the fraction of the work done to compress the spring are to be determined. Assumptions 1 The expansion process is quasi-equilibrium. 2 The spring is linear in the range of interest. Analysis A sketch of the system and the P-V diagram of the process are shown in Fig. 410. (a) The enclosed volume at the final state is 1 2 2 1 0.05 m3 2 2V1 V2 0.1 m3 Then the displacement of the piston (and of the spring) becomes 1 0.1 ¢V A x 0.05 2 m3 0.2 m 0.25 m2 The force applied by the linear spring at the final state is F 1 150 kN> m 2 1 0.2 m 2 kx 30 kN The additional pressure applied by the spring on the gas at this state is F A P 30 kN 0.25 m2 120 kPa Without the spring, the pressure of the gas would remain constant at 200 kPa while the piston is rising. But under the effect of the spring, the pressure rises linearly from 200 kPa to 200 120 320 kPa at the final state. (b) An easy way of finding the work done is to plot the process on a P-V diagram and find the area under the process curve. From Fig. 410 the area under the process curve (a trapezoid) is determined to be W 1 200 area 320 2 kPa 2 3 1 0.1 0.05 2 m3 4 a 1 kJ b 1 kPa # m3 k = 150 kN/m P, kPa 320 II 200 FIGURE 410 Schematic and P-V diagram for Example 44. A = 0.25 m2 P1 = 200 kPa V1 = 0.05 m3 I 0.05 Heat 0.1 V, m3 13 kJ cen84959_ch04.qxd 4/25/05 3:35 PM Page 173 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, 1 23 Wspring Discussion 4 2 200 2 kPa 4 1 0.05 m3 2 a 1 kJ b 1 kPa # m3 3 kJ This result could also be obtained from 1 2 2 k 1 x2 Wspring 1 320 x2 2 1 1 2 1 150 kN> m 2 3 1 0.2 m 2 2 02 4 a 1 kJ b 1 kN # m 3 kJ ENERGY BALANCE FOR CLOSED SYSTEMS Energy balance for any system undergoing any kind of process was expressed as (see Chap. 2) dE system> dt . E out Rate of net energy transfer by heat, work, and mass (411) 1 kW 2 SEE TUTORIAL CH. 4, SEC. 2 ON THE DVD. (412) Change in internal, kinetic, potential, etc., energies or, in the rate form, as . E in 1 kJ 2 ¢ E system E out E in Net energy transfer by heat, work, and mass INTERACTIVE TUTORIAL 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 1 dE> dt 2 ¢ t ¢E 1 kJ 2 (413) The energy balance can be expressed on a per unit mass basis as ein eout 1 kJ> kg 2 ¢ esystem (414) which is obtained by dividing all the quantities in Eq. 411 by the mass m of the system. Energy balance can also be expressed in the differential form as dE in dE out dE system or dein deout desystem P (415) 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 1 for a cycle 2 Qnet = Wnet V (416) That is, the net work output during a cycle is equal to net heat input (Fig. 411). FIGURE 411 For a cycle E 0, thus Q W. cen84959_ch04.qxd 4/25/05 2:48 PM Page 174 174 | Thermodynamics General Q W = E Stationary systems Q W = U Per unit mass q w = e Differential form δq δw = de FIGURE 412 Various forms of the first-law relation for closed systems. EXPERIMENT The energy balance (or the first-law) 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 (417) 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 first-law relation for closed systems are given in Fig. 412. The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, and this should be taken as sufficient proof. Note that if it were possible to prove the first law on the basis of other physical principles, the first law then would be a consequence of those 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 first-law 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 first-law point of view, heat and work are not different at all. From the second-law point of view, however, heat and work are very different, as is discussed in later chapters. EXAMPLE 45 Electric Heating of a Gas at Constant Pressure A pistoncylinder device contains 25 g of saturated water vapor that is maintained at a constant pressure of 300 kPa. A resistance heater within the cylinder is turned on and passes a current of 0.2 A for 5 min from a 120-V 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 first-law 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 pistoncylinder device expands at con- Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4175. © 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. cen84959_ch04.qxd 4/25/05 2:48 PM Page 175 Chapter 4 | 175 P, kPa H 2O m = 25 g P1 = P2 = 300 kPa 2 1 300 0.2 A 120 V EXPERIMENT Sat. vapor 5 min v Qout = 3.7 kJ FIGURE 413 Schematic and P-v diagram for Example 45. Analysis We take the contents of the cylinder, including the resistance wires, as the system (Fig. 413). This is a closed system since no mass crosses the system boundary during the process. We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work Wb. The pressure remains constant during the process and thus P2 P1. Also, heat is lost from the system and electrical work We is done on the system. (a) This part of the solution involves a general analysis for a closed system undergoing a quasi-equilibrium constant-pressure 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 Q Wother © Ronald Mullisen ¢ E system Net energy transfer by heat, work, and mass Q Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4176. Change in internal, kinetic, potential, etc., energies W ¢U Wb U2 EXPERIMENT 0 0 ¢ KE ¢ PE ¡ ¡ U1 For a constant-pressure process, the boundary work is given as Wb P0(V2 V1). Substituting this into the preceding relation gives Q Wother P0 1 V2 V1 2 U2 Wother 1 U2 P2V2 2 However, P0 Also H P2 U P1 S Q PV, and thus Q Wother H2 H1 1 kJ 2 U1 1 U1 P1V1 2 (418) which is the desired relation (Fig. 414). This equation is very convenient to use in the analysis of closed systems undergoing a constant-pressure 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 end-of-chapter problem 4177. © Ronald Mullisen cen84959_ch04.qxd 4/25/05 2:48 PM Page 176 176 | Thermodynamics EXPERIMENT (b) The only other form of work in this case is the electrical work, which can be determined from 1 120 V 2 1 0.2 A 2 1 300 s 2 a VI ¢ t We P1 300 kPa f sat. vapor State 1: h1 1 kJ> s 1000 VA b 2724.9 kJ> kg h g @ 300 kPa 7.2 kJ 1 Table A5 2 The enthalpy at the final state can be determined directly from Eq. 418 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 constant-pressure expansion or compression process: E out ¢ E system Net energy transfer by heat, work, and mass We,in © Ronald Mullisen Q out We,in 7.2 kJ E in Use actual data from the experiment shown here to verify the first law of thermodynamics. See end-of-chapter problem 4178. Change in internal, kinetic, potential, etc., energies Wb ¢U Q out ¢H 3.7 kJ h2 m 1 h2 1 0.025 kg 2 1 h 2 2864.9 kJ> kg h1 2 1 since P 2724.9 2 kJ> kg constant 2 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 h2 300 kPa f 2864.9 kJ> kg T2 200°C 1 Table A6 2 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 first-law relation. Therefore, in problems of this kind, the potential energy term is always neglected. Q W other = H FIGURE 414 For a closed system undergoing a quasi-equilibrium, P constant process, U Wb H. EXAMPLE 46 Unrestrained Expansion of Water A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains 5 kg of water at 200 kPa and 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 177 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. 415). This is a closed system since no mass crosses the system boundary during the process. We observe that the water fills the entire tank when the partition is removed (possibly as a liquidvapor mixture). (a) Initially the water in the tank exists as a compressed liquid since its pressure (200 kPa) is greater than the saturation pressure at 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 1 Table A4 2 Then the initial volume of the water is V1 mv1 1 5 kg 2 1 0.001 m3> kg 2 0.005 m3 The total volume of the tank is twice this amount: 1 2 2 1 0.005 m3 2 Vtank 0.01 m3 (b) At the final state, the specific volume of the water is v2 V2 m 0.01 m3 5 kg 0.002 m3> 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: 0.001003 m3> kg vf and 43.340 m3> kg vg 1 Table A4 2 Since vf v2 vg, the water is a saturated liquidvapor mixture at the final state, and thus the pressure is the saturation pressure at 25°C: P2 Psat @ 25°C 1 Table A4 2 3.1698 kPa 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 415 Schematic and P-v diagram for Example 46. | 177 cen84959_ch04.qxd 4/25/05 2:48 PM Page 178 178 | 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 ¢ E system H 2O Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies Heat FIGURE 416 Expansion against a vacuum involves no work and thus no energy transfer. m 1 u2 ¢U Qin u1 2 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. 416). Then W 0 since the system does not involve any other forms of work. (Can you reach the same conclusion by choosing the water as our system?) Initially, u1 104.83 kJ> kg uf @ 25°C 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 m = 1 kg T = 1°C Specific heat = 5 kJ/kg ·°C 5 kJ FIGURE 418 Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 3 ON THE DVD. vfg 0.002 43.34 0.001 0.001 2.3 10 5 Then u2 uf x2ufg 104.83 kJ> kg 41.8 kJ FIGURE 417 It takes different amounts of energy to raise the temperature of different substances by the same amount. vf v2 x2 104.88 kJ> kg Substituting yields Qin 1 5 kg 2 3 1 104.88 1 2.3 10 5 2 1 2304.3 kJ> kg 2 104.83 2 kJkg 4 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. 417). 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. 418). In general, this energy depends on how the process is executed. In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume cv and specific heat at constant pressure cp. Physically, the specific heat at constant volume cv can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. The energy required to cen84959_ch04.qxd 4/20/05 5:10 PM Page 179 Chapter 4 do the same as the pressure is maintained constant is the specific heat at constant pressure cp. This is illustrated in Fig. 419. The specific heat at constant 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 constant-volume process (and thus no expansion or compression eout esystem work is involved). The conservation of energy principle ein 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 left-hand 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 419 Constant-volume and constantpressure specific heats cv and cp (values given are for helium gas). or cv a 0u b 0T v (419) Similarly, an expression for the specific heat at constant pressure cp can be obtained by considering a constant-pressure expansion or compression process. It yields cp a 0h b 0T p (420) 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. 419 and 420. First, these equations are property relations and as such are independent of the type of processes. They are valid for any substance undergoing any process. The only relevance cv has to a constant-volume process is that cv happens to be the energy transferred to a system during a constant-volume 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 constant-pressure 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. 419 and 420 is that cv is related to the changes in internal energy and cp to the changes in enthalpy. In fact, it would be more proper to define cv as the change in the internal energy of a substance per unit change in temperature at constant (( 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 420 Formal definitions of cv and cp. cen84959_ch04.qxd 4/25/05 2:48 PM Page 180 180 | Thermodynamics AIR AIR m = 1 kg m = 1 kg 300 301 K 1000 1001 K 0.718 kJ 0.855 kJ FIGURE 421 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 some times given on a molar basis. They are then denoted by cv and cp and have the unit kJ/kmol · °C or kJ/kmol · K. 44 INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES We defined an ideal gas as a gas whose temperature, pressure, and specific volume are related by Pv RT It has been demonstrated mathematically (Chap. 12) and experimentally (Joule, 1843) that for an ideal gas the internal energy is a function of the temperature only. That is, u Thermometer WATER AIR (high pressure) Evacuated FIGURE 422 Schematic of the experimental apparatus used by Joule. u 1T 2 (421) In his classical experiment, Joule submerged two tanks connected with a pipe and a valve in a water bath, as shown in Fig. 422. Initially, one tank contained air at a high pressure and the other tank was evacuated. When 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 ideal-gas 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 Pv u Pv f RT h u 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 1T 2 (422) cen84959_ch04.qxd 4/20/05 5:10 PM Page 181 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 420 can be replaced by ordinary derivatives. Then the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as du cv 1 T 2 dT cp 1 T 2 dT (424) and 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: 2 ¢u u2 181 u( T ) h( T ) cv (T ) c p( T ) (423) dh u= h= cv = cp = | u1 1 cv 1 T 2 dT 1 kJ> kg 2 (425) cp 1 T 2 dT 1 kJ> kg 2 FIGURE 423 For ideal gases, u, h, cv, and cp vary with temperature only. (426) and 2 ¢h h2 h1 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 ideal-gas behavior, and therefore their specific heats depend on temperature only. The specific heats of real gases at low pressures are called ideal-gas specific heats, or zero-pressure specific heats, and are often denoted cp0 and cv 0. Accurate analytical expressions for ideal-gas specific heats, based on direct measurements or calculations from statistical behavior of molecules, are available and are given as third-degree polynomials in the appendix (Table A2c) for several gases. A plot of cp0(T) data for some common gases is given in Fig. 4 24. The use of ideal-gas 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 ideal-gas behavior significantly. The integrations in Eqs. 4 25 and 426 are straightforward but rather time-consuming and thus impractical. To avoid these laborious calculations, u and h data for a number of gases have been tabulated over small temperature intervals. These tables are obtained by choosing an arbitrary reference point and performing the integrations in Eqs. 425 and 426 by treating state 1 as the reference state. In the ideal-gas tables given in the appendix, zero kelvin is chosen as the reference state, and both the enthalpy and the internal energy are assigned zero values at that state (Fig. 425). The choice of the reference state has no effect on u or h calculations. The u and h data are given in kJ/kg for air (Table A17) 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 Cp 0 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 424 Ideal-gas constant-pressure specific heats for some gases (see Table A2c for cp equations). cen84959_ch04.qxd 4/20/05 5:10 PM Page 182 182 | 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 u2 cp Approximation 2 cp,avg 1 T1 T avg T2 cv,avg 1 T2 T1 2 1 kJ> kg 2 (427) h1 cp,avg 1 T2 T1 2 1 kJ> kg 2 (428) and FIGURE 425 In the preparation of ideal-gas tables, 0 K is chosen as the reference temperature. Actual u1 h2 0 . . 300.19 310.24 . . T FIGURE 426 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 A2b. 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. 426. If the final temperature T2 is not known, the specific heats may be evaluated at T1 or at the anticipated average temperature. Then T2 can be determined by using these specific heat values. The value of T2 can be refined, if necessary, by evaluating the specific heats at the new average temperature. Another way of determining the average specific heats is to evaluate them at T1 and T2 and then take their average. Usually both methods give 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 ideal-gas 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 constant-volume specific heat cv in an equation should not lead one to believe that this equation is valid for a constant-volume process only. On the contrary, the relation u cv,avg T is valid for any ideal gas undergoing any process (Fig. 427). 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 183 Chapter 4 | 183 Replacing dh by cp dT and du by cv dT and dividing the resulting expression by dT, we obtain cv cp 1 kJ> kg # K 2 R (429) 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. 429). cp cv 1 kJ> kmol # K 2 Ru cp (431) cv 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 47 Evaluation of the AIR P = constant T1 = 20 °C T2 = 30 °C Q2 (430) At this point, we introduce another ideal-gas 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 427 The relation u cv T is valid for any kind of process, constant-volume or not. 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 A17), (b) the functional form of the specific heat (Table A2c), and (c) the average specific heat value (Table A2b). u = u 2 u1 (table) 2 u = cv (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 A17 and take the difference: u1 u2 Thus, ¢u u2 u1 214.07 kJ> kg u @ 300 K u @ 600 K 1 434.78 434.78 kJ> kg 214.07 2 kJ> kg 220.71 kJ> kg (b) The c p(T) of air is given in Table A2c in the form of a third-degree polynomial expressed as cp 1 T 2 a bT cT 2 dT 3 FIGURE 428 Three ways of calculating u. cen84959_ch04.qxd 4/20/05 5:10 PM Page 184 184 | Thermodynamics where a 28.11, b 0.1967 10 2, c d 1.966 10 9. From Eq. 430, AIR at 300 K cv = 0.718 kJ/kg · K R = 0.287 kJ/kg . K { cp 1a Ru Ru 2 bT 10 5, and cT 2 dT 3 From Eq. 425, or cv = 20.80 kJ/kmol . K Ru = 8.314 kJ/kmol . K cv 1 T 2 cp = 1.005 kJ/kg . K 0.4802 { 2 cp = 29.114 kJ/kmol . K ¢u 1 T2 cv 1 T 2 dT T1 3 1a Ru 2 bT dT 3 4 dT cT 2 Performing the integration and substituting the values, we obtain FIGURE 429 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 unit-mass basis is determined by dividing this value by the molar mass of air (Table A1): ¢u 6447 kJ> kmol ¢u M 28.97 kg> kmol 222.5 kJ> kg which differs from the tabulated value by 0.8 percent. (c) The average value of the constant-volume specific heat cv,avg is determined from Table A2b at the average temperature of (T1 T2)/2 450 K to be cv,avg Thus, ¢u cv,avg 1 T2 220 kJ> kg cv @ 450 K T1 2 0.733 kJ> kg # K 1 0.733 kJ> kg # K 2 3 1 600 300 2 K 4 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 48 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 critical-point 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 185 Chapter 4 | Analysis We take the contents of the tank as the system (Fig. 430). This is a closed system since no mass crosses the system boundary during the process. We observe that there is shaft work done on the system. (a) The amount of paddle-wheel work done on the system is Wsh 1 0.02 hp 2 1 0.5 h 2 a # Wsh ¢ t 2545 Btu> h 1 hp b 25.45 Btu Under the stated assumptions and observations, the energy balance on the system can be expressed as ¢ E system E out E in Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies Wsh,in ¢U m 1u2 u1 2 mcv,avg 1 T2 T1 2 As we pointed out earlier, the ideal-gas specific heats of monatomic gases (helium being one of them) are constant. The cv value of helium is determined from Table A2Ea to be cv 0.753 Btu/lbm · °F. Substituting this and other known quantities into the above equation, we obtain 1 1.5 lbm 2 1 0.753 Btu> lbm # °F 2 1 T2 25.45 Btu T2 80°F 2 102.5°F (b) The final pressure is determined from the ideal-gas relation P1V1 T1 P2V2 T2 where V1 and V2 are identical and cancel out. Then the final pressure becomes 50 psia 1 80 460 2 R P2 1 102.5 P2 460 2 R 52.1 psia Discussion Note that the pressure in the ideal-gas relation is always the absolute pressure. P, psia P2 He m = 1.5 lbm T1 = 80°F P1 = 50 psia W sh 2 50 1 V2 = V1 V FIGURE 430 Schematic and P-V diagram for Example 48. 185 cen84959_ch04.qxd 4/20/05 5:10 PM Page 186 186 | Thermodynamics EXAMPLE 49 Heating of a Gas by a Resistance Heater A pistoncylinder 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 120-V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of nitrogen. Solution Nitrogen gas in a pistoncylinder device is heated by an electric resistance heater. Nitrogen expands at constant pressure while some heat is lost. The final temperature of nitrogen is to be determined. Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical-point 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. 431). This is a closed system since no mass crosses the system boundary during the process. We observe that a pistoncylinder device typically involves a moving boundary and thus boundary work, Wb. Also, heat is lost from the system and electrical work We is done on the system. First, let us determine the electrical work done on the nitrogen: 1 120 V 2 1 2 A 2 1 5 VI ¢ t We 60 s 2 a 1 kJ> s 1000 VA b 72 kJ The mass of nitrogen is determined from the ideal-gas relation: 1 400 kPa 2 1 0.5 m3 2 P1V1 RT1 m 1 0.297 kPa # m3> kg # K 2 1 300 K 2 2.245 kg 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 We,in ¢U Q out ¢H m 1 h2 h1 2 mcp 1 T2 T1 2 since U Wb H for a closed system undergoing a quasi-equilibrium expansion or compression process at constant pressure. From Table A2a, 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 T2 1 2.245 kg 2 1 1.039 kJ> kg # K 2 1 T2 27°C 2 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 187 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 V, m3 0.5 FIGURE 431 Schematic and P-V diagram for Example 49. EXAMPLE 410 Heating of a Gas at Constant Pressure A pistoncylinder 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. 432, and the enclosed volume is 400 L. The mass of the piston is such that a 350-kPa 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 pistoncylinder device with a set of stops is heated until its volume is doubled. The final temperature, work done, and the total heat transfer are to be determined. Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical-point 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. 432). 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, 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 ideal-gas relation between states 1 and 3 in the following form: P1V1 T1 1 150 kPa 2 1 V1 2 P3V3 ¡ T3 300 K T3 1 350 kPa 2 1 2V1 2 T3 1400 K | 187 cen84959_ch04.qxd 4/25/05 2:48 PM Page 188 188 | Thermodynamics P, kPa 2 350 3 AIR A V 1 = 400 L P 1 = 150 kPa FIGURE 432 Schematic and P-V diagram for Example 410. EXPERIMENT 1 150 T1 = 27°C Q 0.8 0.4 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 P-V diagram, shown in Fig. 432: A 1 V2 V1 2 P2 1 0.4 m3 2 1 350 kPa 2 140 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 13) can be expressed as ¢ E system Net energy transfer by heat, work, and mass © Ronald Mullisen Change in internal, kinetic, potential, etc., energies Q in EXPERIMENT E out E in Use actual data from the experiment shown here to obtain the specific heat of aluminum. See end-of-chapter problem 4179. Wb,out ¢U m 1u3 u1 2 The mass of the system can be determined from the ideal-gas relation: 1 150 kPa 2 1 0.4 m3 2 P1V1 RT1 m 1 0.287 kPa # m3> kg # K 2 1 300 K 2 0.697 kg The internal energies are determined from the air table (Table A17) to be u1 u @ 300 K u3 u @ 1400 K Thus, Use actual data from the experiment shown here to obtain the specific heat of aluminum. See end-of-chapter problem 4180. © Ronald Mullisen Q in 140 kJ Q in Discussion 214.07 kJ> kg 1113.52 kJ> kg 1 0.697 kg 2 3 1 1113.52 214.07 2 kJ> kg 4 767 kJ The positive sign verifies that heat is transferred to the system. cen84959_ch04.qxd 4/25/05 3:41 PM Page 189 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. 433). Therefore, liquids and solids can be approximated as incompressible substances without sacrificing much in accuracy. The constant-volume 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 liquid-in-glass thermometers. It can be mathematically shown that (see Chap. 12) the constant-volume and constant-pressure specific heats are identical for incompressible substances (Fig. 434). Therefore, for solids and liquids, the subscripts on cp and cv can be dropped, and both specific heats can be represented by a single symbol c. That is, cv cp c (432) | 189 INTERACTIVE TUTORIAL SEE TUTORIAL CH. 4, SEC. 5 ON THE DVD. LIQUID vl = constant SOLID vs = constant FIGURE 433 The specific volumes of incompressible substances remain constant during a process. This result could also be deduced from the physical definitions of constantvolume and constant-pressure specific heats. Specific heat values for several common liquids and solids are given in Table A3. Internal Energy Changes Like those of ideal gases, the specific heats of incompressible substances depend on temperature only. Thus, the partial differentials in the defining equation of cv can be replaced by ordinary differentials, which yield c 1 T 2 dT cv dT du ˛˛ 2 u2 u1 1 c 1 T 2 dT 1 kJ> kg 2 (434) The variation of specific heat c with temperature should be known before this integration can be carried out. For small temperature intervals, a c value at the average temperature can be used and treated as a constant, yielding cavg 1 T2 ¢u T1 2 1 kJ> kg 2 (435) Enthalpy Changes du ¢u v ¢P v dP 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 dh P dv du Integrating, ¢h cavg ¢ T v ¢P = 0.45 kJ/kg . °C (433) ˛ The change in internal energy between states 1 and 2 is then obtained by integration: ¢u IRON 25°C c = cv = cp v dP 1 kJ> kg 2 (436) (437) FIGURE 434 The cv and cp values of incompressible substances are identical and are denoted by c. cen84959_ch04.qxd 4/20/05 5:10 PM Page 190 190 | Thermodynamics For solids, the term v P is insignificant and thus h liquids, two special cases are commonly encountered: u cavg T. For 1. Constant-pressure processes, as in heaters ( P 0): h u cavg T 2. Constant-temperature 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 hf @ T vf @ T 1 P Psat @ T 2 (438) as discussed in Chap. 3. This is an improvement over the assumption that the enthalpy of the compressed liquid could be taken as hf at the given 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. 438 may overcorrect the enthalpy and result in a larger error than the approximation h hf @ T.) EXAMPLE 411 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. 438. 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 T 15 MPa f 100°C h 430.39 kJ> kg 1 Table A7 2 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. 438, h@P,T hf @ T vf @ T 1 P 1 419.17 kJ> kg 2 434.07 kJ> kg Psat @ T 2 1 0.001 m3 kg 2 3 1 15,000 101.42 2 kPa 4 a 1 kJ b 1 kPa # m3 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 191 Chapter 4 EXAMPLE 412 | Cooling of an Iron Block by Water A 50-kg 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 well-insulated and thus there is no heat transfer. Analysis We take the entire contents of the tank as the system (Fig. 435). This is a closed system since no mass crosses the system boundary during the process. We 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 WATER 25°C IRON m = 50 kg 80°C 0.5 m 3 ¢ E system FIGURE 435 Schematic for Example 412. Net energy transfer by heat, work, and mass 0 Change in internal, kinetic, potential, etc., energies ¢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 3 mc 1 T2 ¢ Usys T1 2 4 iron ¢ Uiron 3 mc 1 T2 ¢ Uwater T1 2 4 water 0 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 v 0.5 m3 0.001 m3> kg 500 kg The specific heats of iron and liquid water are determined from Table A3 to be ciron 0.45 kJ/kg · °C and cwater 4.18 kJ/kg · °C. Substituting these values into the energy equation, we obtain 1 50 kg 2 1 0.45 kJ> kg # °C 2 1 T2 80°C 2 T2 1 500 kg 2 1 4.18 kJ> kg # °C 2 1 T2 25°C 2 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. 0 191 cen84959_ch04.qxd 4/20/05 5:10 PM Page 192 192 | Thermodynamics EXAMPLE 413 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. 436). 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 436 Schematic for Example 413. E in E out ¢ E system Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies 0 0 ¢ Uaffected tissue ¢ KEhand 1 mc ¢ T 2 affected tissue 3 m 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 1 mc ¢ T 2 affected tissue mhand 2 1 0.15 kg 2 1 3.8 kJ> kg # °C 2 1 1.8°C 2 1000 m2> s2 a b B 1.2 kg 1 kJ> kg 41.4 m> s 1 or 149 km> h 2 Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering. cen84959_ch04.qxd 4/20/05 5:10 PM Page 193 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, 1.8-m2 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. 437). 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 1-kW electrical resistance heater (Fig. 438). The fraction of sensible heat varies from about 40 percent in the case of heavy work to about 70 percent in the case of light work. The rest of the energy is rejected from the body by perspiration in the form of latent heat. The basal metabolic rate varies with sex, body size, general health 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 437 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 438 Two fast-dancing people supply more energy to a room than a 1-kW electric resistance heater. cen84959_ch04.qxd 4/20/05 5:10 PM Page 194 194 | Thermodynamics A 300-W refrigerator A 300-W resistance heater A 300-W fan A 300-W TV Two people, each dissipating 150 W Three light bulbs, 100 W each A 100-W computer with a 200-W 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 300-W TV set or three 100-W light bulbs will produce the same heating effect in a room as a 300-W resistance heater (Fig. 439). This is a consequence of the conservation of energy principle, which requires that the energy input into a system must equal the energy output when the total energy content of a system remains constant during a process. Food and Exercise Solar energy 300 W FIGURE 439 Some arrangements that supply a room the same amount of energy as a 300-W electric resistance heater. Mixer and motor Electrical switch Thermometer Water Bomb (combustion chamber) Insulation Food sample FIGURE 440 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 well-insulated rigid tank (Fig. 440). 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 195 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. 441) 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 1 Calorie 2 1000 calories 1 kcal 1 kilocalorie 2 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 | 195 FIGURE 441 Evaluating the calorie content of one serving of chocolate chip cookies (values are for Chips Ahoy cookies made by Nabisco). © Vol. 30/PhotoDisc cen84959_ch04.qxd 4/20/05 5:10 PM Page 196 196 | 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 41 and 42. 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 42 during some activities are for a 68-kg 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 68-kg bicyclist is listed in Table 42 to be 639 Calories/h. Then the rate of energy consumption by a 50-kg bicyclist is 1 50 kg 2 639 Cal> h 68 kg 470 Cal> h For a 100-kg 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 41 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 1 slice, 23 cm 8 diameter) Pizza (large, cheese, one 1 slice) 8 Calories 76 136 65 300 350 cen84959_ch04.qxd 4/20/05 5:10 PM Page 197 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 fuel-efficient car needing less fuel for traveling a given distance. It is interesting that we want our cars to be fuel efficient but we do not want the same high efficiency for our bodies. One thing that frustrates the dieters is that the body interprets dieting as starvation and starts using the energy reserves of the body more stringently. Shifting from a normal 2000-Calorie daily diet to an 800-Calorie 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 low-calorie 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. 442). 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 | 197 TABLE 42 Approximate energy consumption of a 68-kg adult during some activities (1 Calorie 4.1868 kJ 3.968 Btu) Activity Basal metabolism Basketball Bicycling (21 km/h) Cross-country 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 442 The body tends to keep the body fat level at a set point by speeding up metabolism when a person splurges and by slowing it down when the person starves. cen84959_ch04.qxd 4/20/05 5:10 PM Page 198 198 | Thermodynamics TABLE 43 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 91119 97127 103136 111146 118156 125165 133175 140185 148195 156205 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 4053 4356 4660 4964 5268 5572 5877 6281 6586 6990 *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 dont impose on it. Dont expect to see this unscientific diet advertised anywhere since there is nothing to be sold and thus no money to be made. Also, it is not as easy as it sounds since food is at the center stage of most leisure activities in social life, and eating and drinking have become synonymous with having a good time. However, it is comforting to know that the human body is quite forgiving of occasional impositions. Being overweight is associated with a long list of health risks from high blood pressure to some forms of cancer, especially for people who have a weight-related medical condition such as diabetes, hypertension, and heart disease. Therefore, people often wonder if their weight is in the proper range. Well, the answer to this question is not written in stone, but if you cannot see your toes or you can pinch your love handles more than an inch, you dont need an expert to tell you that you went over your range. On the other hand, some people who are obsessed with the weight issue try to lose more weight even though they are actually underweight. Therefore, it is useful to have a scientific criterion to determine physical fitness. The range of healthy weight for adults is usually expressed in terms of the body mass index (BMI), defined, in SI units, as BMI W 1 kg 2 H 2 1 m2 2 with 19 BMI 6 19 BMI 25 BMI 7 25 underweight healthy weight overweight (439) where W is the weight (actually, the mass) of the person in kg and H is the height in m. Therefore, a BMI of 25 is the upper limit for the healthy weight and a person with a BMI of 27 is 8 percent overweight. It can be shown that the formula 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 43 in both SI and English units. cen84959_ch04.qxd 4/20/05 5:10 PM Page 199 Chapter 4 EXAMPLE 414 | Burning Off Lunch Calories A 90-kg man had two hamburgers, a regular serving of french fries, and a 200-ml Coke for lunch (Fig. 443). Determine how long it will take for him to burn the lunch calories off (a) by watching TV and (b) by fast swimming. What would your answers be for a 45-kg 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 41 and 42 are applicable for food and exercise. Analysis (a) We take the human body as our system and treat it as a closed system whose energy content remains unchanged during the process. Then the conservation of energy principle requires that the energy input into the body must be equal to the energy output. The net energy input in this case is the metabolizable energy content of the food eaten. It is determined from Table 41 to be Ein 2 Ehamburger 2 275 887 Cal 250 Efries FIGURE 443 A typical lunch discussed in Example 414. © Vol. 30/PhotoDisc Ecola 87 The rate of energy output for a 68-kg man watching TV is given in Table 42 to be 72 Calories/h. For a 90-kg man it becomes E out 1 90 kg 2 72 Cal> h 68 kg 95.3 Cal> h Therefore, it will take ¢t 887 Cal 95.3 Cal> h 9.3 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 45-kg man is half as large as the 90-kg 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 415 Losing Weight by Switching to Fat-Free 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 long-term effects are unknown. A 1-oz (28.3-g) serving of regular potato chips has 10 g of fat and 150 Calories, whereas 1 oz of the so-called fat-free 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 fat-free chips (Fig. 444). FIGURE 444 Schematic for Example 415. 199 cen84959_ch04.qxd 4/20/05 5:10 PM Page 200 200 | Thermodynamics Solution A person switches from regular potato chips to fat-free 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 fat-free chips consumes 75 fewer Calories a day. Then the annual reduction in calories consumed becomes Ereduced 1 75 Cal> day 2 1 365 day> year 2 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 fat-free chips will lose mfat Ereduced Energy content of fat 27,375 Cal 4.1868 kJ a b 33,100 kJ> kg 1 Cal 3.46 kg (about 7.6 pounds) of body fat that year. SUMMARY V1 2 (P1 P2 P0 V2 V1 mRT0 ln Rate of net energy transfer by heat, work, and mass 1 kW 2 Rate of change in internal, kinetic, potential, etc., energies Q W ¢U ¢ KE ¢ PE W Wother 1 kJ 2 Wb where P2V2 P1V1 Wb 1 n 1 2 (PV n 1n (4) Isothernal process of an ideal gas P1V1 ln It can also be expressed in the rate form as . . dE system> dt E in Eout constant) (3) Polytropic process Wb Change in internal, kinetic, potential, etc., energies (2) Isobaric process P0 1 V2 Net energy transfer by heat, work, and mass 1 kJ 2 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 1 Wb ¢ E system P dV E out 2 Wb (1) General E in 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 P-V diagram, the area under the process curve represents the boundary work for a quasi-equilibrium process. Various forms of boundary work are expressed as follows: V2 V1 (PV constant) mRT0 ¢U ¢ KE 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 ¢ PE m 1u2 1 2 2m 1 V 2 mg 1 z 2 For a constant-pressure process, Wb Q Wother ¢H ¢ KE u1 2 V22 1 z1 2 U ¢ PE H. Thus, 1 kJ 2 cen84959_ch04.qxd 4/20/05 5:10 PM Page 201 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 constant-volume process and the specific heat at constant pressure cp for a constantpressure process. They are defined as a cv 0u b 0T v and cp a 2 u2 u1 1 2 ¢h h2 h1 1 201 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 constant-pressure and constant-volume specific heats are identical and denoted by c: 0h b 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 | cv 1 T 2 dT cv,avg 1 T2 T1 2 cp 1 T 2 dT cp,avg 1 T2 T1 2 cp cv c 1 kJ> kg # K 2 The u and h of imcompressible substances are given by 2 ¢u 1 ¢h c 1 T 2 dT ¢u cavg 1 T2 v¢P T1 2 1 kJ> kg 2 1 kJ> kg 2 For ideal gases, cv and cp are related by cp cv R 1 kJ> kg # K 2 REFERENCES AND SUGGESTED READINGS 1. ASHRAE Handbook of Fundamentals. SI version. Atlanta, GA: American Society of Heating, Refrigerating, and Air-Conditioning 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 P-v diagram, what does the area under the process curve represent? 4 2C Is the boundary work associated with constant-volume 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 CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 45 A pistoncylinder 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. 46 A pistoncylinder 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 P46 Q cen84959_ch04.qxd 4/20/05 5:10 PM Page 202 202 | Thermodynamics 4 7 A pistoncylinder 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 P-V 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 pistoncylinder device initially contains 200 L of saturated liquid refrigerant-134a. 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 R-134a P = const. FIGURE P49 P = aV + b FIGURE P414 415E 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 416 During some actual expansion and compression processes in pistoncylinder devices, the gases C, have been observed to satisfy the relationship PV n 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 P-v diagram. 417 4 11E A frictionless pistoncylinder 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. 418 A frictionless pistoncylinder device contains 2 kg of nitrogen at 100 kPa and 300 K. Nitrogen is now compressed constant until it slowly according to the relation PV1.4 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 gas-tight, frictionless pistoncylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the 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 quasi-equilibrium Reconsider Prob. 416. Using the EES (or other) software, plot the process described in the problem on a P-V 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 P418 cen84959_ch04.qxd 4/20/05 5:10 PM Page 203 Chapter 4 4 19 The equation of state of a gas is given as v (P RuT, where the units of v and P are 10/ v 2) 3/kmol and kPa, respectively. Now 0.5 kmol of this gas is m expanded in a quasi-equilibrium 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 20 Reconsider Prob. 4 19. Using the integration feature of the EES software, calculate the work done, and compare your result with the hand-calculated result obtained in Prob. 4 19. Plot the process described in the problem on a P-v diagram. 4 21 Carbon dioxide contained in a pistoncylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volume are related by P aV 2, 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 pistoncylinder 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 cross-sectional 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 P-V diagram. 4 23 A pistoncylinder device contains 50 kg of water at 250 kPa and 25°C. The cross-sectional 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 P-V diagram. Answers: (a) 450 kPa, 147.9°C, (b) 44.5 kJ | 203 424 Reconsider Prob. 423. 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. 425 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. 426 A pistoncylinder 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 P426 427 A pistoncylinder 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 428 A 0.5-m3 rigid tank contains refrigerant-134a 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 P-v diagram with respect to saturation lines. 2 A = 0.1 m H 2O m = 50 kg FIGURE P423 429E A 20-ft3 rigid tank initially contains saturated refrigerant-134a vapor at 160 psia. As a result of heat transfer from the refrigerant, the pressure drops to 50 psia. Show the process on a P-v 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 204 204 | Thermodynamics 4 30 A well-insulated rigid tank contains 5 kg of a saturated liquidvapor mixture of water at l00 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 110-V 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 T-v diagram with respect to saturation lines. H 2O V = constant We FIGURE P430 434 A pistoncylinder 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 T-v diagram with respect to saturation lines. Answer: 1173 kJ 435E A pistoncylinder 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 T-v diagram with respect to saturation lines. 436 An insulated pistoncylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constantpressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-v 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 P432 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 P436 437 A pistoncylinder 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 T-v 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. 438 A pistoncylinder 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 P-v 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 205 Chapter 4 205 442 A 30-L electrical radiator containing heating oil is placed in a 50-m3 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 well-sealed so that there are no air leaks. H2 O 200 kPa 200°C Q | FIGURE P438 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 pistoncylinder 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 P-v 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 2-kg steam at 1 MPa and 300°C while Tank B contains 3-kg saturated liquidvapor 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 P441 Room 10°C Q Radiator FIGURE P442 Specific Heats, u, and h of Ideal Gases 443C Is the relation u mcv,avg T restricted to constantvolume processes only, or can it be used for any kind of process of an ideal gas? 444C Is the relation h mcp,avg T restricted to constantpressure processes only, or can it be used for any kind of process of an ideal gas? _ _ 445C Show that for an ideal gas cp cv Ru. 446C 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. 447C In the relation u mcv T, what is the correct unit of cv kJ/kg · °C or kJ/kg · K? 448C 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? 449C 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? 450C 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? 451 Determine the kJ/kg, as it is heated empirical specific heat (Table A2c), (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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 206 206 | Thermodynamics (Table A2b), and (c) the cp value at room temperature (Table A2a). 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 A2Ec), (b) the cp value at the average temperature (Table A2Eb), and (c) the cp value at room temperature (Table A2Ea). 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 A2c), (b) the cv value at the average temperature (Table A2b), and (c) the cv value at room temperature (Table A2a). at a rate of 10,000 kJ/h, and a 100-W 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. 459 A student living in a 4-m 6-m 6-m dormitory room turns on her 150-W 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 Closed-System Energy Analysis: Ideal Gases 4 54C Is it possible to compress an ideal gas isothermally in an adiabatic pistoncylinder 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 3-m3 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 P459 4 57 A 4-m 5-m 6-m 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 460E A 10-ft3 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 paddle-wheel work done. Neglect the energy stored in the paddle wheel. 4 58 A 4-m 5-m 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat 5000 kJ/h 461 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 P458 Evacuated FIGURE P461 cen84959_ch04.qxd 4/20/05 5:10 PM Page 207 Chapter 4 4 62 A pistoncylinder 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 pistoncylinder 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. | 207 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. 469 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 pistoncylinder 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 pistoncylinder 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 P469 Answer: 0.235 kWh 470E A pistoncylinder 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 P465 4 66 An insulated pistoncylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa and 27°C. An electric switch is turned on, and a 110-V 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 pistoncylinder 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 one-half. 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 P-V diagram, and investigate the effect of the polytropic exponent n on the boundary work and heat 471 A pistoncylinder device contains 4 kg of argon at 250 kPa and 35°C. During a quasi-equilibrium, isothermal expansion process, 15 kJ of boundary work is done by the system, and 3 kJ of paddle-wheel work is done on the system. Determine the heat transfer for this process. 472 A pistoncylinder 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 P-v diagram. Answers: 516 kJ, 2674 kJ 473 A pistoncylinder 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 P-v diagram. Closed-System Energy Analysis: Solids and Liquids 474 In a manufacturing facility, 5-cm-diameter 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 208 208 | Thermodynamics 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 478 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. 479 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 P474 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 12-fluid-oz 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 1000-W iron whose base plate is made of 0.5-cm-thick aluminum alloy 2024-T6 (r 2770 kg/m3 and 875 J/kg · °C). The base plate has a surface area of 0.03 cp 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 P479 480 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 5-min operating period. What would your answer be if the device were attached to a 0.2-kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal. 481 Reconsider Prob. 480. 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. 482 An ordinary egg can be approximated as a 5.5-cmdiameter sphere. The egg is initially at a uniform temperature of 8°C and is dropped into boiling water at 97°C. Taking the 3.32 properties of the egg to be r 1020 kg/m3 and cp 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 P477 © Vol. 58/PhotoDisc 483E ln a production facility, 1.2-in-thick 2-ft 2-ft 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 209 Chapter 4 Furnace, 1300°F | 209 494 A 55-kg man gives in to temptation and eats an entire 1-L 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 495 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 P483E 4 84 Long cylindrical steel rods (r 7833 kg/m3 and cp 0.465 kJ/kg · °C) of 10-cm diameter are heat-treated 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 2-kW 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 80-kg 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 68-kg woman is planning to bicycle for an hour. If she is to meet her entire energy needs while bicycling by eating 30-g chocolate candy bars, determine how many candy bars she needs to take with her. 496 Consider two identical 50-kg 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 497 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? 498 A 60-kg man used to have an apple every day after dinner without losing or gaining any weight. He now eats a 200-ml 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: 0.087-kg gain 499 The average specific heat of the human body is 3.6 kJ/kg · °C. If the body temperature of an 80-kg 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. 4100E Alcohol provides 7 Calories per gram, but it provides no essential nutrients. A 1.5 ounce serving of 80-proof 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. 4101 A 12-oz serving of a regular beer contains 13 g of alcohol and 13 g of carbohydrates, and thus 150 Calories. A 12-oz 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 P4101 cen84959_ch04.qxd 4/20/05 5:10 PM Page 210 210 | Thermodynamics 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 12-oz can of (a) regular beer and (b) light beer on a treadmill. 4 102 A 5-oz serving of a Bloody Mary contains 14 g of alcohol and 5 g of carbohydrates, and thus 116 Calories. A 2.5-oz 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 cross-country 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 cross-country ski machine. 4 103E A 176-pound man and a 132-pound 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 150-pound person goes to Hardees for dinner and orders a regular roast beef (270 Cal) and a big roast beef (410 Cal) sandwich together with a 12-oz can of Pepsi (150 Cal). A 150-pound 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 McDonalds 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. 4109 The body mass index (BMI) of a 1.7-m tall woman who normally has 3 large slices of cheese pizza and a 400-ml Coke for lunch is 30. She now decides to change her lunch to 2 slices of pizza and a 200-ml 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 4110 Consider a pistoncylinder 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. 4111 In solar-heated 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 5-m3 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. 4112 A pistoncylinder 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 100-kg 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 1 kg 2 H 2 1 m2 2 FIGURE P4112 Q cen84959_ch04.qxd 4/20/05 5:10 PM Page 211 Chapter 4 is done during this process, determine the gas constant and the molar mass of the gas. Also, determine the constantvolume and constant-pressure 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 pistoncylinder 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 P-v diagram. Answers: (a) 150 kJ, (b) 90 kJ | 211 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 P-v diagram. 4116E 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 4117E Reconsider Prob. 4116E. Using the integration feature of the EES software, determine the work done. Compare the result with your handcalculated result. 4118 A mass of 12 kg of saturated refrigerant-134a vapor is contained in a pistoncylinder device at 240 kPa. Now 300 kJ of heat is transferred to the refrigerant at constant pressure while a 110-V 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 T-v diagram with respect to the saturation lines. Answer: 12.8 A AIR P = 200 kPa 1 V1 = 0.2 m3 R-134a P = constant FIGURE P4114 We 4 115 A mass of 5 kg of saturated liquidvapor mixture of water is contained in a pistoncylinder 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 P4115 Q FIGURE P4118 4119 A mass of 0.2 kg of saturated refrigerant-134a is contained in a pistoncylinder 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 P-v 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. 4120 A pistoncylinder 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 212 212 | Thermodynamics 4124 One ton (1000 kg) of liquid water at 80°C is brought into a well-insulated and well-sealed 4-m 5-m 6-m 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 P4120 4 121 A frictionless pistoncylinder 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 thermostat-controlled 15-kW back-up 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, 4125 A 4-m 5-m 6-m 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 24-h 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. 4126 The energy content of a certain food is to be determined in a bomb calorimeter that contains 3 kg of water by burning a 2-g 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 P4126 22ºC Water 80°C Pump FIGURE P4122 4 123 An 1800-W 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. 4127 A 68-kg 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. 4128 A 0.2-L 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 213 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. | 213 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 pistoncylinder device initially contains 0.01 m3 of saturated liquidvapor 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 pistoncylinder 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 P4133 4134 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 P4134 4135 0.05 m 3 100 kPa Steam FIGURE P4132 4 133 Water is boiled at sea level in a coffee maker equipped with an immersion-type 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. 4134. 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. 4136 A rigid tank containing 0.4 m3 of air at 400 kPa and 30°C is connected by a valve to a pistoncylinder 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 214 214 | Thermodynamics 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 P4138 4139 Repeat Prob. 4138 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 P4136 4 137 A well-insulated 4-m 4-m 5-m 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 120-W 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. 4140 Reconsider Prob. 4139. 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. 4141 An insulated rigid tank initially contains 1.4-kg 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 P4141 FIGURE P4137 4 138 Consider a well-insulated 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? 4142 A vertical 12-cm diameter pistoncylinder 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. 4143 A pistoncylinder device initially contains 0.15-kg 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- cen84959_ch04.qxd 4/20/05 5:10 PM Page 215 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 P4144 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 closed-system conservation of energy relation in this case reduces to Wout m(u1 u2). Then the explosive energy Eexp becomes Eexp m 1 u1 u2 2 where the subscripts 1 and 2 refer to the state of the fluid before and after the explosion, respectively. The specific a1 T2 b T1 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 1 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 fa , ,T,T b m3 m3 1 2 SIDE 1 Mass = m1 Temperature = T1 v1 P1 k FIGURE P4143 T3 u2 u1 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 P4145 4146 Using the relations in Prob. 4145, 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 4147 A room is filled with saturated steam at 100°C. Now a 5-kg 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 4148 A frictionless pistoncylinder 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 cen84959_ch04.qxd 4/20/05 5:10 PM Page 216 216 | Thermodynamics 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 3-m3 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 0.8-m3 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 well-sealed 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 120-W 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 2-kW 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 250-W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120-W TV, a 1kW electric resistance heater, and a 50-W 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 pistoncylinder device contains 5 kg of air at 400 kPa and 30°C. During a quasi-equilibium isothermal expansion process, 15 kJ of boundary work is done by the system, and 3 kJ of paddle-wheel 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 4156 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 4157 A 6-pack 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 4158 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 4159 A 2-kW electric resistance heater submerged in 5-kg 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 4160 3 kg of liquid water initially at 12°C is to be heated at 95°C in a teapot equipped with a 1200-W 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 4161 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. cen84959_ch04.qxd 4/20/05 5:10 PM Page 217 Chapter 4 | 217 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 10 4)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 10 4)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 pistoncylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant 3 T1. Then the piston is held stationary pressure until T2 4 1 while the gas is further cooled to T3 T1, where all temper2 atures 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 pistoncylinder 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 4167 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. 4168 Find out how the specific heats of gases, liquids, and solids are determined in national laboratories. Describe the experimental apparatus and the procedures used. 4169 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? 4170 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? 4171 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? 4172 It is claimed that fruits and vegetables are cooled by 6°C for each percentage point of weight loss as moisture cen84959_ch04.qxd 4/20/05 5:10 PM Page 218 218 | Thermodynamics 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 polyc, where p is pressure, v is specific tropic relation pv n 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 write-up, and the data provided on the DVD accompanying this book. 4 175 First Law of ThermodynamicsLead 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 write-up, and the data provided on the DVD accompanying this book. 4 176 First Law of ThermodynamicsFriction 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 write-up, and the data provided on the DVD accompanying this book. 4 177 First Law of ThermodynamicsCopper 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 write-up, and the data provided on the DVD accompanying this book. 4178 First Law of ThermodynamicsBicycle Braking Experiment The first law of thermodynamics is verified yet againthis time with a bicycle. A bicycle front caliper brake is removed and replaced with a lever-mounted, 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 coast-down data read into a cassette audio recorder by the bicycle rider. Verify the first law of thermodynamics using the video clip, the complete write-up, and the data provided on the DVD accompanying this book. 4179 Specific Heat of AluminumElectric Calorimeter Experiment The specific heat of aluminum is obtained with an electric calorimeter. The design consists of two individual calorimeterseach an assembly of 13 aluminum plates with electric resistance heater wires laced in-between the plates. The exterior surfaces of both calorimeters and the surrounding insulation are identical. However, the interior plates are differentone calorimeter has solid interior plates and the other has perforated interior plates. By initially adjusting the electrical power into each calorimeter the temperature-versustime 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 write-up, and the data provided on the DVD accompanying this book. 4180 Specific Heat of AluminumTransient Cooling Experiment The specific heat of aluminum is obtained with an entirely different experiment than the one described in Prob. 4179. 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 write-up, and the data provided on the DVD accompanying this book. cen84959_ch05.qxd 4/25/05 3:00 PM Page 219 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 steady-flow 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 steady-flow processes and analyze the common steady-flow devices such as nozzles, diffusers, compressors, turbines, throttling devices, mixing chambers, and heat exchangers. Finally, we apply the energy balance to general unsteady-flow 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 unsteady-flow 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 steady-flow devices such as nozzles, compressors, turbines, throttling valves, mixers, heaters, and heat exchangers. Apply the energy balance to general unsteady-flow processes with particular emphasis on the uniform-flow process as the model for commonly encountered charging and discharging processes. | 219 cen84959_ch05.qxd 4/25/05 3:00 PM Page 220 220 | Thermodynamics 51 INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 1 ON THE DVD. 2 kg H2 16 kg O2 18 kg H2O FIGURE 51 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 vinegar-and-oil 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. 51). 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 well-known formula proposed by Albert Einstein (18791955): E mc2 (51) 108 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 10 10 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. 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. 52) # dm Control surface FIGURE 52 The normal velocity Vn for a surface is the component of velocity perpendicular to the surface. rVn dA c (52) 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 1 r 2 r 2 2 but dm mtotal (total mass flow rate 2 1 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 cen84959_ch05.qxd 4/25/05 3:00 PM Page 221 Chapter 5 | 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 cross-sectional area of a pipe or duct is obtained by integration: # m # dm rVn dA c Ac Ac 1 kg> s 2 (53) While Eq. 53 is always valid (in fact it is exact), it is not always practical for engineering analyses because of the integral. We would like instead to express mass flow rate in terms of average values over a cross section of the pipe. In a general compressible flow, both 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. 53. 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 no-slip 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. 53), Vavg Average velocity: 1 Ac Vn dA c (54) 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. 53 becomes # m rVavg A c 1 kg> s 2 (55) For compressible flow, we can think of r as the bulk average density over the cross section, and then Eq. 55 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 cross-sectional area normal to the flow direction. The volume of the fluid flowing through a cross section per unit time is . called the volume flow rate V (Fig. 54) and is given by # V Vn dA c Vavg A c VA c Ac 1 m3> s 2 (56) An early form of Eq. 56 was published in 1628 by the Italian monk Benedetto Castelli (circa 15771644). Note that most fluid mechanics text. . books 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 # m # rV # V v (57) 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 53 The average velocity Vavg is defined as the average speed through a cross section. Ac Vavg V = Vavg Ac Cross section FIGURE 54 The volume flow rate is the volume of fluid flowing through a cross section per unit time. cen84959_ch05.qxd 4/25/05 3:00 PM Page 222 222 | 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 b the CV during ¢ t a Total mass leaving b the CV during ¢ t a or mout min FIGURE 55 Conservation of mass principle for an ordinary bathtub. Net change in mass b within the CV during ¢ t 1 kg 2 ¢ mCV (58) where mCV mfinal minitial is the change in the mass of the control volume during the process (Fig. 55). It can also be expressed in rate form as # min dmCV> dt # mout 1 kg> s 2 (59) . . 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 58 and 59 are often referred to as the mass balance and are applicable to any control volume undergoing any kind of process. Consider a control volume of arbitrary shape, as shown in Fig. 56. The mass of a differential volume dV within the control volume is dm 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 (510) CV Then the time rate of change of the amount of mass within the control volume can be expressed as dmCV dt Rate of change of mass within the CV: dV n dm dA Control volume (CV) u V Control surface (CS) FIGURE 56 The differential control volume dV and the differential control surface dA used in the derivation of the conservation of mass relation. d dt r dV (511) 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. 56. 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 (512) 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 # Differential mass flow rate: dm rVn dA r 1 V cos u 2 dA r 1 V # n 2 dA (513) S S cen84959_ch05.qxd 4/25/05 3:00 PM Page 223 Chapter 5 | 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: r 1 V # n 2 dA S # dm rVn dA CS CS CS S (514) 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. 514 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. . . mout min 0, the conservation of Rearranging Eq. 59 as dmCV/dt mass relation for a fixed control volume can then be expressed as d dt General conservation of mass: r 1 V # n 2 dA S r dV CV CS S 0 (515) It states that the time rate of change of mass within the control volume plus the net mass flow rate through the control surface is equal to zero. Splitting the surface integral in Eq. 515 into two partsone for the outgoing flow streams (positive) and one for the incoming streams (negative) the general conservation of mass relation can also be expressed as d dt a r dV out CV rVn dA a in A rVn dA 0 (516) 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. 516 can also be expressed as d dt r dV CV # am in # am out or dmCV dt # am in # am (517) out Equations 515 and 516 are also valid for moving or deforming control vol umes provided that the absolute velocity V is replaced by the relative velocity Vr , which is the fluid velocity relative to the control surface. Mass Balance for Steady-Flow Processes ˙ m1 = 2 kg/s During a steady-flow 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 steady-flow 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 steady-flow system with multiple inlets and outlets can be expressed in rate form as (Fig. 57) Steady flow: # am in # am out 1 kg> s 2 (518) ˙ m2 = 3 kg/s CV ˙˙m m3 = m1 + ˙ 2 = 5 kg/s FIGURE 57 Conservation of mass principle for a two-inletone-outlet steady-flow system. cen84959_ch05.qxd 4/25/05 3:00 PM Page 224 224 | 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. 518 reduces, for single-stream steady-flow systems, to Steady flow (single stream): # m1 # m2 S r 1V1A 1 r 2V2 A 2 (519) Special Case: Incompressible Flow The conservation of mass relations can be simplified even further when the fluid is incompressible, which is usually the case for liquids. Canceling the density from both sides of the general steady-flow relation gives ˙ m2 = 2 kg/s ˙ V2 = 0.8 m3/s Steady, incompressible flow: # aV in ˙ m1 = 2 kg/s ˙ V1 = 1.4 m3/s FIGURE 58 During a steady-flow process, volume flow rates are not necessarily conserved although mass flow rates are. out 1 m3> s 2 (520) For single-stream steady-flow systems it becomes Steady, incompressible flow (single stream): Air compressor # aV # V1 # V2 S V1A 1 V2 A 2 (521) 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 steady-flow 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. 58). 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 (constant-density) 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 51 Water Flow through a Garden Hose Nozzle Nozzle Garden hose Bucket A garden hose attached with a nozzle is used to fill a 10-gal bucket. The inner diameter of the hose is 2 cm, and it reduces to 0.8 cm at the nozzle exit (Fig. 59). 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 59 Schematic for Example 51. 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 225 Chapter 5 | 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 # V # m V ¢t # rV 10 gal 3.7854 L a b 50 s 1 gal 1 1 kg> L 2 1 0.757 L> s 2 0.757 L/s 0.757 kg/s (b) The cross-sectional area of the nozzle exit is pr 2 e Ae p 1 0.4 cm 2 2 0.5027 cm2 0.5027 10 4 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 Ae 0.757 L/s 1 m3 a b 4 2 1000 L 0.5027 10 m 15.1 m/s 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 52 Discharge of Water from a Tank A 4-ft-high, 3-ft-diameter 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. 510). 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 unsteady-flow 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 # min # mout dmCV dt r 22ghAjet . During this process no mass enters the control volume (min mass flow rate of discharged water can be expressed as # mout (rVA)out (1) 0), and the (2) Air Water h0 h2 h 0 Djet Dtank FIGURE 510 Schematic for Example 52. cen84959_ch05.qxd 4/25/05 3:00 PM Page 226 226 | Thermodynamics where Ajet pD 2 /4 is the cross-sectional area of the jet, which is constant. jet Noting that the density of water is constant, the mass of water in the tank at any time is mCV r 22ghAjet rAtankh rV (3) r 22gh(pD2 /4) jet D 2 /4 tank 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(rAtankh) dt 22gh r(pD2 /4) dh tank dt Canceling the densities and other common terms and separating the variables give D2 tank dt Integrating from t t dt 0 D 2 22g jet 24 ft D2 jet 2h 0 at which h h2 D2 tank 22 ft 232.2/2 ft/s h0 dh dh h0 to t St 2h 0 2h 2 2g/2 t at which h a h2 gives D tank 2 b D jet Substituting, the time of discharge is t 2 a 3 12 in 2 b 0.5 in 757 s 12.6 min 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 511 Schematic for flow work. 52 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. 511. 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 cross-sectional area of the fluid element is A (Fig. 512), the force applied on the fluid element by the imaginary piston is F PA (522) cen84959_ch05.qxd 4/25/05 3:00 PM Page 227 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 1 kJ 2 PV wflow Pv 1 kJ> kg 2 (524) The flow work relation is the same whether the fluid is pushed into or out of the control volume (Fig. 513). 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. 514). On a unit-mass basis, it is expressed as e u ke pe u V2 2 gz 1 kJ> kg 2 (525) where V is the velocity and z is the elevation of the system relative to some external reference point. Flow Flow energy energy Kinetic Kinetic energy energy Non lowing Non f lowing fluid fluid e=u+ Internal Internal energy energy V2 + gz 2 Potential Potential energy energy F P FIGURE 512 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 Flowing Flowing fluid fluid CV (a) Before entering wflow Total Energy of a Flowing Fluid 227 A (523) 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 513 Flow work is the energy needed to push a fluid into or out of a control volume, and it is equal to Pv. Kinetic Kinetic energy energy 2 θ = Pv + u + V + gz 2 Internal Internal energy energy Potential Potential energy energy FIGURE 514 The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid. cen84959_ch05.qxd 4/25/05 3:00 PM Page 228 228 | Thermodynamics The fluid entering or leaving a control volume possesses an additional form of energythe flow energy Pv, as already discussed. Then the total energy of a flowing fluid on a unit-mass basis (denoted by u) becomes Pv u e 1u Pv pe 2 ke (526) But the combination Pv u has been previously defined as the enthalpy h. So the relation in Eq. 526 reduces to u h ke pe V2 2 h 1 kJ> kg 2 gz (527) By using the enthalpy instead of the internal energy to represent the energy of a flowing fluid, one does not need to be concerned about the flow work. The energy associated with pushing the fluid into or out of the control volume is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy. From now on, the energy of a fluid stream flowing into or out of a control volume is represented by Eq. 527, and no reference will be made to flow work or flow energy. Energy Transport by Mass ˙ mi,kg/s θ i,kJ/kg CV ˙ m i θi (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. 515). That is, Rate of energy transport: FIGURE 515 . The product miui is the energy transported into control volume by mass per unit time. # E mass mah mu E mass Amount of energy transport: # mu # mah V2 2 V2 2 gz b gz b 1 kJ 2 1 kW 2 (528) (529) 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 u i dmi mi mi a hi V i2 2 gz i b dmi (530) Most flows encountered in practice can be approximated as being steady and one-dimensional, and thus the simple relations in Eqs. 528 and 529 can be used to represent the energy transported by a fluid stream. cen84959_ch05.qxd 4/25/05 3:00 PM Page 229 Chapter 5 EXAMPLE 53 Energy Transport by Mass Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa (Fig. 516). 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 cross-sectional 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 start-up 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 A5). 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 # m V ¢ V liquid 0.6 L 1 m3 a b 0.001053 m3> kg 1000 L vf 0.570 kg 0.570 kg m 0.0142 kg> min 2.37 10 4 kg/s ¢t 40 min # # m vg 1 2.37 10 4 kg> s 2 1 1.1594 m3> kg 2 m 34.3 m/s 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 u Pv h h u ke pe 2693.1 h 2519.2 173.9 kJ/kg 2693.1 kJ/kg Note that the kinetic energy in this case is ke V 2/2 (34.3 m/s)2/2 2/s2 588 m 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 # mu 1 2.37 10 4 kg/s 2 1 2693.1 kJ/kg 2 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 516 Schematic for Example 53. | 229 cen84959_ch05.qxd 4/25/05 3:00 PM Page 230 230 | Thermodynamics 53 INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 3 ON THE DVD. FIGURE 517 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 518 Under steady-flow conditions, the mass and energy contents of a control volume remain constant. ENERGY ANALYSIS OF STEADY-FLOW 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 start-up period is completed and steady operation is established, and they are classified as steady-flow devices (Fig. 517). Processes involving such devices can be represented reasonably well by a somewhat idealized process, called the steady-flow 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 steady-flow 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. 518). As a result, the boundary work is zero for steady-flow 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. 519). As an added simplification, the fluid properties at an opening are usually considered to be uniform (at some average value) over the cross section. Thus, the fluid properties at an inlet or exit may be specified by the average single values. Also, the heat and work interactions between a steady-flow system and its surroundings do not change with time. Thus, the power delivered by a system and the rate of heat transfer to or from a system remain constant during a steady-flow process. The mass balance for a general steady-flow system was given in Sec. 51 as # am in ˙ m2 h2 ˙ m1 h1 Control volume ˙ m3 h3 FIGURE 519 Under steady-flow conditions, the fluid properties at an inlet or exit remain constant (do not change with time). # am out 1 kg> s 2 (531) The mass balance for a single-stream (one-inlet and one-outlet) steady-flow system was given as # m1 # m2 S r 1V1A 1 r 2V2 A 2 (532) 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 cross-sectional area normal to flow direction. During a steady-flow process, the total energy content of a control volume constant), and thus the change in the total energy remains constant (ECV 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 steady-flow process to cen84959_ch05.qxd 4/25/05 3:00 PM Page 231 # E in Chapter 5 0 (steady) dE system> dt ¡ 0 # E out | 231 (533) Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies or 1 kW 2 . E out . E in Energy balance: Rate of net energy transfer in by heat, work, and mass (534) Rate of net energy transfer out by heat, work, and mass Noting that energy can be transferred by heat, work, and mass only, the energy balance in Eq. 534 for a general steady-flow system can also be written more explicitly as # Q in # Win # a mu # Q out # Wout # a mu in (535) out or # Q in 2 V # gz b a mah 2 in 123 # Win # Q out V2 2 # a mah # Wout out gz b 123 for each inlet ˙˙ m2 = m 1 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 steady-flow systems first appeared in 1859 in a German thermodynamics book written by Gustav Zeuner. Consider, for example, an ordinary electric hot-water heater under steady operation, as shown in Fig. 520. A cold-water stream with a mass flow rate . m is continuously flowing into the water heater, and a hot-water 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 first-law or energy balance relation in that case for a general steady-flow system becomes # Q # W V2 2 # a mah gz b V2 2 # a mah out 123 in gz b (537) 123 for each inlet for each exit . . Obtaining a negative quantity for Q or W simply means that the assumed direction is wrong and should be reversed. For single-stream devices, the steady-flow energy balance equation becomes # Q # W # m c h2 h1 V2 2 V2 1 2 g 1 z2 z1 2 d Heat ˙ loss Qout (536) (538) Electric heating element ˙ Win Hot water out CV (Hot-water tank) ˙ m1 Cold water in FIGURE 520 A water heater in steady operation. cen84959_ch05.qxd 4/25/05 3:00 PM Page 232 232 | Thermodynamics . Dividing Eq. 538 by m gives the energy balance on a unit-mass basis as q w h2 V2 2 h1 V2 1 2 g 1 z2 z1 2 (539) .. .. where q Q/m and w W/m are the heat transfer and work done per unit mass of the working fluid, respectively. When the fluid experiences negligible changes in its kinetic and potential energies (that is, ke 0, pe 0), the energy balance equation is reduced further to q ˙ We CV ˙ Wsh FIGURE 521 Under steady operation, shaft work and electrical work are the only forms of work a simple compressible system may involve. J N. m kg m kg kg s2 ( m ( kg m s 2 2 2 tu (Also, lBbm 25,037 fst ( 2 FIGURE 522 The units m2/s2 and J/kg are equivalent. w h2 h1 (540) 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 steady-flow 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. 521). Many steady-flow 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. 2 ke (V 2 V 2)/2. The unit of kinetic energy is m2/s2, which is equivalent 1 to J/kg (Fig. 522). 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 steady-flow 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. 523). 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 233 Chapter 5 54 SOME STEADY-FLOW 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. 524). Therefore, these devices can be conveniently analyzed as steady-flow devices. In this section, some common steady-flow 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. 5-Stage LPC Bleed Low Pressure Air Collector Compressor (LPC) Cold End Drive Flange 14-Stage High Pressure Compressor Combustor Fuel System Manifolds 2-Stage 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 523 At very high velocities, even small changes in velocities can cause significant changes in the kinetic energy of the fluid. 5-Stage Low Pressure Turbine Hot End Drive Flange FIGURE 524 A modern land-based 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 cross-sectional 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. 525). Therefore, the kinetic energy changes must be accounted for in analyzing the flow through these devices ( ke 0). | cen84959_ch05.qxd 4/25/05 3:00 PM Page 234 234 | Thermodynamics EXAMPLE 54 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 525 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 steady-flow 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 critical-point 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. 526). 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 ideal-gas relation at the inlet conditions: P1 = 80 kPa T1 = 10°C V1 = 200 m/s A1 = 0.4 m2 0.287 kPa # m3/kg # K 2 1 283 K 2 RT1 P1 v1 1.015 m3/kg 80 kPa Then, AIR m=? # m T2 = ? 1 V1A 1 v1 1 1 200 m/s 2 1 0.4 m2 2 1.015 m3/kg 78.8 kg/s Since the flow is steady, the mass flow rate through the entire diffuser remains constant at this value. FIGURE 526 Schematic for Example 54. (b) Under stated assumptions and observations, the energy balance for this steady-flow system can be expressed in the rate form as # E in dE system> dt ¡ # E out 0 (steady) 0 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies # Ein # m a h1 2 V1 b 2 h2 # Eout # m a h2 h1 V2 2 b 2 V2 2 # 1 since Q # 0, W 0, and ¢ pe 02 V2 1 2 The exit velocity of a diffuser is usually small compared with the inlet velocity (V2 V1); thus, the kinetic energy at the exit can be neglected. The enthalpy of air at the diffuser inlet is determined from the air table (Table A17) to be h1 h @ 283 K 283.14 kJ/kg cen84959_ch05.qxd 4/25/05 3:00 PM Page 235 Chapter 5 | 235 Substituting, we get h2 283.14 kJ/kg 0 1 200 m/s 2 2 2 a 1 kJ/kg 1000 m2/s2 b 303.14 kJ/kg From Table A17, 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 55 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 steady-flow 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. 526A). 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 T1 250 psia f 700°F v1 h1 10 lbm> s V1 1 VA v1 1 1 1 1 V1 2 1 0.2 ft2 2 2.6883 ft3> lbm 134.4 ft/s (b) Under stated assumptions and observations, the energy balance for this steady-flow system can be expressed in the rate form as # E in 0 (steady) dE system> dt ¡ 0 # E out Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies # Ein # m a h1 # Eout 2 V1 b 2 # Qout # m a h2 P2 = 200 psia V2 = 900 ft/s FIGURE 526A Schematic for Example 55. Then, # m STEAM m = 10 lbm/s P1 = 250 psia T1 = 700°F A1 = 0.2 ft2 1 Table A6E 2 2.6883 ft3/lbm 1371.4 Btu/lbm qout = 1.2 Btu/lbm V2 2 b 2 # 1 since W 0, and ¢ pe 02 cen84959_ch05.qxd 4/25/05 3:00 PM Page 236 236 | Thermodynamics . Dividing by the mass flow rate m and substituting, h2 is determined to be h2 h1 qout 1 1371.4 V2 2 V2 1 2 1.2 2 Btu/lbm 1 900 ft/s 2 2 2 1 134.4 ft/s 2 2 a 1 Btu/lbm b 25,037 ft2/s2 1354.4 Btu/lbm Then, P2 h2 200 psia f 1354.4 Btu/lbm T2 662.0°F 1 Table A6E 2 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 negligi. ble (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 56 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 237 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 steady-flow 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 critical-point values. 3 The kinetic and potential energy changes are zero, ke pe 0. Analysis We take the compressor as the system (Fig. 527). This is a control volume since mass crosses the system boundary during the process. We . . . observe that there is only one inlet and one exit and thus 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 steady-flow system can be expressed in the rate form as # E in dE system> dt ¡ # E out 237 qout = 16 kJ/kg P2 = 600 kPa T2 = 400 K AIR m = 0.02 kg/s ˙ ˙ Win = ? P1 = 100 kPa T1 = 280 K FIGURE 527 Schematic for Example 56. Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies # Eout # # Qout mh2 # # mqout m 1 h2 # Ein # Win 0 (steady) 0 | # mh1 # Win 1 since ¢ ke ¢ pe h1 2 02 The enthalpy of an ideal gas depends on temperature only, and the enthalpies of the air at the specified temperatures are determined from the air table (Table A17) to be 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 1 0.02 kg/s 2 1 16 kJ/kg 2 1 0.02 kg/s 2 1 400.98 280.13 2 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 57 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. 528. (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 steady-flow 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 528 Schematic for Example 57. cen84959_ch05.qxd 4/25/05 3:00 PM Page 238 238 | 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 2 MPa f 400 C P1 T1 h1 3248.4 kJ/kg (Table A6) At the turbine exit, we obviously have a saturated liquidvapor mixture at 15-kPa pressure. The enthalpy at this state is h2 hf x2hfg Then ¢h ¢ ke ¢ pe h2 h1 V2 2 V2 1 2 g 1z2 [225.94 1 2361.01 1 180 m> s 2 2 z1 2 (0.9)(2372.3)] kJ/kg 3248.4 2 kJ> kg 2 1 50 m> s 2 2 1 9.81 m> s2 2 3 1 6 2361.01 kJ/kg 887.39 kJ/kg a 1 kJ> kg 1000 m2> s2 10 2 m 4 a 1 kJ> kg b 1000 m2> s2 14.95 kJ/kg b 0.04 kJ/kg (b) The energy balance for this steady-flow system can be expressed in the rate form as # E in dE system> dt ¡ # E out Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies # Ein # m a h1 2 V1 2 0 (steady) 0 # Eout gz1 b # Wout # m a h2 V2 2 2 gz2 b # 1 since Q 02 . 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 1 h2 h1 2 [ 887.39 V2 2 V2 1 2 14.95 g 1z2 0.04] kJ/kg z1 2 d 1 ¢h ¢ ke ¢ pe 2 872.48 kJ/kg (c) The required mass flow rate for a 5-MW power output is # m # Wout wout 5000 kJ/s 872.48 kJ/kg 5.73 kg/s Discussion Two observations can be made from these results. First, the change in potential energy is insignificant in comparison to the changes in enthalpy and kinetic energy. This is typical for most engineering devices. Second, as a result of low 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 239 Chapter 5 | 239 3 Throttling Valves Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. Some familiar examples are ordinary adjustable valves, capillary tubes, and porous plugs (Fig. 529). 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 air-conditioning applications. The magnitude of the temperature drop (or, sometimes, the temperature rise) during a throttling process is governed by a property called the Joule-Thomson 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 single-stream steady-flow device reduces to h2 h1 1 kJ> kg 2 (a) An adjustable valve (b) A porous plug (c) A capillary tube FIGURE 529 Throttling valves are devices that cause large pressure drops in the fluid. (541) That is, enthalpy values at the inlet and exit of a throttling valve are the same. For this reason, a throttling valve is sometimes called an isenthalpic device. Note, however, that for throttling devices with large exposed surface areas such as capillary tubes, heat transfer may be significant. To gain some insight into how throttling affects fluid properties, let us express Eq. 541 as follows: u1 P1v1 u2 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. 530). EXAMPLE 58 Expansion of Refrigerant-134a in a Refrigerator Refrigerant-134a 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 Refrigerant-134a 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 530 The temperature of an ideal gas does not change during a throttling (h constant) process since h h(T ). cen84959_ch05.qxd 4/25/05 3:00 PM Page 240 240 | 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 531 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 flow-restricting 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. 531). At inlet: At exit: P1 0.8 MPa f sat. liquid P2 (h2 T1 h1 0.12 MPa h1) Tsat @ 0.8 MPa 31.31 C hf @ 0.8 MPa 95.47 kJ/kg ¡ hf hg (Table A12) 22.49 kJ/kg Tsat 236.97 kJ/kg 22.32 C Obviously hf h2 hg; thus, the refrigerant exists as a saturated mixture at the exit state. The quality at this state is x2 hf h2 h fg 95.47 236.97 22.49 22.49 0.340 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 1 22.32 31.31 2 °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 T-elbow FIGURE 532 The T-elbow of an ordinary shower serves as the mixing chamber for the hot- and the cold-water 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 T-elbow or a Y-elbow in a shower, for example, serves as the mixing chamber for the cold- and hot-water streams (Fig. 532). 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 241 Chapter 5 EXAMPLE 59 · m1 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. Solution In a shower, cold water is mixed with hot water at a specified 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 steady-flow 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. 533). 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 steady-flow system can be expressed in the rate form as follows: # min Energy balance: # mout # min # E in # E out Mixing chamber P = 20 psia T3 = 110°F · m3 T2 = 50°F · m2 FIGURE 533 Schematic for Example 59. 0 (steady) dmsystem> dt ¡ 0 # # # # mout S m1 m2 m3 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 # m1h1 241 T1 = 140°F Mixing of Hot and Cold Waters in a Shower Mass balance: | # Eout # # m3h3 1 since Q # m2h2 # 0, W 0, ke pe 02 T Combining the mass and energy balances, # m2h2 # (m1 # m2)h3 t. # m1h1 P = co ns . Dividing this equation by m2 yields yh1 h2 (y 1)h3 .. m1/m2 is the desired mass flow rate ratio. where y 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. 534). A compressed liquid can be approximated as a saturated liquid at the given temperature. Thus, h1 h2 h3 h f @ 140°F h3 h1 Compressed liquid states 107.99 Btu/lbm v h f @ 50°F 18.07 Btu/lbm h f @ 110°F 78.02 Btu/lbm FIGURE 534 A substance exists as a compressed liquid at temperatures below the saturation temperatures at the given pressure. Solving for y and substituting yields y Tsat h2 h3 78.02 107.99 18.07 78.02 2.0 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. cen84959_ch05.qxd 4/25/05 3:00 PM Page 242 242 | Thermodynamics 4b Heat Exchangers Fluid B 70°C Heat Fluid A 20°C 50°C Heat 35°C FIGURE 535 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 double-tube (also called tubeand-shell) heat exchanger, shown in Fig. 535. 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 direct-contact 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. 536). 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 510 Cooling of Refrigerant-134a by Water Refrigerant-134a 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 536 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 cen84959_ch05.qxd 4/25/05 3:01 PM Page 243 Chapter 5 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 steady-flow 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. 537). 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 multiple-stream steady-flow 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 steady-flow system can be expressed in the rate form as follows: # min Mass balance: # mout for each fluid stream since there is no mixing. Thus, # m1 # m3 # Energy balance: E in # m2 # m4 # mw # mR dE system> dt ¡ # E out Rate of change in internal, kinetic, potential, etc., energies # Eout # Ein # m3h3 # m 2h 2 0 Rate of net energy transfer by heat, work, and mass # m 1h 1 0 (steady) # 1 since Q # m4h4 # 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 h1 h f @ 15° C 62.982 kJ/kg h2 h f @ 25° C 104.83 kJ/kg (Table A4) The refrigerant enters the condenser as a superheated vapor and leaves as a compressed liquid at 35°C. From refrigerant-134a tables, P3 T3 1 MPa f 70 C h3 303.85 kJ/kg P4 T4 1 MPa f 35 C h4 hf @ 35 C (Table A13) 100.87 kJ/kg (Table A11) 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 Refrigerant-134a is cooled by water in a condenser. The mass flow | R-134a 3 70°C 1MPa 4 35°C 2 25°C FIGURE 537 Schematic for Example 510. cen84959_ch05.qxd 4/25/05 3:01 PM Page 244 244 | Thermodynamics Substituting, we find # mw 1 62.982 . . Qw,in = QR,out R-134a 104.83 2 kJ/kg # mw 1 6 kg/min 2 [ 1 100.87 303.85 2 kJ/kg] 29.1 kg/min (b) To determine the heat transfer from the refrigerant to the water, we have to choose a control volume whose boundary lies on the path of heat transfer. We can choose the volume occupied by either fluid as our control volume. For no particular reason, we choose the volume occupied by the water. All the assumptions stated earlier apply, except that the heat transfer is no longer zero. Then assuming heat to be transferred to water, the energy balance for this single-stream steady-flow system reduces to # E in dE system> dt ¡ # E out 0 (steady) 0 Control volume boundary FIGURE 538 In a heat exchanger, the heat transfer depends on the choice of the control volume. Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc., energies # Qw, in # Ein # mwh1 # Eout # mwh2 Rearranging and substituting, # Q w, in # mw 1 h2 h1 2 1 29.1 kg/min 2 [ 1 104.83 62.982 2 kJ/kg] 1218 kJ/min . Surroundings 20°C Qout Discussion Had we chosen the volume occupied by the refrigerant as . the control volume (Fig. 538), 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 539 Heat losses from a hot fluid flowing through an uninsulated pipe or duct to the cooler environment may be very significant. ˙ We Control volume ˙ Wsh FIGURE 540 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 steady-flow conditions and thus can be analyzed as a steady-flow process. This, of course, excludes the transient start-up and shut-down 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. 539). 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. 540). Of these, fan work is usually small and often neglected in energy analysis. cen84959_ch05.qxd 4/25/05 3:01 PM Page 245 Chapter 5 | 245 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 cross-sectional 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 511 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 15-kW 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 steady-flow 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 critical-point 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. 541). 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 air-conditioning applications, h can be replaced by cp T where cp 1.005 kJ/kg · °Cthe value at room temperaturewith negligible error (Fig. 542). Then the energy balance for this steady-flow system can be expressed in the rate form as # E in dE system> dt ¡ # E out 0 (steady) 0 T2 = ? ˙ We, in = 15 kW T1 = 17°C P1 = 100 kPa ˙ V1 = 150 m3/min FIGURE 541 Schematic for Example 511. Rate of net energy transfer by heat, work, and mass ˙ Qout = 200 W Rate of change in internal, kinetic, potential, etc., energies # E in # mh1 # Q out # We,in # We,in # E out # # Q out mh 2 1 since ¢ ke # mcp 1 T2 T1 2 ¢ pe 02 From the ideal-gas relation, the specific volume of air at the inlet of the duct is v1 RT1 P1 1 0.287 kPa # m3/kg # K 2 1 290 K 2 100 kPa # V1 v1 150 m3/min 1 min b a 60 s 0.832 m3/kg h = 1.005 T (kJ/kg) 0.832 m3/kg The mass flow rate of the air through the duct is determined from # m AIR 20 to 70°C 3.0 kg/s FIGURE 542 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. cen84959_ch05.qxd 4/25/05 3:01 PM Page 246 246 | Thermodynamics Substituting the known quantities, the exit temperature of the air is determined to be 1 15 kJ/s 2 1 0.2 kJ/s 2 T2 Discussion of air. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 5, SEC. 5 ON THE DVD. Supply line Control volume CV boundary FIGURE 543 Charging of a rigid tank from a supply line is an unsteady-flow process since it involves changes within the control volume. CV boundary Control volume 55 1 3 kg/s 2 1 1.005 kJ/kg # °C 2 1 T2 17 2 °C 21.9°C Note that heat loss from the duct reduces the exit temperature ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES During a steady-flow 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 unsteady-flow, or transientflow, processes. The steady-flow relations developed earlier are obviously not applicable to these processes. When an unsteady-flow 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 unsteady-flow processes are the charging of rigid vessels from supply lines (Fig. 543), 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 steady-flow processes, unsteady-flow 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 unsteady-flow 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 unsteady-flow systems is that steady-flow systems are fixed in space, size, and shape. Unsteady-flow systems, however, are not (Fig. 544). They are usually stationary; that is, they are fixed in space, but they may involve moving boundaries and thus boundary work. The mass balance for any system undergoing any process can be expressed as (see Sec. 51) min mout ¢ msystem 1 kg 2 (542) 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 544 The shape and size of a control volume may change during an unsteady-flow process. mi me 1 m2 m1 2 CV (543) where i inlet, e exit, 1 initial state, and 2 final state of the control volume. Often one or more terms in the equation above are zero. For exam- cen84959_ch05.qxd 4/25/05 3:01 PM Page 247 Chapter 5 | 247 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 unsteady-flow 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 unsteady-flow process, we must keep track of the energy content of the control volume as well as the energies of the incoming and outgoing flow streams. The general energy balance was given earlier as Energy balance: E in E out 1 kJ 2 ¢ E system Net energy transfer by heat, work, and mass (544) Change in internal, kinetic, potential, etc., energies The general unsteady-flow process, in general, is difficult to analyze because the properties of the mass at the inlets and exits may change during a process. Most unsteady-flow processes, however, can be represented reasonably well by the uniform-flow 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 steady-flow systems, the state of an unsteady-flow 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 uniform-flow system can be expressed explicitly as a Q in Win a mu b in a Q out Wout a mu b out 1 m2e2 m1e1 2 system W a mh out a mh in 1 m 2u 2 m 1u 1 2 system W Closed system Closed Q W = U Closed FIGURE 545 The energy equation of a uniform-flow system reduces to that of a closed system when all the inlets and exits are closed. (545) 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 Q Wb Moving boundary We (546) 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. 545). Also note that an unsteady-flow system may involve boundary work as well as electrical and shaft work (Fig. 546). Although both the steady-flow and uniform-flow processes are somewhat idealized, many actual processes can be approximated reasonably well by Wsh FIGURE 546 A uniform-flow system may involve electrical, shaft, and boundary work all at once. cen84959_ch05.qxd 4/25/05 3:01 PM Page 248 248 | 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 512 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 uniform-flow 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. 547). This is a control volume since mass crosses the system boundary during the process. We observe that this is an unsteady-flow 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 uniform-flow 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 547 Schematic for Example 512. P2 = 1 MPa (b) The closed-system equivalence cen84959_ch05.qxd 4/25/05 3:01 PM Page 249 Chapter 5 min mout Energy balance: ¢ msystem E in Mass balance: mi E out S 0 m1 m2 m2u2 Q 249 ¢ E system Net energy transfer by heat, work, and mass mihi m2 | Change in internal, kinetic, potential, etc., energies (since W Q 0, ke pe 0, m1 0) Combining the mass and energy balances gives hi u2 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 Ti 1 MPa f 300 C hi 3051.6 kJ/kg (Table A6) which is equal to u2. Since we now know two properties at the final state, it is fixed and the temperature at this state is determined from the same table to be P2 u2 1 MPa f 3051.6 kJ/kg T2 456.1°C 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 P v. Part of the energy represented by enthalpy is the flow energy P v, 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. 548). 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. 547b. 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 2 Wb,in Pi dV 1 Pi 1 V2 V1 2 Pi 3 Vtank 1 Vtank 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 out E in Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc., energies Wb,in miPivi u2 ¢U m 2u 2 ui miu i Pivi hi Steam Ti = 300°C T2 = 456.1°C FIGURE 548 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. cen84959_ch05.qxd 4/25/05 3:01 PM Page 250 250 | 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 uniform-flow analysis. Once again, the temperature rise is caused by the so-called flow energy or flow work, which is the energy required to move the fluid during flow. EXAMPLE 513 System boundary H2O m1 = 1 kg V=6L P = 75 kPa (gage) Vapor Liquid ˙ Qin = 500 W FIGURE 549 Schematic for Example 513. 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 uniform-flow 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. 549). This is a control volume since mass crosses the system boundary during the process. We observe that this is an unsteady-flow 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 FIGURE 550 As long as there is liquid in a pressure cooker, the saturation conditions exist and the temperature remains constant at the saturation temperature. Pgage Patm 75 100 175 kPa Since saturation conditions exist in the cooker at all times (Fig. 550), the cooking temperature must be the saturation temperature corresponding to this pressure. From Table A5, it is T Tsat @ 175 kPa 116.04°C which is about 16°C higher than the ordinary cooking temperature. cen84959_ch05.qxd 4/25/05 3:01 PM Page 251 Chapter 5 | 251 (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 uniform-flow system can be expressed as Mass balance: min mout ¢ msystem S me E in E out Energy balance: 1 m2 m 1 2 CV (m2u2 m2 2 CV mehe 1 m1 me ¢ E system Net energy transfer by heat, work, and mass Qin or Change in internal, kinetic, potential, etc., energies m1u1)CV (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 1 0.5 kJ/s 2 1 30 # Q in ¢ t Q in 60 s 2 900 kJ Steam leaves the pressure cooker as saturated vapor at 175 kPa at all times (Fig. 551). Thus, hg @ 175 kPa he P Sat. vapor 2700.2 kJ/kg The initial internal energy is found after the quality is determined: v1 x1 0.006 m3 1 kg V m1 vf v1 vfg Thus, u1 uf x1u fg 486.82 0.006 m3/kg 0.006 1.004 0.001 0.001 Sat. liquid 0.00499 1 0.00499 2 1 2037.7 2 kJ/kg 497 kJ/kg and U1 m1u1 (1 kg)(497 kJ/kg) The mass of the system at the final state is m2 into the energy equation yields Q in a m1 V bh v2 e a V u v2 2 497 kJ V/v2. Substituting this m 1u 1 b 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 u2 uf x2u fg 486.82 x2 1 1.004 0.001 2 m3/kg x2 1 2037.7 2 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 551 In a pressure cooker, the enthalpy of the exiting steam is hg @ 175 kPa (enthalpy of the saturated vapor at the given pressure). cen84959_ch05.qxd 4/25/05 3:01 PM Page 252 252 | Thermodynamics Thus, v2 1 0.009 2 1 1.004 0.001 0.001 2 m3/kg 0.010 m3/kg and 0.006 m3 0.01 m3/kg V v2 m2 0.6 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 # Q # W dE sys dt or # Q # W d dt re dV (547) sys . . . . where Q Qnet,in Qin Qout is the net rate of. heat transfer to the system . . . (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 unit-mass basis as e u ke pe u V2 2 gz (548) Note that total energy is a property, and its value does not change unless the state of the system changes. An energy interaction is heat if its driving force is a temperature 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 (549) where Wshaft is the work transmitted by a rotating shaft, Wpressure is the work done by the pressure forces on the control surface, Wviscous is the work done *This section can be skipped without a loss in continuity. cen84959_ch05.qxd 4/25/05 3:01 PM Page 253 Chapter 5 | 253 by the normal and shear components of viscous forces on the control surface, and Wother is the work done by other forces such as electric, magnetic, and surface tension, which are insignificant for simple compressible systems and are not considered in this text. We do not consider Wviscous either since it is usually small relative to other terms in control volume analysis. But it should be kept in mind that the work done by shear forces as the blades shear through the fluid may need to be considered in a refined analysis of turbomachinery. Work Done by Pressure Forces Consider a gas being compressed in the pistoncylinder device shown in Fig. 552a. When the piston moves down a differential distance ds under the influence of the pressure force PA, where A is the cross-sectional 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. 552b. 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 # dWpressure P dA 1 V # n 2 S P dA Vn S (550) 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 P 1V # n 2 dA S A S SS P r 1V # n 2 dA r A (551) 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 (552) S A S Then the rate form of the conservation of energy relation for a closed system becomes # Q net,in # Wshaft,net out # Wpressure,net out dE sys dt (553) To obtain a relation for the conservation of energy for a control volume, we apply the Reynolds transport theorem by replacing the extensive property B with total energy E, and its associated intensive property b with total P ds A Vpiston System (gas in cylinder) (a) dV P dm dA System n u V System boundary, A (b) FIGURE 552 The pressure force acting on (a) the moving boundary of a system in a pistoncylinder device, and (b) the differential surface area of a system of arbitrary shape. cen84959_ch05.qxd 4/25/05 3:01 PM Page 254 254 | Thermodynamics energy per unit mass e, which is e 553). This yields dBsys dt dE sys d = dt br dV + CV B=E br (Vr · n ) dA dt b=e dt er dV + CV er (Vr · n ) dA CS FIGURE 553 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). # Wshaft,net out Qnet,in In energyin m in , energyin Out mout , energyout Fixed control volume In CV CS gz (Fig. er 1 Vr # n 2 A S er dV V 2/2 u S (554) Out mout , energyout Out Wshaft, net,in mout , energyout FIGURE 554 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. # Wpressure,net out d dt er 1 Vr # n 2 dA S er dV CV CS S (555) which can be stated as The net rate of energy ° transfer into a CV by ¢ heat and work transfer The time rate of °change of the energy¢ content of the CV The net flow rate of °energy out of the control¢ surface by mass flow 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 control volume. Again noting that n is the outer normal of the dA, the quantity V r · n and thus mass flow is positive for outflow and negative for inflow. Substituting the surface integral for the rate of pressure work from Eq. 551 into Eq. 555 and combining it with the surface integral on the right give # Q net,in m in , pe Substituting the left-hand side of Eq. 553 into Eq. 554, the general form of the energy equation that applies to fixed, moving, or deforming control volumes becomes b=e d = ke CS # Q net,in dEsys d dt dt u d dt # Wshaft,net out er dV CV CS a P r e b r 1 Vr # n 2 dA (556) S S 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 no-slip 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 SS # r 1 Vr # n 2 dAc is the mass flow rate across an inlet integral. Noting that m 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. 554) # Q net,in # Wshaft,net out d dt er dV CV #P a ma r out eb #P a ma r in eb (557) cen84959_ch05.qxd 4/25/05 3:01 PM Page 255 Chapter 5 | 255 where e u V 2/2 gz is the total energy per unit mass for both the control volume and flow streams. Then, # Q net,in # Wshaft,net out d dt #P a ma r er dV u out CV #P a ma r V2 2 u in V2 2 gz b gz b (558) or # Q net,in # Wshaft,net out d dt er dV CV # a mah in # a mah out V2 2 gz b V2 2 gz b (559) 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 h ke pe h V2 2 gz 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 mh and amount and rate of energy transport become Emass . . 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 density of fluid, V average fluid velocity norwhere r cross-sectional area normal to flow direcmal to A, and A tion. 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: steady-flow processes and cen84959_ch05.qxd 4/25/05 3:01 PM Page 256 256 | Thermodynamics unsteady-flow processes. During a steady-flow 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 steady-flow 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 steady-flow processes are expressed as # am in # Q # am out # W # a mah gz b V2 2 # a mah gz b for each exit for each inlet Q These are the most general forms of the equations for steadyflow processes. For single-stream (one-inletone-exit) systems such as nozzles, diffusers, turbines, compressors, and pumps, they simplify to # m1 # Q # W 1 # m2 S VA v1 1 1 # m c h2 h1 V2 2 mout min ¢ msystem in out V2 2 In these relations, subscripts 1 and 2 denote the inlet and exit states, respectively. Most unsteady-flow 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 uniform-flow system are expressed as 1 VA v2 2 2 V2 1 2 g 1z2 z1 2 d W a mh out a mh in 1 m 2u 2 m1u 1 2 system Qnet,in Qin Qout is the net heat input and where Q Wnet,out Wout Win is the net work output. W When solving thermodynamic problems, it is recommended that the general form of the energy balance Eout Esystem be used for all problems, and simplify Ein 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 Air-Conditioning 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 51C 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 CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 52C Define mass and volume flow rates. How are they related to each other? 53C Does the amount of mass entering a control volume have to be equal to the amount of mass leaving during an unsteady-flow process? 54C When is the flow through a control volume steady? 55C 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? 56E A garden hose attached with a nozzle is used to fill a 20-gal bucket. The inner diameter of the hose is 1 in and it cen84959_ch05.qxd 4/25/05 3:01 PM Page 257 Chapter 5 | 257 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. 57 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 58 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 P512 1.05 kg/m3 1.20 kg/m3 FIGURE P58 513 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. 59E Air whose density is 0.078 lbm/ft3 enters the duct of an air-conditioning 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. 510 A 1-m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure 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 511 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. 512 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 P513 514 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 2.7-m-high, 200-m2 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. cen84959_ch05.qxd 4/27/05 5:42 PM Page 258 258 | Thermodynamics 515 Air enters a 28-cm 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 P515 516 Refrigerant-134a enters a 28-cm 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. 517 Consider a 300-L 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 2-cm 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 20-minute 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. 522 Refrigerant-134a 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. 523 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 524 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. Steady-Flow Energy Balance: Nozzles and Diffusers 525C How is a steady-flow system characterized? 526C Can a steady-flow system involve boundary work? 527C 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? 528C 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 P517 Flow Work and Energy Transfer by Mass 518C What are the different mechanisms for transferring energy to or from a control volume? 529C 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? 530 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 519C What is flow energy? Do fluids at rest possess any flow energy? 520C How do the energies of a flowing fluid and a fluid at rest compare? Name the specific forms of energy associated with each case. 521E 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 P530 cen84959_ch05.qxd 4/26/05 12:29 PM Page 259 Chapter 5 Reconsider Prob. 530. 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. | 259 531 P1 = 13 psia T1 = 20°F V1 = 600 ft/s 532 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. 533E 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 534 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. 535 Air at 600 kPa and 500 K enters an adiabatic nozzle that has an inlet-to-exit area ratio of 2:1 with a velocity of AIR P2 = 14.5 psia V2 < < V1 A2 = 5A1 FIGURE P537E leaves at 100 kPa and 450 m/s. The inlet area of the nozzle is 40 cm2. Determine (a) the inlet velocity and (b) the exit temperature. Answers: (a) 60.8 m/s, (b) 685.8 K 539 Refrigerant-134a at 700 kPa and 120°C enters an adiabatic nozzle steadily with a velocity of 20 m/s and leaves at 400 kPa and 30°C. Determine (a) the exit velocity and (b) the ratio of the inlet to exit area A1/A2. 540 Air at 80 kPa, 27°C, and 220 m/s enters a diffuser at a rate of 2.5 kg/s and leaves at 42°C. The exit area of the diffuser is 400 cm2. The air is estimated to lose heat at a rate of 18 kJ/s during this process. Determine (a) the exit velocity and (b) the exit pressure of the air. Answers: (a) 62.0 m/s, (b) 91.1 kPa 541 Nitrogen gas at 60 kPa and 7°C enters an adiabatic diffuser steadily with a velocity of 200 m/s and leaves at 85 kPa and 22°C. Determine (a) the exit velocity of the nitrogen and (b) the ratio of the inlet to exit area A1/A2. P1 = 3 MPa T1 = 400°C V1 = 40 m/s 542 STEAM P2 = 2.5 MPa V2 = 300 m/s FIGURE P534 120 m/s and leaves with a velocity of 380 m/s. Determine (a) the exit temperature and (b) the exit pressure of the air. Answers: (a) 436.5 K, (b) 330.8 kPa 536 Air at 80 kPa and 127°C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the airstream is decreased from 230 to 30 m/s as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the exit area of the diffuser. 537E Air at 13 psia and 20°F enters an adiabatic diffuser steadily with a velocity of 600 ft/s and leaves with a low velocity at a pressure of 14.5 psia. The exit area of the diffuser is 5 times the inlet area. Determine (a) the exit temperature and (b) the exit velocity of the air. 538 Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa and 500°C with a mass flow rate of 6000 kg/h and Reconsider Prob. 541. Using EES (or other) software, investigate the effect of the inlet velocity on the exit velocity and the ratio of the inlet-to-exit area. Let the inlet velocity vary from 180 to 260 m/s. Plot the final results against the inlet velocity, and discuss the results. 543 Refrigerant-134a enters a diffuser steadily as saturated vapor at 800 kPa with a velocity of 120 m/s, and it leaves at 900 kPa and 40°C. The refrigerant is gaining heat at a rate of 2 kJ/s as it passes through the diffuser. If the exit area is 80 percent greater than the inlet area, determine (a) the exit velocity and (b) the mass flow rate of the refrigerant. Answers: (a) 60.8 m/s, (b) 1.308 kg/s 544 Steam enters a nozzle at 400°C and 800 kPa with a velocity of 10 m/s, and leaves at 300°C and 200 kPa while losing heat at a rate of 25 kW. For an inlet area of 800 cm2, determine the velocity and the volume flow rate of the steam at the nozzle exit. Answers: 606 m/s, 2.74 m3/s 400°C 800 kPa 10 m/s 300°C 200 kPa Steam Q FIGURE P544 cen84959_ch05.qxd 4/25/05 3:01 PM Page 260 260 | Thermodynamics Turbines and Compressors 545C Consider an adiabatic turbine operating steadily. Does the work output of the turbine have to be equal to the decrease in the energy of the steam flowing through it? 546C Consider an air compressor operating steadily. How would you compare the volume flow rates of the air at the compressor inlet and exit? 547C Will the temperature of air rise as it is compressed by an adiabatic compressor? Why? 548C Somebody proposes the following system to cool a house in the summer: Compress the regular outdoor air, let it cool back to the outdoor temperature, pass it through a turbine, and discharge the cold air leaving the turbine into the house. From a thermodynamic point of view, is the proposed system sound? 549 Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit conditions are 10 kPa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the change in kinetic energy, (b) the power output, and (c) the turbine inlet area. Answers: (a) 1.95 kJ/kg, (b) 10.2 552E Steam flows steadily through a turbine at a rate of 45,000 lbm/h, entering at 1000 psia and 900°F and leaving at 5 psia as saturated vapor. If the power generated by the turbine is 4 MW, determine the rate of heat loss from the steam. 553 Steam enters an adiabatic turbine at 8 MPa and 500°C at a rate of 3 kg/s and leaves at 20 kPa. If the power output of the turbine is 2.5 MW, determine the temperature of the steam at the turbine exit. Neglect kinetic energy changes. Answer: 60.1°C 554 Argon gas enters an adiabatic turbine steadily at 900 kPa and 450°C with a velocity of 80 m/s and leaves at 150 kPa with a velocity of 150 m/s. The inlet area of the turbine is 60 cm2. If the power output of the turbine is 250 kW, determine the exit temperature of the argon. 555E Air flows steadily through an adiabatic turbine, entering at 150 psia, 900°F, and 350 ft/s and leaving at 20 psia, A1 = 60 cm2 P1 = 900 kPa T 1 = 450°C V2 = 80 m/s MW, (c) 0.00447 m2 550 Reconsider Prob. 549. Using EES (or other) software, investigate the effect of the turbine exit ARGON P1 = 10 MPa T1 = 450° C V1 = 80 m/s 250 kW P2 = 150 kPa V2 = 150 m/s STEAM m = 12 kg/s ˙ FIGURE P554 · Wout P 2 = 10 kPa x 2 = 0.92 V2 = 50 m/s FIGURE P549 pressure on the power output of the turbine. Let the exit pressure vary from 10 to 200 kPa. Plot the power output against the exit pressure, and discuss the results. 551 Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes in kinetic and potential energies, determine the mass flow rate required for a power output of 5 MW. Answer: 4.852 kg/s 300°F, and 700 ft/s. The inlet area of the turbine is 0.1 ft2. Determine (a) the mass flow rate of the air and (b) the power output of the turbine. 556 Refrigerant-134a enters an adiabatic compressor as saturated vapor at 24°C and leaves at 0.8 MPa and 60°C. The mass flow rate of the refrigerant is 1.2 kg/s. Determine (a) the power input to the compressor and (b) the volume flow rate of the refrigerant at the compressor inlet. 557 Air enters the compressor of a gas-turbine plant at ambient conditions of 100 kPa and 25°C with a low velocity and exits at 1 MPa and 347°C with a velocity of 90 m/s. The compressor is cooled at a rate of 1500 kJ/min, and the power input to the compressor is 250 kW. Determine the mass flow rate of air through the compressor. 558E Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by cen84959_ch05.qxd 4/26/05 12:29 PM Page 261 Chapter 5 circulating water through the compressor casing. The volume flow rate of the air at the inlet conditions is 5000 ft3/min, and the power input to the compressor is 700 hp. Determine (a) the mass flow rate of the air and (b) the temperature at the compressor exit. Answers: (a) 6.36 lbm/s, (b) 801 R | 261 566 Refrigerant-134a is throttled from the saturated liquid state at 700 kPa to a pressure of 160 kPa. Determine the temperature drop during this process and the final specific volume of the refrigerant. Answers: 42.3°C, 0.0344 m3/kg P1 = 700 kPa Sat. liquid 559E Reconsider Prob. 558E. Using EES (or other) software, investigate the effect of the rate of cooling of the compressor on the exit temperature of air. Let the cooling rate vary from 0 to 100 Btu/lbm. Plot the air exit temperature against the rate of cooling, and discuss the results. 560 Helium is to be compressed from 120 kPa and 310 K to 700 kPa and 430 K. A heat loss of 20 kJ/kg occurs during the compression process. Neglecting kinetic energy changes, determine the power input required for a mass flow rate of 90 kg/min. R-134a P2 = 160 kPa 20 kJ/kg P2 = 700 kPa T2 = 430 K He · m = 90 kg/min · W in FIGURE P566 567 Refrigerant-134a at 800 kPa and 25°C is throttled to a temperature of 20°C. Determine the pressure and the internal energy of the refrigerant at the final state. Answers: 133 kPa, 80.7 kJ/kg 568 A well-insulated valve is used to throttle steam from 8 MPa and 500°C to 6 MPa. Determine the final temperature of the steam. Answer: 490.1°C 569 P1 = 120 kPa T1 = 310 K FIGURE P560 561 Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 0.5 kg/s and leaves at 600 kPa and 450 K. Neglecting kinetic energy changes, determine (a) the volume flow rate of the carbon dioxide at the compressor inlet and (b) the power input to the compressor. Reconsider Prob. 568. Using EES (or other) software, investigate the effect of the exit pressure of steam on the exit temperature after throttling. Let the exit pressure vary from 6 to 1 MPa. Plot the exit temperature of steam against the exit pressure, and discuss the results. 570E Air at 200 psia and 90°F is throttled to the atmospheric pressure of 14.7 psia. Determine the final temperature of the air. 571 Carbon dioxide gas enters a throttling valve at 5 MPa and 100°C and leaves at 100 kPa. Determine the temperature change during this process if CO2 is assumed to be (a) an ideal gas and (b) a real gas. Answers: (a) 0.28 m3/s, (b) 68.8 kW Throttling Valves 562C Why are throttling devices commonly used in refrigeration and air-conditioning applications? 563C During a throttling process, the temperature of a fluid drops from 30 to 20°C. Can this process occur adiabatically? 564C Would you expect the temperature of air to drop as it undergoes a steady-flow throttling process? Explain. 565C Would you expect the temperature of a liquid to change as it is throttled? Explain. CO2 5 MPa 100°C 100 kPa FIGURE P571 Mixing Chambers and Heat Exchangers 572C When two fluid streams are mixed in a mixing chamber, can the mixture temperature be lower than the temperature of both streams? Explain. 573C Consider a steady-flow mixing process. Under what conditions will the energy transported into the control volume cen84959_ch05.qxd 4/26/05 12:29 PM Page 262 262 | Thermodynamics by the incoming streams be equal to the energy transported out of it by the outgoing stream? 574C Consider a steady-flow heat exchanger involving two different fluid streams. Under what conditions will the amount of heat lost by one fluid be equal to the amount of heat gained by the other? 575 A hot-water stream at 80°C enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold water at 20°C. If it is desired that the mixture leave the chamber at 42°C, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa. Answer: 0.865 kg/s . T1 = 8 m 1 = 0° 0.5 C kg /s =2 T2 .2 m H2 O (P = 250 kPa) 0° C T3 = 42°C 578E Water at 50°F and 50 psia is heated in a chamber by mixing it with saturated water vapor at 50 psia. If both streams enter the mixing chamber at the same mass flow rate, determine the temperature and the quality of the exiting stream. Answers: 281°F, 0.374 579 A stream of refrigerant-134a at 1 MPa and 12°C is mixed with another stream at 1 MPa and 60°C. If the mass flow rate of the cold stream is twice that of the hot one, determine the temperature and the quality of the exit stream. 580 Reconsider Prob. 579. Using EES (or other) software, investigate the effect of the mass flow rate of the cold stream of R-134a on the temperature and the quality of the exit stream. Let the ratio of the mass flow rate of the cold stream to that of the hot stream vary from 1 to 4. Plot the mixture temperature and quality against the cold-tohot mass flow rate ratio, and discuss the results. 581 Refrigerant-134a at 1 MPa and 90°C is to be cooled to 1 MPa and 30°C in a condenser by air. The air enters at 100 kPa and 27°C with a volume flow rate of 600 m3/min and leaves at 95 kPa and 60°C. Determine the mass flow rate of the refrigerant. Answer: 100 kg/min FIGURE P575 576 Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 1.8 kg/s. If the mixture leaves the mixing chamber at 60°C, determine the mass flow rate of the superheated steam required. R-134a AIR V3 = 600 m3/min P3 = 100 kPa T3 = 27°C P1 = 1 MPa T1 = 90°C Answer: 0.107 kg/s 577 In steam power plants, open feedwater heaters are frequently utilized to heat the feedwater by mixing it with steam bled off the turbine at some intermediate stage. Consider an open feedwater heater that operates at a pressure of 1000 kPa. Feedwater at 50°C and 1000 kPa is to be heated with superheated steam at 200°C and 1000 kPa. In an ideal feedwater heater, the mixture leaves the heater as saturated liquid at the feedwater pressure. Determine the ratio of the mass flow rates of the feedwater and the superheated vapor for this case. Answer: 3.73 . T1 = 5 m 1 =2 0 °C °C 00 H2O Sat. (P = 1000 kPa) liquid T2 .2 m P4 = 95 kPa T4 = 60°C FIGURE P581 582E Air enters the evaporator section of a window air conditioner at 14.7 psia and 90°F with a volume flow rate of 200 ft3/min. Refrigerant-134a at 20 psia with a quality of 30 percent enters the evaporator at a rate of 4 lbm/min and leaves as saturated vapor at the same pressure. Determine (a) the exit temperature of the air and (b) the rate of heat transfer from the air. 583 Refrigerant-134a at 700 kPa, 70°C, and 8 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the same pressure. The cooling water enters the condenser at 300 kPa and 15°C and leaves at 25°C at the same pressure. Determine the mass flow rate of the cooling water required to cool the refrigerant. Answer: 42.0 kg/min 584E FIGURE P577 P2 = 1 MPa T2 = 30°C In a steam heating system, air is heated by being passed over some tubes through which steam flows steadily. Steam enters the heat exchanger at 30 psia and 400°F at a rate of 15 lbm/min and leaves at 25 psia cen84959_ch05.qxd 4/25/05 3:01 PM Page 263 Chapter 5 and 212°F. Air enters at 14.7 psia and 80°F and leaves at 130°F. Determine the volume flow rate of air at the inlet. 585 Steam enters the condenser of a steam power plant at 20 kPa and a quality of 95 percent with a mass flow rate of 20,000 kg/h. It is to be cooled by water from a nearby river by circulating the water through the tubes within the condenser. To prevent thermal pollution, the river water is not allowed to experience a temperature rise above 10°C. If the steam is to leave the condenser as saturated liquid at 20 kPa, determine the mass flow rate of the cooling water required. Answer: 297.7 kg/s | 263 from a nearby lake, which enters the tubes of the condenser at 18°C at a rate of 101 kg/s and leaves at 27°C. Determine the rate of condensation of the steam in the condenser. Answer: 1.60 kg/s 587 Reconsider Prob. 586. Using EES (or other) software, investigate the effect of the inlet temperature of cooling water on the rate of condensation of steam. Let the inlet temperature vary from 10 to 20°C, and assume the exit temperature to remain constant. Plot the rate of condensation of steam against the inlet temperature of the cooling water, and discuss the results. 588 A heat exchanger is to heat water (cp 4.18 kJ/kg · °C) from 25 to 60°C at a rate of 0.2 kg/s. The heating is to be 4.31 kJ/kg · °C) accomplished by geothermal water (cp available at 140°C at a mass flow rate of 0.3 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of geothermal water. m3 = 20,000 kg/h P3 = 20 kPa x3 = 0.95 Steam Water T1 589 A heat exchanger is to cool ethylene glycol (cp 2.56 kJ/kg · °C) flowing at a rate of 2 kg/s from 80°C to 4.18 kJ/kg · °C) that enters at 20°C and 40°C by water (cp leaves at 55°C. Determine (a) the rate of heat transfer and (b) the mass flow rate of water. 590 T1 + 10 °C P4 = 20 kPa Sat. liquid FIGURE P585 586 Steam is to be condensed in the condenser of a steam power plant at a temperature of 50°C with cooling water Reconsider Prob. 589. Using EES (or other) software, investigate the effect of the inlet temperature of cooling water on the mass flow rate of water. Let the inlet temperature vary from 10 to 40°C, and assume the exit temperature to remain constant. Plot the mass flow rate of water against the inlet temperature, and discuss the results. 591 A thin-walled double-pipe counter-flow heat exchanger is used to cool oil (cp 2.20 kJ/kg · °C) from 150 to 40°C at 4.18 kJ/kg · °C) that enters at a rate of 2 kg/s by water (cp 22°C at a rate of 1.5 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of water. Hot oil 2 kg/s 150°C Steam 50°C Cold water Cooling water 18°C 1.5 kg/s 22°C 40°C FIGURE P591 27°C 50°C FIGURE P586 592 Cold water (cp 4.18 kJ/kg · °C) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at 15°C at a rate of 0.60 kg/s and is heated to 45°C by hot 4.19 kJ/kg · °C) that enters at 100°C at a rate of water (cp 3 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of the hot water. 593 Air (cp 1.005 kJ/kg · °C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters cen84959_ch05.qxd 4/25/05 3:01 PM Page 264 264 | Thermodynamics the furnace. Air enters the heat exchanger at 95 kPa and 20°C at a rate of 0.8 m3/s. The combustion gases (cp 1.10 kJ/kg · °C) enter at 180°C at a rate of 1.1 kg/s and leave at 95°C. Determine the rate of heat transfer to the air and its outlet temperature. Air 95 kPa 20°C 0.8 m3/s 597 Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at 2 MPa pressure. The exhaust gases enter the heat exchanger at 400°C at a rate of 32 kg/min while water enters at 15°C. The heat exchanger is not well insulated, and it is estimated that 10 percent of heat given up by the exhaust gases is lost to the surroundings. If the mass flow rate of the exhaust gases is 15 times that of the water, determine (a) the temperature of the exhaust gases at the heat exchanger exit and (b) the rate of heat transfer to the water. Use the constant specific heat properties of air for the exhaust gases. Exh. gas 400°C Heat exchanger Exhaust gases 1.1 kg/s 95°C 2 MPa sat. vap. FIGURE P593 595E Steam is to be condensed on the shell side of a heat exchanger at 85°F. Cooling water enters the tubes at 60°F at a rate of 138 lbm/s and leaves at 73°F. Assuming the heat exchanger to be well-insulated, determine the rate of heat transfer in the heat exchanger and the rate of condensation of the steam. 596 An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 5°C and 105 kPa at a rate of 1.25 m3/s while warm air enters at 34°C and 105 kPa. The air leaves the room at 24°C. The ratio of the mass flow rates of the hot to cold air streams is 1.6. Using variable specific heats, determine (a) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room. Cold air 5°C Room FIGURE P596 Water 15°C FIGURE P597 594 A well-insulated shell-and-tube heat exchanger is 4.18 kJ/kg · °C) in the tubes from used to heat water (cp 20 to 70°C at a rate of 4.5 kg/s. Heat is supplied by hot oil 2.30 kJ/kg · °C) that enters the shell side at 170°C at a (cp rate of 10 kg/s. Determine the rate of heat transfer in the heat exchanger and the exit temperature of oil. Warm air 34°C Q 24°C Pipe and Duct Flow 598 A desktop computer is to be cooled by a fan. The electronic components of the computer consume 60 W of power under full-load conditions. The computer is to operate in environments at temperatures up to 45°C and at elevations up to 3400 m where the average atmospheric pressure is 66.63 kPa. The exit temperature of air is not to exceed 60°C to meet the reliability requirements. Also, the average velocity of air is not to exceed 110 m/min at the exit of the computer case where the fan is installed to keep the noise level down. Determine the flow rate of the fan that needs to be installed and the diameter of the casing of the fan. 599 Repeat Prob. 598 for a computer that consumes 100 W of power. 5100E Water enters the tubes of a cold plate at 95°F with an average velocity of 60 ft/min and leaves at 105°F. The diameter of the tubes is 0.25 in. Assuming 15 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation, and the remaining 85 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. Answer: 263 W 5101 A sealed electronic box is to be cooled by tap water flowing through the channels on two of its sides. It is specified that the temperature rise of the water not exceed 4°C. The power dissipation of the box is 2 kW, which is removed entirely by water. If the box operates 24 hours a day, 365 days a year, determine the mass flow rate of water flowing through the box and the amount of cooling water used per year. 5102 Repeat Prob. 5101 for a power dissipation of 4 kW. cen84959_ch05.qxd 4/26/05 12:29 PM Page 265 Chapter 5 5103 A long roll of 2-m-wide and 0.5-cm-thick 1-Mn manganese steel plate (r 7854 kg/m3 and cp 0.434 kJ/kg · °C) coming off a furnace at 820°C is to be quenched in an oil bath at 45°C to a temperature of 51.1°C. If the metal sheet is moving at a steady velocity of 10 m/min, determine the required rate of heat removal from the oil to keep its temperature constant at 45°C. Answer: 4368 kW Furnace Steel plate 10 m/min Oil bath, 45°C FIGURE P5103 5104 Reconsider Prob. 5103. Using EES (or other) software, investigate the effect of the moving velocity of the steel plate on the rate of heat transfer from the oil bath. Let the velocity vary from 5 to 50 m/min. Plot the rate of heat transfer against the plate velocity, and discuss the results. | 265 400 Btu/h per ft length of the tube, determine the required length of the parabolic collector to meet the hot-water requirements of this house. 5108 Consider a hollow-core printed circuit board 12 cm high and 18 cm long, dissipating a total of 20 W. The width of the air gap in the middle of the PCB is 0.25 cm. If the cooling air enters the 12-cm-wide core at 32°C and 1 atm at a rate of 0.8 L/s, determine the average temperature at which the air leaves the hollow core. Answer: 53.4°C 5109 A computer cooled by a fan contains eight PCBs, each dissipating 10 W power. The height of the PCBs is 12 cm and the length is 18 cm. The cooling air is supplied by a 25-W fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed 10°C, determine (a) the flow rate of the air that the fan needs to deliver and (b) the fraction of the temperature rise of air that is due to the heat generated by the fan and its motor. Answers: (a) 0.0104 kg/s, (b) 24 percent Air outlet 5105 The components of an electronic system dissipating 180 W are located in a 1.4-m-long hori20 cm. The zontal duct whose cross section is 20 cm components in the duct are cooled by forced air that enters the duct at 30°C and 1 atm at a rate of 0.6 m3/min and leaves at 40°C. Determine the rate of heat transfer from the outer surfaces of the duct to the ambient. Answer: 63 W Natural convection Air inlet 40°C 25°C PCB, 10 W FIGURE P5109 180 W 30°C 0.6 m3/min 1.4 m FIGURE P5105 5110 Hot water at 90°C enters a 15-m section of a cast iron pipe whose inner diameter is 4 cm at an average velocity of 0.8 m/s. The outer surface of the pipe is exposed to the cold air at 10°C in a basement. If water leaves the basement at 88°C, determine the rate of heat loss from the water. 5111 5106 Repeat Prob. 5105 for a circular horizontal duct of diameter 10 cm. 5107E The hot-water needs of a household are to be met by heating water at 55°F to 180°F by a parabolic solar collector at a rate of 4 lbm/s. Water flows through a 1.25-in-diameter thin aluminum tube whose outer surface is black-anodized in order to maximize its solar absorption ability. The centerline of the tube coincides with the focal line of the collector, and a glass sleeve is placed outside the tube to minimize the heat losses. If solar energy is transferred to water at a net rate of Reconsider Prob. 5110. Using EES (or other) software, investigate the effect of the inner pipe diameter on the rate of heat loss. Let the pipe diameter vary from 1.5 to 7.5 cm. Plot the rate of heat loss against the diameter, and discuss the results. 5112 A 5-m 6-m 8-m room is to be heated by an electric resistance heater placed in a short duct in the room. Initially, the room is at 15°C, and the local atmospheric pressure is 98 kPa. The room is losing heat steadily to the outside at a rate of 200 kJ/min. A 200-W fan circulates the air steadily through the duct and the electric heater at an average cen84959_ch05.qxd 4/25/05 3:01 PM Page 266 266 | Thermodynamics mass flow rate of 50 kg/min. The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 15 min for the room air to reach an average temperature of 25°C, find (a) the power rating of the electric heater and (b) the temperature rise that the air experiences each time it passes through the heater. 5113 A house has an electric heating system that consists of a 300-W fan and an electric resistance heating element placed in a duct. Air flows steadily through the duct at a rate of 0.6 kg/s and experiences a temperature rise of 7°C. The rate of heat loss from the air in the duct is estimated to be 300 W. Determine the power rating of the electric resistance heating element. Answer: 4.22 kW 5114 A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. Air enters a 1200-W hair dryer at 100 kPa and 22°C and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglecting the power consumed by the fan and the heat losses through the walls of the hair dryer, determine (a) the volume flow rate of air at the inlet and (b) the velocity of the air at the exit. Answers: (a) 0.0404 m3/s, (b) 7.31 m/s T 2 = 47°C P1 = 100 kPa T 1 = 22° C A 2 = 60 cm 2 · We = 1200 W FIGURE P5114 Reconsider Prob. 5114. Using EES (or other) software, investigate the effect of the exit cross-sectional area of the hair dryer on the exit velocity. Let the exit area vary from 25 to 75 cm2. Plot the exit velocity against the exit cross-sectional area, and discuss the results. Include the effect of the flow kinetic energy in the analysis. the heater steadily at 20°C and leaves at 75°C, determine the mass flow rate of water. 5119 Steam enters a long, horizontal pipe with an inlet diameter of D1 12 cm at 1 MPa and 300°C with a velocity of 2 m/s. Farther downstream, the conditions are 800 kPa and 10 cm. Determine (a) the 250°C, and the diameter is D2 mass flow rate of the steam and (b) the rate of heat transfer. Answers: (a) 0.0877 kg/s, (b) 8.87 kJ/s 5120 Steam enters an insulated pipe at 200 kPa and 200°C and leaves at 150 kPa and 150°C. The inlet-to-outlet diameter ratio for the pipe is D1/D2 1.80. Determine the inlet and exit velocities of the steam. D1 200 kPa 200°C Steam FIGURE P5120 Charging and Discharging Processes 5121 A balloon that initially contains 50 m3 of steam at 100 kPa and 150°C is connected by a valve to a large reservoir that supplies steam at 150 kPa and 200°C. Now the valve is opened, and steam is allowed to enter the balloon until the pressure equilibrium with the steam at the supply line is reached. The material of the balloon is such that its volume increases linearly with pressure. Heat transfer also takes place between the balloon and the surroundings, and the mass of the steam in the balloon doubles at the end of the process. Determine the final temperature and the boundary work during this process. 5115 Steam 150 kPa 200°C 5116 The ducts of an air heating system pass through an unheated area. As a result of heat losses, the temperature of the air in the duct drops by 4°C. If the mass flow rate of air is 120 kg/min, determine the rate of heat loss from the air to the cold environment. 5117E Air enters the duct of an air-conditioning system at 15 psia and 50°F at a volume flow rate of 450 ft3/min. The diameter of the duct is 10 in, and heat is transferred to the air in the duct from the surroundings at a rate of 2 Btu/s. Determine (a) the velocity of the air at the duct inlet and (b) the temperature of the air at the exit. 5118 Water is heated in an insulated, constant-diameter tube by a 7-kW electric resistance heater. If the water enters D2 150 kPa 150°C Steam 50 m3 100 kPa 150°C FIGURE P5121 cen84959_ch05.qxd 4/25/05 3:01 PM Page 267 Chapter 5 5122 A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 4 MPa. Now the valve is opened, and steam is allowed to flow into the tank until the pressure reaches 4 MPa, at which point the valve is closed. If the final temperature of the steam in the tank is 550°C, determine the temperature of the steam in the supply line and the flow work per unit mass of the steam. Air 100 kPa 17°C 8L evacuated FIGURE P5125 5126 An insulated rigid tank is initially evacuated. A valve is opened, and atmospheric air at 95 kPa and 17°C enters the tank until the pressure in the tank reaches 95 kPa, at which point the valve is closed. Determine the final temperature of the air in the tank. Assume constant specific heats. Answer: 406 K FIGURE P5123 5124 A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries helium at 200 kPa and 120°C. Now the valve is opened, and helium is allowed to flow into the tank until the pressure reaches 200 kPa, at which point the valve is closed. Determine the flow work of the helium in the supply line and the final temperature of the helium in the tank. Answers: 816 kJ/kg, 655 K 200 kPa, 120°C Helium 267 AIR 5123 A vertical pistoncylinder device initially contains 0.25 m3 of air at 600 kPa and 300°C. A valve connected to the cylinder is now opened, and air is allowed to escape until three-quarters of the mass leave the cylinder at which point the volume is 0.05 m3. Determine the final temperature in the cylinder and the boundary work during this process. Air 0.25 m3 600 kPa 300°C | Initially evacuated FIGURE P5124 5125 Consider an 8-L evacuated rigid bottle that is surrounded by the atmosphere at 100 kPa and 17°C. A valve at the neck of the bottle is now opened and the atmospheric air is allowed to flow into the bottle. The air trapped in the bottle eventually reaches thermal equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle. The valve remains open during the process so that the trapped air also reaches mechanical equilibrium with the atmosphere. Determine the net heat transfer through the wall of the bottle 0.8 kJ during this filling process. Answer: Qout 5127 A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flowing in the supply line at 600 kPa and 22°C. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77°C. Determine (a) the mass of air that has entered the tank and (b) the amount 339 kJ of heat transfer. Answers: (a) 9.58 kg, (b) Qout P i = 600 kPa T i = 22° C V = 2 m3 Qout P1 = 100 kPa T1 = 22° C FIGURE P5127 5128 A 0.2-m3 rigid tank initially contains refrigerant-134a at 8°C. At this state, 70 percent of the mass is in the vapor phase, and the rest is in the liquid phase. The tank is connected by a valve to a supply line where refrigerant at 1 MPa and 100°C flows steadily. Now the valve is opened slightly, and the refrigerant is allowed to enter the tank. When the pressure in the tank reaches 800 kPa, the entire refrigerant in the cen84959_ch05.qxd 4/25/05 3:01 PM Page 268 268 | Thermodynamics tank exists in the vapor phase only. At this point the valve is closed. Determine (a) the final temperature in the tank, (b) the mass of refrigerant that has entered the tank, and (c) the heat transfer between the system and the surroundings. 3-ft3 5129E A rigid tank initially contains saturated water vapor at 300°F. The tank is connected by a valve to a supply line that carries steam at 200 psia and 400°F. Now the valve is opened, and steam is allowed to enter the tank. Heat transfer takes place with the surroundings such that the temperature in the tank remains constant at 300°F at all times. The valve is closed when it is observed that one-half of the volume of the tank is occupied by liquid water. Find (a) the final pressure in the tank, (b) the amount of steam that has entered the tank, and (c) the amount of heat transfer. Answers: observed that the tank contains saturated liquid at 1.2 MPa. Determine (a) the mass of the refrigerant that has entered the tank and (b) the amount of heat transfer. Answers: (a) 128.4 kg, (b) 1057 kJ 5133 A 0.3-m3 rigid tank is filled with saturated liquid water at 200°C. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank. Heat is transferred to the water such that the temperature in the tank remains constant. Determine the amount of heat that must be transferred by the time one-half of the total mass has been withdrawn. H 2O V = 0.3 m3 T = 200°C Sat. liquid (a) 67.03. psia, (b) 85.74 lbm, (c ) 80,900 Btu 5130 A vertical pistoncylinder device initially contains 0.01 m3 of steam at 200°C. The mass of the frictionless piston is such that it maintains a constant pressure of 500 kPa inside. Now steam at 1 MPa and 350°C is allowed to enter the cylinder from a supply line until the volume inside doubles. Neglecting any heat transfer that may have taken place during the process, determine (a) the final temperature of the steam in the cylinder and (b) the amount of mass that has entered. Answers: (a) 261.7°C, (b) 0.0176 kg 5131 An insulated, vertical pistoncylinder device initially contains 10 kg of water, 6 kg of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and 350°C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine (a) the final temperature in the cylinder and (b) the mass of the steam that has entered. Answers: (a) 120.2°C, (b) 19.07 kg Qin 1 me = m1 2 FIGURE P5133 5134 A 0.12-m3 rigid tank contains saturated refrigerant134a at 800 kPa. Initially, 25 percent of the volume is occupied by liquid and the rest by vapor. A valve at the bottom of the tank is now opened, and liquid is withdrawn from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when no liquid is left in the tank and vapor starts to come out. Determine the total heat transfer for this process. Answer: 201.2 kJ 5135E A 4-ft3 rigid tank contains saturated refrigerant134a at 100 psia. Initially, 20 percent of the volume is occupied by liquid and the rest by vapor. A valve at the top of the tank is now opened, and vapor is allowed to escape slowly from the tank. Heat is transferred to the refrigerant such that the pressure inside the tank remains constant. The valve is closed when the last drop of liquid in the tank is vaporized. Determine the total heat transfer for this process. P = 200 kPa m1 = 10 kg H 2O P i = 0.5 MPa T i = 350°C FIGURE P5131 0.12-m3 5132 A rigid tank initially contains refrigerant134a at 1 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant-134a at 1.2 MPa and 36°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is R-134a Sat. vapor P = 100 psia V = 4 ft3 FIGURE P5135E Qin cen84959_ch05.qxd 4/25/05 3:01 PM Page 269 Chapter 5 5136 A 0.2-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated. The regulator keeps the steam pressure constant by letting out some steam, but the temperature inside rises. Determine the amount of heat transferred when the steam temperature reaches 500°C. 5137 A 4-L pressure cooker has an operating pressure of 175 kPa. Initially, one-half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for 1 h, determine the highest rate of heat transfer allowed. | 269 constant by an electric resistance heater placed in the tank. Determine the electrical work done during this process. 5140 A vertical pistoncylinder device initially contains 0.2 m3 of air at 20°C. The mass of the piston is such that it maintains a constant pressure of 300 kPa inside. Now a valve connected to the cylinder is opened, and air is allowed to escape until the volume inside the cylinder is decreased by one-half. Heat transfer takes place during the process so that the temperature of the air in the cylinder remains constant. Determine (a) the amount of air that has left the cylinder and (b) the amount of heat transfer. Answers: (a) 0.357 kg, (b) 0 5141 A balloon initially contains 65 m3 of helium gas at atmospheric conditions of 100 kPa and 22°C. The balloon is connected by a valve to a large reservoir that supplies helium gas at 150 kPa and 25°C. Now the valve is opened, and helium is allowed to enter the balloon until pressure equilibrium with the helium at the supply line is reached. The material of the balloon is such that its volume increases linearly with pressure. If no heat transfer takes place during this process, determine the final temperature in the balloon. Answer: 334 K Ti = 25° C Pi = 150 kPa V=4L (P = 175 kPa) · Q in He P1 = 100 kPa T1 = 22° C FIGURE P5137 5138 An insulated 0.08-m3 tank contains helium at 2 MPa and 80°C. A valve is now opened, allowing some helium to escape. The valve is closed when one-half of the initial mass has escaped. Determine the final temperature and pressure in the tank. Answers: 225 K, 637 kPa 5139E An insulated 60-ft3 rigid tank contains air at 75 psia and 120°F. A valve connected to the tank is now opened, and air is allowed to escape until the pressure inside drops to 30 psia. The air temperature during this process is maintained FIGURE P5141 5142 An insulated vertical pistoncylinder device initially contains 0.8 m3 of refrigerant-134a at 1.2 MPa and 120°C. A linear spring at this point applies full force to the piston. A valve connected to the cylinder is now opened, and refrigerant AIR V = 60 ft3 P = 75 psia T = 120°F FIGURE P5139E We,in R-134a 0.8 m3 1.2 MPa 120°C FIGURE P5142 cen84959_ch05.qxd 4/25/05 3:01 PM Page 270 270 | Thermodynamics is allowed to escape. The spring unwinds as the piston moves down, and the pressure and volume drop to 0.6 MPa and 0.5 m3 at the end of the process. Determine (a) the amount of refrigerant that has escaped and (b) the final temperature of the refrigerant. 5143 A 2-m3 rigid insulated tank initially containing saturated water vapor at 1 MPa is connected through a valve to a supply line that carries steam at 400°C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be 300°C. Determine the mass of the steam that has entered and the pressure of the steam in the supply line. 400°C Steam 0.015, and the discharge velocity is expressed as 2gz V where z is the water height above the B 1.5 fL> 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. to be f 5146 Underground water is being pumped into a pool 4 m while water is discharged whose cross section is 3 m through a 5-cm-diameter 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. 5147 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. 5148 Air at 4.18 kg/m3 enters a nozzle that has an inlet-toexit 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 P5143 5144 A pistoncylinder 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. 5149 The air in a 6-m 5-m 4-m 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. 5150 A long roll of 1-m-wide and 0.5-cm-thick 1-Mn man7854 kg/m3) coming off a furnace is ganese steel plate (r 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 P5144 Review Problems 5145 A D0 10-m-diameter tank is initially filled with 10-cm-diameter valve water 2 m above the center of a D near the bottom. The tank surface is open to the atmosphere, 100-m-long pipe conand the tank drains through a L nected to the valve. The friction factor of the pipe is given FIGURE P5150 5151E 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 cen84959_ch05.qxd 4/25/05 3:01 PM Page 271 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 5152 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. 1.8. Using conThe diameter ratio for the pipe is D1/D2 stant specific heats for air, determine the inlet and exit velocities of the air. Answers: 28.6 m/s, 120 m/s 5153 In a single-flash 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. 5156 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. 5157 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? 5158 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 P5153 5154 The hot-water needs of a household are met by a 60L electric water heater whose heaters are rated at 1.6 kW. The hot-water 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 P5158 5159 Water is boiled at 100°C electrically by a 3-kW 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 P5154 5155 In a gas-fired boiler, water is boiled at 150°C by hot gases flowing through a stainless steel pipe submerged in FIGURE P5159 5160 Two streams of the same ideal gas having different mass flow rates and temperatures are mixed in a steady-flow, adiabatic mixing device. Assuming constant specific heats, cen84959_ch05.qxd 4/25/05 3:01 PM Page 272 272 | Thermodynamics find the simplest expression for the mixture temperature written in the form # # m1 m2 T3 f a # , # , T1, T2 b m3 m3 m1, T1 m2, T2 Mixing device m3, T3 FIGURE P5160 5161 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. 5162 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. 5163 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. 5164 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 5165 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 continuous-flow-type 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. 5166 Repeat Prob. 5165 assuming heat gain of the chiller is negligible. 5167 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 natural-gas-fired 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 nat105,500 kJ), determine ural gas is $1.10/therm (1 therm 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 P5167 FIGURE P5163 5168E 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 cen84959_ch05.qxd 4/25/05 3:01 PM Page 273 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. 5169 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 P5169 5170 A glass bottle washing facility uses a well-agitated hot-water 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. Make-up 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. 5170 for a water bath temperature of 5172 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. 5173 Repeat Prob. 5172 for a copper wire (r 0.383 kJ/kg · °C). kg/m3 and cp 8950 5174 Steam at 40°C condenses on the outside of a 5-mlong, 3-cm-diameter 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 P5174 15 kJ/kg Dough 5171 50°C. 273 25°C Qout 15°C | Tair = 30°C 10 m/min Aluminum wire FIGURE P5172 5175E 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. 5176 Saturated refrigerant-134a vapor at 34°C is to be condensed as it flows in a 1-cm-diameter 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? 5177E 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 9-ft-high, 3000-ft2 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. 5178 Determine the rate of sensible heat loss from a building due to infiltration if the outdoor air at 5°C and 90 kPa cen84959_ch05.qxd 4/25/05 3:01 PM Page 274 274 | Thermodynamics enters the building at a rate of 35 L/s when the indoors is maintained at 20°C. 5179 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 low-flow 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 T-elbow 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 T-elbow and (b) the amount of electricity that will be saved per year, in kWh, by replacing the standard shower heads by the low-flow ones. 5180 Reconsider Prob. 5179. 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. 5181 A fan is powered by a 0.5-hp 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. 5182 An air-conditioning 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. 5184 An adiabatic air compressor is to be powered by a direct-coupled 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 P5184 5185 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 P5182 5183 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 P5185 cen84959_ch05.qxd 4/26/05 12:29 PM Page 275 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 10-min shower as a result of installing this heat exchanger. 5186 Reconsider Prob. 5185. 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. 5187 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. 5188 Reconsider Prob. 5187. 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. 5189E Refrigerant-134a 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 5191 It is proposed to have a water heater that consists of an insulated pipe of 5-cm 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. 5192 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. 5193 A building with an internal volume of 400 m3 is to be heated by a 30-kW 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 250-W 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 R-134a 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 P5193 FIGURE P5189E 5190 In large gas-turbine 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 5194 An insulated vertical pistoncylinder 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 cen84959_ch05.qxd 4/25/05 3:01 PM Page 276 276 | Thermodynamics opened, and air from the high-pressure 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. 5197 The pump of a water distribution system is powered by a 15-kW 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 P5194 5195 A pistoncylinder device initially contains 2 kg of refrigerant-134a 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 500-kPa pressure is required to move it. A valve at the bottom of the tank is opened, and R-134a 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 5196 A pistoncylinder 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 P5197 5198 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. 5199 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 P5196 Turbine Exhaust gases 400°C 120 kPa 50°C 100 kPa Air Compressor 130 kPa 30°C Cold air Aftercooler FIGURE P5199 40°C cen84959_ch05.qxd 4/25/05 3:01 PM Page 277 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 5200 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 5201 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 5202 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 5203 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 5204 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 5205 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 5206 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 5207 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 5208 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 5209 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 5210 Refrigerant-134a 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 5211 Refrigerant-134a 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 cen84959_ch05.qxd 4/25/05 3:01 PM Page 278 278 | Thermodynamics 5212 Refrigerant-134a 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 5213 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 5214 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/kg (c) 0.375 m 5215 Air is to be heated steadily by an 8-kW 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 5216 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 5217 Design a 1200-W electric hair dryer such that the air temperature and velocity in the dryer will not exceed 50°C and 3 m/s, respectively. 5218 Design a scalding unit for slaughtered chickens to loosen their feathers before they are routed to feather-picking 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 slow-moving conveyor. The chicken is expected to leave the tank 15 percent heavier as a result of the water that sticks to its surface. The center-tocenter 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 well-insulated. 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. cen84959_ch06.qxd 3/31/05 3:51 PM Page 279 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 KelvinPlanck and Clausius statements of the second law of thermodynamics. Discuss the concepts of perpetual-motion 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. | 279 cen84959_ch06.qxd 4/25/05 3:10 PM Page 280 280 | Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 1 ON THE DVD. HOT COFFEE Heat FIGURE 61 A cup of hot coffee does not get hotter in a cooler room. Heat I=0 FIGURE 62 Transferring heat to a wire will not generate electricity. Heat FIGURE 63 Transferring heat to a paddle wheel will not cause it to rotate. 61 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. 61). 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 processthe hot coffee getting even hotter in a cooler room as a result of heat transfer from the room air. We all know that this process never takes place. Yet, doing so would not violate the first law as long as the amount of energy lost by the air is equal to the amount gained by the coffee. As another familiar example, consider the heating of a room by the passage of electric current through a resistor (Fig. 62). Again, the first law dictates that the amount of electric energy supplied to the resistance wires be equal to the amount of energy transferred to the room air as heat. Now let us 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 paddle-wheel mechanism that is operated by the fall of a mass (Fig. 63). The paddle wheel rotates as the mass falls and stirs a fluid within an insulated container. As a result, the potential energy of the mass decreases, and the internal energy of the fluid increases in accordance with the conservation of energy principle. However, the reverse process, raising the mass by transferring heat from the fluid to the paddle wheel, does not occur in nature, although doing so would not violate the first law of thermodynamics. It is clear from these arguments that processes proceed in a certain direction and not in the reverse direction (Fig. 64). The first law places no restriction on the direction of a process, but satisfying the first law does not ensure that the process can actually occur. This inadequacy of the first law to identify whether a process can take place is remedied by introducing another general principle, the second law of thermodynamics. We show later in this chapter that the reverse processes 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. 65). 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 cen84959_ch06.qxd 4/25/05 3:10 PM Page 281 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 high-temperature 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. 62 | 281 ONE WAY FIGURE 64 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. 66). The atmosphere, for example, does not warm up as a result of heat losses from residential buildings in winter. Likewise, megajoules of waste energy dumped in large rivers by power plants do not cause any significant change in water temperature. A two-phase 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. 67). Thermal energy reservoirs are often referred to as heat reservoirs since they supply or absorb energy in the form of heat. Heat transfer from industrial sources to the environment is of major 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 65 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 66 Bodies with relatively large thermal masses can be modeled as thermal energy reservoirs. Thermal energy SOURCE HEAT HEAT Thermal energy SINK FIGURE 67 A source supplies energy in the form of heat, and a sink absorbs it. cen84959_ch06.qxd 4/25/05 3:10 PM Page 282 282 | Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 3 ON THE DVD. Work No work Heat Heat WATER WATER FIGURE 68 Work can always be converted to heat directly and completely, but the reverse is not true. High-temperature SOURCE Qin HEAT ENGINE Wnet,out Qout Low-temperature SINK FIGURE 69 Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink. 63 HEAT ENGINES As pointed out earlier, work can easily be converted to other forms of energy, but converting other forms of energy to work is not that easy. The mechanical work done by the shaft shown in Fig. 68, for example, is first converted to the internal energy of the water. This energy may then leave the water as heat. We know from experience that any attempt to reverse this process will fail. That is, transferring heat to the water does not cause the shaft to rotate. From this and other observations, we conclude that work can be converted to heat directly and completely, but converting heat to work requires the use of some special devices. These devices are called heat engines. Heat engines differ considerably from one another, but all can be characterized by the following (Fig. 69): 1. They receive heat from a high-temperature 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 low-temperature 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 air-and-fuel mixture at the end of the cycle. The work-producing device that best fits into the definition of a heat engine is the steam power plant, which is an external-combustion engine. That is, combustion takes place outside the engine, and the thermal energy released during this process is transferred to the steam as heat. The schematic of a basic steam power plant is shown in Fig. 610. This is a rather simplified diagram, and the discussion of actual steam power plants is given in later chapters. The various quantities shown on this figure are as follows: Qin Qout Wout Win amount of heat supplied to steam in boiler from a high-temperature source (furnace) amount of heat rejected from steam in condenser to a lowtemperature sink (the atmosphere, a river, etc.) amount of work delivered by steam as it expands in turbine 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. cen84959_ch06.qxd 3/31/05 3:51 PM Page 283 Chapter 6 | 283 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 610 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. 611): Wnet,out Wout Win 1 kJ 2 (61) The net work can also be determined from the heat transfer data alone. The four components of the steam power plant involve mass flow in and out, and therefore 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. 610; 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 1 kJ 2 Win FIGURE 611 A portion of the work output of a heat engine is consumed internally to maintain continuous operation. SOURCE Heat input 100 kJ 100 kJ (62) 1 Thermal Efficiency In Eq. 62, Qout represents the magnitude of the energy wasted in order to complete the cycle. But Qout is never zero; thus, the net work output of a heat engine is always less than the amount of heat input. That is, only part of the heat transferred to the heat engine is converted to work. The fraction of the heat input that is converted to net work output is a measure of the performance of a heat engine and is called the thermal efficiency hth (Fig. 612). 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 Wnet,out Net work output Total heat input (63) 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 612 Some heat engines perform better than others (convert more of the heat they receive to work). cen84959_ch06.qxd 3/31/05 3:51 PM Page 284 284 | Thermodynamics or Wnet,out hth (64) Qin It can also be expressed as h th High-temperature reservoir at TH QH QL HE QL magnitude of heat transfer between the cyclic device and the hightemperature medium at temperature TH magnitude of heat transfer between the cyclic device and the lowtemperature medium at temperature TL Notice that both QL and QH are defined as magnitudes and therefore are positive quantities. The direction of QH and QL is easily determined by inspection. Then the net work output and thermal efficiency relations for any heat engine (shown in Fig. 613) can also be expressed as Wnet,out Low-temperature reservoir at TL FIGURE 613 Schematic of a heat engine. QH = 100 MJ Wnet,out = 55 MJ HE QL = 45 MJ The atmosphere FIGURE 614 Even the most efficient heat engines reject almost one-half of the energy they receive as waste heat. QH QL and hth Wnet,out QH or h th Furnace (65) since Wnet,out Qin Qout. Cyclic devices of practical interest such as heat engines, refrigerators, and heat pumps operate between a high-temperature medium (or reservoir) at temperature TH and a low-temperature 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 Wnet,out Q out Q in 1 1 QL QH (66) The thermal efficiency of a heat engine is always less than unity since both QL and QH are defined as positive quantities. Thermal efficiency is a measure of how efficiently a heat engine converts the heat that it receives to work. Heat engines are built for the purpose of converting heat to work, and engineers are constantly trying to improve the efficiencies of these devices since increased efficiency means less fuel consumption and thus lower fuel bills and less pollution. The thermal efficiencies of work-producing devices are relatively low. Ordinary spark-ignition 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 gas-turbine plants and as high as 60 percent for large combined gas-steam power plants. Thus, even with the most efficient heat engines available today, almost one-half of the energy supplied ends up in the rivers, lakes, or the atmosphere as waste or useless energy (Fig. 614). cen84959_ch06.qxd 3/31/05 3:51 PM Page 285 Chapter 6 | 285 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. 615 that is used to lift weights. It consists of a pistoncylinder device with two sets of stops. The working fluid is the gas contained within the cylinder. Initially, the gas 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 quasi-equilibrium 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 low-temperature 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 low-temperature 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 low-temperature 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 615 A heat-engine cycle cannot be completed without rejecting some heat to a low-temperature sink. cen84959_ch06.qxd 3/31/05 3:51 PM Page 286 286 | 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 KelvinPlanck expression of the second law of thermodynamics discussed later in this section. EXAMPLE 61 Net Power Production of a Heat Engine Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine. 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. 616. The furnace serves as the high-temperature reservoir for this heat engine and the river as the low-temperature reservoir. The given quantities can be expressed as # QH · QL = 50 MW 80 MW and # QL 50 MW The net power output of this heat engine is # Wnet,out RIVER FIGURE 616 Schematic for Example 61. # QH # QL 1 80 50 2 MW 30 MW Then the thermal efficiency is easily determined to be hth # Wnet,out # QH 30 MW 80 MW 0.375 1 or 37.5% 2 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 617 Schematic for Example 62. EXAMPLE 62 Fuel Consumption Rate of a Car A car engine with a power output of 65 hp has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 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. 617. The car engine is powered by converting 24 percent of the chemical energy released during the combustion process to work. The amount of energy input required to produce a power output of 65 hp is determined from the definition of thermal efficiency to be # QH Wnet,out h th 65 hp 2545 Btu> h a b 0.24 1 hp 689,270 Btu> h cen84959_ch06.qxd 4/25/05 3:10 PM Page 287 Chapter 6 | 287 To supply energy at this rate, the engine must burn fuel at a rate of # m 689,270 Btu> h 19,000 Btu> lbm 36.3 lbm/h 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: KelvinPlanck Statement We have demonstrated earlier with reference to the heat engine shown in Fig. 615 that, even under ideal conditions, a heat engine must reject some heat to a low-temperature reservoir in order to complete the cycle. That is, no heat engine can convert all the heat it receives to useful work. This limitation on the thermal efficiency of heat engines forms the basis for the KelvinPlanck 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 low-temperature sink as well as a high-temperature source to keep operating. The KelvinPlanck statement can also be expressed as no heat engine can have a thermal efficiency of 100 percent (Fig. 618), or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. Note that the impossibility of having a 100 percent efficient heat engine is not due to friction or other dissipative effects. It is a limitation that applies to both the idealized and the actual heat engines. Later in this chapter, we develop a relation for the maximum thermal efficiency of a heat engine. We also demonstrate that this maximum value depends on the reservoir temperatures only. 64 REFRIGERATORS AND HEAT PUMPS We all know from experience that heat is transferred in the direction of decreasing temperature, that is, from high-temperature 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 low-temperature medium to a high-temperature 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 vapor-compression refrigeration cycle, which involves four main components: a compressor, a condenser, an expansion valve, and an evaporator, as shown in Fig. 619. · Wnet,out = 100 kW HEAT ENGINE · QL = 0 FIGURE 618 A heat engine that violates the KelvinPlanck statement of the second law. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 4 ON THE DVD. cen84959_ch06.qxd 3/31/05 3:51 PM Page 288 288 | 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 619 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 620 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 low-temperature 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. 620. 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 Required input QL Wnet,in (67) . 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 1 kJ 2 (68) cen84959_ch06.qxd 3/31/05 5:53 PM Page 289 Chapter 6 Then the COP relation becomes COPR QL QH QL 1 Q H> Q L | 289 Warm heated space at TH > TL 1 (69) 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 termthe coefficient of performanceis 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 low-temperature medium to a high-temperature one is the heat pump, shown schematically in Fig. 621. 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 higher-temperature medium is merely a necessary part of the operation, not the purpose. The objective of a heat pump, however, is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low-temperature source, such as well water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a house (Fig. 622). An ordinary refrigerator that is placed in the window of a house with its door open to the cold outside air in winter will function as a heat pump since it will try to cool the outside by absorbing heat from it and rejecting this heat into the house through the coils behind it (Fig. 623). The measure of performance of a heat pump is also expressed in terms of the coefficient of performance COPHP, defined as Desired output Required input COPHP Cold environment at TL FIGURE 621 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 QH Wnet,in (610) 1 Q L >Q H (611) COP = 3.5 HP which can also be expressed as COPHP QH QH QL 1 Cold outdoors at 4°C A comparison of Eqs. 67 and 610 reveals that COPHP COPR 1 QL = 5 kJ (612) 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 622 The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors. cen84959_ch06.qxd 3/31/05 3:51 PM Page 290 290 | Thermodynamics FIGURE 623 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 air-source heat pumps. The COP of such heat pumps is about 3.0 at design conditions. Air-source 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 ground-source) 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 air-source heat pumps). The COP of ground-source 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 air-conditioning unit can be used as a heat pump in winter by installing it backwards as shown in Fig. 623. In this mode, the unit absorbs heat from the cold outside and delivers it to the room. Air-conditioning 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 Btus 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 high-efficiency heat pump manufactured by the Trane Company using a reciprocating variable-speed 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. Variable-speed 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 air-conditioning 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 2.63.0 for cutting and preparation rooms; 2.32.6 for meat, deli, dairy, and produce; 1.21.5 for frozen foods; and 1.01.2 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. cen84959_ch06.qxd 3/31/05 3:51 PM Page 291 Chapter 6 EXAMPLE 63 Heat Rejection by a Refrigerator | 291 Kitchen The food compartment of a refrigerator, shown in Fig. 624, 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 COPR # QL # Wnet,in 360 kJ> min 2 kW a 1 kW b 60 kJ> min 3 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, # QH # QL # Wnet,in 360 kJ> min 1 2 kW 2 a 60 kJ> min 1 kW b · QL = 360 kJ/min Food compartment 4°C FIGURE 624 Schematic for Example 63. 480 kJ/min Discussion Notice that both the energy removed from the refrigerated space as heat and the energy supplied to the refrigerator as electrical work eventually show up in the room air and become part of the internal energy of the air. This demonstrates that energy can change from one form to another, can move from one place to another, but is never destroyed during a process. EXAMPLE 64 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. 625, is determined from the definition of the coefficient of performance to be # Wnet,in # QH COPHP 80,000 kJ> h 2.5 32,000 kJ/h 1 or 8.9 kW 2 (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 · QL = ? Outdoor air at 2°C FIGURE 625 Schematic for Example 64. cen84959_ch06.qxd 3/31/05 3:51 PM Page 292 292 | 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 1 80,000 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 32,000-kJ/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 lawthe KelvinPlanck 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 626 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 lower-temperature body to a higher-temperature 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. 626). 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 KelvinPlanck 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 KelvinPlanck 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 KelvinPlanck statement also violates the Clausius statement, and vice versa. This can be demonstrated as follows. cen84959_ch06.qxd 4/25/05 3:10 PM Page 293 Chapter 6 High-temperature reservoir at TH QH HEAT ENGINE η th = 100% QL REFRIGERATOR REFRIGERATOR QL = QH 293 High-temperature reservoir at TH QH + QL Wnet | QL Low-temperature reservoir at TL Low-temperature reservoir at TL (a) A refrigerator that is powered by a 100 percent efficient heat engine (b) The equivalent refrigerator FIGURE 627 Proof that the violation of the KelvinPlanck statement leads to the violation of the Clausius statement. Consider the heat-engine-refrigerator combination shown in Fig. 627a, operating between the same two reservoirs. The heat engine is assumed to have, in violation of the KelvinPlanck statement, a thermal efficiency of 100 percent, and therefore it converts all the heat QH it receives to work W. This work is now supplied to a refrigerator that removes heat in the amount of QL from the low-temperature reservoir and rejects heat in the amount of QH to the high-temperature reservoir. During this process, the highQL temperature reservoir receives a net amount of heat QL (the difference between QL QH and QH). Thus, the combination of these two devices can be viewed as a refrigerator, as shown in Fig. 627b, that transfers heat in an amount of QL from a cooler body to a warmer one without requiring any input from outside. This is clearly a violation of the Clausius statement. Therefore, a violation of the KelvinPlanck statement results in the violation of the Clausius statement. It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the KelvinPlanck statement. Therefore, the Clausius and the KelvinPlanck statements are two equivalent expressions of the second law of thermodynamics. 65 PERPETUAL-MOTION 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 perpetual-motion machine, and despite numerous attempts, no perpetual-motion 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 perpetual-motion machine of the first kind (PMM1), and a device that violates the second law of thermodynamics is called a perpetual-motion machine of the second kind (PMM2). INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 5 ON THE DVD. cen84959_ch06.qxd 3/31/05 3:51 PM Page 294 294 | Thermodynamics · Wnet,out System boundary BOILER Resistance heater TURBINE PUMP FIGURE 628 A perpetual-motion machine that violates the first law of thermodynamics (PMM1). GENERATOR CONDENSER · Qout Consider the steam power plant shown in Fig. 628. It is proposed to heat the steam by resistance heaters placed inside the boiler, instead of by the energy supplied from fossil or nuclear fuels. Part of the electricity generated by the plant is to be used to power the resistors as well as the pump. The rest of the electric energy is to be supplied to the electric network as the net work output. The inventor claims that once the system is started, this power plant will produce electricity indefinitely without requiring any energy input from the outside. Well, here is an invention that could solve the worlds energy problemif it works, of course. A careful examination of this invention reveals that the system enclosed by the shaded area is continuously supplying energy to the . . outside at a rate of Qout Wnet,out without receiving any energy. That is, this . . 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 one-half 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. 629. 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 cen84959_ch06.qxd 3/31/05 3:51 PM Page 295 Chapter 6 System boundary | · Qin BOILER · Wnet,out PUMP TURBINE FIGURE 629 A perpetual-motion 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 perpetual-motion 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 perpetual-motion 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 hydropneumatic-pulsating-vacu-engine, 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 perpetual-motion 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 battery-operated. However, it did not convince the inventor that his machine will not work. The proposers of perpetual-motion 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. 295 cen84959_ch06.qxd 4/25/05 3:10 PM Page 296 296 | Thermodynamics INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 6 ON THE DVD. (a) Frictionless pendulum (b) Quasi-equilibrium expansion and compression of a gas FIGURE 630 Two familiar reversible processes. 66 REVERSIBLE AND IRREVERSIBLE PROCESSES The second law of thermodynamics states that no heat engine can have an efficiency of 100 percent. Then one may ask, What is the highest efficiency that a heat engine can possibly have? Before we can answer this question, we need to define an idealized process first, which is called the reversible process. The processes that were discussed at the beginning of this chapter occurred in a certain direction. Once having taken place, these processes cannot reverse themselves spontaneously and restore the system to its initial state. For this reason, they are classified as irreversible processes. Once a cup of hot coffee cools, it will not heat up by retrieving the heat it lost from the surroundings. If it could, the surroundings, as well as the system (coffee), would be restored to their original condition, and this would be a reversible process. A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings (Fig. 630). 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 work-producing devices such as car engines and gas or steam turbines deliver the most work, and work-consuming devices such as compressors, fans, and pumps consume the least work when reversible processes are used instead of irreversible ones (Fig. 631). 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 cen84959_ch06.qxd 3/31/05 3:51 PM Page 297 Chapter 6 Expansion Compression Expansion | 297 Compression Pressure distribution Water Water (a) Slow (reversible) process Water Water (b) Fast (irreversible) process FIGURE 631 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 work-producing device or the less work required by a work-consuming device. The concept of reversible processes leads to the definition of the secondlaw efficiency for actual processes, which is the degree of 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 second-law 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. 632), 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 632 Friction renders a process irreversible. cen84959_ch06.qxd 3/31/05 3:51 PM Page 298 298 | Thermodynamics (a) Fast compression (b) Fast expansion 700 kPa 50 kPa (c) Unrestrained expansion FIGURE 633 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. 633. 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. 634). 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 634 (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, cen84959_ch06.qxd 4/25/05 3:10 PM Page 299 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 quasi-equilibrium 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. 635). A totally reversible process involves no heat transfer through a finite temperature difference, no nonquasi-equilibrium changes, and no friction or other dissipative effects. As an example, consider the transfer of heat to two identical systems that are undergoing a constant-pressure (thus constant-temperature) phasechange process, as shown in Fig. 636. 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. 67 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 heat-engine 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 No irreversibilities outside the system | 299 No irreversibilities inside the system FIGURE 635 A reversible process involves no internal and external irreversibilities. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 7 ON THE DVD. Boundary at 20°C Heat Thermal energy reservoir at 20.000 ...1°C Thermal energy reservoir at 30°C (a) Totally reversible (b) Internally reversible FIGURE 636 Totally and interally reversible heat transfer processes. cen84959_ch06.qxd 3/31/05 3:51 PM Page 300 300 | Thermodynamics (2) TH = const. (1) Energy source at TH QH (a) Process 1-2 Insulation (2) (3) TH TL (b) Process 2-3 (3) TL = const. (4) Energy sink at TL QL (c) Process 3-4 Insulation (1) (4) TH TL (d ) Process 4-1 FIGURE 637 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 processestwo isothermal and two adiabaticand it can be executed either in a closed or a steady-flow system. Consider a closed system that consists of a gas contained in an adiabatic pistoncylinder device, as shown in Fig. 637. 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 1-2, 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 2-3, 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 3-4, 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 cen84959_ch06.qxd 4/25/05 3:10 PM Page 301 Chapter 6 | 301 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 4-1, temperature rises from TL to TH). State 4 is such that when the low-temperature 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 P-V diagram of this cycle is shown in Fig. 638. Remembering that on a P-V diagram the area under the process curve represents the boundary work for quasi-equilibrium (internally reversible) processes, we see that the area under curve 1-2-3 is the work done by the gas during the expansion part of the cycle, and the area under curve 3-4-1 is the work done on the gas during the compression part of the cycle. The area enclosed by the path of the cycle (area 1-2-3-4-1) 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 3-2. 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 steady-flow 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 638 P-V diagram of the Carnot cycle. P 1 QH The Reversed Carnot Cycle The Carnot heat-engine 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 low-temperature reservoir, heat in the amount of QH is rejected to a high-temperature reservoir, and a work input of Wnet,in is required to accomplish all this. The P-V 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. 639. 68 THE CARNOT PRINCIPLES The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the KelvinPlanck and Clausius statements. A heat engine cannot operate by exchanging heat with a single reservoir, and a 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 639 P-V diagram of the reversed Carnot cycle. INTERACTIVE TUTORIAL SEE TUTORIAL CH. 6, SEC. 8 ON THE DVD. cen84959_ch06.qxd 3/31/05 3:51 PM Page 302 302 | Thermodynamics High-temperature reservoir at TH 1 Irrev. HE 2 Rev. HE ηth,1 < ηth,2 3 Rev. HE ηth,2 = ηth,3 Low-temperature reservoir at TL FIGURE 640 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. 640), 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. 641. 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 high-temperature 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 Wrev have an engine that produces a net work in the amount of Wirrev High-temperature reservoir at TH QH QH Wirrev Irreversible HE Reversible HE (or R) QL,irrev < QL,rev (assumed) QL,rev Low-temperature reservoir at TL FIGURE 641 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 Low-temperature reservoir at TL (b) The equivalent combined system cen84959_ch06.qxd 4/25/05 3:10 PM Page 303 Chapter 6 while exchanging heat with a single reservoira violation of the Kelvin Planck statement of the second law. Therefore, our initial assumption that hth,rev is incorrect. Then we conclude that no heat engine can be hth,irrev 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. 69 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 68 states that all reversible heat engines have the same thermal efficiency when operating between the same two reservoirs (Fig. 642). 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, g 1 TH, TL 2 High-temperature reservoir at TH = 1000 K Another reversible HE ηth,B A reversible HE ηth,A ηth,A = ηth,B = 70% Low-temperature reservoir at TL = 300 K 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 QL f 1 TH, TL 2 303 FIGURE 642 All reversible heat engines operating between the same two reservoirs have the same efficiency (the second Carnot principle). THE THERMODYNAMIC TEMPERATURE SCALE hth,rev | WC (613) since hth 1 QL/QH. In these relations TH and TL are the temperatures of the high- and low-temperature reservoirs, respectively. The functional form of f (TH, TL) can be developed with the help of the three reversible heat engines shown in Fig. 643. Engines A and C are supplied with the same amount of heat Q1 from the high-temperature reservoir at T1. Engine C rejects Q3 to the low-temperature 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 643 The arrangement of heat engines used to develop the thermodynamic temperature scale. cen84959_ch06.qxd 3/31/05 3:51 PM Page 304 304 | 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. 613 to all three engines separately, we obtain Q1 Q2 f 1 T1, T2 2 , f 1 T2, T3 2 , Q2 Q3 and Q1 Q3 f 1 T1, T3 2 Now consider the identity Q1 Q3 Q1 Q2 Q2 Q3 which corresponds to f 1 T1, T3 2 f 1 T1, T2 2 # f 1 T2, T3 2 A careful examination of this equation reveals that the left-hand side is a function of T1 and T3, and therefore the right-hand side must also be a function of T1 and T3 only, and not T2. That is, the value of the product on the right-hand 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 1 T1, T2 2 f 1 T1 2 f 1 T2 2 and f 1 T2, T3 2 f 1 T2 2 f 1 T3 2 so that f(T2) will cancel from the product of f (T1, T2) and f (T2, T3), yielding f 1 T1 2 f 1 T1, T3 2 Q1 Q3 f 1 T3 2 (614) This relation is much more specific than Eq. 613 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. 614 can be written as High-temperature reservoir at TH QL Wnet QH QL f 1 TH 2 f 1 TL 2 (615) This is the only requirement that the second law places on the ratio of heat transfers to and from the reversible heat engines. Several functions 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. 644) QH Reversible heat engine or refrigerator QH QL = TH TL Low-temperature reservoir at TL FIGURE 644 For reversible cycles, the heat transfer ratio QH /QL can be replaced by the absolute temperature ratio TH /TL. a QH b QL rev TH TL (616) This temperature scale is called the Kelvin scale, and the temperatures on this scale are called absolute temperatures. On the Kelvin scale, the temperature ratios depend on the ratios of heat transfer between a reversible heat engine and the reservoirs and are independent of the physical properties of any substance. On this scale, temperatures vary between zero and infinity. The thermodynamic temperature scale is not completely defined by Eq. 616 since it gives us only a ratio of absolute temperatures. We also need to know the magnitude of a kelvin. At the International Conference on cen84959_ch06.qxd 4/25/05 3:10 PM Page 305 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. 645). The magnitude of a kelvin is defined as 1/273.16 of the temperature interval between absolute zero and the triple-point 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 °C 2 T 1K 2 273.15 (617) Even though the thermodynamic temperature scale is defined with the help of the reversible heat engines, it is not possible, nor is it practical, to actually operate such an engine to determine numerical values on the absolute temperature scale. Absolute temperatures can be measured accurately by other means, such as the constant-volume ideal-gas thermometer together with extrapolation techniques as discussed in Chap. 1. The validity of Eq. 616 can be demonstrated from physical considerations for a reversible cycle using an ideal gas as the working fluid. 610 THE CARNOT HEAT ENGINE The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat engine. The thermal efficiency of any heat engine, reversible or irreversible, is given by Eq. 66 as QL QH 1 hth where QH is heat transferred to the heat engine from a high-temperature reservoir at TH, and QL is heat rejected to a low-temperature 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. 616. Then the efficiency of a Carnot engine, or any reversible heat engine, becomes hth,rev 1 TL TH (618) This relation is often referred to as the Carnot efficiency, since the Carnot heat engine is the best known reversible engine. This is the highest efficiency a heat engine operating between the two thermal energy reservoirs at temperatures TL and TH can have (Fig. 646). 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. 618 are absolute temperatures. Using °C or °F for temperatures in this relation gives results grossly in error. The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows (Fig. 647): hth 6 hth,rev hth,rev 7 hth,rev irreversible heat engine reversible heat engine impossible heat engine (619) | 305 Heat reservoir T QH W Carnot HE QL 273.16 K (assigned) Water at triple point QH T = 273.16 QL FIGURE 645 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. High-temperature reservoir at TH = 1000 K QH Carnot HE ηth = 70% Wnet,out QL Low-temperature reservoir at TL = 300 K FIGURE 646 The Carnot heat engine is the most efficient of all heat engines operating between the same high- and lowtemperature reservoirs. cen84959_ch06.qxd 3/31/05 3:51 PM Page 306 306 | Thermodynamics High-temperature reservoir at TH = 1000 K Rev. HE ηth = 70% FIGURE 647 No heat engine can have a higher efficiency than a reversible heat engine operating between the same high- and low-temperature reservoirs. High-temperature reservoir at TH = 652°C Impossible HE ηth = 80% Irrev. HE ηth = 45% Low-temperature reservoir at TL = 300 K Most work-producing 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 limitsbecause this is the true theoretical upper limit for the efficiency, not 100 percent. The maximum efficiency of a steam power plant operating between TH 1000 K and TL 300 K is 70 percent, as determined from Eq. 618. Compared with this value, an actual efficiency of 40 percent does not seem so bad, even though there is still plenty of room for improvement. It is obvious from Eq. 618 that the efficiency of a Carnot heat engine increases as TH is increased, or as TL is decreased. This is to be expected since as TL decreases, so does the amount of heat rejected, and as TL approaches zero, the Carnot efficiency approaches unity. This is also true for actual heat engines. The thermal efficiency of actual heat engines can be maximized by supplying heat to the engine at the highest possible temperature (limited by material strength) and rejecting heat from the engine at the lowest possible temperature (limited by the temperature of the cooling medium such as rivers, lakes, or the atmosphere). QH = 500 kJ Wnet,out Carnot HE EXAMPLE 65 Analysis of a Carnot Heat Engine A Carnot heat engine, shown in Fig. 648, receives 500 kJ of heat per cycle from a high-temperature source at 652°C and rejects heat to a low-temperature 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 Low-temperature reservoir at TL = 30°C FIGURE 648 Schematic for Example 65. 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. 618 to be hth,C hth,rev 1 TL TH 1 1 30 1 652 273 2 K 273 2 K 0.672 cen84959_ch06.qxd 3/31/05 3:51 PM Page 307 Chapter 6 | 307 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. 616 to be QL,rev TL Q TH H,rev 1 30 1 652 273 2 K 273 2 K 1 500 kJ 2 164 kJ Discussion Note that this Carnot heat engine rejects to a low-temperature sink 164 kJ of the 500 kJ of heat it receives during each cycle. The Quality of Energy The Carnot heat engine in Example 65 receives heat from a source at 925 K and converts 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. 618 and is listed in Fig. 649. 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. 649 that more of the high-temperature thermal energy can be converted to work. Therefore, the higher the temperature, the higher the quality of the energy (Fig. 650). 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 high-temperature 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. High-temperature reservoir at TH TH, K Rev. HE ηth ηth, % 925 800 700 500 350 67.2 62.1 56.7 39.4 13.4 Low-temperature reservoir at TL = 303 K FIGURE 649 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 650 The higher the temperature of the thermal energy, the higher its quality. cen84959_ch06.qxd 3/31/05 3:51 PM Page 308 308 | Thermodynamics 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 high-quality energy can be more valuable than three units of lower-quality 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 second-law 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 first-law point of view, which considers quantity only, it is obvious that Andy has more friends. However, from the second-law 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 moregram for gram, calorie for caloriethan Americans do, but they weigh less, with less body fat. Studies indicate that the metabolism rates and hormone levels change noticeably in the mid-30s. 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 calorie-counting technique (the first law) leaves many questions on dieting unanswered. Obviously, more research focused on the second-law effects of dieting is needed before we can fully understand the weight-gain and weight-loss process. cen84959_ch06.qxd 4/25/05 3:10 PM Page 309 Chapter 6 | 309 It is tempting to judge things on the basis of their quantity instead of their quality since assessing quality is much more difficult than assessing quantity. However, assessments made on the basis of quantity only (the first law) may be grossly inadequate and misleading. 611 THE CARNOT REFRIGERATOR AND HEAT PUMP 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. 69 and 611 as COPR 1 QH> QL 1 and COPHP 1 SEE TUTORIAL CH. 6, SEC. 11 ON THE DVD. 1 QL > QH where QL is the amount of heat absorbed from the low-temperature medium and QH is the amount of heat rejected to the high-temperature 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 low-temperature reservoirs, as expressed by Eq. 616. Then the COP relations for reversible refrigerators and heat pumps become COPR,rev and COPHP,rev 1 TH> TL 1 (620) 1 1 TL > TH (621) These are the highest coefficients of performance that a refrigerator or a heat pump operating between the temperature limits of TL and TH can have. All actual refrigerators or heat pumps operating between these temperature limits (TL and TH) have lower coefficients of performance (Fig. 651). 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 651 No refrigerator can have a higher COP than a reversible refrigerator operating between the same temperature limits. cen84959_ch06.qxd 3/31/05 3:51 PM Page 310 310 | Thermodynamics The coefficients of performance of actual and reversible refrigerators operating between the same temperature limits can be compared as follows: COPR Warm environment at TH = 75°F Refrigerator COP = 13.5 FIGURE 652 Schematic for Example 66. 135,000 kJ/h Heat loss House 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. 652) can be evaluated by comparing it with a reversible refrigerator operating between the same temperature limits: COPR,rev TH = 21°C 1 75 · Wnet,in = ? HP · QL Cold outside air TL = 5°C FIGURE 653 Schematic for Example 67. (622) A Questionable Claim for a Refrigerator COPR,max · QH irreversible refrigerator reversible refrigerator impossible refrigerator A similar relation can be obtained for heat pumps by replacing all COPRs in Eq. 622 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 lower-temperature 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 66 Cool refrigerated space at TL = 35°F 6 COPR,rev COPR,rev 7 COPR,rev 1 TH> TL 1 460 R 2 > 1 35 1 460 R 2 1 12.4 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 67 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. 653. 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. cen84959_ch06.qxd 4/19/05 10:01 AM Page 311 Chapter 6 | 311 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 TL > TH 1 15 1 273 K 2> 1 21 273 K 2 11.3 Then the required power input to this reversible heat pump becomes # Wnet,in QH COPHP 37.5 kW 11.3 3.32 kW 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 one-to-one 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 refrigerator-freezer since it has a freezer compartment to make ice and to store frozen food. Todays refrigerators use much less energy as a result of using smaller and higher-efficiency motors and compressors, better insulation materials, larger coil surface areas, and better door seals (Fig. 654). 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 25-year-old, 18-ft3 refrigerator with a new energy-efficient 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 vapor-compression 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 654 Todays refrigerators are much more efficient because of the improvements in technology and manufacturing. cen84959_ch06.qxd 3/31/05 3:51 PM Page 312 312 | Thermodynamics TABLE 61 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 Vapor-compression Absorption refrigeration Thermoelectric refrigeration 1.3 0.4 0.1 efficient, and they have found limited use in some specialized applications (Table 61). 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 special-purpose compartments. Practically all full-size refrigerators have a large air-tight 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 chilled-water 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 fiber-glass insulation (thermal conductivity k 0.0320.040 W/m · °C) to expanded-in-place 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. 655), 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 cen84959_ch06.qxd 3/31/05 3:51 PM Page 313 Chapter 6 Steel shell | 313 Steel or plastic liner Thermal Insulation 6% Defrost heater 6% Fan motor 6% External heater 52% Wall insulation FIGURE 655 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. 656). 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 force-cool the condensers of large or built-in 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 one-third 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 656 Putting hot foods into the refrigerator without cooling them first not only wastes energy but also could spoil the foods nearby. cen84959_ch06.qxd 3/31/05 3:51 PM Page 314 314 | Thermodynamics Cabinet Warm air Refrigerator Coils Cool air FIGURE 657 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 build-up 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 no-frost refrigerators by supplying heat to the evaporator by a 300-W to 1000-W 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 no-frost 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 frost-free. 7. Use the power-saver 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 air-conditioning. The moisture formation on the surfaces is undesirable since it may cause the painted finish of the outer surface to deteriorate and it may wet the kitchen floor. About 10 percent of the total energy consumed by the refrigerator can be saved by turning this heater off and keeping it off unless there is visible condensation on the outer surfaces. 8. Do not block the 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. 657). 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 trouble-free life of the device. EXAMPLE 68 FIGURE 658 Schematic for Example 68. Malfunction of a Refrigerator Light Switch The interior lighting of refrigerators is provided by incandescent lamps whose switches are actuated by the opening of the refrigerator door. Consider a refrigerator whose 40-W lightbulb remains on continuously as a result of a malfunction of the switch (Fig. 658). 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. cen84959_ch06.qxd 3/31/05 3:51 PM Page 315 Chapter 6 | 315 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 # Wrefrig COPR 40 W 1.3 30.8 W Therefore, the total additional power consumed by the refrigerator is # Wlight # Wtotal,additional # Wrefrig 40 30.8 70.8 W The total number of hours in a year is 1 365 days> yr 2 1 24 h> day 2 Annual hours 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 1 20 times> day 2 1 30 s> time 2 1 1 h> 3600 s 2 1 365 days> yr 2 61 h> yr Then the additional hours the light remains on as a result of the malfunction becomes Additional operating hours Annual hours 8760 61 Normal operating hours 8699 h> yr Therefore, the additional electric power consumption and its cost per year are Additional power consumption # Wtotal,additional 1 Additional operating hours 2 1 0.0708 kW 2 1 8699 h> yr 2 616 kWh/yr and Additional power cost 1 Additional power consumption 2 1 Unit cost 2 1 616 kWh> yr 2 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 cen84959_ch06.qxd 4/19/05 10:01 AM Page 316 316 | Thermodynamics 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 low-temperature media and reject it to higher-temperature ones. The performance of a refrigerator or a heat pump is expressed in terms of the coefficient of performance, which is defined as COPR QL Wnet,in COPHP QH Wnet,in 1 QH> QL 1 1 1 QL > QH The KelvinPlanck 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 perpetual-motion 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, non-quasi-equilibrium 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 low-temperature reservoirs by a QH b QL rev TH 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 low-temperature 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 and COPHP,rev 1 1 TL > TH 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. 193208. 4. K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGraw-Hill, 1999. PROBLEMS* Second Law of Thermodynamics and Thermal Energy Reservoirs 61C A mechanic claims to have developed a car engine that runs on water instead of gasoline. What is your response to this claim? 62C Describe an imaginary process that 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 CD-EES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computer-EES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. cen84959_ch06.qxd 4/20/05 4:40 PM Page 317 Chapter 6 63C Describe an imaginary process that satisfies the second law but violates the first law of thermodynamics. 64C Describe an imaginary process that violates both the first and the second laws of thermodynamics. 65C An experimentalist claims to have raised the temperature of a small amount of water to 150°C by transferring heat from high-pressure steam at 120°C. Is this a reasonable claim? Why? Assume no refrigerator or heat pump is used in the process. 66C What is a thermal energy reservoir? Give some examples. 67C 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. 68C Consider the energy generated by a TV set. What is a suitable choice for a thermal energy reservoir? Heat Engines and Thermal Efficiency 69C Is it possible for a heat engine to operate without rejecting any waste heat to a low-temperature reservoir? Explain. 610C What are the characteristics of all heat engines? 611C 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. 612C Baseboard heaters are basically electric resistance heaters and are frequently used in space heating. A home owner claims that her 5-year-old baseboard heaters have a conversion efficiency of 100 percent. Is this claim in violation of any thermodynamic laws? Explain. 613C What is the KelvinPlanck expression of the second law of thermodynamics? 614C 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. 615C In the absence of any friction and other irreversibilities, can a heat engine have an efficiency of 100 percent? Explain. 616C Are the efficiencies of all the work-producing devices, including the hydroelectric power plants, limited by the KelvinPlanck statement of the second law? Explain. | 317 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 619E 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. 620 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 621 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 622E 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. 623 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 coal-fired power plants in the United States that year. 624 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 coal-fired power plants, which cost $1300 per kW to construct and have an efficiency of 34 percent. Another possibility is to use the clean-burning 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. 617 A 600-MW 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? 625 Reconsider Prob. 624. Using EES (or other) software, investigate the price of coal for varying simple payback periods, plant construction costs, and operating efficiency. 618 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- 626 Repeat Prob. 624 for a simple payback period of three years instead of five years. 627E An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate cen84959_ch06.qxd 3/31/05 3:51 PM Page 318 318 | Thermodynamics 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 40-in-diameter 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. 628 A coal-burning steam power plant produces a net power of 300 MW with an overall therma