Baixe Thermodynamics 2011 Gregory Nellis e outras Manuais, Projetos, Pesquisas em PDF para Engenharia Elétrica, somente na Docsity! THERMODYNAMICS Sanford Klein feto a NDT more information - www.combridge.org/9780521 195706 Thermodynamics SANFORD KLEIN University of Wisconsin, Madison GREGORY NELLIS University of Wisconsin, Madison cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Tokyo, Mexico City Cambridge University Press 32 Avenue of the Americas, New York, NY 10013-2473, USA www.cambridge.org Information on this title: www.cambridge.org/9780521195706 C© Sanford Klein and Gregory Nellis 2012 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2012 Printed in the United States of America A catalog record for this publication is available from the British Library. Library of Congress Cataloging in Publication data Klein, Sanford A., 1950– Thermodynamics / Sanford Klein, Gregory Nellis. p. cm. Includes bibliographical references and index. ISBN 978-0-521-19570-6 (hardback) 1. Thermodynamics. 2. Engineering – Problems, exercises, etc. I. Nellis, Gregory. II. Title. QC311.15.K58 2011 536′.7–dc22 2011001982 ISBN 978-0-521-19570-6 Hardback Additional resources for this publication at www.cambridge.org/kleinandnellis Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party Internet Web sites referred to in this publication and does not guarantee that any content on such Web sites is, or will remain, accurate or appropriate. CONTENTS Preface page xv Acknowledgments xvii Nomenclature xix 1 BASIC CONCEPTS 1 1.1 Overview 1 1.2 Thermodynamic Systems 3 1.3 States and Properties 4 1.3.1 State of a System 4 1.3.2 Measurable and Derived Properties 4 1.3.3 Intensive and Extensive Properties 5 1.3.4 Internal and External Properties 5 1.4 Balances 6 1.5 Introduction to EES (Engineering Equation Solver) 8 1.6 Dimensions and Units 11 1.6.1 The SI and English Unit Systems 11 EXAMPLE 1.6-1: WEIGHT ON MARS 14 1.6.2 Working with Units in EES 14 EXAMPLE 1.6-2: POWER REQUIRED BY A VEHICLE 15 1.7 Specific Volume, Pressure, and Temperature 24 1.7.1 Specific Volume 24 1.7.2 Pressure 24 1.7.3 Temperature 26 References 28 Problems 28 2 THERMODYNAMIC PROPERTIES 34 2.1 Equilibrium and State Properties 34 2.2 General Behavior of Fluids 36 2.3 Property Tables 41 2.3.1 Saturated Liquid and Vapor 41 EXAMPLE 2.3-1: PRODUCTION OF A VACUUM BY CONDENSATION 45 2.3.2 Superheated Vapor 47 Interpolation 49 2.3.3 Compressed Liquid 50 2.4 EES Fluid Property Data 51 2.4.1 Thermodynamic Property Functions 51 v viii Contents EXAMPLE 6.6-3: INTERCOOLED COMPRESSION 278 6.6.3 Pump Efficiency 287 EXAMPLE 6.6-4: SOLAR POWERED LIVESTOCK PUMP 289 6.6.4 Nozzle Efficiency 292 EXAMPLE 6.6-5: JET-POWERED WAGON 294 6.6.5 Diffuser Efficiency 300 EXAMPLE 6.6-6: DIFFUSER IN A GAS TURBINE ENGINE 302 6.6.6 Heat Exchanger Effectiveness 305 EXAMPLE 6.6-7: ARGON REFRIGERATION CYCLE 308 Heat Exchangers with Constant Specific Heat Capacity 312 EXAMPLE 6.6-8: ENERGY RECOVERY HEAT EXCHANGER 316 References 322 Problems 322 7 EXERGY 350 7.1 Definition of Exergy and Second Law Efficiency 350 7.2 Exergy of Heat 351 EXAMPLE 7.2-1: SECOND LAW EFFICIENCY 353 7.3 Exergy of a Flow Stream 355 EXAMPLE 7.3-1: HEATING SYSTEM 358 7.4 Exergy of a System 361 EXAMPLE 7.4-1: COMPRESSED AIR POWER SYSTEM 364 7.5 Exergy Balance 367 EXAMPLE 7.5-1: EXERGY ANALYSIS OF A COMMERCIAL LAUNDRY FACILITY 369 7.6 Relation Between Exergy Destruction and Entropy Generation∗ (E1) 378 Problems 379 8 POWER CYCLES 385 8.1 The Carnot Cycle 385 8.2 The Rankine Cycle 388 8.2.1 The Ideal Rankine Cycle 388 Effect of Boiler Pressure 395 Effect of Heat Source Temperature 397 Effect of Heat Sink Temperature 397 8.2.2 The Non-Ideal Rankine Cycle 399 8.2.3 Modifications to the Rankine Cycle 405 Reheat 405 Regeneration 410 EXAMPLE 8.2-1: SOLAR TROUGH POWER PLANT 413 8.3 The Gas Turbine Cycle 426 8.3.1 The Basic Gas Turbine Cycle 427 Effect of Air-Fuel Ratio 433 Effect of Pressure Ratio and Turbine Inlet Temperature 434 Effect of Compressor and Turbine Efficiencies 437 8.3.2 Modifications to the Gas Turbine Cycle 437 Reheat and Intercooling 437 EXAMPLE 8.3-1: OPTIMAL INTERCOOLING PRESSURE 439 Recuperation 442 ∗ Section can be found on the Web site that accompanies this book (www.cambridge.org/kleinandnellis). Contents ix EXAMPLE 8.3-2: GAS TURBINE ENGINE FOR SHIP PROPULSION 443 8.3.3 The Gas Turbine Engines for Propulsion 452 Turbojet Engine 452 EXAMPLE 8.3-3: TURBOJET ENGINE 454 Turbofan Engine 458 EXAMPLE 8.3-4: TURBOFAN ENGINE 460 Turboprop Engine 467 8.3.4 The Combined Cycle and Cogeneration 467 8.4 Reciprocating Internal Combustion Engines 468 8.4.1 The Spark-Ignition Reciprocating Internal Combustion Engine 468 Spark-Ignition, Four-Stroke Engine Cycle 469 Simple Model of Spark-Ignition, Four-Stroke Engine 472 Octane Number of Gasoline 477 EXAMPLE 8.4-1: POLYTROPIC MODEL WITH RESIDUAL COMBUSTION GAS 479 Spark-Ignition, Two-Stroke Internal Combustion Engine 488 8.4.2 The Compression-Ignition Reciprocating Internal Combustion Engine 491 EXAMPLE 8.4-2: TURBOCHARGED DIESEL ENGINE 493 8.5 The Stirling Engine 501 8.5.1 The Stirling Engine Cycle 502 8.5.2 Simple Model of the Ideal Stirling Engine Cycle∗ (E2) 504 8.6 Tradeoffs Between Power and Efficiency 505 8.6.1 The Heat Transfer Limited Carnot Cycle 505 8.6.2 Carnot Cycle using Fluid Streams as the Heat Source and Heat Sink∗ (E3) 511 8.6.3 Internal Irreversibilities∗ (E4) 511 8.6.4 Application to other Cycles 511 References 512 Problems 512 9 REFRIGERATION AND HEAT PUMP CYCLES 529 9.1 The Carnot Cycle 529 9.2 The Vapor Compression Cycle 532 9.2.1 The Ideal Vapor Compression Cycle 532 Effect of Refrigeration Temperature 538 9.2.2 The Non-Ideal Vapor Compression Cycle 540 EXAMPLE 9.2-1: INDUSTRIAL FREEZER 542 EXAMPLE 9.2-2: INDUSTRIAL FREEZER DESIGN 545 9.2.3 Refrigerants 550 Desirable Refrigerant Properties 550 Positive Evaporator Gage Pressure 551 Moderate Condensing Pressure 551 Appropriate Triple Point and Critical Point Temperatures 551 High Density/Low Specific Volume at the Compressor Inlet 553 High Latent Heat (Specific Enthalpy Change) of Vaporization 553 High Dielectric Strength 553 Compatibility with Lubricants 553 Non-Toxic 554 Non-Flammable 554 ∗ Section can be found on the Web site that accompanies this book (www.cambridge.org/kleinandnellis). x Contents Inertness and Stability 554 Refrigerant Naming Convention 554 Ozone Depletion and Global Warming Potential 556 9.2.4 Vapor Compression Cycle Modifications 557 Liquid-Suction Heat Exchanger 559 EXAMPLE 9.2-3: REFRIGERATION CYCLE WITH A LIQUID-SUCTION HEAT EXCHANGER 560 Liquid Overfed Evaporator 564 Intercooled Cycle 567 Economized Cycle 568 Flash-Intercooled Cycle 571 EXAMPLE 9.2-4: FLASH INTERCOOLED CYCLE FOR A BLAST FREEZER 571 EXAMPLE 9.2-5: CASCADE CYCLE FOR A BLAST FREEZER 578 9.3 Heat Pumps 584 EXAMPLE 9.3-1: HEATING SEASON PERFORMANCE FACTOR 588 9.4 The Absorption Cycle 598 9.4.1 The Basic Absorption Cycle 598 9.4.2 Absorption Cycle Working Fluids∗ (E6) 601 9.5 Recuperative Cryogenic Cooling Cycles 601 9.5.1 The Reverse Brayton Cycle 603 9.5.2 The Joule-Thomson Cycle 611 9.5.3 Liquefaction Cycles∗ (E7) 614 9.6 Regenerative Cryogenic Cooling Cycles∗ (E8) 614 References 614 Problems 615 10 PROPERTY RELATIONS FOR PURE FLUIDS 629 10.1 Equations of State for Pressure, Volume, and Temperature 629 10.1.1 Compressibility Factor and Reduced Properties 630 10.1.2 Characteristics of the Equation of State 633 Limiting Ideal Gas Behavior 633 The Boyle Isotherm 633 Critical Point Behavior 634 10.1.3 Two-Parameter Equations of State 637 The van der Waals Equation of State 637 EXAMPLE 10.1-1: APPLICATION OF THE VAN DER WAALS EQUATION OF STATE 641 The Dieterici Equation of State 646 EXAMPLE 10.1-2: DIETERICI EQUATION OF STATE 646 The Redlich-Kwong Equation of State 649 The Redlich-Kwong-Soave (RKS) Equation of State 650 The Peng-Robinson (PR) Equation of State 651 EXAMPLE 10.1-3: PENG-ROBINSON EQUATION OF STATE 653 10.1.4 Multiple Parameter Equations of State 656 10.2 Application of Fundamental Property Relations 657 10.2.1 The Fundamental Property Relations 658 10.2.2 Complete Equations of State 659 EXAMPLE 10.2-1: USING A COMPLETE EQUATION OF STATE 660 EXAMPLE 10.2-2: THE REDUCED HELMHOLTZ EQUATION OF STATE 661 ∗ Section can be found on the Web site that accompanies this book (www.cambridge.org/kleinandnellis). Contents xiii 13.3.1 Enthalpy of Formation 864 13.3.2 Heating Values 866 EXAMPLE 13.3-1: HEATING VALUE OF A PRODUCER GAS 871 13.3.3 Enthalpy and Internal Energy as a Function of Temperature 873 EXAMPLE 13.3-2: PROPANE HEATER 875 13.3.4 Use of EES for Determining Properties 879 EXAMPLE 13.3-3: FURNACE EFFICIENCY 882 13.3.5 Adiabatic Reactions 889 EXAMPLE 13.3-4: DETERMINATION OF THE EXPLOSION PRESSURE OF METHANE 894 13.4 Entropy Considerations 898 EXAMPLE 13.4-1: PERFORMANCE OF A GAS TURBINE ENGINE 901 13.5 Exergy of Fuels∗ (E18) 907 References 907 Problems 908 14 CHEMICAL EQUILIBRIUM 922 14.1 Criterion for Chemical Equilibrium 922 14.2 Reaction Coordinates 924 EXAMPLE 14.2-1: SIMULTANEOUS CHEMICAL REACTIONS 927 14.3 The Law of Mass Action 931 14.3.1 The Criterion of Equilibrium in terms of Chemical Potentials 931 14.3.2 Chemical Potentials for an Ideal Gas Mixture 933 14.3.3 Equilibrium Constant and the Law of Mass Action for Ideal Gas Mixtures 933 EXAMPLE 14.3-1: REFORMATION OF METHANE 935 14.3.4 Equilibrium Constant and the Law of Mass Action for an Ideal Solution 938 EXAMPLE 14.3-2: AMMONIA SYNTHESIS 939 14.4 Alternative Methods for Chemical Equilibrium Problems 943 14.4.1 Direct Minimization of Gibbs Free Energy 944 EXAMPLE 14.4-1: REFORMATION OF METHANE (REVISITED) 945 14.4.2 Lagrange Method of Undetermined Multipliers 949 EXAMPLE 14.4-2: REFORMATION OF METHANE (REVISITED AGAIN) 951 14.5 Heterogeneous Reactions∗ (E19) 953 14.6 Adiabatic Reactions 954 EXAMPLE 14.6-1: ADIABATIC COMBUSTION OF HYDROGEN 954 EXAMPLE 14.6-2: ADIABATIC COMBUSTION OF ACETYLENE 960 Reference 967 Problems 967 15 STATISTICAL THERMODYNAMICS 972 15.1 A Brief Review of Quantum Theory History 973 15.1.1 Electromagnetic Radiation 973 15.1.2 Extension to Particles 975 15.2 The Wave Equation and Degeneracy for a Monatomic Ideal Gas 976 15.2.1 Probability of Finding a Particle 976 15.2.2 Application of a Wave Equation 976 15.2.3 Degeneracy 979 15.3 The Equilibrium Distribution 979 15.3.1 Macrostates and Thermodynamic Probability 980 ∗ Section can be found on the Web site that accompanies this book (www.cambridge.org/kleinandnellis). xiv Contents 15.3.2 Identification of the Most Probable Macrostate 982 15.3.3 The Significance of β 985 15.3.4 Boltzmann’s Law 987 15.4 Properties and the Partition Function 989 15.4.1 Definition of the Partition Function 989 15.4.2 Internal Energy from the Partition Function 990 15.4.3 Entropy from the Partition Function 991 15.4.4 Pressure from the Partition Function 992 15.5 Partition Function for an Monatomic Ideal Gas 993 15.5.1 Pressure for a Monatomic Ideal Gas 994 15.5.2 Internal Energy for a Monatomic Ideal Gas 995 15.5.3 Entropy for a Monatomic Ideal Gas 995 EXAMPLE 15.5-1: CALCULATION OF ABSOLUTE ENTROPY VALUES 997 15.6 Extension to More Complex Particles 998 15.7 Heat and Work from a Statistical Thermodynamics Perspective 1001 References 1004 Problems 1005 16 COMPRESSIBLE FLOW∗ (E20) 1009 Problems 1009 Appendices A: Unit Conversions and Useful Information 1015 B: Property Tables for Water 1019 C: Property Tables for R134a 1031 D: Ideal Gas & Incompressible Substances 1037 E: Ideal Gas Properties of Air 1039 F: Ideal Gas Properties of Common Combustion Gases 1045 G: Numerical Solution to ODEs 1056 H: Introduction to Maple∗ (E26) 1057 Index 1059 ∗ Section can be found on the Web site that accompanies this book (www.cambridge.org/kleinandnellis). PREFACE Thermodynamics is a mature science. Many excellent engineering textbooks have been written on the subject, which leads to the question: Why yet another textbook on classical thermodynamics? There is a simple answer to this question: this book is different. The objective of this book is to provide engineers with the concepts, tools, and experience needed to solve practical real-world energy problems. With this in mind, the focus of this effort has been to integrate a computer tool with thermodynamic concepts in order to allow engineering students and practicing engineers to tackle problems that they would otherwise not be able to solve. It is generally acknowledged that students need to solve problems in order to inte- grate concepts and skills. The effort required to solve a thermodynamics problem can be broken into two parts. First, it is necessary to identify the fundamental relationships that describe the problem. The set of equations that leads to a useful solution to a problem results from application of appropriate balances and rate relations, simplified with justified assumptions. Identifying the necessary equations is the conceptual part of the problem, and no computer program can provide this capability in general. Proper application of the First and Second Laws of Thermodynamics is at the heart of this process. The ability to identify the appropriate equations does not come easily to most thermodynamics students. This is an area in which problem-solving experience is helpful. A distinguishing feature of this textbook is that it presents detailed examples and discus- sion that explain how to apply thermodynamics concepts identify a set of equations that will provide solutions to non-trivial problems. Once the appropriate equations have been identified, they must be solved. In our experience, much of the time and effort required to solve thermodynamics problems results from looking up property information in tables and solving the appropriate equations. Though necessary for obtaining a solution, these tasks contribute little to the learning process. For example, once the student is familiar with the use of property tables, further use of the tables does not contribute to the student’s grasp of the subject – nor does doing the tedious algebra that is required to solve a large set of equations. Practical problems that focus on real engineering issues tend to be more interesting to students, but also more mathematically complex. The time and effort required to do problems without computing tools may actually detract from learning the subject matter by forcing the student to focus on the mathematical complexity of the problem rather than on the underlying concepts. The motivation for writing this book is a result of our experience in teaching mechan- ical engineering thermodynamics in a manner that is tightly integrated with the EES (Engineering Equation Solver) program. EES eliminates much of the mathematical complexity involved in solving thermodynamics problems by providing a large bank of high-accuracy property data and the capability to solve large sets of simultaneous alge- braic and differential equations. EES also provides the capability to check equations for unit consistency; do parametric studies; produce high-quality plots; and apply numeri- cal integration, optimization, and uncertainty analyses. Using EES, students can easily xv NOMENCLATURE a specific Helmholtz free energy (J/kg) parameter in an equation of state A area (m2) Helmholtz free energy (J) amplitude of wave A∗ critical area (m2) Ac cross-sectional area (m2) Af frontal area (m2) AF air-fuel ratio (-) As surface area (m2) b parameter in an equation of state (m3/kg) B parameter defined in Eq. (10-71) parameter defined in Eq. (15-106) BPR bypass ratio (-) bwr back work ratio (-) c specific heat capacity (J/kg-K) speed of sound (m/s) speed of light in a vacuum (3×108 m/s) c̄ molar specific heat of an incompressible substance (J/kmol-K) C number of chemical species in a mixture (-) Ċ capacitance rate (product of mass flow rate and specific heat) (W/K) CD drag coefficient (-) COP coefficient of performance (-) cP specific heat capacity at constant pressure (J/kg-K) c̄P molar specific heat capacity at constant pressure (J/kmol-K) CR capacitance ratio (-) CR compression ratio (-) cv specific heat capacity at constant volume (J/kg-K) c̄v molar specific heat capacity at constant volume (J/kmol-K) D diameter (m) d⇀x differential displacement vector (m) E energy (J) voltage (Volt) number of elements (-) Ė rate of energy transfer (W) Ē intensity of radiation (W/m2) Eb,λ blackbody spectral emissive power (W/m2-μm) EER energy efficiency rating (Btu/W-hr) ei,j number of moles of element j per mole of substance i (-) Ej number of moles of element j (kmol) xix xx Nomenclature f frequency (Hz) fugacity (Pa) partition function (-) fraction of flow (-) F force (N) number of intensive properties (for Gibbs phase rule) ⇀ F force vector (N) fi fugacity of pure fluid i (Pa) f̂ i partial fugacity of component i in a mixture (Pa) g gravitational acceleration (m/s2) specific Gibbs free energy (J/kg) G Gibbs free energy (J) ḡ molar specific Gibbs free energy (J/kmol) gi degeneracy of energy level i (-) h specific enthalpy (J/kg) Planck’s constant (6.625×10−34 J/s) h̄ molar specific enthalpy (J/kmol) H enthalpy (J) Ḣ enthalpy flow rate (W) hav enthalpy of an air-water vapor mixture per kg dry air (J/kga) hconv convective heat transfer coefficient (W/m2-K) HC heat of combustion (J/kg) h̄dep molar specific enthalpy departure (J/kmol or J/kg) h̄dep,i molar specific enthalpy departure of component i in a mixture (J/kmol) hfg specific enthalpy of vaporization (J/kg) h̄fg molar specific enthalpy of vaporization (J/kmol) h̄form molar specific enthalpy of formation (J/kmol) HHV higher heating value (J/kmol) h̄i molar specific enthalpy of component i in a mixture (J/kmol) HSPF heating season performance factor (Btu/W-hr) h̄std standardized molar specific enthalpy (J/kmol) HV heating value (J/kmol) i current (Amp) ī unit vector in the x-direction I moment of inertia (kg-m2) j̄ unit vector in the y-direction k thermal conductivity (W/m-K) ratio of specific heat capacities, cP/cv (-) Boltzmann’s constant (1.3805×10−23 J/K) k̄ unit vector in the z-direction K spring constant (N/m) KE kinetic energy (J) kij binary mixing parameter (-) Kj equilibrium constant for reaction j (-) KP coefficient of pressure recovery (-) KT isothermal compressibility (1/Pa) L the dimension length L distance or length (m) LHV lower heating value (J/kmol or J/kg) m mass (kg) parameter in the RKS and PR equations of state (-) ṁ mass flow rate (kg/s) Nomenclature xxiii Xdes exergy destroyed (J) Xf exergy associated with a mass transfer (J) xi mole fraction of component i in the liquid phase of a mixture (-) XQ exergy associated with a heat transfer (J) Xs exergy of a system (J) x f specific exergy of a flowing substance (J/kg) xs specific exergy of a system (J/kg) Ẋdes rate of exergy destruction, also called the irreversibility rate (W) Ẋf rate of exergy flow with mass flow (W) ẊQ rate of exergy flow with heat (W) yi mole fraction of component i in a mixture (-) mole fraction of component i in the vapor phase of a mixture (-) z elevation in a gravitational field (m) Z compressibility factor (-) Zcrit critical compressibility factor (-) zi total mole fraction of component i in a mixture (-) Zi compressibility factor for component i in a mixture (-) Greek Symbols α reduced Helmholtz free energy (-) parameter in the RK or PR equation of state (-) β parameter defined in Eq. (15-101) (1/J) δ uncertainty in some measurement reduced density (-) differential amount change of some property of a system Goj standard state Gibbs free energy change of reaction (J) hfg latent heat of vaporization (J/kg) h̄mix molar specific enthalpy change of mixing (J/kmol) P pressure drop (Pa) s̄mix molar specific entropy change of mixing (J/kmol-K) T approach temperature difference (K) v̄mix molar specific volume change of mixing (m3/kmol) Vmix volume change of mixing (m3) ε Lennard-Jones energy potential (J) emissivity (-) effectiveness of a heat exchanger (-) reaction coordinate or degree of reaction (kmol) εi energy associated with a energy level i (J) ε j reaction coordinate for reaction j (kmol) φ fugacity coefficient (-) relative humidity (-) γ surface tension (N/m) η efficiency (-) η2 Second Law efficiency (-) λ wavelength (μm) undetermined multiplier number of phases (-) θ angle (radian) ρ density (kg/m3) xxiv Nomenclature σ Lennard-Jones length potential (m) Stefan-Boltzmann constant (5.67×10−8 W/m2-K4) τ torque (N-m) inverse reduced temperature (-) μ viscosity (Pa-s) μf,i chemical potential of component i in the liquid phase of a mixture (J/kmol) μg,i chemical potential of component i in the vapor phase of a mixture (J/kmol) μJT Joule-Thomson coefficient (K/Pa) νi stoichiometric coefficient for component i νi, j stoichiometric coefficient for component i in reaction j ψ constraint function that evaluates to zero ω angular velocity (rad/s) acentric factor (-) humidity ratio (kgv/kga) ωeff effective acentric factor of a mixture (-) thermodynamic probability (-) Superscripts ∗ quantity evaluated at location of critical area o under conditions where fluid behaves as an ideal gas (i.e., at low pressure) Subscripts a dry air act actual amb ambient as adiabatic saturation atm atmospheric av psychrometric property defined on a per mass of dry air basis avg average b boundary boiler blackbody B Boyle isotherm BDC bottom dead center BE Bose-Einstein model c compressor C cold fluid in a heat exchanger or cold thermal reservoir comp compression process cond condenser crit critical, related to the critical point CTHB cold-to-hot blow process cv associated with a control volume cyl cylinder d diffuser downstream of a normal shock drag des destroyed within system dp dew point ec evaporative cooler Nomenclature xxv evap evaporator exp expansion process f saturated liquid fuel furnace FD Fermi-Dirac model g saturated vapor gage gage gen generated within system generator in an absorption cycle H hot fluid in a heat exchanger or hot thermal reservoir heat pump HTCB hot-to-cold blow process hf heat transfer fluid hx heat exchanger i the ith component in a mixture IC based on incompressible model in in, entering a system ini initial, at time = 0 load refrigeration or building load max maximum or maximum possible min minimum or minimum possible mix associated with a mixture mp maximum power MB Maxwell-Boltzmann model n nozzle net net output nom nominal value o overall stagnation initial p pump piston propulsive P related to an isobaric process, at constant pressure associated with the products of a reaction pure associated with a pure substance out out, leaving a system R refrigeration cycle associated with the reactants of a reaction Rankine Rankine cycle r reduced ref reference res residual component rev reversible rh reheat cycle or reheater s associated with a reversible device sat saturated sc subcool sh superheat sur surroundings 1 Basic Concepts 1.1 Overview Thermodynamics is unquestionably the most powerful and most elegant of the engi- neering sciences. Its power arises from the fact that it can be applied to any discipline, technology, application, or process. The origins of thermodynamics can be traced to the development of the steam engine in the 1700’s, and thermodynamic principles do govern the performance of these types of machines. However, the power of thermodynamics lies in its generality. Thermodynamics is used to understand the energy exchanges accom- panying a wide range of mechanical, chemical, and biological processes that bear little resemblance to the engines that gave birth to the discipline. Thermodynamics has even been used to study the energy exchanges that are involved in nuclear phenomena and it has been helpful in identifying sub-atomic particles. The elegance of thermodynamics is the simplicity of its basic postulates. There are two primary ‘laws’ of thermodynamics, the First Law and the Second Law, and they always apply with no exceptions. No other engineering science achieves such a broad range of applicability based on such a simple set of postulates. So, what is thermodynamics? We can begin to answer this question by dissecting the word into its roots: ‘thermo’ and ‘dynamics’. The term ‘thermo’ originates from a Greek word meaning warm or hot, which is related to temperature. This suggests a concept that is related to temperature and referred to as heat. The concept of heat will receive much attention in this text. ‘Dynamics’ suggests motion or movement. Thus the term ‘thermodynamics’ may be loosely interpreted as ‘heat motion’. This interpretation of the word reflects the origins of the science. Thermodynamics was developed in order to explain how heat, usually generated from combusting a fuel, can be provided to a machine in order to generate mechanical power or ‘motion’. However, as noted above, thermodynamics has since matured into a more general science that can be applied to a wide range of situations, including those for which heat is not involved at all. The term thermodynamics is sometimes criticized because the science of thermodynamics is ordinarily limited to systems that are in equilibrium. Systems in equilibrium are not ‘dynamic’. This fact has prompted some to suggest that the science would be better named ‘thermostatics’ (Tribus, 1961). Perhaps the best definition of thermodynamics is this: Thermodynamics is the science that studies the conversion of energy from one form to another. This definition captures the generality of the science. The definition also introduces a new concept – energy. Thermodynamics involves a number of concepts that may be new to you, such as heat and energy, and these terms must each be carefully defined. As you read this, it may seem that heat and energy are familiar words and therefore no further definition of these concepts is necessary. However, the common understanding of these terms differs from the formal definitions that are needed in order to apply the laws of thermodynamics. 1 2 Basic Concepts The First Law of Thermodynamics states that energy is conserved in all processes (in the absence of nuclear reactions). If energy is conserved (i.e., it is not generated or destroyed) then the amount of energy that is available must be constant. But if the amount of energy is constant then why do we hear on the news that the world is experiencing an energy shortage? How could we be ‘running out of energy’? Why do we receive monthly ‘energy’ bills? The answer to these questions lies in the difference between the term energy as it is commonly used and the formal, thermodynamic definition of energy. These differences between common vernacular and precise thermodynamic definitions are a source of confusion. The term energy that is used in everyday conversations should be thought of as ‘the capacity to do work’. This definition is not consistent with the thermodynamic definition of energy, but rather refers to a different thermodynamic concept that is referred to as exergy and is studied in Chapter 7. The thermodynamic definition of energy is not as satisfying. Energy is not really ‘something’; rather, it is a property of matter. We cannot see, smell, taste, hear or feel energy. We can measure it, but only indirectly. Hopefully, the thermodynamic concept of energy will become clearer as you progress through this book. The First Law of Thermodynamics is concerned with the conservation of energy. However, energy has both quantity and quality. The quality of energy is not conserved and the Second Law of Thermodynamics can be interpreted as a system for assigning quality to energy. Although energy is conserved, the quality of energy is always reduced during energy transformation processes. Lower quality energy is less useful to us in the sense that its capability for doing work has been diminished. The quality of energy is continuously degraded by all real processes; this observation can be expressed in lay terms as ‘running out of energy’. The Second Law is responsible for the directional nature of all real processes. That is, processes can occur in only one direction and will not spontaneously reverse themselves because doing so would require a spontaneous increase in the quality of energy. The Second Law explains why heat flows from hot to cold and why objects at different temperatures will eventually come to the same temperature. The Second Law explains why gases mix and things break. It can be used to explain why we age and why time moves forward. The Second Law of Thermodynamics is likely the most famous law in all of the physical sciences. Our society is now facing some very challenging problems. Some of these problems are related to the diminishing supply of petroleum, coal, natural gas, and the other combustible materials that provide the energy (the common definition rather than the thermodynamic definition) that powers our world. Even if these fuels were inexhaustible (which they are not), combustion of carbon-based fuels necessarily produces carbon dioxide, which has been linked to global warming and other climate change phenomena. What alternatives exist to provide the power that we need? Hydrogen-powered fuel cells, biomass, nuclear power plants, solar and wind energy systems have all been mentioned in the popular media as potential solutions. Which one of these alternatives is actually best? What role can each of them play in terms of displacing our current energy supplies? These are huge questions. The solution to our energy problem will likely be one of the biggest challenges facing our species this century. In one sense this is alarming, but it is also very exciting. You are reading this book because you have either a professional or personal interest in the subject of thermodynamics. Thermodynamics plays a major role in addressing these energy-related questions. It is clear that the demand for people who are well-educated in thermodynamics and capable of applying the discipline to a wide range of problems will only increase. 1.3 States and Properties 5 measured. Derived properties include internal energy, enthalpy, entropy, and other related thermodynamic properties that will be defined in this text. It is not always clear whether a property is measurable or derived. For example, temperature is normally considered to be a measurable property. But how does one actually measure temperature? The common thermometer consists of a precision bore within a transparent glass enclosure that is filled with a liquid that expands when its tem- perature is increased. By observing the height of the liquid in the bore, we can measure the volume of the fluid and infer the temperature. There are many other ways to measure temperature. For example, thermistors relate the electrical resistance of a material to its temperature. Thermocouples are junctions between two dissimilar metals that gen- erate a voltage potential that is a function of temperature. In each of these instruments, however, something (e.g., volume, resistance, or voltage) is directly measured and tem- perature is then inferred from this measurement. Although we do not directly measure temperature, it is still considered to be a measurable property. 1.3.3 Intensive and Extensive Properties Thermodynamic properties are also classified as being either intensive or extensive. Inten- sive properties are independent of the amount of mass in the system whereas the values of extensive properties depend directly on the amount of mass. Temperature and pres- sure, for example, are intensive properties. If you were told the temperature or pressure of a system and nothing more, you would have no idea of the size of the system. Volume and energy are extensive properties. The greater the volume of a system, the more mass it must have. Extensive properties are linearly related to the system mass. A specific property is defined as the ratio of an extensive property of a system to the mass of the system. Thus specific volume, v, is the ratio of volume (an extensive property) to mass: v = V m (1-1) where V is the volume of the system and m is the mass of the system. The inverse of specific volume is density: ρ = 1 v = m V (1-2) Specific volume and density are both intensive properties. We will encounter several other extensive properties, including internal energy (U), enthalpy (H), and entropy (S). The corresponding specific properties are specific internal energy (u), specific enthalpy (h), and specific entropy (s): u = U m (1-3) h = H m (1-4) s = S m (1-5) 1.3.4 Internal and External Properties Properties can also be classified as being either internal or external. The value of an internal property depends on the nature of the matter that composes the system. External 6 Basic Concepts properties are independent of the nature of the matter within the system. Examples of external properties include the velocity of the system (Ṽ) and its elevation in a gravitational field (z). These properties do not depend on whether we are talking about a system composed of helium or one composed of steel. For example, in Section 3.2.2 we will see that a system with a mass m = 1 kg that is elevated a distance z = 1 m in Earth’s gravitational field (g = 9.81 m/s2) will have a potential energy PE = m g z = 9.81 J regardless of the type of matter that the system is composed of. Internal properties depend on the nature of the matter in the system and, as a consequence, they depend upon each other. That is, internal properties are functionally related to one another. The interdependence of internal properties is of fundamental importance because it allows us to completely fix the state of a system by specifying only a few internal properties. The values of other internal properties can be found by employing the rela- tionships that exist between these properties. It will be shown in Chapter 2 that only two internal intensive properties are required to fix the state of a system containing a pure substance that consists of only one phase (i.e., solid, liquid, or vapor). For example, if the temperature and pressure of water vapor are specified, then the density, specific internal energy, and specific enthalpy all have fixed values. Any other intensive property of the water vapor could also be determined. It is only necessary to know the temperature and pressure of a single phase pure substance in order to determine the specific heat capacity at constant pressure, the magnetic moment, the surface tension, the speed of sound, the electrical resistivity, and many other properties. You likely have already employed a property relationship in your chemistry class by using the absolute temperature (T) and absolute pressure (P) of a gas in order to calculate its specific volume (v) using the ideal gas law: v = R T P (1-6) where the parameter R is the ideal gas constant. The ideal gas law will be discussed in Section 2.5. It does not apply under all conditions. The accuracy of Eq. (1-6) is reduced as the pressure is increased or as the temperature is decreased. Under some conditions, the ideal gas law may not be sufficiently accurate to be of any use at all. However, this complication does not change the fact that the specific volume is fixed at some value (i.e., it is not an independent variable) when the temperature and pressure are specified. If the ideal gas law is not applicable, a more complicated relation between specific volume, temperature and pressure may be needed, as discussed in Chapter 10. The properties of many substances have been measured and the relationships between internal properties can be expressed using tables, charts, equations, and computer programs, as described in Chapter 2. 1.4 Balances Balances are the basic tool of engineering. Once a system has been carefully defined, it is possible to apply a balance to the system. A balance is simply a mathematical statement of what we know to be true. Any number of quantities can be balanced for any arbitrary system. The general balance equation, written for a finite period of time and some arbitrary quantity is: In + Generated = Out + Destroyed + Stored (1-7) where In is the amount entering the system by crossing its boundary, Generated is the amount generated within the system, Out is the amount leaving the system by crossing its boundary, Destroyed is the amount destroyed within the system, and Stored is the 1.4 Balances 7 amount stored in the system (i.e., the change in the quantity during the period of time). A balance can also be written on a rate basis, in which case the rate of each of the terms in Eq. (1-7) must be balanced at a particular instant in time. It is important to emphasize that the balance provided by Eq. (1-7) makes no sense until you have carefully defined a system and its boundaries. Every month, most of us define a system that is referred to as our household and carry out a money balance on this system: Din + Dgen = Dout + Ddes + D (1-8) where D indicates the amount of money. The variable Din is the amount of money that enters your household (from wages and other forms of income), Ddes is the amount of money that is destroyed, Dout is the amount of money that leaves your household (expenses), and Dgen is the amount of money that is created in your household. The term D in Eq. (1-8) is the amount of money stored in your household, i.e., the change in the amount of money contained within your household during the time period of interest. A positive value of D indicates that you managed to save some money during the month while a negative value indicates that you had to dip into your savings. For most of us, Ddes will be zero every month as we do not often literally burn up or destroy currency. Also, Dgen will be zero as we cannot (legally) generate money. Thus, balancing our monthly finances will result in the following equation: Din = Dout + D (1-9) Equation (1-9) shows that, at least on a personal level, money is a conserved quantity; that is, it is neither destroyed nor produced. Other quantities are not conserved. For example, we could define a system around the borders of the United States and balance people for a year: Pin + Pgen = Pout + Pdes + P (1-10) In Eq. (1-10), Pin is the number of people that enter the U.S. by crossing its borders (immigration) and Pout is the number of people leaving the U.S. by crossing its borders (emigration). The quantity P is the change in the population of the U.S. during the year (i.e., the number of people at the end of the year less the number of people at the beginning of the year). People are, as you know, not a conserved quantity; they are both destroyed and generated. The quantity Pdes is the number of people that die and Pgen is the number of babies born within the borders of the U.S during the year. Equation (1-10) must be satisfied as it is simply a mathematical statement of what we know to be true. Mass is a conserved quantity since it cannot be generated or destroyed (in the absence of nuclear reactions). Therefore, a mass balance on a system for a finite period of time leads to: min = mout + m (1-11) where min is the amount of mass that enters the system by crossing its boundary and mout is the amount of mass that leaves the system by crossing its boundary. Note that the quantities min and mout must be zero for a closed system. The quantity m is the amount of mass stored in the system; this is the mass in the system at the end of time period less the mass in the system at the beginning of the time period. A mass balance written on a rate basis at a particular instant in time is: ṁin = ṁout + dmdt (1-12) 10 Basic Concepts Equations Window “this is a comment” {this is also a comment} Figure 1-4: Equations window with equations and comments. practice to enter the comments immediately to the right of each equation; this process is facilitated by pressing the tab key. Information within comments is ignored by EES and comments may span as many lines as needed. EES will display the comments in blue. It is sometimes difficult to interpret equations that have been entered in text for- mat in the Equations window, particularly when many nested sets of parentheses or operations are employed. Therefore, EES provides a Formatted Equations window that displays the equations that are entered in the Equations window using a mathematical notation. Select Formatted Equations from the Windows menu in order to access the Formatted Equations window (Figure 1-5). Notice that the comments that are entered in the Equations window within quotes are also displayed in the Formatted Equations window whereas comments entered in curly braces are not displayed. Normally, com- ments within quotes are used to document the equations whereas curly braces are used to “comment out” text that you do not wish EES to use at this time. “Commenting out” a set of equations is a convenient way to remove these equations temporarily. To accomplish this, highlight the equation(s) to be removed and right-click. Select Com- ment {} from the pop-up menu that appears. To re-instate the equation(s), highlight them again, right-click, and select Undo Comment {}. The equations in the Formatted Equations window can be copied and pasted, for example into a report documenting the model. Highlight the equation(s) of interest and right-click on the selection. Notice that it is possible to copy the equation as a picture that can be pasted into a word processor. The Professional version of EES can also copy the equation as a LaTeX object or a MathType R© equation. Select Solve from the Calculate menu. A dialog window will appear indicating the progress of the solution. Click the Continue button when the calculations are completed in order to display the Solution window that contains the solution to the set of equations (Figure 1-6). EES can solve thousands of equations very quickly, which makes it a very powerful tool. The Equations window allows a free form input. As you can see in Figure 1-4, the position of variables within the equation does not matter and it is not necessary to isolate the unknown variable on the left side of an equal sign, as is required in formal programming languages. This capability is convenient because in many problems, it is not possible to isolate the unknown variable. Also, the order in which the equations are entered does not matter. Before the equations are solved, EES will (internally) rearrange them into an order that leads to the most efficient solution process, regardless Formatted Equations this is a comment Figure 1-5: Equations displayed in the Formatted Equations window. 1.6 Dimensions and Units 11 Solution Unit Settings: SI K Pa J mass radFigure 1-6: Solution window. of the order that they are entered in the Equations window. Although EES allows you to enter equations in any order, it is still recommended that you enter your equations in an organized manner that progresses logically from the known information to the desired results. Also, it is best to enter and solve a few equations at a time, rather than entering all of the equations needed to solve a problem at once. This strategy allows you to efficiently debug your program because problems are naturally isolated to the last group of equations that were entered. These best practices for using EES will be demonstrated in the example problems presented throughout this textbook. 1.6 Dimensions and Units Most of the variables used in models of thermodynamic processes and systems represent physical quantities and therefore they are dimensional. It is necessary to know both the value of the variable as well as its associated units. Calculation errors that result from incorrectly converting between units are common and frustrating. Careful attention to units is essential. This section reviews the dimensions and units of the fundamental quantities used in thermodynamic analyses. One of the most basic and powerful features of the EES program is its ability to keep track of units, convert between units, and check equations for unit consistency. 1.6.1 The SI and English Unit Systems Dimensions are the fundamental measures of a physical quantity. Dimensions are cat- egorized as being either primary or secondary. Primary dimensions for the quantities encountered in most thermodynamic models are usually chosen to be mass, length, time, and temperature, although other choices are equally valid. Secondary dimensions are combinations of the primary dimensions and result from definitions or physical laws. Examples of secondary dimensions are length/time for velocity and mass-length/time2 for force. Note that in this text, when dealing with units, the dash symbol (-) will indicate multiplication on one side of the divisor. Therefore, the unit N-m/kg should be read as N m kg and the unit J/kg-K should be read as J kgK . This convention is consistent with how units are entered in the EES program. The scale for a dimension is its units. Many different units can be used to express a primary dimension. For example, the possible units for the dimension length include inches, feet, yards, miles, meters, centimeters, furlongs, and many others. Secondary dimensions can also be expressed in many different units. Energy, for example, is a sec- ondary dimension that is expressed in terms of primary dimensions as mass-length2/time2. The units of energy include ft-lbf, Btu (British thermal units), calories, Joules, and many others. Units are commonly categorized as belonging to the English or SI (Systems Interna- tional) unit system. A list of the standard units used for the primary dimensions in each system is provided in Table 1-1. Some secondary dimensions and their typical units in each system are provided in Table 1-2. 12 Basic Concepts Table 1-1: Standard units for primary dimensions in the SI and English unit systems. SI Unit System English Unit System Physical Quantity Unit Symbol Unit Symbol mass (M) kilogram kg pound-mass lbm length (L) meter m foot ft time (t) second s second s temperature (T) Kelvin K Rankine R degree Celsius ◦C degree Fahrenheit ◦F Why is there more than one unit for each dimension? There is no single answer to this question. Convenience certainly provides one explanation. It is possible but not convenient to express the distance between New York and California in inches or feet, but miles is a more convenient measure. Different societies have historically adopted different units for the same dimension and these conventions tend to persist for a long time. In at least one case, the primary units associated with a secondary dimension remained undiscovered until after the dimension itself had already achieved widespread use. Heat and work were for a long time considered to be unrelated quantities. The sci- ence of calorimetry was developed to measure heat and used as its fundamental unit the British Thermal Unit (Btu) or calorie. Work was understood to be a separate quantity that resulted in lifting a weight; therefore, the unit of work was taken to be ft-lbf. Joule demonstrated in the late 1800’s that many of the effects produced by heat could also be produced by work. At that time it became clear that heat and work both refer to the transfer of energy and therefore could share a common unit. Whatever the reasons, physical quantities can be expressed using many different units and this fact introduces the possibility for unit conversion errors. The practicing engineer must be able to deal with and convert between a variety of units and unit systems. Instruments will report measurements in a variety of units. For example, a vacuum gage may report pressure measurements in units of torr while a water manometer will naturally lead to a pressure measurement in units of inches of water. Engineers must communicate the results of their analyses to a variety of audiences, some of whom are most comfortable thinking in terms of a specific set of units. For example, cooling power Table 1-2: Units of some secondary dimensions in the SI and English unit systems. SI Unit System English Unit System Physical Quantity Dimensions Unit Symbol Unit Symbol force M-L/t2 Newton N pound-force lbf energy M-L2/t2 Joule J British Thermal Unit Btu power M-L2/t3 Watt (J/s) W horsepower hp pressure M/L-t2 Pascal (N/m2) Pa pound/inch2 psi bar bar atmosphere atm 1.6 Dimensions and Units 15 EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL E EXAMPLE 1.6-2: POWER REQUIRED BY A VEHICLE The two major forces opposing the motion of a vehicle on a level road are the rolling resistance of the tires (Fr) and the aerodynamic drag on the car (Fd). The rolling resistance is the product of the dimensionless rolling resistance coefficient, f = 0.02, and the force exerted by the vehicle on the road (i.e., its weight, W ). Fr = f W (1) The aerodynamic drag is expressed in terms of a dimensionless drag coefficient (Cd) according to: Fd = Af Cd 12ρ Ṽ 2 (2) where Af is the frontal area of the vehicle, ρ = 0.075 lbm/ft3 is the density of air and Ṽ is the speed of the vehicle. The Toyota Prius has a drag coefficient of Cd = 0.29, a frontal area of Af = 21.2 ft2 and a curb weight of W = 2930 lbf. a) Determine the power required by a Prius (in hp) traveling at a velocity of Ṽ = 65 mph. It is good form to put the problem inputs at the top of the Equations window. Each input is entered and immediately annotated. We’ll start with the density of air. rho=0.075 [lbm/ftˆ3] “density of air” Note that the units of the numerical constant, 0.075, are entered in square brackets immediately after the constant is typed in EES. If the EES code is solved at this point (select Solve from the Calculate menu) you will see that the units of the variable rho are indicated next to its value in the Solutions window (Figure 1). The ˆ symbol in the unit designation for rho is used to raise a unit to a power. EES will also accept the unit lbm/ft3, but the ˆ symbol helps EES recognize that the 3 should be a superscript in the formatted output. Also note that the unit for pound-mass is represented in EES as either lb_m or lbm. The underscore causes EES to recognize that the m should be a subscript, but either designation is acceptable. Solution Unit Settings: SI K Pa J mass rad [lbm/ft 3]Figure 1: Solution window. The unit designations that are recognized by EES can be viewed by selecting Unit Conversion Info from the Options menu. Select the dimension from the list on the left and EES will display the defined units associated with that dimension in the list on the right, as shown in Figure 2 for the dimension mass. As discussed in Section 1.6.1, the inputs to the problem should be converted to SI units. EES provides the function convert in order to easily convert from one 16 Basic Concepts EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL E lbm kg Mass Unit Conversion Information Figure 2: Unit Conversion Information dialog. unit to another without resorting to the use of tables of unit conversion factors (like those included in Appendix A). The convert function accepts two arguments. The first argument indicates the unit(s) that you wish to convert from, and the second argument is the unit(s) that you wish to convert to. Both sets of units must have the same dimensions (i.e., they must be dimensionally consistent). So, for example, the conversion factor that is required to convert from feet to meters is obtained from the EES code convert(ft,m). According to Appendix A, the function convert(ft,m) will return the conversion factor 0.30480 m/ft. In our case, we wish to convert the density of air from the English units that it was provided in, lbm/ft3, to SI units, kg/m3. We can do this conversion by revising the information in the Equations window as follows: rho=0.075 [lbm/ftˆ3]*convert(lbm/ftˆ3,kg/mˆ3) “density of air” Select Solve from the Equations menu (or press the F2 key, the shortcut for Solve) and you will see the Solution window (Figure 3). Solution Unit Settings: SI K Pa J mass rad [lbm/ft 3] Figure 3: Solution window. Notice the warning message indicating that EES has detected a unit problem. EES checks the units of your equations each time you solve. You can have EES check units at any time by selecting Check Units from the Calculate menu (or by pressing F8). A window will appear that provides a list of all of the equations that 1.6 Dimensions and Units 17 EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL Econtain unit errors as well as some description of the unit inconsistency that was detected (Figure 4). Check Units The units of rho [lbm/ft^3] and 0.075 * 16.0184635 [(lbm/ft^3) * ((kg/m^3)/(lbm/ft^3))] differ by a factor of 16.02. Figure 4: Check Units dialog. The units of the variable rho were previously set to lbm/ft3; these units are no longer consistent with the equation that is used to specify the variable. This is most evident by examining the Formatted Equations window (Figure 5). Formatted Equations [lbm/ft 3] density of air Figure 5: Formatted Equations window. Figure 5 shows that the units of the variable rho are clearly kg/m3 and therefore the units assigned to this variable should be updated to reflect the unit conversion that was carried out. There are several ways to set the units of a variable. One way is to highlight the variable in the Equations window and right-click on it; select Variable Info from the pop-up menu that appears (Figure 6). rho Equations Window [lbm/ft ^3] “density of air” Variable Info Shift+Ctrl+V Figure 6: Setting the units of a variable from the Equations window. A dialog box will appear that allows you to set the units of the variable rho, as shown in Figure 7. 20 Basic Concepts EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL E generate a Parametric table. Select New Parametric Table from the Tables menu in order to bring up the New Parametric Table dialog (Figure 10). vel W_dot hp New Parametric Table Figure 10: New Parametric Table dialog. The dialog asks for the number of runs that should be placed in the table and the name of the table. A list of all of the variables used by the EES code is shown on the left. Highlight the variable that will be varied in order to make the plot as well as any variables that will be examined parametrically and select the Add button. In order to make the plot requested by the problem statement, the variables vel and W_dot_hp should be placed in the table. Select OK and a Parametric table will be created with a column for each of the selected variables (Figure 11). Parametric Table [m/s] [hp] Figure 11: Parametric table, empty. When Solve Table is selected from the Calculate menu, EES will sequentially work with each row of the table. If a value is set in one of the columns of the table for that row, then EES will assign that value to the corresponding variable in the Equations window and solve the resulting set of equations. The values of the variables in any of the other columns in the table will be calculated. Entered values are, by default, displayed in black type whereas calculated variables are displayed in blue after the calculations are completed. This process is repeated for each row in the table. It is possible to specify the value for velocity in each run manually, but it is much more convenient to do this automatically. Right-click on the heading 1.6 Dimensions and Units 21 EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL Eof the column corresponding to the variable vel and select Alter Values in order to bring up the dialog shown in Figure 12. vel: Column 1 Figure 12: Alter Values dialog. Fill in the dialog so that the variable vel is varied from Ṽ = 0 to 50 m/s in equal increments (Figure 13). Parametric Table [m/s] [hp] Figure 13: Parametric table with velocity values set. Select Solve Table from the Calculate menu (or the corresponding shortcut key, F3) and you will receive an error indicating that the value of the variable vel has already been set. It is not possible for EES to set the value of velocity to 0 m/s in Run 1 of the table when the value of the velocity has already been set to 29.06 m/s (corresponding to 65 mph) in the Equations window. In order to proceed, it is necessary to remove the line of code in the Equations window that specifies velocity. One convenient method for temporarily removing the line of code is to highlight the appropriate line in the Equations window and right-click on it. Select Comment {} from the pop-up menu that appears and the line will be “commented out”; that is, it will be surrounded by curly brackets so that it is read as a comment, having no effect on the equation set. {vel=65 [mph]*convert(mph,m/s) “velocity of car”} It is easy to re-activate (or “uncomment”) the line of EES code by highlighting it again, right-clicking, and selecting Undo Comment {}. For now, leave the line 22 Basic Concepts EX AM PL E 1. 6- 2: PO W ER RE QU IR ED BY A VE HI CL E “commented out” and solve the Parametric table. The column corresponding to the variable W_dot_hp should be filled in (Figure 14). Parametric Table [m/s] [hp] 0 2.021 4.515 7.956 12.82 19.57 28.69 40.65 55.93 74.99 Figure 14: Parametric table, after it is solved. Data contained in a Parametric table (or any other type of table in the EES program) can be plotted. Select New Plot Window from the Plots menu and then select X-Y plot in order to bring up the New Plot Setup dialog shown in Figure 15. W_dot_hp New Plot Setup vel Figure 15: New Plot Setup dialog. The right side of the dialog is used to specify the source of the data (the table in which the data are stored). There is only one table in our EES code (Table 1, a Parametric table). Select the X- and Y-Axis variables and then hit OK. The plot, with some formatting, is shown in Figure 16. 1.7 Specific Volume, Pressure, and Temperature 25 tank at pressure P L atmospheric pressure, Patm Patm Ac P Ac A Lc r g Figure 1-7: Manometer. balance on the liquid column is shown in Figure 1-7 and provided below: P Ac︸︷︷︸ upward pressure force on bottom of column = Ac Lρ g︸︷︷︸ weight of column + Patm Ac︸︷︷︸ downward pressure force on top of column (1-21) where Ac is the cross-sectional area of the tube, ρ is the density of the fluid in the tube, and g is the acceleration of gravity. Equation (1-21) can be rearranged: P − Patm = Lρ g (1-22) Both ρ and g are known and therefore measurement of L is sufficient to provide the pressure relative to the atmospheric pressure (P – Patm). Manometers are in such common use that units for pressure include inches of water (inH2O) and inches of mercury (inHg); these units are included in Table A-4 and are recognized by EES. Many other measurement devices also measure pressure relative to atmospheric pressure. Various types of elastic pressure transducers determine pressure by measuring the displacement caused when atmospheric pressure is applied to one side of a mechanical element (e.g., a membrane) while the pressure of interest is applied to the other. The result is clearly a direct measurement of the pressure relative to the atmospheric pressure (P – Patm) rather than the pressure alone. The term gage pressure refers to the pressure measured with respect to atmospheric pressure: Pgage = P − Patm (1-23) The absolute pressure (P) can be obtained from a gage pressure measurement by adding the atmospheric pressure. Unless otherwise stated, all pressures used within this text will be absolute pressure. Gage pressures are sometimes reported by adding the letter g to the end of the unit and absolute pressures by adding the letter a. For example, 5 psig should be read as 5 pounds per square inch gage pressure and 19.7 psia should be read as 19.7 pounds per square inch absolute pressure; these two pressures are the same provided that atmospheric pressure is 14.7 psi. There are specialized instruments that measure the absolute atmospheric pressure. For example, a barometer is an inverted glass tube that has been evacuated and placed in a pool of liquid, as shown in Figure 1-8. The height of the liquid in the tube relative to the pool of liquid (L) is proportional to the absolute pressure above the pool (i.e., the local atmospheric pressure). A force balance on the column of liquid is shown in Figure 1-8: Patm Ac︸︷︷︸ upward pressure force on bottom of column = Ac Lρ g︸︷︷︸ weight of column (1-24) 26 Basic Concepts atmospheric pressure, Patm evacuated Patm Ac Ac L r gL Figure 1-8: Barometer. Equation (1-24) can be rearranged: Patm = Lρ g (1-25) Most laboratories include a barometer so that gage pressure measurements can be accu- rately converted to absolute pressure. 1.7.3 Temperature The precise meaning of temperature is not easily stated. From a molecular viewpoint, the temperature of a substance is related to the average energy or velocity of the individual molecules of the substance. When high velocity (i.e., hot) molecules are in the presence of low velocity (i.e., cold) molecules, then collisions between the molecules tend to cause an energy transfer that we call heat from the hot molecules to the cold molecules. Most of us are familiar with either the Fahrenheit or Celsius temperature scales. Depending on what country you live in, the weather forecasts are provided in these scales. However, these are not absolute temperature scales in the same way that gage pressure is not absolute pressure. Zero gage pressure (e.g., 0 psig) does not correspond to zero force per unit area but rather to atmospheric pressure. Zero degrees Fahrenheit (or Celsius) does not correspond to an absence of molecular motion (i.e., absolute zero). The different temperature scales are of interest because they are in common use and their origin reflects the problems that are encountered when setting up a unit system. How might you set up a scale for temperature? At a minimum, it is necessary to select two fixed or reference temperatures. One point fixes the zero for the scale and the second allows the degree size to be specified. In the early 1700’s, Anders Celsius in Sweden chose the freezing and boiling points of pure water as the fixed points for a temperature scale (although he apparently did not initially choose the zero point to be the freezing point of water, as we do today). Around the same time, Daniel Fahrenheit in Holland developed a scale in which the zero point was defined as the freezing point of a salt- water solution. There are different stories about how the second point was selected. One explanation is that he used normal human body temperature as the second point. Normal human body temperature is now accepted to be 98.6◦F but, for some reason, he initially assigned this temperature to 96◦F. An alternative story suggests that he used the body temperature of a cow (approximately 100◦F) as the second fixed point (which is perhaps more understandable than using the body temperature of a human). Yet another story suggests that he assigned 212◦F to be the boiling point of water in order to fix the scale. It is not clear why 212◦F would be chosen for this point as opposed to a more round number. 1.7 Specific Volume, Pressure, and Temperature 27 0 100 200 300 400 500 600 -300 -250 -200 -150 -100 -50 0 50 100 Pressure (kPa) T e m p e ra tu re ( °C ) data best fit line to data 1 g of air 3 g of air 5 g of air -273.15°C Figure 1-9: Temperature as a function of pressure in a 1 liter container containing various masses of air. The Kelvin and Rankine scales are absolute temperature scales. They have the same degree size as the Celsius and Fahrenheit scales, respectively, but because they are absolute scales, the zero point corresponds to absolutely no molecular motion (i.e., absolute zero – an unattainable condition). The zero point for these scales was first identified from observations of the pressure of a gas in a fixed volume container, as reported by Boyle and Charles in the 1600’s. Recall that pressure is the manifestation of molecular motion through the force exerted by collisions of molecules with a surface. Therefore, pressure is related to temperature. Researchers noticed that the pressure and temperature of a gas are linearly related to one another under some conditions (i.e., conditions where the ideal gas law applies). For example, Figure 1-9 illustrates measurements of temperature as a function of pressure for 3 g of air contained in a 1 liter rigid volume. Extrapolation of these data to zero pressure (corresponding to some theoretical condition where there is no molecular motion – i.e., absolute zero) leads to an ultimate lower limit on temperature of –273.15◦C. Figure 1-9 shows that this lower limit on the temperature applies regardless of the mass of air that is contained in the tank (provided, of course, that the ideal gas law con- tinues to hold). Data taken in this manner for other gases also extrapolates to –273.15◦C. All of this evidence suggests that the absolute temperature scale corresponding to the Celsius scale should be defined based on adding 273.15 to the value of the temperature in ◦C. The Kelvin temperature scale is therefore defined according to: T [K] = T [◦C] + 273.15 (1-26) A similar experiment accomplished using temperature measurements taken on the Fahrenheit scale would suggest that -459.67◦F is the lowest possible temperature. There- fore, the Rankine temperature scale is an absolute temperature defined according to: T [R] = T [◦F] + 459.67 (1-27) Equations for converting between any of the temperature scales are summarized in Table A-3 in Appendix A. Most unit conversions require only a multiplicative correction that can be provided by the convert function in EES, as discussed in Example 1.6-2. However, 30 Basic Concepts relation between resistance (R, ohms) and temperature (T, K) for a thermistor is given by: R = Ro exp [ α ( 1 T − 1 To )] where Ro is the resistance in ohms at temperature To in K and α is a material constant. For a particular resistor, it is known that Ro = 2.6 ohm with To = 298.15 K (25◦C). A calibration test indicates that R1 = 0.72 ohm at T1 = 60◦C. a) Determine the value of α. b) Prepare a plot of R versus T (in ◦C) for temperatures between 0◦ and 100◦C. Indicate the range over which this instrument will work best. C. Dimensions and Units 1.C-1 The damage that a bullet does to a target is largely dictated by the kinetic energy of the bullet. A 0.22 caliber bullet is fired from a handgun with a muzzle velocity of approximately Ṽ = 1060 ft/s and has a mass of m = 40 grains. A 0.357 magnum bullet is fired with a muzzle velocity of approximately Ṽ = 1450 ft/s and has a mass of m = 125 grains. Grains is the typical unit that is used to report the mass of a bullet, there are 7000 grains in a lbm. The kinetic energy of an object (EK) is given by: EK = m Ṽ 2 2 a) Determine the kinetic energy of the 0.22 and 0.357 caliber bullets (in lbf-ft) as they are fired. 1.C-3 Figure 1.C-3 illustrates a spring-loaded pressure relief valve. spring K= 5000 N/m x = 0.01 m atmospheric pressure, Patm = 100 kPa disk D = 1 cm m = 34 g internal pressure Figure 1.C-3: Spring-loaded valve. When the valve is seated as shown, the spring is compressed by x = 0.01 m and pushes down on a disk having a D = 1 cm diameter. The disk has a mass of m = 34 grams. One side of the valve is exposed to atmospheric pressure at Patm = 100 kPa and the other is exposed to an elevated internal pressure. The spring constant is K = 5000 N/m. a) Determine the pressure at which the valve opens (in kPa and lbf/in2). b) Generate a plot showing how the opening pressure and spring constant are related. 1.C-5 One of the main purposes of a seat belt is to ensure that the passenger stops with the car during a crash rather than flying freely, only to be stopped more quickly Problems 31 by a hard object. Assume that a vehicle traveling at Ṽini = 30 mph comes to a halt in a distance of L = 0.75 ft during a crash. Assume that the vehicle and passenger experience a constant rate of deceleration during the crash. a) The stopping distance of a passenger that is not wearing a seat belt is estimated to be 20% of the stopping distance of the vehicle. Calculate the force experi- enced by a m = 180 lbm passenger (in lbf and N) if he is not wearing a seat belt during the crash. b) Calculate the force experienced by a m = 180 lbm passenger (in lbf and N) if he is wearing a seat belt that does not stretch during the crash. c) Many seat belts are designed to stretch during a crash in order to increase the stopping distance of the passenger relative to that of the vehicle. Calculate the force experienced by the passenger (in lbf and N) if his seat belt stretches by Lstretch = 0.5 ft during the crash. d) Plot the force experienced by the passenger as a function of the stretch in the seat belt. 1.C-6 Some hybrid and electric car manufacturers have begun to advertise that their cars are “green” because they can incorporate solar photovoltaic panels on their roof and use the power generated by the panel to directly power the wheels. For example, a solar array can be installed on the roof of the Fisker Karma car (see www.fiskerautomotive.com). In this problem we will assess the value of a solar panel installed on the roof of a car. Assume that the panel is L = 5 ft long and W = 4 ft wide. On a very sunny day (depending on your location), the rate of solar energy per area hitting the roof of the car is sf = 750 W/m2. Assume that the panel’s efficiency relative to converting solar energy to electrical energy is ηp = 10% (0.10). The cruising power required by the car is Ẇcar = 20 hp. a) Estimate the rate of electrical power produced by the panel. b) Calculate the fraction of the power required by the car that is produced by the solar panel. c) If the car (and therefore the panel) sits in the sun for timesit = 6 hours during a typical day then determine the total amount of electrical energy produced by the panel and stored in the car’s battery. d) If the car is driven for timedrive = 30 minutes during a typical day then determine the total amount of energy required by the car. e) Calculate the fraction of the energy required by the car that is produced by the solar panel during a typical day. f) Create a plot showing the fraction of energy required by the car that is pro- duced by the solar panel, your answer from part (e), as a function of the panel efficiency; note that available photovoltaic panels operate at efficiencies less than 20%. D. Pressure, Volume and Temperature 1.D-2 A manometer is a device that is commonly used to measure gage pressure and a barometer is used to measure absolute pressure. In your lab there is a mercury barometer that is used to measure ambient pressure and a water manometer that is used to measure the gage pressure in a tank, as shown in Figure 1.D-2. The temperature in the tank is also measured. 32 Basic Concepts zw = 25.3 inch water manometer tank, T 92 F= ° mercury barometer evacuated, P = 0 zHg = 29.5 inch Patm Patm Figure 1.D-2: Mercury barometer and water manometer. The mercury barometer consists of a tube that is open at one end and contains a column of mercury. The closed end of the tube is evacuated. Therefore, the height of the mercury column can be related to the absolute pressure that is applied to the open end using a force balance. The height of the mercury column is zHg = 29.5 inch and the density of mercury is ρHg = 13,530 kg/m3. The water manometer is a U-shaped tube that is open at both ends. One end of the tube is exposed to the pressure in the tank and the other end is exposed to local atmospheric pressure. Therefore, the height of the column of water can be related to the pressure difference between the tank and ambient (i.e., the gage pressure). The height of the water column is zw = 25.3 inch and the density of water is ρw = 996.6 kg/m3. The temperature in the tank is 92◦F. a) Determine the gage pressure in the tank in Pa and psi. b) Determine the absolute pressure in the tank in Pa and psi. c) Determine the temperature in the tank in ◦C, K, and R. 1.D-4 You are working on an experimental fission reactor that is fueled by deuterium. The deuterium gas is very valuable and is delivered in V = 1 liter tanks, as shown in Figure 1.D-4. You would like to transfer as much of the deuterium that is initially in the tank to your experiment as possible. tank, V= 1 liter Tini = 20°C Pfill,gage = 15 psig to experiment at Pexp,gage = 5 psig Figure 1.D-4: Deuterium tank connected to your experiment. The initial temperature of the tank is Tini = 20◦C and the initial pressure is Pfill,gage = 15 psig (i.e., pounds per square inch gage pressure). The experiment is always maintained at pressure Pexp,gage = 5 psig (also a gage pressure). The atmospheric pressure is Patm = 1 atm. You should model deuterium as an ideal gas for this problem; the ideal gas law relates pressure, temperature, and specific volume according to: v = R T P where R = 2064 N-m/kg-K is the ideal gas constant for deuterium and T and P are the absolute temperature and pressure. 2.1 Equilibrium and State Properties 35 water vapor Figure 2-1: Water vapor enclosed in a piston-cylinder device. There is only one chemical species present in the system (i.e., water) and therefore C = 1. There is only one phase present (vapor) so = 1. Applying the phase rule, Eq. (2-1), to this system leads to: F = C − + 2 = 1 − 1 + 2 = 2 (2-2) Therefore, only two internal intensive properties are required to completely specify the internal intensive state of this system. The state is fixed if, for example, we specify the temperature and pressure of the water. The word ‘fixed’ here means that we are no longer free to arbitrarily specify the values of any other internal intensive property. For example, if the temperature and pressure of the water are known, all other specific internal properties, such as the specific volume, specific internal energy, and specific entropy, have fixed values. Suppose that the piston cylinder device contains a mixture of liquid water in equi- librium with water vapor, as shown in Figure 2-2. In this case, the number of phases is = 2 but the number of chemical species remains C = 1. Applying the phase rule to this situation: F = C − + 2 = 1 − 2 + 2 = 1 (2-3) indicates that it is only necessary to specify a single internal intensive property in order to fix the intensive state of both the liquid and vapor phases in the apparatus. For example, if we specify that the temperature is 100◦C, then every other internal intensive property is fixed. The pressure in the piston-cylinder device must be 101.325 kPa. We know this to be true because of careful experiments that have repeatedly been conducted in order to measure this property. The specific volumes of the liquid and vapor phases are also fixed as are the specific internal energy, specific enthalpy, and specific entropy of each phase. One purpose of this chapter is to explain how properties can be obtained from exist- ing data sources once the state of the substance has been fixed. The internal properties of many engineering substances of general interest have been measured and are recorded in tables, equations, or graphs. These databases can be used to retrieve the relationships between the internal properties for these substances. Under some conditions, the behav- ior of the substance is so simple that these databases are not necessary and very simple models (e.g., the ideal gas or incompressible substance models) can be applied with little loss of accuracy. water vapor liquid water Figure 2-2: Water in liquid and vapor phases enclosed in a piston-cylinder device. 36 Thermodynamic Properties 1 Pa 1 kPa 1 MPa 1 GPa 0 100 200 300 400 500 600 700 800 900 1000 Temperature (K) 1 atm critical point g f d e liquid supercritical vapor solid triple point cba Figure 2-3: Pressure-temperature diagram for water. Some of the properties that are tabulated in these databases (e.g., specific internal energy or specific entropy) are not directly measurable. If that is the case, how were these values determined and recorded? Internal properties are functionally related by the First and Second Laws of Thermodynamics. Therefore it is possible to infer the value of non- measurable properties from the values of other properties that are measurable. For example, we will see in Chapter 10 that the specific internal energy, specific enthalpy and specific entropy of a substance can be derived provided that the specific heat capacity and specific volume of the substance (both of which are measurable properties) are accurately known as a function of temperature and pressure. 2.2 General Behavior of Fluids All pure substances exhibit qualitatively similar behavior and therefore it is only neces- sary to completely understand the behavior of one fluid in order to have a firm grasp of the general behavior of all fluids. Water is a fluid that we are familiar with, so we will examine its behavior in some detail in order to develop our understanding of fluids in general. Figure 2-3 illustrates the phase of water in the parameter space of pressure and temperature. The phase rule dictates that two independent properties (e.g., pressure and temperature) are required to specify the state of a single phase (i.e., = 1), pure (C = 1) fluid. Therefore, single phase liquid water can exist over a range of both tem- perature and pressure; this area is labeled ‘liquid’ in Figure 2-3. The regions of pressure and temperature where single phase water vapor and single phase solid water (i.e., ice) exist are also labeled in Figure 2-3. At a specified pressure (e.g., 1 atm or 101.325 kPa) and a sufficiently low temperature water exists as a solid, as indicated by point a in Figure 2-3. As the temperature of water is increased at constant pressure from its value at point a, the water remains a solid until it reaches its freezing point at one atmosphere (which lies on the line separating the solid and liquid phases) at which point it transitions from a solid to a liquid. Figure 2-3 shows that the freezing point of water at atmospheric pressure is 273.15 K (0◦C). Eventually, the water completely melts and becomes a single phase liquid, as indicated by point b. Further heating at constant pressure causes the liquid temperature to rise until eventually 2.2 General Behavior of Fluids 37 the water reaches its boiling temperature and begins to evaporate (i.e., to turn from liquid to vapor) when it reaches the line separating the liquid and vapor phases. With additional heating, the water becomes a single phase vapor, as indicated by point c. The normal boiling point is defined as the boiling point at atmospheric pressure; the normal boiling point of water is 373.15 K (100◦C). At very low pressures, the solid phase of water will transition directly to vapor phase without passing through a liquid phase. This process is called sublimation and is shown in Figure 2-3 by the process of moving from point d to point e. A familiar substance that sublimes at atmospheric pressure is carbon dioxide, which is also called dry ice. An interesting and unusual characteristic of water is that the saturation line delin- eating its solid phase from its liquid phase actually has a slightly negative slope (i.e., the freezing temperature of water decreases slightly with increasing pressure). This charac- teristic is a result of the fact that solid water (ice) is less dense than liquid water. This is the reason ice chunks float rather than sink. One consequence of this behavior is that it is possible to melt ice at constant temperature by increasing the pressure that is applied to it; this process is shown in Figure 2-3 by the process of moving from state f to state g. This characteristic of water is the reason that ice skates work. Your body weight is concentrated on the very small area associated with the metal skate blade, substantially increasing the pressure exerted on the ice below the blade. The increase in pressure causes the ice under the blade to melt, forming a thin film of liquid water that lubricates the blade. Relatively few substances exhibit this characteristic; some examples include bismuth, gallium, antimony, silicon and acetic acid. Notice that the information presented in Figure 2-3 is consistent with the phase rule. When water exists in a single phase, both temperature and pressure may vary independently and so single phase water exists over regions in the pressure-temperature diagram. However, there are two phases present during a transition process (i.e., sublima- tion, melting, or evaporation). During evaporation water is a mixture of liquid and vapor, both phases existing at the same temperature and pressure. The phase rule requires that a single intensive property (e.g., either temperature or pressure, but not both) fixes the state of a two-phase system. The phase transitions in Figure 2-3 appear as lines because the temperature at which a phase transition occurs is uniquely defined once the pressure is given (or vice versa). When a mixture of two or more phases is present, the substance is said to be in a saturated condition. The term saturated applies to the coexistence of any phases, but it will most often be used in this textbook to refer to a liquid and vapor mixture. It is possible for a pure fluid to exist in three phases simultaneously. The phase rule, Eq. (2-1), indicates that the number of degrees of freedom (F) for a pure fluid in a saturated state for which three phases (solid, liquid, and vapor) are present is zero. Zero degrees of freedom implies that there is only one such state and it is impossible to independently specify the value of any property. The single condition at which solid, liquid and vapor phases coexist is called the triple point. Figure 2-3 shows that the triple point temperature of water is 273.16 K and the triple point pressure of water is 611.7 Pa. The triple point state is often used as a reference condition because it is uniquely and precisely defined for any fluid. One last feature of Figure 2-3 that should be examined is the critical point. The existence of separate liquid and vapor phases is observable if the fluid is placed in a transparent tube under appropriate conditions. The liquid phase has higher density (lower specific volume) and will therefore tend to pool at the bottom of the tube (assum- ing that the experiment is carried out in a gravitational field). The boundary between the liquid and vapor is called a meniscus; the meniscus can have a concave or convex shape, depending on the interaction of the fluid with the tube wall. However, at a sufficiently high 40 Thermodynamic Properties Figure 2-6: Temperature as a function of specific volume for water measured at several different pressures. Upon heating, the initial behavior of the experiment is essentially the same as it was when the pressure was 101.325 kPa. However, when the pressure is held at 500 kPa, the water does not begin to evaporate until the temperature reaches 151.8◦C; the saturation temperature of water at 500 kPa is 151.8◦C. The specific volume of the saturated liquid water at 500 kPa, state f,2, is only slightly larger than it was at 101.325 kPa, state f,1. The specific volume of saturated vapor at 500 kPa, state g,2, is substantially less than it was at 101.325 kPa, state g,1. The specific volumes of saturated liquid and saturated vapor approach one another as the pressure increases. The temperature as a function of specific volume measured at P = 2 MPa is also shown in Figure 2-6 in order to demonstrate this behavior. If we repeat this experiment for many different pressures then it is possible to generate the property plot that is referred to as a temperature-specific volume (or T-v) diagram; the T-v diagram for water is shown in Figure 2-7. Each line of constant pressure is referred to as an isobar. The locus of the states at which saturated liquid exists (e.g., states f,1 and f,2 in Figure 2-6) forms one side of the vapor dome. The locus of the states at which saturated vapor exists (e.g., states g,1 and g,2) forms the other side. 5x10-4 10-2 10-1 100 101 0 50 100 150 200 250 300 350 400 450 500 Specific volume (m3/kg) T em pe ra tu re ( °C ) 101.325 kPa 500 kPa 2 MPa 5 MPa 10 MPa 22.064 MPa 50 MPa crit sat'd liquid line sat'd vapor line co m p re ss ed li q u id superheated vapor saturated liquid and vapor supercritical Figure 2-7: Temperature–specific volume (T-v) diagram for water. 2.3 Property Tables 41 The top of the vapor dome is terminated by the critical point, state crit in Figure 2-7. All states that lie within the vapor dome consist of both liquid and vapor phases. The left boundary of the vapor dome is the saturated liquid line. States on this line consist entirely of saturated liquid that is ready to evaporate. The right boundary of the vapor dome is the saturated vapor line. All states falling on this line consist entirely of water vapor that is ready to condense. Several isobars are displayed in Figure 2-7 in order to show their trajectory. Note that the isobars are horizontal when they pass through the vapor dome. An important implication of the behavior in the saturated region is that it is not possible to determine the relative amounts of liquid and vapor for a state that lies within the vapor dome by specifying only the temperature (or only the pressure). All states between saturated liquid and saturated vapor at a given pressure exhibit the same temperature. States that lie to the right of the vapor dome (i.e., states with specific volume greater than the specific volume of saturated vapor at the same temperature) are referred to as superheated vapor. At the other extreme, states that lie to the left of the vapor dome (i.e., states with specific volume less than the specific volume of the saturated liquid at the same temperature) are referred to as compressed liquid. Notice that the specific volume of compressed liquid is smaller than that of saturated liquid at the same temperature, but only by a small amount. Liquids are nearly incompressible. Unless an enormous pressure is exerted on the liquid, its specific volume cannot be changed significantly (i.e., it cannot be compressed). Compressed liquid states are also called subcooled liquid. The isobar that passes through state crit in Figure 2-7 is the critical pressure, which corresponds to Pcrit = 22.064 MPa for water. The critical temperature for water is Tcrit = 374.0◦C and the critical specific volume for water is vcrit = 0.003106 m3/kg. A fluid that is at a pressure or temperature above its critical pressure or temperature cannot exist in two phases. Figure 2-8 illustrates, qualitatively, the three-dimensional surface that is formed when pressure is plotted as a function of temperature and specific volume. Notice that the P-T and T-v plots, shown in Figure 2-3 and Figure 2-7 for water, are projections of the P-v-T surface viewed from different directions. 2.3 Property Tables Fluid property data of the type represented in Figure 2-7 have traditionally been made available in tables. The tables developed for water are commonly referred to as steam tables, but of course these tables provide information for liquid water as well as for steam. A detailed set of property tables for water is contained in Appendix B as Tables B-1 through B-4. Data tables for refrigerant R134a are provided in Appendix C. Section 2.4 describes how the property data for these substances and many others can be obtained using built-in property functions in EES. These computer functions are more convenient to use than property tables and therefore most of the examples in this book will rely on the property data obtained using EES. 2.3.1 Saturated Liquid and Vapor Tables B-1 and B-2 in Appendix B provide property data for water that lie on the vapor dome (i.e., the saturated liquid and saturated vapor lines). Recall that the number of degrees of freedom for a pure substances existing in two phases is one; therefore, the specification of a single intensive variable fixes the intensive state of both phases. Table B-1 provides information at regular intervals of temperature whereas Table B-2 uses regular intervals of pressure. A small excerpt of Table B-1 is provided as Table 2-1. The 42 Thermodynamic Properties P v T P v P T v T triple point triple point line critical point critical point critical point critical point triple point line triple point line Figure 2-8: Pressure as a function of specific volume and temperature for a pure substance that contracts upon freezing, based on Myers (2007). first column in Table 2-1 is the temperature of interest while the second column is the saturation pressure that corresponds to this temperature; notice in Figure 2-7 that there is a unique saturation pressure corresponding to each temperature. After pressure, the table provides four sets of two columns. Each set of columns corresponds to a different property: specific volume (v), specific internal energy (u), specific enthalpy (h), and specific entropy (s). Within each set, the first column corresponds to the value of the property for the saturated liquid (designated by the subscript f) and the second column is the value for the saturated vapor (designated by the subscript g). We have thus far only discussed specific volume; however, it will also be necessary to know the values of u, h, and s in order to apply the laws of thermodynamics. Table B-2 provides exactly the same information as Table B-1, except that the order of columns 1 and 2 is switched and the data are provided for equal increments of pressure. Note that the value of the specific volume of saturated liquid reported in these tables is multiplied by 1000 in order to minimize the space required by the column for vf. Therefore, the specific volume of saturated liquid water at 8◦C is 0.0010002 m3/kg (and not 1.0002 m3/kg). The data in Tables B-1 and B-2 provide property information for states that are entirely saturated liquid (subscript f) and entirely saturated vapor (subscript g). How- ever, a state that lies under the vapor dome will consist of both liquid and vapor phases that are in equilibrium. For example, Figure 2-9 illustrates a tank with volume V. The water contained in the tank is in a two-phase state at T = 20◦C. According to Table 2-1, the pressure in the tank must be 2.3392 kPa, which is the saturation pressure of water at 20◦C. The liquid contained in the tank in Figure 2-9 is saturated liquid at 20◦C and 2.3392 kPa; the properties of the liquid phase are vf = 0.0010018 m3/kg, uf = 83.913 kJ/kg, hf = 83.915 kJ/kg and sf = 0.29649 kJ/kg-K. The vapor contained in the tank is saturated vapor at 20◦C and 2.3392 kPa; the properties of the vapor phase are vg = 57.762 m3/kg, ug = 2402.3 kJ/kg, hg = 2537.4 kJ/kg, and 2.3 Property Tables 45 EX AM PL E 2. 3- 1: PR OD UC TI ON OF A VA CU UM BY CO ND EN SA TI ON 0.0005 0.01 0.1 1 10 0 50 100 150 200 250 300 350 400 450 500 Specific volume (m3/kg) T em pe ra tu re ( °C ) 101.325 kPa 500 kPa 2 MPa 5 MPa 10 MPa 22.064 MPa 50 MPa crit x = 0 x = 0.4 x = 0.2 x = 0.6 x = 0.8 x = 1 Figure 2-10: A T-v diagram for water showing lines of constant quality. Quality is only relevant for states under the vapor dome and therefore its value must lie between zero and one. The quality is equal to zero for saturated liquid and equal to one for saturated vapor. Equation (2-15) shows that the quality associated with a state represents the fractional progression of the state across the vapor dome along an isobar. Figure 2-10 illustrates a T-v diagram for water that includes lines of constant quality in addition to isobars. Because the abscissa is logarithmic, the lines of constant quality appear to be compressed towards the right side of the vapor dome. EXAMPLE 2.3–1: PRODUCTION OF A VACUUM BY CONDENSATION In 1601, Giambattista della Porta described a method by which a vacuum (i.e., a pressure lower than the pressure of the atmosphere) can be produced through the condensation of steam. A small amount of liquid water at T1 = 20◦C and P1 = 1 atm is placed in a rigid container with total volume V = 0.35 liter. The water is heated until it boils. The boiling continues for some time in order to allow the water vapor gen- erated by the boiling process to push all of the air out of the container. Then the con- tainer is sealed and weighed. The total mass of water in the container is m1 = 1.25 g. a) What is the volume of liquid water and the volume of water vapor in the container at the time that it is sealed? The specific volume of the water in the container at the time that it is sealed is: v1 = Vm1 = 0.35 liter ∣∣∣∣ 1.25 g ∥∥∥∥ 1000 g1 kg ∣∣∣∣ 0.001 m31 liter = 0.28 m3/kg The pressure and specific volume together specify state 1. Whenever you solve a problem, it is useful to sketch a T-v diagram (or some other type of property diagram) and qualitatively locate each state that is involved in the problem on the diagram. This exercise will usually help you to better understand the problem. Figure 1 illustrates a qualitative sketch of a T-v diagram. State 1 lies at the intersection of the isobar P1 = 101.325 kPa and the line of constant specific vol- ume (also called an isochor) v1 = 0.28 m3/kg. On a T-v diagram, an isochor is a vertical line (and an isotherm is a horizontal line). According to Table B-1 in Appendix B, at 1 atm (101.3 kPa, corresponding to a saturation tempera- ture of 100◦C) the specific volume of saturated liquid is vf,1 = 0.0010435 m3/kg 46 Thermodynamic Properties EX AM PL E 2. 3- 1: PR OD UC TI ON OF A VA CU UM BY CO ND EN SA TI ON and the specific volume of saturated vapor is vg,1 = 1.6720 m3/kg. Because vf,1 < v1 < vg,1, state 1 must lie within the vapor dome as shown in Figure 1. T v P = 101.3 kPa T = 20°C 2 1 v = 0.28 m3/kgvf,1 vf , 2 vg,1 vg, 2 Figure 1: Qualitative sketch of the T-v diagram. The quality of the water in the container at the time that it is sealed is: x1 = ( v1 − vf,1 )( vg,1 − vf,1 ) = (0.28 − 0.0010435) (1.6720 − 0.0010435) = 0.1669 The mass of vapor in the container is obtained from Eq. (2-6): mg,1 = x1 m1 = 0.1669 ∣∣∣∣ 1.25 g = 0.2087 g and the mass of liquid in the container is: mf,1 = (1 − x1) m1 = (1 − 0.1669) ∣∣∣∣ 1.25 g = 1.041 g The volume occupied by the vapor is the product of the mass of the vapor and its specific volume: Vg,1 = mg,1 vg,1 = 0.2087 g ∣∣∣∣ 1.6720 m3kg ∥∥∥∥ 1 kg1000 g ∣∣∣∣ liter0.001 m3 = 0.349 liter The volume of liquid is calculated according to: Vf,1 = mf,1 vf,1 = 1.041 g ∣∣∣∣ 0.0010435 m3kg ∥∥∥∥ 1 kg1000 g ∣∣∣∣ liter0.001 m3 = 0.0011 liter b) The sealed container is allowed to cool to T2 = 20◦C. What is the resulting pressure in the container? What is the quality of the water in the container? Assume that the container volume does not change. In this process, both the volume and the mass of water in the container remain constant. Therefore, the specific volume at state 2 is equal to its value at state 1, v2 = v1 = 0.28 m3/kg. State 2 is specified by the intersection of an isochor, v2 = 0.28 m3/kg, and an isotherm, T2 = 20◦C, as shown in Figure 1. The specific volume falls between the values of saturated liquid water at 20◦C, vf,2 = 0.0010018 m3/kg, and saturated water vapor at 20◦C, vg,2 = 57.762 m3/kg (both of these values are provided in Table B-1). Therefore, state 2 also lies inside the vapor dome (as shown in 2.3 Property Tables 47 EX AM PL E 2. 3- 1: PR OD UC TI ON OF A VA CU UM BY CO ND EN SA TI ON Figure 1) and the pressure in the container must be P2 = 2.3392 kPa (the saturation pressure at 20◦C). The quality at state 2 is: x2 = ( v2 − vf,2 )( vg,2 − vf,2 ) = (0.28 − 0.0010018) (57.762 − 0.0010018) = 0.00483 Note that the pressure in the container is very much lower than atmospheric and, unless the container is very sturdy, it will likely be crushed by the force exerted from the atmospheric pressure. c) What are the volumes of vapor and liquid in the container at the final state? The mass of vapor is calculated according to: mg,2 = x2 m2 = 0.00483 ∣∣∣∣ 1.25 g = 0.006038 g and the mass of liquid is: mf,2 = (1 − x2) m2 = (1 − 0.00483) ∣∣∣∣ 1.25 g = 1.244 g The volume of vapor is: Vg,2 = mg,2 vg,2 = 0.006038 g ∣∣∣∣ 57.762 m3kg ∥∥∥∥ 1 kg1000 g ∣∣∣∣ liter0.001 m3 = 0.349 liter The volume of liquid is: Vf,2 = mf,2 vf,2 = 1.244 g ∣∣∣∣ 0.0010018 m3kg ∥∥∥∥ 1 kg1000 g ∣∣∣∣ liter0.001 m3 = 0.0012 liter Notice that the specific volume of saturated water vapor at low temperature is orders of magnitude larger than the specific volume of saturated liquid water at the same temperature. As a result, even though the water in the container has a very small quality (only 0.483% of the mass is vapor) the vapor takes up most of the volume (99.7% of the volume is vapor). 2.3.2 Superheated Vapor The term superheated vapor is used to describe a state for which the specific volume is larger than the specific volume of saturated vapor at the same temperature. The superheated vapor region lies to the right of the saturation dome on a T-v diagram, as shown in Figure 2-7. Saturated vapor cannot exist for temperatures above the critical temperature; states that are above the critical point are referred to as supercritical. The properties of water in the superheated vapor and supercritical regions are provided in Table B-3 of Appendix B. These states are single phase ( = 1) for a pure fluid (C = 1); therefore, according to phase rule in Eq. (2-1), there are two degrees of freedom (F = 2). The data in Table B-3 are presented as an array of information in the dimensions of pressure and temperature. An excerpt of Table B-3 is shown in Table 2-2. The individual columns in Table 2-2 correspond to different temperatures and each set of four rows corresponds to a different pressure. The first row in each set is the specific volume, the second is the specific internal energy, the third is the specific enthalpy, and the fourth is the specific entropy. Therefore, it is possible to obtain the properties v, u, h, and s at a given pressure and temperature. For example, the specific volume of superheated water vapor at T = 300◦C and P = 40 kPa is v = 6.6067 m3/kg. 50 Thermodynamic Properties already been completed for 20 kPa. At 40 kPa, the interpolated value of specific volume at 115◦C is obtained from: v = 4.2799 m3/kg + (115 ◦C − 100◦C) (150◦C − 100◦C) ( 4.8662 m3/kg − 4.2799 m3/kg) = 4.4558 m3/kg (2-26) The other interpolated properties at T = 115◦C and P = 40 kPa are u = 2534.6 kJ/kg, h = 2712.9 kJ/kg, and s = 7.8744 kJ/kg-K. Finally, the interpolated values corresponding to 115◦C at pressures of 20 kPa and 40 kPa are interpolated for a pressure of 32 kPa. Applying Eq. (2-22) for specific volume provides: v − v at 115 ◦ C and 20 kPa︷︸︸︷ 8.9344 m3/kg 4.4558 m3/kg︸︷︷︸ v at 115 ◦ C and 40 kPa − 8.9344 m3/kg︸︷︷︸ v at 115 ◦ C and 20 kPa = 32 kPa − 20 kPa 40 kPa − 20 kPa (2-27) or v = 8.9344 m3/kg + (32 kPa − 20 kPa)(40 kPa − 20 kPa) ( 4.4558 m3/kg − 8.9344 m3/kg) = 6.2472 m3/kg (2-28) Repeating the process for the other properties results in u = 2535.3 kJ/kg, h = 2713.7 kJ/kg, and s = 8.0042 kJ/kg-K. The accepted values at this state obtained using EES are: v = 5.5763 m3/kg, u = 2535.4 kJ/kg, h = 2713.8 kJ/kg, and s = 7.9826 kJ/kg-K. Notice that the interpolated and accepted values of u, h, and s agree quite well. However, the interpolated and accepted values of specific volume differ by 12%. This discrepancy occurs because specific volume varies approximately according to the inverse of pressure and the table entries are too far apart to allow linear interpolation to provide an accurate answer. The specific volume estimated from interpolation could be improved if the interpolation were carried out using the inverse of pressure, as suggested by the ideal gas law. In this case, Eq. (2-22) is applied by replacing X with the inverse of pressure: v − 8.9344 m3/kg 4.4558 m3/kg − 8.9344 m3/kg = 1/32 kPa − 1/20 kPa 1/40 kPa − 1/20 kPa (2-29) Solving Eq. (2-29) results in v = 5.5754 m3/kg, which is within 0.014% of the accepted value. 2.3.3 Compressed Liquid The term compressed liquid is used to describe a state for which the specific volume is less than the specific volume of saturated liquid at the same temperature. The compressed liquid region lies to the left of the saturation dome on a T-v diagram, as shown in Figure 2-7. Compressed liquid property data for water are provided in Table B-4 in Appendix B. The compressed liquid table is arranged in a manner that is similar to the superheated vapor table, Table B-3 (or Table 2-2); data are provided over a matrix of temperatures and pressures. Most liquids at a temperature much lower than their critical temperature are nearly incompressible; that is, the specific volume of the liquid at a specified temperature is essentially constant regardless of pressure. When plotted to scale on a temperature- specific volume plot, each isobar nearly collapses onto a single, nearly vertical line that coincides with the saturated liquid side of the vapor dome. This characteristic behavior is highlighted in Figure 2-11. Notice that even at relatively high temperatures, the various isobars tend to collapse onto the saturated liquid line of the vapor dome; 2.4 EES Fluid Property Data 51 5x10-4 10-2 10-1 100 101 0 50 100 150 200 250 300 350 400 450 500 Specific volume (m3/kg) T em pe ra tu re ( °C ) 101.325 kPa 500 kPa 2 MPa 5 MPa 10 MPa 22.064 MPa 50 MPa incompressible Figure 2-11: Temperature-specific volume plot for water showing the incompressible liquid behavior. therefore, the specific volume of a compressed liquid depends slightly on temperature, but is almost completely independent of pressure. One useful result of this behavior is that the properties of a substance in the compressed liquid region can usually be estimated fairly accurately using the properties of saturated liquid evaluated at the same temperature. 2.4 EES Fluid Property Data You do not have to solve many thermodynamics problems before the limitations asso- ciated with the use of the property tables become evident. Looking up property values in tables usually requires either single or double interpolation. The process is time- consuming and likely to introduce math errors. It is not easy or even practical to carry out the parametric studies that are required for optimization or design using property tables. Fortunately, computer programs are available that automate the process of deter- mining property values. EES is the most convenient program available in this regard. This section will describe how EES can be used to provide the property information for many substances. 2.4.1 Thermodynamic Property Functions EES provides a number of built-in thermodynamic and transport property functions. In this textbook we will focus on the thermodynamic property functions. The transport property functions that provide properties such as viscosity and thermal conductivity are useful for problems that you may encounter in a Fluid Dynamics or Heat Transfer class. The thermodynamic property functions that are of greatest interest are listed in Table 2-3, together with the thermodynamic property that is returned by the function and the possible units of the returned value. It is essential that you pay close attention to the units of the variables that are used to access the built-in thermodynamic property functions in EES. The units that EES will expect in the call to the thermodynamic property functions can be specified with the Unit System menu command in the Options menu. The Unit System dialog, shown in Figure 2-12, allows either SI or English units to be selected (but it does not allow a combination of SI and English units to be used). Within each unit system, there are options for the temperature scale, the energy units and the pressure units. Specific property values can be defined on either a mass or molar basis. The molar 52 Thermodynamic Properties Table 2-3: Commonly used EES thermodynamic property functions and their units. Function Name Returns SI Units English Units Enthalpy specific enthalpy J/kg, kJ/kg Btu/lbm Entropy specific entropy J/kg-K, kJ/kg-K Btu/lbm-R IntEnergy specific internal energy J/kg, kJ/kg Btu/lbm Pressure absolute pressure Pa, kPa, bar, MPa psia, atm Quality quality none none Temperature temperature ◦C, K ◦F, R Volume specific volume m3/kg ft3/lbm Density density kg/m3 lbm/ft3 Phase$ phase, e.g, ‘superheated’ none none MolarMass1 molecular weight kg/kmol lbm/lbmol T_sat2 saturation temperature ◦C, K ◦F, R P_sat3 saturation pressure Pa, kPa, bar, MPa psia, atm T_crit1 critical temperature ◦C, K ◦F, R P_crit1 critical pressure Pa, kPa, bar, MPa psia, atm v_crit1 critical specific volume m3/kg ft3/lbm 1. The functions MolarMass, T_crit, P_crit, and v_crit require only the name of the fluid. 2. The function T_sat requires the name of the fluid and the pressure. 3. The function P_sat requires the name of the fluid and the temperature. basis is particularly convenient for combustion and chemical equilibrium calculations, as discussed in Chapters 13 and 14. The unit system that EES uses can alternatively be specified using the $UnitSystem directive, which is a command that is entered directly in the Equations window. For example, the following line: $UnitSystem SI, K, Pa, J, Mass, Radian entered at the top of the Equations window will specify that the SI unit system should be used with the Kelvin temperature scale and units of J and Pa for energy and pressure, respectively. It also specifies that specific property values must be entered and returned on a mass basis and angles must be provided to trigonometric functions in radians rather than in degrees. These specifications are consistent with the choices that would otherwise be made in the Unit System dialog shown in Figure 2-12. The entries in the $UnitSystem directive over-ride any entries made in the Unit System dialog. The advantage of using the $UnitSystem directive is that the units associated with the EES solution are visible and will not change if the EES code is copied and pasted to a different EES file that may have a different default set of unit specifications. Unit setting are saved with other information when the EES file is saved. All property values that are returned by the EES built-in property functions will have the units that are specified with the Unit System dialog or the $UnitSystem directive. Further, any values provided to the built-in property functions will be interpreted as if 2.4 EES Fluid Property Data 55 EX AM PL E 2. 4- 1: TH ER M OS TA TI C EX PA NS IO N VA LV E Table 2-4: Property designations used in the EES thermodynamic property functions. Property Designator Property H or h specific enthalpy P or p pressure S or s specific entropy T or t temperature U or u specific internal energy V or v specific volume X or x quality which produces v = 5.576 m3/kg, u = 2.535×106 J/kg, h = 2.713×106 J/kg, and s = 7981 J/kg-K; these values agree with the results of the interpolation. Notice that the functions volume, intenergy, enthalpy, and entropy all required that temperature be provided in units of K and pressure in units of Pa based on the settings in the $UnitSystem directive. The units of the variables v, u, h, and s returned by the thermodynamic functions are m3/kg, J/kg, J/kg, and J/kg-K, respectively, consistent with the unit system settings. EES knows what the units of these variables should be, based on the unit system that has been set, and will check to ensure that the units entered for each variable are consistent. EXAMPLE 2.4–1: THERMOSTATIC EXPANSION VALVE A thermostatic expansion valve is a passive device that is used to control the flow of refrigerant to an evaporator (a heat exchanger that allows the refrigerant to accept a heat transfer from a refrigerated space), as shown in Figure 1. liquid line bulb evaporator evaporator outlet, Tevap,out valve seat spring head space valve body diaphragm Figure 1: Thermostatic expansion valve controlling the flow of a refrigerant to an evaporator. 56 Thermodynamic Properties EX AM PL E 2. 4- 1: TH ER M OS TA TI C EX PA NS IO N VA LV E Saturated (or nearly saturated) refrigerant enters the thermostatic expansion valve through the liquid line. The flow rate of refrigerant is controlled by the posi- tion of the valve seat. The refrigerant subsequently flows through the evaporator at nearly constant pressure. The heat transfer causes the refrigerant to completely evaporate and so it leaves the evaporator as superheated vapor at temperature Tevap,out. The job of the thermostatic expansion valve is to regulate the refrigerant flow so that it leaves at a precisely controlled temperature even if the load on the evaporator (i.e., the heat transfer to the refrigerant) changes. If the load on the evaporator decreases then Tevap,out will tend to decrease and the thermostatic valve should act to reduce the refrigerant flow rate. Alternatively, if the load increases then Tevap,out will tend to increase and this should result in the thermostatic valve increasing the flow rate. A bulb filled with refrigerant is attached to the evaporator outlet. The bulb is connected, through a tube, to the head space of the valve. The pressure in the bulb and the head space are the same; therefore, an increase in Tevap,out tends to increase the pressure in the head space of the valve which causes the diaphragm in the thermostatic expansion valve to move downward and open the valve. If Tevap,out decreases then the opposite happens; the head space pressure decreases which, along with the force exerted by the spring, causes the diaphragm to move up thereby closing the valve. Figure 2 provides a simplified schematic of the thermostatic expansion valve and bulb. The bulb volume is Vbulb = 10 cm3 and the head space volume is Vhead = 2 cm3, the volume of the connecting tube can be neglected. The temperature of the refrigerant contained in the head space is equal to the evaporator inlet tem- perature, Tevap = –5◦C. Under nominal operating conditions, the temperature of the refrigerant leaving the evaporator (and therefore the temperature of the refrig- erant contained in the bulb) is Tevap,out,nom = –2◦C and the bulb and head space pressure are both Pbulb,nom = 700 kPa. The refrigerant is R22. (The naming code for refrigerants is discussed in Chapter 9.) liquid line valve seat spring K = 7.2x104 N/m head space Vhead = 2 cm 3 Tevap = -5°C Pbulb,nom = 700 kPa valve body diaphragm A = 1 cm2 to evaporator bulb Vbulb = 10 cm 3 Tevap,out,nom = -2°C Pbulb,nom = 700 kPa Figure 2: Schematic of thermostatic expansion valve and bulb. a) What is the total mass of refrigerant used to charge the bulb and the head space? 2.4 EES Fluid Property Data 57 EX AM PL E 2. 4- 1: TH ER M OS TA TI C EX PA NS IO N VA LV EThe inputs are entered in EES and the unit system is specified using the $UnitSystem directive. $UnitSystem SI, K, Pa, J, Mass, Radian “Inputs” Vol_bulb=10 [cmˆ3]∗convert(cmˆ3,mˆ3) “Volume of bulb” Vol_head=2 [cmˆ3]∗convert(cmˆ3,mˆ3) “Volume of head space” T_evap=converttemp(C,K,-5 [C]) “Temperature of refrigerant in head space” T_evap_out_nom=converttemp(C,K,-2 [C]) “Nominal evaporator outlet temperature” P_bulb_nom=700 [kPa]∗convert(kPa,Pa) “Nominal bulb pressure” We will start by identifying the state of the refrigerant in the bulb and the state of the refrigerant in the head space. The pressure and temperature are known for both of these spaces. The saturation temperature of the refrigerant at the nominal bulb pressure (Tsat) is obtained using the T_sat function; note that the T_sat function requires the name of the fluid and the pressure of interest. T_sat=T_sat(R22,P=P_bulb_nom) “saturation temperature at bulb pressure” T_sat_C=converttemp(K,C,T_sat) “in C” Solving provides Tsat = 10.91◦C. Because both Tevap and Tevap,out,nom are both less than Tsat, we know that the state of the refrigerant in the bulb and the state of the refrigerant in the head space both lie in the compressed liquid region. The specific volume of the refrigerant in the bulb under nominal operating conditions (vbulb,nom) is obtained using the thermodynamic function Volume by specifying the state with the pressure (Pbulb,nom) and temperature (Tevap,out,nom). The mass of refrigerant in the bulb is: mbulb,nom = Vbulb vbulb,nom v_bulb_nom=volume(R22,T=T_evap_out_nom,P=P_bulb_nom) “specific volume of refrigerant in bulb” m_bulb_nom=Vol_bulb/v_bulb_nom “mass of refrigerant in bulb” The specific volume of the refrigerant in the head space under nominal operating conditions (vhead,nom) is obtained using the Volume function by specifying the state with the pressure (Pbulb,nom) and temperature (Tevap). The mass of refrigerant in the head space is: mhead,nom = Vhead vhead,nom v_head_nom=volume(R22,T=T_evap,P=P_bulb_nom) “specific volume of refrigerant in head space” m_head_nom=Vol_head/v_head_nom “mass of refrigerant in head space” The total mass of refrigerant used to charge the bulb and the head space is: m = mbulb,nom + mhead,nom m=m_bulb_nom+m_head_nom “total mass of refrigerant” which leads to m = 0.01549 kg (15.49 g). 60 Thermodynamic Properties Arrays Table 100000 100000 100000 Figure 2-14: Arrays window. If you solve the problem (select Solve from the Calculate menu or press F2) you will find that the Solution window is empty. By default, the arrays are displayed in a separate window called the Arrays window that can be accessed by selecting Arrays from the Windows menu (Figure 2-14). Each element of array P is shown in the Arrays window. The units for all of the elements of P can be set at once by right-clicking on the column heading and entering the units in the dialog that appears (Figure 2-15). The temperatures of the propane at the three states are T1 = 300 K, T2 = 350 K, and T3 = 375 K. These values are entered into the array T: T[1]=300 [K] T[2]=350 [K] T[3]=375 [K] The elements of an array can be manipulated like any other variable in EES. For example, we can use the Volume function to obtain the specific volume of each state: v[1]=Volume(Propane,T=T[1],P=P[1]) v[2]=Volume(Propane,T=T[2],P=P[2]) v[3]=Volume(Propane,T=T[3],P=P[3]) The phase associated with each state can be found using the Phase$ function: Phase$[1]=Phase$(Propane,T=T[1],P=P[1]) Phase$[2]=Phase$(Propane,T=T[2],P=P[2]) Phase$[3]=Phase$(Propane,T=T[3],P=P[3]) Format Arrays Window Column 1 Figure 2-15: Enter units for array. 2.4 EES Fluid Property Data 61 Arrays Table [Pa] 100000 300 350 375 0.7017 0.6536 0.5568 100000 100000 [K] [m3/kg] Figure 2-16: Arrays window. The Phase$ function returns a string that describes the state (e.g., ‘liquid’ or ‘superheated’) and therefore the array used to store the output of the Phase$ function must contain string variables. Variable names that end (before the array index) with the $ character indicate that the variable will store a character string rather than a number. After setting the units for the arrays T and v, the Arrays window should appear as shown in Figure 2-16. Figure 2-16 illustrates one of the advantages of using arrays to solve a thermodynamics problem: the information about each state is organized in tabular form. It is possible to quickly examine the properties associated with each state by examining each row of the Arrays table rather than having to gather this information from a list of variables in the Solution window. In Section 2.2, we discussed the importance of property plots. Thus far we have dealt exclusively with T-v plots; however, we will encounter other types of property plots that are also useful. The process of generating property plots can be automated in EES. For example, we can use EES to generate a T-v plot for propane and locate states 1, 2 and 3 on this plot. Select Property Plot from the Plot menu and select Propane from the fluid list. Click the T-v radio button in order to create a temperature-specific volume plot. Up to six lines of constant pressure and six lines of constant specific entropy can optionally be placed on the plot. Since we know that a pressure of interest is 100 kPa, enter that isobar (Figure 2-17). Also, enter isobar pressures of 2.8 MPa, 1.1 MPa, and 350 kPa. Click on [×] in the list on the right to unselect all of the constant specific entropy lines. Property Plot Information Propane Figure 2-17: Property plot information dialog. 62 Thermodynamic Properties 10-3 10-2 10-1 100 101 150 200 250 300 350 400 450 500 550 Specific volume (m3/kg) T e m p e ra tu re ( K ) 2.8 MPa 1.1 MPa 350 kPa 100 kPa x = 0.05 0.1 0.2 0.5 (a) 10 -3 10 -2 10 -1 10 0 101 150 200 250 300 350 400 450 500 550 Specific volume (m3/kg) T e m p e ra tu re ( K ) 2.8 MPa 1.1 MPa 350 kPa 100 kPa x = 0.05 0.1 0.2 0.5 1 2 3 (b) Figure 2-18: (a) T-v plot for propane and (b) with states overlaid. Figure 2-18(a) illustrates the T-v plot that is produced (with some minor editing). Because the properties at each state are stored in arrays rather than as scalar variables, it is possible to overlay the states onto the T-v plot. Select Overlay Plot from the Plots menu. Data should be plotted from the Arrays table. Select v[i] as the X-Axis variable and T[i] as the Y-axis variable. Check the Show array indices selection, as shown in Figure 2-19, in order to identify each state with its number on the property plot. The result is shown in Figure 2-18(b). It is possible to setup the plot so that the states are automatically updated each time the program is run. Double-click on the plot in order to access a list of all of the data that are being presented. The last item in the list should be the data from the Arrays table. If the Automatic update selection is checked then the states will automatically re-position 2.4 EES Fluid Property Data 65 EX AM PL E 2. 4- 2: LI QU ID OX YG EN TA NKSolving provides T1 = 90.2 K, m1 = 68.5 kg, x1 = 3.26 × 10−4, and v1 = 0.00948 m3/kg. b) The engine requires liquid oxygen at a pressure of Pengine = 25 atm for tburn = 120 s. The obvious method for providing oxygen to the engine is to pull liquid from the bottom of the tank using a pump that discharges to the engine at Pengine, as shown in Figure 2. mout m2 - m1 Pengine pump Figure 2: Liquid oxygen provided to the engine with a pump. The emptying process ends when the liquid in the tank is completely removed. Assume that the tank pressure remains at P = 1 atm during this process. Deter- mine the mass of oxygen provided to the engine. What fraction of the mass of oxygen that is initially contained in the tank is provided to the engine? When the tank is empty of liquid, the quality has reached unity; x2 = 1. The pressure is also given, P2 = 1 atm. The state of the oxygen that remains in the tank at state 2 is specified by the quality and pressure. Therefore, the specific volume (v2) and temperature (T2) are obtained using EES’ internal property routines. P_engine_atm=25 [atm] “engine pressure, in atm” P_engine=P_engine_atm∗convert(atm,Pa) “engine pressure” t_burn = 120 [s] “burn time” x[2]=1 [-] “quality when tank is empty of liquid” P[2]=P[1] “pressure when tank is empty of liquid” v[2]=volume(Oxygen,x=x[2],P=P[2]) “specific volume when tank is empty of liquid” T[2]=temperature(Oxygen,x=x[2],P=P[2]) “temperature when tank is empty of liquid” The mass of oxygen that remains in the tank is: m2 = V v2 Mass is a conserved quantity (it is neither produced nor destroyed). Therefore, for any system it is possible to write a mass balance: In = Out + Stored A mass balance on a system that is defined as the internal volume of the tank is shown in Figure 2 and provides: 0 = mout + (m2 − m1) 66 Thermodynamic Properties EX AM PL E 2. 4- 2: LI QU ID OX YG EN TA NK The fraction of oxygen that is initially contained in the tank (m1) that is provided to the engine (mout) is: fpump = moutm1 m[2]=Vol/v[2] “mass of oxygen in the tank when it is empty of liquid” 0=m_out+m[2]-m[1] “mass balance” f_pump=m_out/m[1] “fraction of oxygen in tank that is provided to the engine” which leads to mout = 68.2 kg and fpump = 0.9958 (i.e., 99.58% of the oxygen initially in the tank is consumed by the engine). Figure 3 illustrates the state of the oxygen initially in the tank, state 1, and the state of the oxygen in the tank when the liquid is gone, labeled state 2p, on a T-v diagram for oxygen generated by selecting Property Plot from the Plots menu and following the procedure discussed in Section 2.4.2. 10-4 10-3 10-2 10-1 100 101 102 50 75 100 125 150 175 200 Specific volume (m /kg)3 T em pe ra tu re ( K ) P = 25 atm 1 atm 1 2p 2h 3h Figure 3: A T-v diagram for oxygen showing the states associated with the problem. c) What is the volumetric flow rate of oxygen at the suction port of the pump? Assume that the mass flow rate of liquid oxygen is constant during the emptying process. The mass flow rate is given by: ṁ = mout tburn The oxygen entering the pump is always saturated liquid at pressure P1 and there- fore it has specific volume vf,1. The volumetric flow rate at the pump inlet is: V̇ = ṁvf,1 m_dot=m_out/t_burn “mass flow rate at pump suction port” V_dot=m_dot∗v_f[1] “volumetric flow rate at pump suction port” V_dot_cfm=V_dot∗convert(mˆ3/s,ftˆ3/min) “in cfm” which results in V̇ = 1.055 cfm. 2.4 EES Fluid Property Data 67 EX AM PL E 2. 4- 2: LI QU ID OX YG EN TA NKd) The pump shown in Figure 2 has some serious drawbacks. The temperature of the oxygen passing through the pump is very low (around 90.2 K according to Figure 3) and therefore the pump needs to be a specialized piece of cryogenic equipment that is expensive and potentially unreliable. An alternative method of providing oxygen to the engine is shown in Figure 4. Pengine resistive heater valve Figure 4: Liquid oxygen tank emptied with a heater. A resistive heater is placed in the tank and activated until the pressure in the tank reaches P2 = Pengine. At that time, a valve placed in the bottom of tank opens and allows liquid oxygen to exit at a rate that maintains the pressure in the tank at Pengine. This continues until the tank is emptied of liquid at state 3. Determine the mass of oxygen provided to the engine using the system shown in Figure 3. What fraction of the mass of oxygen initially in the tank is utilized by this system? The lines of EES code corresponding to the analysis of the pumped system shown in Figure 2, are commented out (i.e., highlight these lines of code, right-click, and select Comment). {x[2]=1 [-] “quality when tank is empty of liquid” P[2]=P[1] “pressure when tank is empty of liquid” v[2]=volume(Oxygen,x=x[2],P=P[2]) “specific volume when tank is empty of liquid” T[2]=temperature(Oxygen,x=x[2],P=P[2]) “temperature when tank is empty of liquid” m[2]=Vol/v[2] “mass of oxygen in the tank when it is empty of liquid” 0=m_out+m[2]-m[1] “mass balance” f_pump=m_out/m[1] “fraction of oxygen in tank that is provided to the engine” m_dot=m_out/t_burn “mass flow rate at pump suction port” V_dot=m_dot∗v_f[1] “volumetric flow rate at pump suction port” V_dot_cfm=V_dot∗convert(mˆ3/s,ftˆ3/min) “in cfm”} State 2 is defined to be the state of the oxygen at the time that the valve opens. The pressure at state 2 is specified to be equal to the engine pressure. A mass balance on the tank for the period of time when it is heated from its initial condition, state 1, to the time when the valve opens, state 2, provides: 0 = m2 − m1 Therefore, the mass in the tank does not change. Since neither the mass nor the volume of oxygen in the tank changes, the specific volume at state 2 must be equal to its value at state 1. State 2 is specified by P2 and v2. The temperature of the oxygen at the time that the valve opens (T2) is computed using EES’ internal property function Temperature.