Fundamentals of Engineering Thermodynamics [2 ed.] 812032790X, 9788120327900

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Fundamentals of Engineering Thermodynamics

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Fundamentals of Engineering Thermodynamics SECOND EDITION

E. RATHAKRISH NAN Professor Department of Aerospace Engineering Indian Institute of Technology Kanpur

Prentice-Hall of India Private New Delhi -110 001 2006

Rs. 425.00 FUNDAMENTALS OF ENGINEERING THERMODYNAMICS, 2nd ed. E. Rathakrishnan

© 2005 by Prentice-Hall of India Private Limited. New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher.

ISBN-81-203-2790-X The export rights of this book are vested solely with the publisher. Sixth Printing (Second Edition)

September, 2006

Published by Asoke K. Ghosh, Prentice-Hall of India Private LirrittE),1 M-97, Connaught Circus, New Delhi-110001 and Printed by Mohan Makhijani at Rekha Printers Private Limited, New

Delhi-110020.

Contents Preface Preface to the First Edition

1. BASIC CONCEPTS AND DEFINITIONS Thermodynamics and Energy 1 Classical and Statistical Thermodynamics 2 Applications of Thermodynamics 3 Dimensions and Units 3 Closed and Open Systems 4 Forms of Energy 6 1.6.1 Internal Energy 7 1.7 Properties of a System 8 1.8 State and Equilibrium 9 1.9 Processes and Cycles 10 1.10 The State Postulate 12 1.11 Pressure and Mechanical Equilibrium 12 1.12 Temperature and the Zeroth Law of Thermodynamics Summary 17 Problems 19

1-21

1.1 12 13 1A 1.5 1.6

15

2. ENERGY AND THE FIRST LAW OF THERMODYNAMICS 22-55 2.1 Energy 22 22 Heat 22 2.3 Work 24 2.3.1 Path and Point Functions 25 2.32 Electrical Work 26 2.33 Mechanical Work 27 23.4 Moving Boundary Work 27 2.3.5 Polytropic Process 29 23.6 Gravitational Work 32 23.7 Acceleration Work 33 23.8 Shaft Work 35 2.3.9 Spring Work 36 23.10 Work Done on Elastic Bars 37 23.11 Work Done in Stretching a Liquid Film 37 23.12 Thermodynamic Definition of Work 38 23.13 Nonmechanical Forms of Work 38 v

vi Contents 2.4 2.5 2.6 2.7

The First Law of Thermodynamics 39 Specific Heats 41 The Internal Energy, Enthalpy, and Specific Heats of Ideal Gases 42 The Internal Energy, Enthalpy, and Specific Heats of Solids and•Liquids

46

Summary 49 Problems 51

3. THERMODYNAMIC ANALYSIS OF CONTROL VOLUME 56 -79 3.1 Introduction 56 3.2 Conservation of Energy Principle 57 3.2.1 Flow Work 58 322 Total Energy of a Flowing Fluid 58 33 The Steady-Flow Process 59 3.4 Some Steady-Flow Engineering Devices 60 3.4.1 Nozzles and Diffusers 61 3.4.2 Turbines and Compressors 62 3.4.3 Throttling Valves 63 3.4.4 Mixing Chambers 64 3.4.5 Heat Exchangers 65 3.4.6 Pipe and Duct Flow 65 35 The Unsteady-Flow Process 66 3.6 The Uniform-Flow Process 67 Summary 74 Problems 76

4. THE SECOND LAW OF THERMODYNAMICS 4.1 Introduction 80 42 Thermal Energy Reservoirs 81 43 Heat Engines 81 43.1 Thermal Efficiency 83 4.4 Kelvin—Planck Statement of the Second Law 84 45 Refrigerators, Heat Pumps and Air-conditioners 85 4.5.1 Heat Pumps 87 4.5/ Air-conditioners 89 4.6 Clausius Statement of the Second Law 89 4.7 Equivalence of Kelvin—Planck and Clausius Statements 4.8 Perpetual-Motion Machines 91 4.9 Reversible and Irreversible Processes 92 4.10 The Carrot Cycle 93 4.10.1 The Reversed Carrot Cycle 94 4.11 The Carrot Principles 94 4.12 The Thermodynamic Temperature Scale 97 4.13 The Carnot Heat Engine 99 4.14 The Quality of Energy 100

80 -109

89

Contents Vii

4.15 The Camot Refrigerator and Heat Pump

101

Summary 103 Problems 106

5. ENTROPY 5.1 Introduction 110 5.2 The Clausius Inequality 110 5.3 Entropy 113 5.3.1 Statistical Definition of Entropy 115 5.3.2 Macrostate, Microstate and Probability 115 53.3 Statistical Interpretation of Entropy 117 5.3.4 Entropy Change for Isothermal Heat Transfer Processes 118 53.5 The Increase-of-Entropy Principle 119 5.3.6 The Increase-of-Entropy Principle for Closed Systems 121 53.7 The Increase-of-Entropy Principle for Control Volumes 123 5.3.8 The Uniform Flow Process 124 5.3.9 The Steady Flow Process 124 53.10 The Concept of Lost Work 125 53.11 Causes of Entropy Change 126 53.12 What is Entropy ? 127 5.4 Property Diagrams 131 5.4.1 The T—S Diagram 131 5.4.2 The Isentropic Process 132 5.4.3 The h—s Diagram 133 5.5 The TdS Relations 134 5.6 Entropy Change for Pure Substances 136 5.7 Entropy Change for Solids and Liquids 136 5.7.1 Isentropic Processes for Solids and Liquids 137 5.8 Entropy Change for Ideal Gases 138 5.8.1 Constant Specific Heats 138 5.8.2 Variable Specific Heats 139 5.83 Isentropic Processes of Ideal Gases 141 5.8.4 Entropy Postulations 142 5.8.5 The Nature of Entropy 142 5.9 Reversible Steady-Flow Work 143 5.9.1 Advantage of Reversible Processes 144 5.10 Adiabatic Efficiency of Compressors and Pumps 145 5.10.1 Adiabatic Efficiency of Turbines 145 5.11 Adiabatic Efficiency of Nozzles 147 Summary 147 Problems 149

110-153

vii Cadmus 6. AVAILABILITY, IRREVERSIBILITY AND AVAILABILITY ANALYSIS OF ENGINEERING PROCESSES

154-175

6.1 Introduction 154 6.2 Availability 154 6.3 Reversible Work and Irreversibility 155 6.3.1 Availability Transfer with Heat and Work Interactions 157 6.4 Second-Law Efficiency rill 159 65 Second-Law Analysis of Closed Systems 161 6.5.1 Heat Transfer with Other Systems (or Bodies) 164 6.6 Second-Law Analysis of Steady-Flow Systems 166 6.6.1 Stream Availability 167 6.6.2 Irreversibility Rate 168 6.6.3 Heat Transfer of a Steady-Flow Device with Other System (or Bodies) 168 Second-Law Analysis of Unsteady-Flow Systems 169 6.7 6.7.1 Uniform-Flow Processes 169 6.7.2 Reversible Work 170 6.7.3 General Unsteady-Flow Processes 170 Summary 171 Problems 172

7. PROPERTIES OF PURE SUBSTANCES 7.1 Introduction 176 Phases of a Pure Substance 176 7.2 7.3 Phase-Change Processes 177 7.4 Property Diagrams 178 7.4.1 The T—v Diagram 178 7.4.2 The P—v Diagram 179 7.43 Solid Phase on Process Diagram 180 7.4.4 The P—T Diagram 181 75 The P—v—T Surface 181 7.6 Property Tables 183 7.6.1 Superheated Tables 184 7.6.2 Compressed Liquid Tables 184 7.63 Reference State and Reference Values 184 7.6.4 Saturated Liquid-Vapour Mixture 185 7.7 The Ideal-Gas Equation of State 189 7.8 Compressibility Factor 190 7.8.1 Principle of the Corresponding States 190 7.8.2 Generalized Compressibility Chart 190 7.9 Other Equations of State 192 7.9.1 The van der Waals Equation of State 192 7.9.2 The Beattie-Bridgeman Equation of State 192 7.93 The Bertholet Equation of State 193

176 -204

Contents ix 7.9.4 The Dieterici Equation of State 7.9.5 Virial Equations of State 193

193

Summary 200 Problems 201

8. NONREACTING GAS MIXTURES

205-225

8.1 Introduction 205 8.2 Composition of a Gas Mixture 205 8.2.1 Mass Fraction mf 205 822 Mole Fraction Y 206 8.3 P—v—T Behaviour of Gas Mixtures 207 8.3.1 Dalton's Law of Additive Pressures 208 83.2 Amagat's Law of Additive Volumes 208 8.4 Ideal-Gas Mixtures 208 85 Real-Gas Mixtures 209 8.5.1 Kay's Rule 209 8.6 Properties of Gas Mixtures 211 8.6.1 Ideal-Gas Mixtures 212 8.62 Real-Gas Mixtures 212 Summary 220 Problems 222

9. VAPOUR POWER CYCLES 9.1 Introduction 226 92 The Carnot Vapour Cycle 226 The Rankine Cycle 226 93 9.4 Actual Vapour Power Cycles 231 9.5 Means to Increase the Efficiency of the Rankine Cycle 233 9.5.1 Lowering the Condenser Pressure 233 9.51 Superheating the Steam to High Temperatures 233 933 Increasing the Boiler Pressure 234 9.6 The Ideal Reheat Rankine Cycle 235 9.7 The Ideal Regenerative Rankine Cycle 240 9.7.1 Open Feedwater Heater 241 9.7.2 Closed Feedwater Heater 244 9.73 Comparison of Open and Closed Feedwater Heaters 248 Second-Law Analysis of Vapour Power Cycles 249 9.8 9.9 Cogeneration 251 9.10 Binary Vapour Cycles 255 9.11 Combined Gas-Vapour Power Cycles 255 Summary 258 Problems 259

226-262

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10. GAS POWER CYCLES

263 -316

10.1 Introduction 263 10.2 Analysis of Power Cycles 263 10.3 The Carrot Cycle 264 10.4 Air-Standard Assumptions 267 10.5 An Overview of Reciprocating Engines 271 10.6 The Otto Cycle 272 10.7 The Diesel Cycle 278 10.7.1 The Cutoff Ratio r, 279 10.72 The Dual Cycle 282 10.8 Stirling and Ericsson Cycles 283 10.9 The Brayton Cycle 284 10.9.1 Deviation of Actual Gas-Turbine Cycles from Idealized Cycles 289 10.10 The Brayton Cycle with Regeneration 291 10.11 The Brayton Cycle with Intercooling, Reheating, and Regeneration 294 10.12 Ideal Jet-Propulsion Cycles 297 10.12.1 Propulsive Power 47i, 299 10.122 Modifications of Turbojet Engines 302 10.12.3 Ramjet Engine 302 10.12.4 Scramjet Engine 303 10.12.5 Rocket 303 10.12.6 Specific Impulse 304 10.13 Second-Law Analysis of Gas Power Cycles 304 Summary 308 Problems 310

11. REFRIGERATION CYCLES 11.1 Introduction 317 112 Refrigerators and Heat Pumps 317 11.3 The Reversed Carnot Cycle 318 11.4 The Ideal Vapour-Compression Refrigeration Cycle 319 116 The Actual Vapour-Compression Refrigeration Cycle 322 11.6 The Refrigerant Selection 324 11.7 Heat Pump Systems 324 11.8 Innovative Vapour-Compression Refrigeration Systems 325 11.8.1 Cascade Refrigeration Cycle 325 11.82 Multistage Compression Refrigeration Systems 328 11.8.3 Multipurpose Refrigeration System with a Single Compressor 11.9 Liquefaction of Gases 330 11.10 Gas Refrigeration Cycles 331 11.11 Absorption Refrigeration Systems 333 11.12 Thermoelectric Power Generation 334 11.12.1 Thermocouple 336 Summary 336 Problems 337

317-339

330

Contents XI

12. PSYCHROMETRICS

340-366

12.1 Introduction 340 12.2 Properties of Atmospheric Air 340 12.3 Humidity of Air 341 123.1 Relative Humidity 0 342 12.4 Temperatures of Atmospheric Air 343 12.4.1 Dew-Point Temperature 343 12.42 Adiabatic Saturation Temperature 344 12.43 Wet-Bulb Temperature 345 12.4.4 Dry-Bulb Temperature 346 12.4.5 Psychrometer 346 12.5 Psychrometric Charts 347 12.6 Air-Conditioning Process 349 12.6.1 Simple Heating or Cooling 349 12.62 Heating with Humidification 350 12.6.3 Cooling with Dehumidification 352 12.6.4 Adiabatic Mixing of Air Streams 356 12.6.5 Wet Cooling Towers 359 12.6.6 Natural-Draft Cooling Towers 359 12.6.7 Spray Ponds 360 Summary 362 Problems 363

13. GENERAL THERMODYNAMIC PROPERTY RELATIONS 367-394 13.1 Introduction 367 13.2 Mathematical Preliminaries on Partial Derivatives and Associated Relations 13.2.1 Reciprocity and Cyclic Relations 369 133 The Maxwell Relations 370 133.1 The Gibbs and Helmholtz Relations 370 13.4 The Clapeyron Equation 373 13.5 General Relations for du, dh, ds, C, and Cp 375 13.5.1 Internal Energy Changes du 375 13.5.2 Enthalpy Changes dh 376 13.53 Entropy Changes ds 377 13.5.4 Specific Heats C,, and Cp 377 13.6 The Joule-Thompson Coefficient 382 13.7 Property Relations for Real Gases 384 13.7.1 Enthalpy Changes of Real Gases 384 13.7.2 Internal Energy Changes of Real Gases 386 13.73 Entropy Changes of Real Gases 386 13.8 Fugacity 387 Summary 390 Problems 392

367

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Contests

14. REACTIVE SYSTEMS

395 - 420

14.1 Introduction 395 142 Fuels and Combustion 395 143 Theoretical and Actual Combustion Processes 396 14.4 Enthalpy of Formation and Enthalpy of Reaction 401 14.4.1 Enthalpy of Reaction hR 401 145 First-Law Analysis of Reacting Systems 403 14.5.1 Steady-Flow Systems 403 1452 Closed Systems 404 14.6 Adiabatic Flame Temperature 407 14.7 Entropy Change 409 14.7.1 Third-Law of Thermodynamics 410 14.8 Second-Law Analysis of Reacting Systems 411 Summary 417 Problems 418

15. CHEMICAL AND PHASE EQUILIBRIUM

421-437

15.1 Introduction 421 15.2 Chemical Equilibrium Criterion 421 15.2.1 Relation for Chemical Equilibrium 423 153 The Equilibrium Constant for Ideal-Gas Mixtures 423 153.1 Equilibrium Constant Kp 424 15.4 Simultaneous Reactions 429 15.5 Kp Variation with Temperature 431 15.6 Phase Equilibrium 432 15.6.1 A Single-Component System 433 15.62 The Gibbs Phase Rule 434 15.63 Multicoraponent Systems 434 Summary 435 Problems 436

16. THERMODYNAMICS OF COMPRESSIBLE FLOW 16.1 Introduction 438 16.2 Stagnation Properties 438 16.2.1 Stagnation State 439 16.2.2 Stagnation Pressure P0 440 163 Speed of Sound and Mach Number 441 163.1 Mach Number 444 16.4 One-Dimensional Isentropic Flow 446 16.4.1 Streamtube Area-Velocity Relation 446 16.5 Property Relations for Isentropic Flow of Perfect Gases 448 16.5.1 Critical Properties 449 16.6 Isentropic Flow through Nozzles 450 16.6.1 Convergent Nozzles 451

438-482

Contents Xiii 16.6.2 Area-Mach Number Relation 452 16.63 Converging-Diverging Nozzles 457 16.7 Normal Shocks 459 16.7.1 Properties Relations across the Shock 463 16.8 Flow through Actual Nozzles and Diffusers 466 16.8.1 Velocity Coefficient Cy 467 16.8.2 Discharge Coefficient CD 467 16.83 Diffusers 470 16.8.4 Pressure Recovery Factor Fp 471 16.8.5 Pressure Rise Coefficient CpR 471 16.9 Steam Nozzles 473 Summary 477 Problems 479

17. KINETIC THEORY OF AN IDEAL GAS

483-495

17.1 Introduction 483 172 Equation of State of an Ideal Gas 484 173 Collision Frequency and Mean Free Path 489 17.4 Velocity and Speed Distribution Functions 491 17.4.1 Speed Distribution Function 493 Summary 495

18. ELEMENTS OF STATISTICAL THERMODYNAMICS

496-522

18.1 Introduction 496 182 Counting the Number of Microstates for a Given Macrostate 501 183 The Most Probable Macrostate 503 18.3.1 The Boltzmann Distribution 507 18.3.2 Evaluation of the Lagrangian Constants a and fi 508 18.4 Partition Function 508 18.5 Evaluation of the Partition Function in Terms of T and V 512 18.6 Evaluation of Thermodynamics Properties for a Single Chemical Species 515 18.6.1 Pressure of an Ideal Monatomic Gas 515 18.6.2 Partition Function of an Ideal Monatomic Gas 515 18.7 Equipartition of Energy 518 18.8 Statistical Interpretation of Work and Heat 520 18.9 Disorder and Entropy 521 Summary 522

19. ELEMENTS OF HEAT TRANSFER 19.1 Introduction 523 192 Driving Potential 523 193 Conduction Heat Transfer 524 19.4 Convection Heat Transfer 527 19.5 Radiation Heat Transfer 531

523-645

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Contents

19.6 Conduction Heat Transfer in a Stationary Medium 533 19.6.1 Energy Flux by Conduction 534 19.7 Boundary and Initial Conditions 536 19.8 One-Dimensional, Steady-State Conduction 537 19.8.1 Steady-State Conduction through a Plane Wall 537 19.8.2 Thermal Resistance 540 19.9 Heat Conduction across a Cylindrical Shell 547 19.9.1 Critical Thickness of Insulation 552 19.10 Steady Conduction through a Spherical Shell 553 19.11 Heat Transfer from an Extended Surface 556 19.11.1 Fins 556 19.112 Infmitely Long Fin 560 19.113 Fin Efficiency or Effectiveness of Fin 561 19.12 Unsteady Heat Conduction 564 19.12.1 Transient Conduction in an Infmite Wall 564 19.122 The Dimensionless Groups 565 19.123 Transient Heating of Bodies with Negligible Internal Resistance 567 19.12.4 Lumped System Analysis 569 19.12.5 Penetration Depth—Significance of Thermal Diffusivity 572 1933 Introduction to Convection 579 19.13.1 The Convection Boundary Layers 579 19.132 The Convection Transfer Equations 580 19.14 Approximation and Special Conditions 586 19.15 Convection Heat Transfer 588 19.15.1 Similarity Parameters in Heat Transfer 588 19.152 Boundary Layer Similarity Parameters 589 19.16 Physical Significance of the Dimensionless Parameters 592 19.17 Boundary Layer Concepts 594 19.18 Thermal Boundary Layer for Flow Past a Heated Plate 597 19.18.1 Transition Flow 600 19.19 Free Convection 604 19.19.1 Free Convection Correlations 605 19.192 Combined Natural and Forced Convection 606 1920 Radiation Heat Transfer 610 1920.1 Properties of Surfaces 612 19202 The View or Configuration Factor 618 1921 Blackbody Radiation Exchange 622 19.22 Radiation Exchange between Diffuse, Gray Surfaces in an Enclosure 623 1922.1 Net Radiation Exchange at a Surface 624 19.222 Radiation Exchange between Surfaces 624 19.223 The Two-Surface Enclosure 626 19.22.4 Radiation Shields 632 1922.5 The Radiating Surface 634 1923 Limitations 635 Summary 635 Problems 638

Contents XV

Appendix

647-686

Table 1 Molar mass, gas constant, and critical-point properties 647 Table 2 Ideal-gas specific heats of various common gases at 300 K 648 Table 3 Saturated water—Temperature table 649 Table 4 Saturated water—Pressure table 652 Table 5 Superheated water 655 Table 6 Compressed liquid water 661 Table 7 Saturated ice-water vapour 663 Table 8 Ideal-gas properties of air 664 Table 9 Ideal-gas properties of nitrogen, N2 666 Table 10 Ideal-gas properties of oxygen, 02 668 Table 11 Ideal-gas properties of carbon dioxide, CO2 670 Table 12 Ideal-gas properties of carbon monoxide, CO 672 Table 13 Ideal-gas properties of hydrogen, H2 674 Table 14 Ideal-gas properties of water vapour, H2O 675 Table 15 Ideal-gas properties of monatomic oxygen, 0 677 Table 16 Ideal-gas properties of hydroxyl, OH 677 Table 17 Enthalpy of formation, Gibbs function of formation, and absolute entropy at 25°C, 1 atm 678 Table 18 Enthalpy of combustion and enthalpy of vaporization at 25°C, 1 atm 679 Table 19 Logarithms to base e of the equilibrium constant Kp 680 Nelson—Obert generalized compressibility chart—Low Pressures 681 Nelson—Obert generalized compressibility chart—Intermediate pressures 682 Nelson—Obert generalized compressibility chart—High pressures 683 ASHRAE psychrometric chart at 1-atm total pressure 684 Generalized entropy departure chart 685 Generalized enthalpy departure chart 686

Selected References Index

687 689-696

•

Preface This second edition of the textbook, Fundamentals of Engineering Thermodynamics, continues to introduce the students to physical phenomena of thermodynamic processes in a coherent and lucid manner. The book also elucidates the applications of thermodynamic processes in the simplest and clearest possible manner without the use of complicated mathematics. The text is suitable for a first-course in thermodynamics for engineering students. It is written with sufficient breadth of coverage so that it can also serve in a number of ways for a second course, if desired. It is essential that the engineering student clearly visualizes the physics behind every process. Throughout the book, therefore, considerable emphasis is placed on physical phenomena of thermodynamic processes. The governing principles, the assumptions made in their development, and their limits on applicability, and how we can apply the principles to the solution of practical engineering problems are all stressed. The emphasis is also on teachability for the instructor. A revised solutions manual can be obtained from the publisher, Prentice-Hall of India, by the instructors teaching from this book. A large number of worked examples are presented to demonstrate the application of basic principles. Exercise problems with answers are provided at the end of most chapters to help students rapidly reinforce their understanding of the subject. Some improvements have been made throughout this edition by adding many worked-out examples, elucidating physical explanations, and increasing the number of exercise problems. In this revision, special attention is given to the chapter on Heat Transfer. I appreciate the many comments and suggestions that I received from the readers of this book since publication of the first edition. Further comments and suggestions regarding this new edition are welcomed. I am grateful for the help extended to me by my doctoral students Shibu Clement, Lovaraju and Vinoth and my masters students Vijay Agrawal, Aznit Kumar and Nischal Srivastava in preparing and checking the solutions manual. Finally, it has been a great pleasure working with the staff of Prentice-Hall of India. They all took special care in the release of this second edition. E. Rathakrishnan

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Preface to the First Edition Thermodynamics is the basic science of energy transformations involving work, heat and the properties of matter and has long been considered as an essential part of engineering education all over the world. This book is developed to serve as a text for engineering students for a course in thermodynamics at the undergraduate level. The objective of the book is to help students gain a sound understanding of the fundamentals of thermodynamics, starting with the most elementary ideas of temperature and heat and then moving on to develop the laws of thermodynamics from their experimental and engineering backgrounds. To meet the curriculum requirements of many universities, the text also includes a chapter each on kinetic theory of ideal gases, statistical thermodynamics, and heat transfer. The treatment of the material is designed to emphasize the phenomena which are associated with the various thermodynamic processes, and to enumerate techniques for solving specific classes of thermodynamic problems. A large number of well designed example problems are provided to help the students in mastering the subject matter. In order to aid the learning process, the end-of-chapter problems are developed to illustrate points which are brought out in the text and to extend the material covered in the text by including problems characteristic of engineering design. A solutions manual is available for the benefit of the instructor and can be obtained from the publisher. I wish to thank my colleagues who assisted me in reviewing this text and offered constructive criticism during the course of its development. In particular, I am grateful to Professor S. Elangovan who not only encouraged me to write this book but also provided help in checking the manuscript. I also sincerely thank my undergraduate and graduate students at the Indian Institute of Technology Kanpur who read the manuscript, gave their comments and suggestions, and encouraged me to proceed on. I express my special appreciation to my undergraduate student Sunil Kumar Dwivedi for his excellent help in critically checking the manuscript and giving some extremely useful suggestions. My special thanks go to my doctoral students S. Elangovan, Himanshu Agrawal, K. Srinivasan and Shashi Bhushan Verma for their help with the development of the manuscript, and my graduate students R.B. Sreejith, S. Saravanan, Amith Agrawal, Gaurav Garg and Ramesh who assisted me in the preparation of several illustrations and the solutions manual. I also thank Manoj Kumar for typing the manuscript, A.K. Ganguly for preparing the drawings and Sharad Chauhan for typing the solutions manual. Last but not the least, I am indebted to the Continuing Education Centre of the Indian Institute of Technology Kanpur for the financial support provided to this project. E. Rathakrishnan

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1 Basic Concepts and Definitions 1.1 THERMODYNAMICS AND ENERGY Thermodynamics is a basic science that deals with energy. In other words, thermodynamics may be defined as the study of energy, its forms and transformations, and its interaction with matter. Thermodynamics also deals with various properties of substances and the changes in these properties as a result of energy transformations. The principles of thermodynamics are based on our day-to-day experiences and experimental observations. The name thermodynamics stems from the Greek words therme (heat) and dunamis (power), which terms in early days were used to describe the effort to convert heat into power. But in the present context, the name thermodynamics is used to interpret all aspects of energy and energy transformations, including power production, refrigeration, and relationships among the properties of matter. It has been said that thermodynamics may be defined as the science of energy. Even though it is easy to feel the energy, it is difficult to give a precise definition to it. Energy may be viewed as the capacity to do work or the ability to cause changes. One of the most fundamental laws of nature is the law of conservation of energy. It simply states that during an interaction, energy can change from one form to another but its total amount remains constant. That is, energy is neither created nor destroyed; it can only change forms (r€ First Law of Thermodynamics). This principle is illustrated in Fig. 1.1. Like all sciences, thermodynamics is also based on experimental observations. The findings from these observations have been expressed in the form of some basic laws. • The first law of thermodynamics, for example, is simply an expression of the principle of conservation of energy. • The second law of thermodynamics asserts that processes occur in a certain direction but not in the reverse direction, i.e. processes occur spontaneously only in a particular direction and never in a direction opposite to that. It also asserts that energy has quality as well as quantity. 1

2 Fundamentals of Engineering Thermodynamics PE = 10 units KE = 0 Potential energy

0

PE or 7 units KE = 3 units Kinetic energy

Figure 1.1 Illustration of the first law of thermodynamics. Suppose a cup of hot coffee kept on a dining table. The hot coffee can only eventually cool. On the other hand, a cup of cold coffee left on the same table can never get hot by itself. Thus, left on its own, the heat flow is from higher potential to lower potential and not vice versa. Although the principles of thermodynamics have been in existence since the creation of the universe, thermodynamics did not emerge as a science until about 1700 when first attempts to build a steam engine were made in England by T. Savery and T. Newcomen. These engines were very slow and inefficient, but they opened the way for the development of a new science. Lord Kelvin used the term thermodynamics for the first time in a publication in the year 1849. The first textbook on thermodynamics was written by W. Rankine in 1859. The maximum progress in thermodynamics was made in the early 1900s.

1.2 CLASSICAL AND STATISTICAL THERMODYNAMICS It is well known that a substance consists of a large number of particles called molecules. The properties of the substance naturally depend on the behaviour of these molecules. For example, the pressure of a gas in a container is the result of momentum transfer between the molecules and the walls of the container. But to determine the pressure in the container, we need not know the behaviour of the individual molecules of the gas. This macroscopic approach to the study of thermodynamics which does not require any knowledge of the behaviour of individual particles of the substance is called classical thermodynamics. On the other hand, the study of thermodynamics based on the behaviour of a large number of individual particles, is called statistical thermodynamics. In short, the macroscopic approach and the microscopic approach to the study of energy are, respectively, called the classical thermodynamics and statistical thermodynamics.

Basic Concepts and Definitions 3

1.3 APPLICATIONS OF THERMODYNAMICS All engineering activities involve an interaction between energy and matter. It is difficult even to imagine an area which does not relate to thermodynamics in some respect. We can see the application of thermodynamics in every item, starting from household appliances to high-tech rockets. In fact, the human body itself is an interesting application area of thermodynamics. Engineering thermodynamics fmds applications in diverse areas such as power producing devices, refrigeration and air-conditioning, compressors and turbines, jet engines and rockets, chemical processing such as in oil refineries, and the combustion of hydrocarbon fuels such as coal, oil, and natural gases. Further, the use of passive and active solar energy units, power plants using geothermal sources beneath the ground, wind and tidal power generation units also work on thermodynamics principles. Also, engineering thermodynamics finds applications in space suit of astronauts, in air-separation plants that produce oxygen for use in the steel-manufacturing industry, and in our search for solutions to problems in connection with energy crises, shortage of fresh water, and waste disposal in cities. Thermodynamics is most relevant in our search for a better quality of life. Indeed it is a fascinating subject that challenges our imagination and creative ability.

1.4 DIMENSIONS AND UNITS Any physical quantity can be characterized by dimensions. The magnitudes assigned to the dimensions are called units. Some basic dimensions such as mass M, length L, time t, and temperature T are selected as primary dimensions, while others such as velocity V, energy E, and volume V are expressed in terms of the primary dimensions and are called secondary or derived dimensions. Table 1.1 gives the common systems of units and their symbols, as assigned to primary dimensions. Table 1.1 Common systems of units Symbol Quantity Mass Length Time Temperature

Unit kilogram metre second kelvin

SI

CGS

FPS

MKS

kg m s K

g cm s °C

lb ft s °F

kg m s °C

Throughout this text, we will use the SI system of units. However, other systems of units mentioned in Table 1.1 are equally applicable to all the equations.

Dimensional homogeneity The law of dimensional homogeneity states: An analytically derived equation representing a physical phenomenon must be valid for all systems of units. Thus, the equation for kinetic energy, KE = (÷) mV 2, is properly stated for any system of units. This explains why all natural phenomena proceed completely in accordance with the man-made units, and hence all fundamental equations

4 Fundamentals of Engineering Thermodynamics representing such events should have validity for any system of units. Thus, all equations must be dimensionally homogeneous, and consequently all relations derived from these equations must also be dimensionally homogeneous. For this to occur under all systems of units, it is essential that each grouping in an equation has the same dimensional representation. Now let us examine the dimensional representation of the following equation L = t2 + t where L denotes length and t, time. Changing the units of length from metres to feet will change the value of the left-hand side but not affect the value of right-hand side, thus making the equation invalid in the new system of units. We will consider dimensionally homogeneous equations only, in this book

1.5 CLOSED AND OPEN SYSTEMS A system is defined as a particular quantity of matter or a particular region of space chosen for study. The region outside a system is called its surroundings. The real or imaginary surface that separates the system from its surroundings is called the boundary. The system, the surroundings, and the boundary are illustrated in Fig. 1.2. The boundary of a system can be either stationary or movable. Surroundings

Boundary Figure 1.2 A system, its surroundings and the boundary.

Closed system (Control mass) A closed system consists of a constant amount of mass, and no mass can cross its boundary. That is, no mass can enter or leave a closed system. A closed system is shown in Fig. 1.3. Mass cannot cross the boundary of a closed system, but energy can do so.

mass NO CLOSED SYSTEM m= constant energy YES Figure 1.3

A closed system.

Basic Concepts and Definitions 5

Isolated system It is a special case of a closed system where even energy is not allowed to cross the boundary. Examine the piston-cylinder device shown in Fig. 1.4. Here gas is the system. There is no transfer of mass across its boundary, therefore, it is a closed system. Energy may cross its boundary, And Moving boundary

r

GAS 2 kg 3 m3

Fixed boundary Figure 1.4 A piston-cylinder device. part of the boundary may move. Everything outside the gas, including the piston and the cylinder, is the surroundings. If, as a special case, even energy is not allowed to cross the system boundary, then it may be treated as an isolated system.

Open system (Control volume) It is a properly selected region of space. It usually encloses a device which involves flow of mass, such as a nozzle, diffuser, compressor or turbine. The flow through these devices is best studied by selecting a region within the device as the control volume. Both energy and mass can cross the boundary of a control volume, which is called the control surface, but the shape of the control volume will remain unchanged. An open system is illustrated in Fig. 1.5. Control surface r

L Emass YES CONTROL VOLUME energy YES

Figure 1.5 An open system.

6

Fundamentals of Engineering Thermodynamics

1.6 FORMS OF ENERGY Energy can exist in numerous forms, such as chemical, electrical, kinetic, potential, mechanical, thermal, and nuclear, and their sum constitutes the total energy E of a system. The total energy of a system per unit mass is denoted by e and is defined as

e= — E (kJ/kg) m Thermodynamics provides no information about the absolute value of the total energy of a system. It only deals with the change in value of the total energy of a system, which is of interest in engineering applications. Thus, the total energy of a system can be assigned a value of zero (E = 0) at some convenient reference point. The change in total energy of a system is independent of the reference point selected. For example, the decrease in the potential energy of a falling object depends only on the elevation difference and not on the reference level chosen. In thermodynamic analysis, it is often helpful to consider the various forms of energy that make up the total energy of a system in two groups, namely the macroscopic energy group and the microscopic energy group. The macroscopic energies are those which a system possesses as a whole with respect to some outside reference frame, such as kinetic energy and potential energy. The microscopic forms of energy are those which are related to the molecular structure of a system and the degree of the molecular activity, and they are independent of outside reference frames. The sum of all the microscopic forms of energy is called the internal energy IE of a system, and is denoted by U. The macroscopic energy of a system is related to the motion as well as to the influence of some external effects such as gravity, magnetism, and so on. The energy that a system possesses as a result of its motion relative to some reference frame is called the kinetic energy KE. When all the parts of a system move with the same velocity, the KE is expressed as 1 2 KE = —mV (kJ) (12) where m is the mass of the system. The kinetic energy per unit mass is v2 ke = — (kJ/kg) 2

(1.3)

where V is the velocity of the system relative to some fixed frame of reference. The potential energy PE is the energy that a system possesses as a result of its elevation in a gravitational field. It is expressed as mgz (kJ)

(1.4)

pe = gz (kJ/kg)

(1.5)

PE =

The potential energy per unit mass is where g is the gravitational acceleration and z is the elevation of the centre of gravity of the system with respect to some arbitrarily chosen reference plane. When the external effects such as magnetic, electric, and surface tension are insignificant

Basic Concepts and Definitions 7 for the system under consideration, the total energy E of the system consists of the KE, PE, and 1E and is expressed as E = 1E + KE + PE That is, E = U + —1 ml/ 2 + mgz (Id) (1.6) The energy per unit mass e, also called the specific energy is given by e = u + ke + pe Or

V2 e = u + — + gz (kJ/kg) 2

(1.7)

Most closed systems remain stationary during a process and thus experience no change in their kinetic and potential energies. Such closed systems in which velocity and elevation of the centre of gravity remain unchanged during a process are termed stationary systems. The change in the total energy AE of a stationary system is equal to the change in its internal energy AU.

1.6.1 Internal Energy Internal energy is defined as the sum of all the microscopic forms of energy of a system. It is the energy associated with the molecular structure and the molecular activity of the constituent particles of the system. It may be viewed as the sum of the kinetic and potential energies of the molecules. In general, the individual molecules of a system will move around with some velocity, vibrate about each other and rotate about an axis during their random motion. Associated with these motions are the translational, vibrational, and rotational kinetic energies, the sum of which constitutes the kinetic energy of a molecule. The portion of the internal energy of a system, associated with the kinetic energy of the molecules is called the sensible energy. The average velocity and activity of the molecules are proportional to the temperature of the system. Thus, at higher temperatures the molecules will possess higher kinetic energies, and as a result the system will have a higher internal energy. The internal energy is also associated with the intermolecular forces of a system. These are the forces that bind the molecules to each other, and they are the strongest in solids and weakest in gases. When sufficient energy is added to the molecules of a solid or liquid, they overcome the intermolecular forces and break away, turning the system to a gas. This is described as the phasechange process. Because of the added energy, a system in a gas phase is at a higher internal energy level than when it exists in the solid or the liquid phase. The internal energy associated with the phase of a system is called the latent energy. The above changes can occur without a change in the chemical composition of a system. Most thermodynamic problems fall into this category and we need not pay any attention to the forces binding the atoms in a molecule. The internal energy associated with the bonds in a molecule is called the chemical or bond energy. During a chemical reaction, such as a combustion process, some chemical bonds are destroyed while others are formed. As a result, the internal energy changes.

8 Fundamentals of Engineering Thermodynamics There is an enormous amount of internal energy associated with the bonds within the nucleus of the atom itself. This energy is called the nuclear energy and it is released during nuclear reactions. Unless fusion or fission reaction is the process under consideration, we need not be concerned with nuclear energy in thermodynamics.

1.7 PROPERTIES OF A SYSTEM Property is any characteristic associated with a system. Some familiar properties that we often encounter in the analysis of engineering systems are pressure P, temperature T, volume V, and mass m. In addition to these, viscosity, thermal conductivity, modulus of elasticity, thermal expansion coefficient, electric resistivity, and even velocity and elevation can also be treated as properties. We should note that not all properties are independent. Some are defmed in terms of other properties. For instance, density, defmed as "mass per unit volume", is expressed in terms of mass and volume as p = 1. (kg/m3 )

(1.8)

Sometimes the density of a substance is given relative to that of a substance which is better known. Then it is called the specific gravity or relative density, and is defined as "the ratio of the density of a substance to the density of some standard substance at a specified temperature" (usually water at 4°C for which mho = 1000 kg/m3). That is, the specific gravity ps of a substance is given by A

(1.9)

The specific gravity is obviously a dimensionless quantity. A more frequently used property in thermodynamics is the specific volume v. It is the reciprocal of density and is defmed as the "volume per unit mass". V 1 v = — = — (m3/kg) m

(1.10)

EXAMPLE 11 Determine the specific volume and density of 5 kg of a gas contained in a tank of volume 25 m3. Solution By Eq. (1.10), the specific volume is 25 3 v = — = 5 m /kg 5 The density is 1 1 = — = — =0.2 kg/m3 P==S = v 5 At this stage, it is essential to note that in classical thermodynamics, the atomic structure of a substance (thus the spaces between and within the molecules) is disregarded, and the substance is viewed to be continuous, homogeneous matter with no microscopic gaps or holes.

Basic Concepts and Definitions

9

In other words, the substance is treated as continuum. This idealization is valid as long as we work with areas and lengths which are large compared to intermolecular spacings. In situations, like highly rarefied conditions, where the intermolecular spacings can be much larger than any area or length under consideration, the continuum hypothesis is not valid and the problem has to be analysed by microscopic approach.

Intensive and extensive properties Properties are usually classified as intensive and extensive. Intensive properties are those which are independent of the size of a system. For example, pressure, temperature, density, are all intensive properties. Extensive properties are those whose values depend on the size or extent of the system. For example, mass, volume, total energy are some of the extensive properties. Extensive properties per unit mass are called specific properties. For example, specific volume (v = V/m), specific total energy (e = Elm), and specific internal energy (u = Ulm) are specific properties. A specific property is an intensive property. A phase of a system is the collection of all regions of the system that have the same intensive state and chemical composition. Another classification of the properties often made is in terms of intrinsic and extrinsic characteristics. For a system: • An intrinsic quantity is that whose value depends on the nature of the substance in the system. For example, pressure, temperature, and internal energy are intrinsic quantities. • An extrinsic quantity is that whose value is independent of the nature of the substance in the system. For example, translational velocity, rotational energy, and electric field strength are extrinsic quantities.

1.8 STATE AND EQUILIBRIUM All the properties of a system, which are not undergoing any change, can be computed or measured throughout the entire system. These give us a set of properties that completely describe the condition or state in which all the properties of the system have fixed values. Even if the value of one of the properties changes, the state of the system will change to a different one. Thermodynamics deals with equilibrium states. The word equilibrium implies a state of balance. In an equilibrium state, there are no unbalanced potentials or driving forces within the system. A system is said to be in equilibrium when it involves no changes with time. There are many types of equilibrium, and a system is not in thermodynamic equilibrium unless the conditions of all the relevant types of equilibrium are satisfied. For example, a system is in thermal equilibrium if the temperature is the same throughout the entire system. That is, the system involves no temperature differentials, which constitute the driving potentials for heat flow. Mechanical equilibrium is related to pressure, and a system is said to be in mechanical equilibrium if there is no change in pressure at any point in the system with time. However, the pressure may vary within the system with elevation as a result of gravitational effects. But the higher pressure at a bottom layer is balanced by the extra weight it must carry, and, therefore, there is no imbalance of forces. The variation of pressure as a result of gravity in most thermodynamic systems is relatively small and usually disregarded.

10

Fundamentals of Engineering Thermodynamics

If a system involves two phases, it will be in phase equilibrium only when the mass of each phase reaches an equilibrium level and stays there. A system will be in chemical equilibrium only if its chemical composition does not change with time, that is, no net chemical reaction occurs. A system will be in thermodynamic equilibrium only when it satisfies the conditions for all modes of equilibrium.

1.9 PROCESSES AND CYCLES A process may be defined as a change of a system from one equilibrium state to another. The series of states (marked as *) through which a system passes during a process is called the path of the process, as shown in Fig. 1.6. When a process proceeds in such a manner that the system remains infmitesimally close to an equilibrium state at all times, it is called a quasi-static, or quasiequilibrium, process. State 2

i Process ) path

0......er) State 1

Figure 1.6 Process path. A quasi-equilibrium process is a sufficiently slow process which allows the system to adjust itself internally so that all properties in one part of the system do not change any faster than those at other parts. Examine the slow and fast compression processes shown in Fig. 1.7.

Slow compression (quasi-equilibrium)

Fast compression (nonquasi-equilibrium)

(a)

(b) Figure 1.7 Compression processes.

When the piston is moved slowly (Fig. 1.7(a)), the molecules redistribute themselves such that they do not pile up in front of the piston face. As a result, the pressure inside the cylinder will always be uniform and will rise at the same rate at all locations. This is a quasi-equilibrium process, since equilibrium is maintained at all times. It is essential to note that a quasi-equilibrium process is an idealized process. However, many actual processes closely approximate it. When the process is as shown in Fig. 1.7(b), it cannot be approximated to a quasi-equilibrium process since molecules pile up in front of the piston, because of the fast movement of the piston. The quasi-equilibrium processes are of interest to engineers owing to the following reasons:

Basic Concepts and Definitions 11 • They are easy to analyse. • Work producing devices deliver most of the work when they operate on quasi-equilibrium processes. Therefore, the quasi-equilibrium processes serve as standards, against to which actual processes can be compared. Process diagrams plotted with thermodynamic properties as coordinates, as shown in Fig. 1.8, are very useful in visualizing the processes.

P 2 Final state

Process path 1

(2)

Initial state

(1)

Figure 1.8 The process P-V diagram of a compression process. The process path, which indicates a series of equilibrium states and through which the system passes during a process, has significance for quasi-equilibrium processes only. For nonquasi-equilibrium processes, it is not possible to specify the intermediate states and, therefore, identification of a single process path of the entire process is not possible. A nonquasi-equilibrium process is usually denoted by a dashed line between the initial and final states. The prefix iso- is commonly used to designate a process for which a particular property remains constant. Some of these popularly known processes are the following:

Isothermal process It is a process during which the temperature T of the system remains constant.

Isobaric process It is a process during which the pressure P of the system remains constant.

Isochoric process It is a process during which the specific volume v of the system remains constant.

12 Fundamentals of Engineering Thermodynamics

Cycle A system is said to have undergone a cycle if it returns to its initial state at the end of the process. That is, for a cycle the initial and the final states are identical. A typical two-process cycle is shown in Fig. 1.9.

2

1 ►V

Figure 1.9 A two-process cycle.

1.10 THE STATE POSTULATE The state of a system is described by its properties. But from experience it is known that all properties need not be specified in order to describe a state. Specifying a certain number of properties is sufficient to describe a state. The number of propbrties required to fix the state of a system is given by the state postulate. "The state of a simple compressible substance is completely specified by two independent intensive properties." A simple compressible system is that system for which the electrical, gravitational, magnetic, motion, and surface tension effects are absent. The state postulate requires that the two properties specified be independent to fix the state. Two properties are said to be independent if one can be varied while the other is held constant. The temperature T and specific volume v are always independent properties and together they can fix the state of a simple compressible system. The temperature T and pressure P are independent properties for single-phase systems, but are dependent properties for multi-phase systems. For example, at sea level with atmospheric pressure, water boils at 100°C but at a higher altitude where the pressure is lower, water boils at a lower temperature. That is, T =f(P) during a phase-change process; thus T and P are not sufficient to fix the state of a two-phase system.

1.11 PRESSURE AND MECHANICAL EQUILIBRIUM _Pressure is the force exerted by a fluid per unit area. The counterpart of pressure in solids is stress. The pressure in a fluid increases with depth as a result of the weight of the fluid. There is no variation of pressure in the horizontal direction. The pressure in a gas tank may be considered to be uniform since the weight of the gas is too small to cause a significant difference. Pressure has units of N/m2, which is called pascal (Pa). That is, 1 Pa = 1 N/m2. Pressure is also expressed in atmosphere (atm), bar, toff, or pounds per square inch (psi), where

Basic Concepts and Definitions 13 1 bar = 105 Pa 1 atm = 101,325 Pa = 1.01325 bar 1 atm = 14.7 psi The actual pressure at a given position is called the absolute pressure, and it is measured relative to absolute vacuum, that is, absolute zero pressure. The relation between gauge pressure "gauge measured by a pressure gauge and atmospheric pressure pa„,,, and absolute pressure gbs is "gauge = Pabs — Palm For pressures below atmospheric, the gauge pressure would be negative and is termed subatmospheric. It is a common practice to call the subatmospheric gauge pressures as vacuum pressures. When the absolute pressure is zero, then the system is said to be at perfect vacuum. When two systems are separated by a frictionless piston in a cylinder, as shown in Fig. 1.10, the piston will move in the direction of the lower pressure because of the unbalanced force acting on the piston. The property pressure may thus be looked upon as the driving force for volume change. The piston will stop moving when the pressures in systems A and B become equal. This demonstrates that it is essential to have uniform pressure in a system to attain mechanical equilibrium. Direction of piston movement

P A> PB

Frictionless piston Figure 1.10 Pressure as a driving force for volume change.

Liquid manometer The basic principle of liquid manometer is that the pressure difference indicated by it is balanced by the weight of a liquid column. It is used for measurement of small and moderate pressure differences. It is generally made up of a glass or plastic U-tube containing a fluid such as mercury, water or alcohol (Fig. 1.11). The manometer shown in Fig. 1.11 is connected to a gas tank at high pressure. It is seen that the pressure in the tank pushes the manometric liquid to a height h. Thus, the pressure in the tank is given by the force balance expressed as PA/1 = 1::„„ A + W where A is the cross-sectional area of the tube and W is the weight of the fluid column h in the manometer. The weight W given by W = mg = pVg = pAhg

14 Fundamentals of Engineering Thermodynamics Palm

i

h

GAS

A

Figure 1.11

.-. *.! B

A simple U-tube manometer.

where V is the volume of the fluid column h and p is the fluid density. Thus,

PA = Pam, + Pgh

(1.12)

The pressure difference between the tank pressure and the atmospheric pressure can be expressed as AP =

PA — Pat m = Pgh

Barometer The barometer is a device for measuring atmospheric pressure. Thus, the atmospheric pressure is also called barometric pressure. The barometer invented by Toricelli in the seventeenth century measures atmospheric pressure by inverting a mercury filled tube into a mercury container which is often open to atmosphere, as illustrated in Fig. 1.12. The pressure at point 1 can be taken as atmospheric, and the pressure at point 2 can be taken as zero since there is only mercury vapour above point 2 and the pressure it exerts is negligible. Force balance in the vertical direction gives (1.13) where p is the density of mercury and g the local gravitational acceleration.

Basic Concepts

and Definitions 15

/Th

2

h

Palm

Figure 1.12 Toricelli barometer.

1.12 TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS Although it is well known that temperature is a measure of "hotness" or "coldness", it is not easy to formulate an exact definition for it. Based on psychological sensations, we express the level of temperature qualitatively with words like hot, red-hot, warm, cold, and freezing cold. However, we cannot assign numerical values to temperature based on our sensations alone. Furthermore, our senses may be misleading. For example, a metallic chair will make us feel much colder than a wooden chair even when both are at the same temperature. Luckily, several properties of materials change with temperature in a repeatable and predictable way. This forms the basis for accurate measurement of temperature. We know from experience that when a hot body is brought into contact with a cold body, heat is transferred from the hot body to cold body until both bodies attain the same temperature. At that instant of time, the heat transfer stops and the two bodies are said to have reached thermal equilibrium. The equality of temperature is the only requirement for thermal equilibrium. The zeroth law of thermodynamics states that: If two bodies are in thermal equilibrium with a third body. they are also in thermal equilibrium with each other. The zeroth law is illustrated in Fig. 1.13. If T1 = T3 and T2 = T3, then T1 = T,. It may look funny that such an obvious fact

Figure 1.13 Three bodies in contact.

16 Fundamentals of Engineering Thermodynamics is called one of the basic laws of thermodynamics. However, it cannot be concluded from the other laws of thermodynamics, and it serves as a basis for the validity of temperature measurement. By replacing the third body with a thermometer, the zeroth law can be restated as: Two bodies are said to be in thermal equilibrium f both have the same temperature reading even ff they are not in contact.

Temperature scales Temperature scales provide a common basis for temperature measurements. All temperature scales are based on some easily reproducible states, such as the freezing and boiling points of water, also known as the ice point and the steam point. A mixture of ice and water which is in equilibrium with air saturated with vapour at 1 atm pressure is said to be at the ice point. A mixture of liquid water and water vapour (with no air) in equilibrium at 1 atm pressure is said to be at the steam point. The temperature scales in the SI and English system of units, respectively, are the Celsius scale (named after the Swedish astronomer A. Celsius who devised it, formerly called the Centigrade scale) and the Fahrenheit scale (named after the German instrument maker G. Fahrenheit). On the Celsius scale, the ice and steam points are assigned the values 0°C and 100°C, respectively. The corresponding values on the Fahrenheit scale are 32°F and 212°F. These are often referred to as two-point scales, since temperature values are assigned at two different points. A more useful temperature scale in thermodynamics is the absolute temperature scale. As the name implies, there are no negative temperatures on the absolute temperature scale, and the lowest attainable temperature is absolute zero. The absolute temperature scale in the SI system is the Kelvin scale (named after W. Thomson, also known as Lord Kelvin). To measure absolute temperature, a sealed, rigid tank containing gas at low pressure, shown in Fig. 1.14, can be used as a thermometer. This is called a constant-volume gas thermometer It is based on the principle that the temperature of a gas at low pressure is proportional to its pressure. Thus, when the pressure reading is halved, so is the absolute temperature. It can be determined by extrapolation that as the absolute pressure in a constant-volume gas thermometer approaches zero, the absolute temperature reading will also approach zero. The reading of the thermometer with a Celsius scale, however, will approach —273.15°C. Therefore, T(K) = T(°C)+ 273.15

7(°C) T(K) P(kPa)

Ii —273.15 Absolute vacuum V = constant

Figure 1.14 Constant-volume gas thermometer.

(1.14)

Basic Concepts and Definitions 17 In the English system, the absolute temperature scale is the Rankine scale (named after W.J.H. Rankine). It is related to Fahrenheit scale by T(R) = T(°F) + 459.67

(1.15)

The temperature scales in the SI and English systeM of units are related by AR) =1.87(K)

(1.16)

A°F) = 1.871°C) + 32

(1.17)

At the tenth conference on weights and measures in 1954, the Celsius scale was redefined in terms of a single fixed point and the absolute temperature scale. The selected single point is the triple point of water (the state at which all three phases of water coexist in equilibrium), which is assigned the value 0.01°C; the magnitude of the degree is defined from the absolute temperature scale. As before, the boiling point of water at 1 atm pressure is 100°C. Thus, the new Celsius scale is essentially the same as the old one. It will be useful to note that the magnitudes of each division of 1 K and 1°C are identical, as shown in Fig. 1.15.

1K

1°C

1.8 R

1.8 F

Figure 1.15 Magnitudes of various temperature units. Therefore, when we deal with temperature difference AT, the temperature interval on the SI and English scales is the same. Raising the temperature of a substance by 10°C is the same thing as raising it by 10 K. That is, (1.18) ARK) = o77°C) Similarly, AAR) = o71°F) (1.19)

SUMMARY In this introductory chapter, the basic concepts and definitions essential for a sound understanding of thermodynamics have been highlighted. These basic facts and associated results will be used either directly or indirectly in the study of the subject in the subsequent chapters. Thermodynamics may be defined as the study of energy, its forms and transformations, and its interaction with matter. Thermodynamics deals with the conservation of energy from one form to another. The first law of thermodynamics is an expression of the principle of conservation of energy. The second law of thermodynamics asserts that spontaneous processes occur only in a particular direction and never in a direction opposite to that. Further, it ascertains that energy has quality as well as quantity. A macroscopic approach to the study of thermodynamics which does not require any

18 Fundatnentals of Engineering Thermodynamics knowledge of the behaviour of the individual particles of the substance is called classical thermodynamics. The study of thermodynamics based on the behaviour of the individual particles is called statistical thermodynamics. A closed system or control mass consists of a constant amount of mass and no mass can cross its boundary, but energy can cross its boundary. An open system or control volume is a properly selected region of space. Both mass and energy can cross the boundary of a control volume, but the shape of the control volume will remain unchanged. The sum of all forms of energy of a system is called its total energy E. The total energy of a system is made up of two groups: microscopic energy group and macroscopic energy group. The internal energy of a system is the sum of all its microscopic forms of energy. The portion of the internal energy of a system, associated with the kinetic energy of the molecules is called the sensible energy. The internal energy associated with the phase of a system is called the latent energy. The internal energy associated with the bonds in a molecule is called the chemical or bond energy. Property is any characteristic associated with a system. Intensive properties are those which are independent of the size of a system. Extensive properties are those which vary directly with the size or extent of the system. A system is said to be in equilibrium when it involves no changes with time. Thermal equilibrium implies that the temperature is the same throughout the entire system. Mechanical equilibrium ensures that there is no change in pressure at any point in the system with time. If a system involves more than one phase, it is said to be in phase equilibrium when the mass of each phase reaches an equilibrium level and stays there. If there is no change in the chemical composition of a system with time, it is said to be in chemical equilibrium. When a system satisfies the conditions for all modes of equilibrium, it is said to be in thermodynamic equilibrium. Process may be defined as a change of a system from one equilibrium state to another. The states through which a system passes during a process is called the path of the process. A quasiequilibrium process is a slow process which allows the system to adjust itself internally so that all properties in one part of the system do not change any faster than those at other parts. Processes during which T, or P, or v of the system remains constant, are respectively called isothermal. isobaric, and isochoric processes. The state postulate states that: "The state of a simple compressible substance is completely specified by two independent intensive properties." A simple compressible system is that system for which the electrical, gravitational, magnetic, motion, and surface tension effects are absent. Pressure is the force exerted by a fluid per unit area. It is essential to have uniform pressure in a system to attain mechanical equilibrium. Basically, temperature is a measure of "hotness" or "coldness"; the equality of temperature is the only requirement for thermal equilibrium. The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. In the SI system of units, the absolute temperature scale is the Kelvin scale. It is related to the Celsius scale by T(K) = T(°C) + 273.151

In the English system of units, the absolute temperature scale is the Rankine scale. It is related to the Fahrenheit scale by

Basic Concepts and Definitions 19 T(R) = T(°F) +459.67

The temperature scales in the SI and English system of units are related by 7(1t) = 1.87(K) T(°F) = 1.87(°C) 4- 32

PROBLEMS 1.1

A 5 kg mass at rest is dropped from a height of 100 m above the ground level. Determine the maximum velocity the mass can attain. Also, determine the velocity of the mass at 50 m above the ground. Take g = 9.81 m/s2. [Ans. V„.„ = 44.29 m/s, Aso m = 31.32 m/s]

1.2

The total energy of a closed system of 5 kg mass travelling with a speed of 100 m/s at an altitude of 250 m above the ground level is 50 kJ. Determine the internal energy of the system. [Ans. 12.74 kJ]

1.3

If the specific gravity of 5 kg of a fluid is 0.0012, determine its volume. [Ans. 4.167 m3]

1.4 A pressure gauge connected to a gas tank reads 105 kPa at a location where the atmospheric pressure is 101 kPa. Determine the absolute pressure of the gas in the tank. [Ans. 206 kPa] 1.5

If the pressure of a compressed air tank measured by a manometer is 2210 mm of mercury, calculate the tank pressure in kPa. The density of mercury is 13.6 x 103 kg/m3 and the atmospheric pressure is 101 kPa. [Ans. 395.85 kPa]

1.6

In a heating process the temperature of a system goes up by 22.2°C. Express this increase in temperature in kelvin (K). [Ans. 22.2 K]

1.7 If the temperature of a human body shown by a thermometer is 98.6°F, determine the temperature in degree Celsius (°C). [Ans. 37°C] 1.8

One litre of gasoline weighs 7 N at a location where of this gasoline.

g is 9.81 m/s2. Determine the density [Ans. 713.56 kg/m3]

1.9

A cylindrical tank of length 30 m and diameter 1.5 m contains 25 kg of air. Determine the specific volume and density of the air in the tank. [Ans. 2.12 m3/kg, 0.472 kg/m3]

1.10 One kilogram of a liquid of density 1200 kg/m3 is mixed with 2 kg of another liquid of density 2000 kg/m3. If the mixture volume is the sum of the initial volumes of the liquids, determine the mixture density. [Ans. 1636.7 kg/m3]

20 Fundamentals of Engineering Thermodynamics 1.11 Two litres of water at 25°C is mixed with an unknown liquid to form a liquid mixture. The mass and volume of the mixture formed are 4 kg and 4000 cc respectively. If the mixture volume is the sum of the initial volumes, determine the density of the unknown liquid. [Ans. 1000 kg/m3] 1.12 Two bodies one with thermometer reading in Celsius and the other with thermometer reading in kelvin, are in thermal equilibrium with the same system. What is the temperature of the system if both thermometers indicate the same numerical value? [Ans. —136.58°C] 1.13 The pressures of high-vacuum systems are quite often given in torr (1 torr equals 1 mm of mercury). If the pressure of one such system is 10-8 torr, what is the pressure in pascal. [Ans. 1.3332 µPa] 1.14 A barometer reads 101.3 kPa at the base of a mountain. If the barometer reads 80 kPa at the top of the mountain, determine the height of the mountain assuming the average density of air to be 1.21 kg/m3. [Ans. 1794.4 m] 1.15 The pressure difference between the upstream sides of a contraction through which air flows is measured as 1255 mm Hg by a U-tube manometer. What is the pressure difference in pascal at standard sea-level conditions. [Ans. 16665.3 Pa] 1.16 A fluid inside a cylinder fitted with a frictionless piston, as shown in the figure below, is in mechanical equilibrium. The atmospheric pressure is 101,325 Pa and the piston area is 2500 mm2. The linear spring has a constant of 5000 N/m. What is the absolute pressue of the fluid in pascal if the spring is compressed 300 mm beyond its free length?

[Ans. 701.325 kPa] 1.17 A gas flows through a passage of cross-sectional area 8.5 cm2 at the rate of 0.55 kg/s. If the specific volume of the gas is 4.3 m3/kg, determine the flow velocity. [Ans. 278.2 m/s] 1.18 Determine the pressure exerted on a diver at 40 m below the free surface of the sea, if the specific gravity of sea water is 1.03 and the atmospheric pressure is 101 kPa. [Ans. 505.2 kPa] 1.19 A spherical balloon of diameter 10 m is filled with helium gas which weighs one-seventh of air under identical conditions. The buoyancy force, Fb = pairgVi,,,noon, pushes the balloon upward when it is released at sea-level. If the balloon carries a person of mass 90 kg, determine the acceleration of the balloon when it is just released. [Ans. 24.85 m/s2]

Basic Concepts and Definitions

21

1.20 The lower-half of a 6 m high cylindrical tank is filled with water and the upper-half with an oil of specific gravity 0.8. Determine the pressure at the bottom of the tank if the top is open to atmosphere. [Ans. 154.3 kPa] 1.21 A vertical, frictionless piston-cylinder device contains a gas at 620 kPa. The piston area is 50 cm2. Determine the piston mass, if the atmospheric pressure is 101 kPa. [Ans. 264.5 kg] 1.22 A 3 kg ball is thrown upward with a force of 100 N at a location where the gravitational acceleration is 9.8 m/s2. Determine the acceleration of the ball. [Ans. 23.53 m/s2]

CHAPTER

2 Energy and the First Law of Thermodynamics

2.1 ENERGY We know that energy can neither be created nor be destroyed. It can only change forms. This enunciation is called the first law of thermodynamics or the principle of conservation of energy. The first law of thermodynamics states that during an interaction between a system and its surroundings, the amount of energy gained by the system is equal to the amount of energy lost by the surroundings. Energy can cross the boundary of a closed system in two distinct forms, namely heat and work (Fig. 2.1). It is essential to distinguish between these two forms of energy. System boundary

U

Heat

CLOSED SYSTEM

Work

(constant mass) Figure 2.1 Energy transfer across the boundary of a closed system.

2.2 HEAT Heat is defined as the form of energy that is transformed between two systems (or a system and 22

23

Energy and the First Law of Thermodynamics

its surroundings) by virtue of a temperature difference between them. The meaning of the term `heat' in thermodynamics is different from that in day-to-day usage. In day-to-day usage, heat is often used to mean internal energy, for example, the heat content of a fuel. In thermodynamics, heat and internal energy are two different entities. Energy is a property, but heat is not a property. A body contains energy, but not heat. Energy is associated with a state and heat is associated with a process. Heat is energy in transition. It is recognized only as it crosses the boundary of a system. The hot metal piece (Fig. 2.2) contains energy, but this energy is called heat only as it passes through the outer surface of the body (the system boundary) to reach the surrounding air. Once in the surroundings, the heat becomes part of the internal energy of the surroundings. Thus, in thermodynamics heat simply means heat transfer.

Surrounding air 5 kJ thermal energy

System boundary Figure 2.2 Hot metal piece kept in open air. The transfer of heat into a system is called heat addition, and the transfer of heat out of a system is called heat rejection. A process during which there is no heat transfer is called an adiabatic process. Even though there is no heat transfer during an adiabatic process, the energy content and thus the temperature of a system can still be changed by other means such as work. As a form of energy, heat has energy units, kJ being the most common one in use. The amount of heat transferred during a process between two states 1 and 2 is denoted by Q12, or simply Q. Heat transfer per unit mass of a system is denoted by q and defined as q = 2, (kJ/kg) m

(2.1)

The heat transfer rate is denoted by Q and has the units kJ/s, which is equivalent to kW. The universally accepted sign convention for heat is as follows (Fig. 2.3): • Heat transfer to a system is positive. • Heat transfer from a system is negative.

24

Fundamentals of Engineering Thermodynamics Surroundings Heat in SYSTEM Heat out

Figure 23 Sign convention for heat. EXAMPLE 2.1 To a gas tank containing 5 kg of gas, 100 kJ of heat is transferred in 4 seconds. Determine the heat transfer rate Q and heat transfer per unit mass q of the system. Solution Given m = 5 kg, Q = 100 kJ and A t = 4 s Therefore, Q

=

=

t

At

= 100 = 25 kW 4

and Q m

100 5

20 kJ/kg

2.3 WORK Work is an energy interaction between a system and its surroundings. Energy can cross the boundary of a closed system in the form of heat or work. In fact, heat and work are the only two mechanisms by which the energy of a closed system can be changed. Therefore, if the energy crossing the boundary is no: heat, it must be work. It is easy to recognize heat. Its driving force (potential) is a temperature difference between the system and its surroundings. Hence we can simply say that, an energy interaction which is not caused by a temperature difference between the system and its surroundings is work. For example, a movable piston, a rotating shaft, and an electric current through a wire crossing the system boundaries are all associated with work interactions. Like heat, work is also a form of energy, and therefore, has energy units such as kJ. The work done during a process between states 1 and 2 is denoted by W12, or simply W. The work done per unit mass of a system is denoted by w and defined as w = (kJ/kg) m The work done per unit time is called power and is denoted by W, with units kJ/s or kW.

(2.2)

Energy and the First Law of Thermodynamics

25

The commonly used sign convention for work is: • Work done by a system is positive. • Work done on a system is negative. The sign conventions for both heat and work are illustrated in Fig. 2.4.

Surroundings

Figure 2.4 Sign conventions for heat and work. EXAMPLE 2.2 A system does work equivalent to 100 kJ on the surroundings in 4 seconds. If the mass of the system is 5 kg, determine W and W. Solution Given W= 100 kJ, t = 4 s, m = 5 kg Therefore, W =1

4

ki/s = 25 kW

and W 100 m

5

20 kJ/kg

From our discussion on heat and work, it is clear that both heat and work are interactions between a system and its surroundings. They have the following similarities: 1. Both are recognized only as they cross the boundary of a system. That is, both are boundary phenomena. 2. Systems possess energy, but not heat or work. That is, heat and work are transient phenomena. 3. Both are associated with a process, not with a state. Unlike properties, heat and work have no meaning at a state. 4. Both are path functions. That is, their magnitudes depend on the path followed during a process as well as on the end states.

2.3.1 Path and Point Functions Path functions have inexact differentials designated by the symbol 8, the magnitude of a path function depends on the path followed during a process as well as on the end states. Therefore,

26

Fundamentals of Engineering Thermodynamics

a differential amount of heat or work is represented by SQ or SW, instead of dQ or dW. Properties are point functions. That is, they depend on the state only, and not on how a system reaches that state. Property changes are designated by the symbol d. For example, a differential amount of pressure or temperature is represented by dP or dT. A small change in volume is represented by dV and the total volume change during a process between states 1 and 2 is given by f2

dV = V2 - V1 = AV

The total work done during the process 1-2 is given by .1.12 SW = W1 2 and not A W. This is because of the fact that work is not a property, and systems do not possess work at a state.

2.3.2 Electrical Work Consider an insulated electric oven which is being heated through its heating element, as shown in Fig. 2.5. The energy content of the oven increases owing to the heating process. The energy transfer to the oven is caused by negatively charged particles called electrons crossing the System boundary

Heating element Figure 2.5 An electric oven. system boundary and not by a temperature difference between the oven and the surrounding air. The electrons crossing the system boundary do electrical work on the system. In an electric field, electrons in a wire move under the effect of electromagnetic forces, doing work. When N coulombs of electrons move through a potential difference V the electrical work done is kkiec = VN (1d) This can be expressed in rate form as = VI (kW)

;tie,

where is the electric power and I is the number of electrons flowing per unit time, that is, the current. Generally, both V and / vary with time, and the electrical work done during time interval & is given by Wele, = r VI dt (kJ)

(2.3)

Energy and the First Law of Thermodynamics 27 For the case when both V and I remain constant during At, the work becomes Wek, = VIAL (kJ)

2.3.3 Mechanical Work There are several ways of doing work, each in some way related to a force acting through a distance. From elementary mechanics, we know that the work done by a constant force F on a body which is displaced through a distance s in the direction of the force is given by W= Fs (kJ)

(2.4)

If the force F is constant, the work done is obtained by adding (i.e. integrating) the differential amounts of work (force times the differential displacement ds). That is,

2

W = f Fds (kJ)

(2.5)

In many thermodynamic problems, mechanical work is the only form of work involved. It is associated with the movement of the boundary of a system, or with the movement of the entire ystem as a whole.

2.3.4 Moving Boundary Work One form of mechanical work frequently encountered in practice is associated with the compression or expansion of a gas in a piston-cylinder device. This work is called the moving boundary work, or simply the boundary work. Let us consider the moving boundary work for a quasi-equilibrium process, a process during which the system remains in equilibrium at all times. The quasi-equilibrium process is closely approximated by actual engines, especially when the piston moves at low velocities. Examine the gas enclosed in the piston-cylinder arrangement shown in Fig. 2.6. The initial pressure is P, the total volume is V and the cross-sectional area of the piston is A. If the piston

i747.47137:174':

Moving boundary Figure 2.6 A piston-cylinder device. is allowed to move a distance ds in a quasi-equilibrium manner, the differential work done during this process is SW = Fds = PAds = PdV where the subscript b refers to boundary work. The total boundary work done during the entire

28 Fundamentals of Engineering Thermodynamics process as the piston moves is obtained by adding all the differential works from the initial state to final state. Therefore, Wb=

P dV

(2.6)

This integral can be evaluated only if the functional relation between P and V during the process is known. That is, P = f(V) should be known. Note that P = f(V) is simply the equation of the process path on a P—V diagram. Consider the quasi-equilibrium expansion process shown in Fig. 2.7. The differential area dA is equal to PdV, which is the differential work. The total area A under the process curve 1-2 is obtained by adding these differential areas. Therefore, = 1.2 = 1.2 A dA PdV (2.7)

Process path

Vi

,

dV

Figure 2.7 The P—V diagram for a quasi-equilibrium expansion process. Thus, the area under the process curve on a P—V diagram is equal in magnitude to the work done during a quasi-equilibrium expansion or compression process of a closed system. It is essential to note that the boundary work done during a process depends on the path followed as well as on the end states. If work were not a path function, no cyclic devices, such as automobile engines, power plants, etc. would operate as work-producing devices. The work produced by these devices during one part of the cycle would have to be consumed during another part, and there would be no net work output. Examine the P—V diagram of a cyclic process shown in Fig. 2.8. The net work done during the cycle is the difference between the work done by the system and the work done on the system. The cycle shown in Fig. 2.8 produces a net work output because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B). If the relationship between P and V during an expansion or a compression process is given in terms of experimental data instead of a functional form, obviously we cannot perform the integration analytically. But we can always plot the P—V diagram of the process, using these data points, and calculate the area underneath graphically to determine the work done. The use of the relation Wb= .112 P dV is not limited to the quasi-equilibrium processes of gases only. It can be used for solids and liquids as well.

Energy and the First Law of Thermodynamics 29

P

"2

1/1

Figure 2.8 The P—V diagram of a cyclic process.

2.3.5 Polytropic Process A process for which the pressure and volume are related by Pr = c, where n and c are constants, is called a polytropic process. The index n is called the polytropic index for the process. During expansion and compression, the real gases usually follow the polytropic process. A general expression for the work done during a polytropic process can be developed as follows: Examine a piston-cylinder device and the P—V diagram of the process shown in Fig. 2.9.

1 PV" = constant

P2

-.1

V "2 Figure 2.9 A system and its P—V diagram for a polytropic process. The pressure for a polytropic process can be expressed as

p = cir Substituting this relation into Eq. (2.6), we get Wb

2 PdV

=

r cv—

V

-n+I —

dV — c 2

-n+I

—n +1

Or

PV — PV 11 Wb _ 2 2 — n

(2.8)

30

Fundamentals of Engineering Thermodynamics

EXAMPLE 23 A piston-cylinder device with air at an initial temperature of 303 K undergoes an expansion process for which pressure and volume are related as given below: P (bar) V (m3)

1.0 0.1

0379 02

0.144 0.4

0.081 0.6

Calculate the work done by the system. Solution The problem can be solved in the following two ways: 1. Using the relation W = )Pd V, assuming the average pressure between each step to be the representative pressure. 2. Using the relation W =

P2V2 — PV 1 1, for which the polytropic index n should be 1— n

known. Method 1 W = JPd V —

[(I +0.379)(0 2 01)+ (0.379 + 0.144) (0 4 — 0 2) 2

(0.144+ 0.081) (0.6 — 0.4)]105 2 14. 375 kJ Method 2 PV 2 2— PV 11 , it can be seen that the process obeys the 1— n relation PV1.4 = constant. Thus, n = 1.4. Therefore,

On inspection of the relation W —

W^

(0.081 x 105X0.6) — (1 x 105X0.1) — 0.4 12.85 kJ

Note: The first method gives a result which is different from the correct value obtained from the second method. This is because of the crude assumption of constant pressure during each step. The first method in which pressure is assumed to be constant during each step is also called quadrature method or trapezoidal rule. EXAMPLE 2.4 Three grams of nitrogen gas at 6 atm and 160°C in a frictionless piston-cylinder device is expanded adiabatically to double its initial volume, then compressed at constant pressure to its initial volume and then compressed again at constant volume to its initial state. Calculate the net work done on the gas. Draw the P—V diagram for the processes.

Energy and the First Law of Thermodynamics 31

Solution The P-V diagram for the processes is shown in the figure below. p

32 V For nitrogen, the gas constant R =

VI =

8314 - 296.93 J/(kg K). 28

mRTI 0.003 x 296.93(273.15 + 160)

=

6 x 101,325

= 6.346 x 10 1 m3 Therefore, V2 = 2V1 = 12.692 x 10-4 m3 u P2 = (

1.4 1 = 6 x 101,325 (-) , since n = 1.4 for nitrogen. 2

V2

= 230,370 Pa = (2.3037 x 105) Pa The work done during the processes is given by W12

P2 V2 - P V1 _ (23.037 x 12.692) - (60.795 x 6.346) 1-n - 0.4 = 233.54 J

r3 W23 - j, Pd V = -P2 (V3 - V2)

- -2.3037 x 105(6.346 - 12.692)10-4 = 146.19 J = 0, since dV = 0 Thus, the total work done on the gas is W = W12 + W23 + W31 = 233.54 + 146.19 379.73 J

32 Fundamentals of Engineering Thermodynamics EXAMPLE 25 Air in a cylinder at an initial volume of 0.01 m3 and initial pressure of 6 MPa expands following a quasi-static process given by P VIA = constant. If the final volume of the gas is 0.025 m3, determine the work done by the gas. Solution The system in this problem is air inside the cylinder. Since the process is quasi-static, the work done is given by W12 =

1

PdV

Given that PV' A ='constant. Therefore, constant P= v1.4

pv1.4

p v1.4

—

22

v1.4

The work expression becomes W12

2dV = constant ji v1.4

—

constant (V-0. 4 — 1 — 1.4

-0.4 ) vi

— 6 x 106(0.01)" (0.02$-°A — 0.01) -6A — 0.4 46.12 kJ

2.3.6 Gravitational Work The gravitational work can be defined as the work done by or against the gravitational force field. In a gravitational field, the force acting on a body of mass m is F = mg

(2.9)

where g is the acceleration of gravity. Then the work required to raise this body from elevation zi to z2 is

wg = r F dz = mg Jr: dz = mg(z2 — zi ) (kJ)

(2.10)

The potential energy of a system increases when gravitational work is done on it. EXAMPLE 2.6 A hydraulic turbine receives water from a reservoir which is at an elevation of 200 m. Find the maximum mass flow rate of water required to produce a steady turbine output of 30,000 kW. Take g = 9.81 m/s2.

Energy and the First Law of Thermodynamics

33

Solution The gravitational work is given by Wg = mg(z2 —z1) Thus,

g(z2 — g z1 )

=

30,000 x 103 , „,, , 9.81 x 200 -= .11 LYV g/s

The required mass flow rate of water is therefore 15,290 kg/s

2.3.7 Acceleration Work The acceleration work may be defined as the work associated with a change in velocity of a system. By Newton's second law, F = ma where F is the force, m the mass of the system and a the acceleration. But a in terms of velocity V is dV a= — dt Therefore, dV F = m— dt The velocity V in terms of displacement s is = ds — dt or ds = Vdt Thus, acceleration work W, can be written as dV Rio = 12 F ds = I2 (M —(Vdt) i dt 2 = V dV mf = _1 (v22 2 ) (a) m

(2.11)

vi

From Eq. (2.11) it is seen that the work done to accelerate a body is independent of path and is equivalent to the change in the kinetic energy of the body. The acceleration work done per unit time W. is referred to as power rating.

34 Fundamentals of Engineering Thermodynamics EXAMPLE 2.7 Determine the power required to accelerate a 1200 kg car from rest to 70 kmph in 30 s on a level road. Neglect friction, rolling resistance and other resistances. Solution The acceleration work is given by _ w.. _1 m t v22 —Vi_ 2

where Vt and V2 are the initial and final velocities, respectively, and m is the mass of the car. Therefore, WQ

=1(1200)[1 7°'12 02 ] 2 300 6 = 226.85 kJ

The average power required is determined from

: ,774t —

226.85 — 7.56 kW 30 7560 746

10.13 hp

EXAMPLE 2.8 The mass of a car with passengers is 2000 kg. What is the minimum power rating required to accelerate the car from rest to a speed of 90 kmph? Solution The air drag, friction, and rolling resistance can be neglected, since the energy asked for is the minimum energy. The acceleration work is given by

wa

_1 m (V2 _ ..2) 21 2

= - (2000)

2 90 ) — 02 = 625 kJ 3.6

Thus, the required power rating is W, = 625 kW Note:

Since the time to accelerate to 90 kmph is not given, the problem is solved taking t = I second.

EXAMPLE 2.9 The power rating of the engine of a 1200 kg automobile is 70 kW. Determine the time required to accelerate the vehicle from rest to 60 kmph at full power on a level road. Neglect the air drag and friction.

Energy and the First Law of Thermodynamics

35

Solution Acceleration work is given by ,. „ ,„ = 22 rya 2 n"

vi2) = 1200 [( 60 )2

2

3.6

_2 ]

0

= 166.67 kJ

Power rating It = 70 kW. Therefore, the time required is given by At —

W. 141.

166.67 70

2.381 s

2.3.8 Shaft Work In engineering practice, it is common to transmit energy using a rotating shaft. Most often, the torque r applied to the shaft is constant, implying that the force F applied is also constant. For a specific constant torque, the work done during n revolutions is determined as follows: A force F acting through a moment arm r generates a torque r, which is r = Fr Or

F=

r

The force F acts through a distance s which is related to the radius r by s = (2xr)n. Then, the shaft work is determined as r Wsh = Fs = (— ) (2x r n) = 2x nz (kJ)

(2.12)

The power transmitted through the shaft is the shaft work done per unit time, which can be expressed as (2.13) = 2/tri z (kW) where n is the number of revolutions per unit time. EXAMPLE 2.10 Determine the torque applied to the shaft of a turbine which transmits 275 kW and rotates at a rate of 3600 rpm. Solution The shaft work done, given by Eq. (2.13), is Wsh = brirr

Therefore, T = WI, brit

275 x 103 2/03600/60)

729.5Nm

36

Fundamentals of Engineering Thermodynamics

EXAMPLE 2.11 If a shaft transmitting 350 hp rotates at a rate of 3200 rpm, determine the torque applied to the shaft. Solution The shaft work done is given as gish = 21trir Here, IVA = 350 bp = (350 x 746) W = 261.1 kW Therefore,

r—

2p ri

—

261.1x103 2p (3200/60)

779.2 N

2.3.9 Spring Work When the length of a spring changes by a differential amount dx under the influence of a force F, the work done is given by Fdx 8 =

(5W

To determine the total spring work, the functional relation between F and x should be known. For linear elastic springs, the displacement x is proportional to the force applied. That is, F = kx (kN)

(2.14)

where k is the spring constant (kN/m). Therefore, the spring work for a linear elastic spring is given by — 1 k(.4 — x?) ( d) Wig =2

(2.15) , •

where xi and x2 are the initial and final displacements of the spring, respectively. EXAMPLE 2.12 Determine the work required to deflect a linear spring by 300 mm from its rest position. The spring constant is 300 kN/m. Solution The spring work is given by 2

2

Wing = — k(x — x ) 2

1 =-x 300(0.32 — 02 ) = 13.5 kJ 2

Energy and the First Law of Thermodynamics

37

2.3.10 Work Done on Elastic Bars Since solids elongate or contract under the action of a force, they are usually modelled as linear springs. Examine the solid bar under the influence of a force, as shown in Fig. 2.10. As long as the force applied is within the elastic limits, the bar will return to its original length, like a spring, when the force is lifted. Therefore, the equations applicable to linear springs can also be used for elastic solid bars.

Figure 2.10 A solid bar subjected to tension. We can also determine the work associated with the expansion or contraction of an elastic solid bar by replacing pressure P by normal stress an = F/A in the boundary work Eq. (2.6). Thus, Weins „c = r a dV = r a„A dx i

(2.16)

where A is the cross-sectional area of the solid bar.

2.3.11 Work Done in Stretching a Liquid Film Examine the liquid film suspended on a wire frame, as shown in Fig. 2.11. From experience we know that we need to apply some force, even though very small, to stretch this film using the solid wire shown. This force is required to overcome the microscopic forces between molecules at the liquid—air interfaces. These small forces are normal to any line in the surface. The net force generated by these forces per unit length is called surface tension as, having the units N/m. Thus, the work associated with the stretching of a film is called the surface tension work. It can be expressed as If surtace = f6S dA

(kJ)

Wire frame

b

Figure 2.11 Stretching of a liquid film.

(2.17)

38 Fundamentals of Engineering Thermodynamics where dA = 2bdx is the change in the area of the film surface. The force required to stretch the film through a distance dx is given by F = 2bthca5.

2.3.12 Thermodynamic Definition of Work From our discussions so far, we have learnt that work and heat are the only two mechanisms by which the energy of a closed system can be changed. It is, therefore, essential to have a clear understanding of what work interaction is and what heat interaction is. Let us have a close look at the idea of work. The concept of work has its origins in the study of mechanics, where work is defined as the product of a force and the distance through which this force acts. In thermodynamics, the force and the distance involved are sometimes not so easily recognized. To give a broader interpretation, we shall define work as follows. Work is energy transferred, without transfer of mass, across the boundary of a system, and work is said to be done by a system on its surroundings if the sole effect external to the system could be the raising of a weight. To illustrate, let us consider the expansion of a gas inside a cylinder. As we see from Fig. 2.12, through the proper selection of linkages, we can make use of the expansion process to lift a weight. Therefore, work will be done by the gas.

Piston-cy inder arrangement

Figure 2.12 Lifting of a weight due to work done by a gas in an expansion process. It is important to note that work, as defined, is energy in transit. Once this form of energy crosses the boundary of a system, it "disappears" and becomes part of the energy content of the system or the surroundings, as the case may be. Therefore, we cannot speak of a system having a certain amount of work at a given state. In other words, the quantity of work is not a thermodynamic property. Work interaction is possible only when a change of state occurs.

2.3.13 Nonmechanical Forms of Work So far we have seen the mechanical forms of work. But in practice we also encounter some work modes which are not mechanical in nature. For example, electrical work where the force is the voltage (the electrical potential) and the displacement is the electrical charge. and magnetic work where the force and displacements are magnetic field strength and magnetic dipole moment, respectively, are some of the nonmechanical work modes. However, all these work

Energy and the First Law of Thermodynamics

39

modes can be treated in a similar manner by identifying a generalized force F acting in the direction of a generalized displacement x. Then the work associated with the displacement dx under the action of the force F is given by SW = Fdx.

2.4 THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics, also known as the principle of conservation of energy, provides a firm basis for a study of the relationships among the various forms of energy and energy transformations. Based on experimental observations, the first law of thermodynamics states that, energy can neither be created nor be destroyed, it can only change forms. It is not possible to prove the first law mathematically. However, no process in nature is known to have violated the first law of thermodynamics, and this should be considered as sufficient proof. For a closed system or a fixed mass, the first law may be expressed as (Net energy transferred to (or from) the = Net increase (or decrease) in the total system as heat and work energy of the system Or

Q — W = LIE (kJ)

(2.18)

Q — W = AU + AKE + APE

(2.19)

Or

Most closed systems encountered in practice are stationary, that is, they do not involve any change in their velocity or elevation of their centre of gravity during a process. Thus for stationary closed systems, the changes in ICE and PE are negligible, that is, AKE = 0 and APE = 0, and the first law reduces to Q — W = AU (kJ)

(2.20)

EXAMPLE 2.13 The heat transferred into a closed system during a process is 3000 kJ. If the work done by the system is 1600 kJ, determine the change in total energy of the system. Solution The energy conservation equation for the given system is Q— W = AE Therefore, AE = 3000 — 1600 1400 kJ This is the change in the total energy of the system.

Other forms of the first-law relation The first-law relation for closed systems can be written in various forms as follows:

40 Fundamentals of Engineering Thermodynamics For a general system Q — W = AE (kJ) For a stationary system Q— W= AU(kJ) Per unit mass basis for a general system q — w = Ae (kJ/kg) In differential form 8Q — 8 W = dE (Id) 8q — 8w = de (kJ/kg) In rate form Q—

=

dE (kW) dt

For a cyclic process, the initial and final states are identical, therefore, A E =E2-E1 = 0 Then the first law for a cyclic process simplifies to Q — W = 0 (IJ)

(221)

That is, the net heat transfer and the net work done during a cyclic process are equal. As energy quantities, heat and work are not that different. It is, therefore, natural to ask the question, "Why should we keep heat and work as two separate quantities, after all, the change in the energy content of a system is equal to the amount of energy that crosses the system boundary, and it makes no difference whether the energy crosses the boundary as heat or work?" It seems as if the first-law relations would be much simpler if we had just one quantity which we could call energy interaction to represent both heat and work. In fact, from the first law point of view, heat and work are not different at all. But from the second law point of view, heat and work are very much different. EXAMPLE 2.14 A certain quantity of air is compressed inside a cylinder. The change in internal energy of the air is 10 kJ; work required for the compression process is 500 kJ. What is the amount of heat transfer involved? Solution By the first law we have for the given process, Q — W = AU, since AKE = 0 and APE = O. Thus, Q=

+ W= 10 — 500 = — 490 kJ

Note that W here is negative as per our sign convention.

Energy and the First Law of Thermodynamics 41 EXAMPLE 2.15 During a working stroke, an engine rejects 200 kJ/kg of heat of the working substance. The internal energy of the working substance also decreases by 350 kJ/kg. Determine the work done by the engine. Solution Treating the working substance as our system, we can write the energy conservation equation for the process as Q— W = AU On unit mass basis, we have q — w = Au or —200 — w = — 350 Or

w = 150 kJ/kg Therefore, the work done by the engine is 150 kJ/kg

2.5 SPECIFIC HEATS The specific heat of a substance is defined as the energy required to raise the temperature of a unit mass of the substance by one degree. In general, this energy will depend on how the process is executed. In thermodynamics, we are interested in two kinds of specific heats, namely the specific heat at constant volume C„ and the specific heat at constant pressure Cp. Physically, the C,, can be viewed as the energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. The energy required to do the same as the pressure is maintained constant is the specific heat at constant pressure Cp. The Cp is always greater than C, because at constant pressure the system is allowed to expand and the energy for this expansion work must also be supplied to the system. Both C, and C,, can be expressed in terms of thermodynamic properties. First, consider a stationary closed system undergoing a constant-volume process (wb = 0). The first law relation for the process can be expressed in the differential form as 8q — 8w her = du In the above equation, the left-hand side (8q — SW r) represents the amount of energy transferred to the system in the form of heat and/or work. From the definition of C„, this energy must be equal to C, dT, where dT is the differential change in temperature. Thus, C„ dT = du at constant volume or

au

C ---(al,

(222)

42

Fundamentals of Engineering Thermodynamics

Now, let us consider a constant-pressure process. For this process, wb + Au = Oh, where h is called enthalpy per unit mass of the system (the enthalpy will be discussed in Section 2.6). Take the two independent variables to be P and T, so that

h = h(T, P) or

dh=r T ) dT +( al dP OP r a P For a constant-pressure process, we have

dh

ah — ) dT =Cp dr

=( ar

p

Thus,

Oh) OT )13

Cp

(223)

It is essential to note that both Cv and Cp are expressed in terms of other properties, thus they must be properties themselves. Like any other property, the specific heats of a substance depend on the state which, in general, is specified by two independent intensive properties. That is, the energy required to raise the temperature of a substance by one degree will be different at different temperatures and pressures. But this difference is usually not large. The relations C,, = (

ah

T

and C,. =(— ) are property relations and as such are indev

ar

P

pendent of the type of processes. They are valid for any substance undergoing any process. The C„ may also be recognized as the change in the specific internal energy of a substance per unit change in temperature at constant volume. That is, C,, is a measure of the variation of

internal energy of a substance with temperature. The Cp may also be defined as the change in the specific enthalpy of a substance per unit change in temperature at constant pressure. That is, Cp is a measure of the variation of enthalpy of a substance with temperature. Both internal energy and enthalpy of a substance can be changed by the transfer of energy in any form, with heat being just one of them. Therefore, the term specific energy is more appropriate than the term specific heat, which implies that energy is transferred (and even stored) in the form of heat. A common unit for specific heat is kJ/(kg°C) or kJ/(kg K).

2.6 THE INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES An ideal gas is a substance whose temperature, pressure, and specific volume are related by

Pv = RT

(2.24)

Equation (2.24) is the popular ideal-gas state equation. It has been demonstrated mathematically

Energy and the First Law of Thermodynamics

43

and experimentally (Joule. 1843) that for an ideal gas the internal energy is a function of ►he temperature only. That is, u = u(T) (2.25) Using the definition of enthalpy and the ideal-gas state Eq. (2.24), we have h = u + Pv Pv = RT Combining these two relations, we obtain h u + RT Since R is a constant and u = u(T), it follows that the enthalpy of an ideal gas is also a function of temperature only. Thus, h = h(T) (2.26) Since u and h depend only on temperature for an ideal gas, C,, and Cp also depend, at most, on temperature only. Therefore, at a given temperature, u, h, C, and Cp of an ideal gas will have fixed values regardless of the specific volume or pressure. Thus, for an ideal gas the partial derivatives in C,, and C1 relations can be replaced by ordinary derivatives. Then the differential changes in the internal energy and enthalpy of an ideal gas can be expressed as du = Cv(T)dT (2.27) and dh = Cp(T)dT (2.28) The change in u and h for an ideal gas during a process from state 1 to state 2 is determined by integrating Eqs. (2.27) and (2.28), respectively. Thus, we have Au = u2 — u1 = tc(T)dT (kJ/kg)

(2.29)

A h = h2 — h, = fCp(T)dT (kJ/kg)

(2.30)

To carry out these integrations, the C, and Cp relations as functions of T are required to be known. At low pressures all real gases approach ideal-gas behaviour, and therefore, their specific heats depend on temperature only. The specific heats of real gases at low pressure are called ideal-gas specific heats, or zero-pressure specific hews, and are often denoted as Cpo and C',.0. Accurate analytical expressions for ideal-gas specific heats, based on direct measurements or calculations from statistical behaviour of molecules, are available in literature for several gases. A plot of C, (T) data [specific heat with a bar stands for specific heat on molar basis and has units kJ/(kmol K)] for some common gases is given in Fig. 2.13. From Fig. 2.13 it is seen that: • The specific heats of gases with complex molecules (molecules with two or more atoms) are higher and increase with temperature. • The variation of specific heats with temperature is smooth and may be approximated as linear over small temperature intervals (a few hundred degrees or less). • Assuming the linear relationship as above, the specific heat functions can be replaced by the constant average specific heat values. Now the integrations of Eqs. (2.29) and (2.30) can be performed, yielding.

44 Fundamentals of Engineering Thermodynamics

CO2

60

50 Z 0 •E

40

IC.) 30

Ar, He, Ne, Kr, Xe, Rn 20 0 Temp., K Figure 2.13 Variation of E po(T) with temperature. u2 - u1 = G.,(T2 - Ti) (kJ/kg) h2 - hi = Cp.,(T2 - TO (kJ/kg) • The ideal-gas specific heats of monatomic gases such as argon, neon, and helium remain constant over the entire range of temperature. From our above discussion on specific heats, it is evident that:

1

• The Au and Ah relations [Eqs. (2.29) and (2.30)] are not restricted to any one kind of process. They are valid for all processes. • The presence of G in an equation does not mean that the equation is valid for a constantvolume process only. On the contrary, the relation Au = G AT is valid for any ideal gas undergoing any process. • Similar argument as for G and Art is also valid for Cp and Ah.

Specific-heat relations for ideal gases The enthalpy relation for ideal gases is h = u + RT

!

Eaergy and the First Law of Thermodynamics 45 where h is the specific enthalpy and R is the gas constant. In differential form, we have dh = du + RdT R9lacing dh and du, by Cp dT and C,, dT respectively and dividing by dT, we obtain Cp = C,, + R[L1/(kg

(231)

When the specific heats are given on a molar basis, the gas constant R in Eq. (2.31) should be replaced by the universal gas constant R. That is,. Cp

= C, + R. [kJ/(knol K)]

(232)

At this stage, it is essential to note that the gas constant R is different for each gas and is determined from R=

[kJ/(kg K) or kPa m3/(kg K)]

where M is the molar mass (also called the molecular weight). The universal gas constant R. is the same for all substances and its value in several units is kJ/(1cmol K)

8.314

Ru

kPa m3/(kmol K) 8.314 0.08314 bar m3/(kmol K) Btu/(Ibmol R) 1.986 10.73

psia ft3/(Ibmol R)

1545.00

ftlbV(Ibmol R)

The molar mass M can thus be simply defined as the mass of one mole of a substance in grams, or the mass of one kmol in kilograms. At this point, we introduce another ideal-gas property called the specific heats ratio y, defined as Y=

Cp C,,

(233)

The specific heats ratio y also varies with temperature, but this variation is very small. For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heats ratio of about 1.4 at room temperature. EXAMPLE 2.16 Determine the values of Cp and C,, for air at 300 K. Solution At 300 K, for air, y= 1.4. By Eqs. (2.31) and (2.33), we have Cp =

8314 where R= = 287 J/(kg K) R and C,, = R y-1 y —1' 28.97

Thus, Cp =

1. 4 x 287 = 1004.5 J/(kg K) 0.4

46

Fundamentals of Engineering Thermodynamics

and ,.,

287 0.4

ll, = - =

717.5 J/(kg K)

2.7 THE INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS For liquids and solids, the specific volume essentially remains constant during a process. Therefore, they can be approximated as incompressible substances. Hence the energy associated with the volume change, such as the boundary work, is negligible compared to other forms of energy. It can be mathematically shown (Chapter 13) that C„ and Cp are identical for incompressible substances. Therefore, for solids and liquids, both C,, and Cp can be represented by just C without any subscript. That is, (2.34) Cp = c = C The specific heats for solids and liquids depend on temperature only. Thus, the partial differentials in the defmition of C,, [Eq. (2.22)] can be replaced with ordinary differentials, which gives du = C,,dT = C(T)dT

(235)

The change in internal energy between states 1 and 2 is then obtained as 2

Au = u2 — u1 = ji C(T)dT (kJ/kg)

(236)

To perform the integration in Eq. (2.36), the variation of C with temperature should be known. When the temperature interval is small, the C value at the average temperature can be used and treated as a constant, giving Au = C,,,,(T2 — TO (kJ/kg) (237) The change in enthalpy of liquids or solids between states 1 and 2 of a process can be determined from the defmition of enthalpy (h = u + Pv) to be h2 — hi = (u2 — u1 ) + v(P2 — P1 )

(238) I

since v1 = v2 = v for incompressible substances. Equation (2.38) can also be written as Ah = Au + vAP (kJ/kg)

I

(239)

In Eq. (2.39), the term (vAP) is generally small compared to the term Au. EXAMPLE 2.17 A 0.6 kg copper piece at 100°C is dropped in an insulated rigid tank which contains 0.75 kg of liquid water at 25°C. Determine the temperature of the system when thermal equilibrium is established. Take specific heats of copper and water to be 0.393 kJ/(kg K) and 4.184 kJ/(kg K), respectively. Solution Let us take the copper piece and the water as our system. Since the tank is insulated, Q =

Energy and the First Law of Thermodynamics 47

Also, since there is no movement of the system boundary, thus Wb = 0. Further, there is no other work present, therefore, W,hers = 0. Then the energy conservation equation for this process reduces to Q — W = 0 = AU That is, AUK = AU + =0 Or

[mC(T2 70lopper + [mCqi — 7i Myrna! = 0 The specific heats of copper and water are, respectively C

= 0.393 kJ/(kg K)

C,,,„ter = 4.184 kJ/(kg. K) Therefore, 0.6 (0.393XT2 — 100) + 0.75(4.184XT2 — 25) = 0 or

= 30.24 °C The temperature of the system when thermal equilibrium is established is 30.24°C. That is, both water and copper will be at 30.24°C. EXAMPLE 2.18 The temperature of 3.5 kg of gas in a rigid container is increased from 22°C to 39°C by heating it. The heat transferred during the heating process is 25 kJ. The specific heats ratio and the molar mass of the gas are 1.4 and 28, respectively. Calculate the work done and the change in internal energy for the gas, treating the gas to be a perfect gas. Solution The process given is a constant volume process. For the control mass system, which is the mass of the gas in the tank, AKE = 0 and APE = 0. Therefore, the first law becomes Q — W = AU = U2 —

U1 =

MC4T2 — T1 )

where the subscripts 1 and 2 refer to the initial and final states. For perfect gases, we have U = mC,T For the given gas, C„ =

y -1

814 and R = — = 296.93 J/(kg K) 28

Therefore, C, —

296.93 0.4 — 742.325 J/(kg K)

Thus, Q— W = 3.5 x 742.325 (39 — 22) = 44.17 kJ Therefore, W = Q — AU = 25 — 44.17 kJ= —19.17 kJ

48 Fundamentals of Engineering Thermodynamics Change in internal energy, AU = 44.17 kJ Note: Even though the change in volume, dV = 0, the work done is not zero. This implies that the process is irreversible.

EXAMPLE 2.19 In a piston-cylinder device, 300 g of saturated water vapour, maintained at 200 kPa, is heated by a resistance heater installed within the cylinder for 10 min by passing a current of 0.35 ampere from a 220 V source. The heat loss from the system during the heating process is 2.2 Id. Calculate the work done and the final temperature of the steam. Solution The water vapour and the electrical resistance wire form the system. Since there is no mass entering or leaving the system, it is a closed system. Neglecting the changes in PE and KE, the first-law relation for the system can be written as Q — W = AU

where W = Wb + For a constant-pressure process, Wb = P(V2 — V1 ). Therefore, Q — wet. — PC V2 — V 1) = U2 U1

or Q — We„,= (U2 + P2V2) — (U1 + PIVI)

Note that the constant P is expressed as P2 in the first term and P1 in the second term of the RHS of the above relation. This is done to introduce enthalpy since H = U + PV. Thus, Q — Wee, = H2 — H1

That is, for a closed system undergoing a constant-pressure process, the boundary work Wb and the change in internal energy AU in the first-law relation can be combined into a single term, AH. The electrical work given by Eq. (2.3) is Wdec = V/ At = 220 x 0.35 x 600 = 46.2 Id Thus, the work done Wei = — 46.2 kJ The negative sign is introduced to indicate that work is done on the system. Given that, Q = — 2.2 kJ

Therefore, — 2.2 — (— 46.2) = m(h2 — h1) At P1 = 200 kPa, the enthalpy for saturated vapour (from Table 4 of the Appendix) is h1 = 2706.5 kJ/kg Thus, 44 = 0.3(h2 — 2706.5)

Energy and the First Law of Thermodynamics 49 Or h2 = 2853.16 id/kg For P2 = 200 kPa and h2 = 2853.16 id/kg, from Table 5 of the Appendix, we have T2 = 191.47°C EXAMPLE 2.20 Nitrogen at 25 atm and 37°C is contained in a cylinder of volume 10 cm3. The cylinder is placed in a large container of 150 times the cylinder volume. The container is perfectly evacuated and insulated. The nitrogen gas is allowed to discharge and fill the container. Calculate the final pressure after the system has reached equilibrium. Solution For the system, Q = 0 and W = 0. The first-law relation for the system becomes Q — W = 0 = AU= m(u2 — u1) But for N2, being a perfect gas, u = CST. Therefore, m c(T2 — T1 ) = 0 Or

T2 = TI By ideal-gas state equation, we have = PI — ljr

since 7'1 = T2

2

Or (

P2 = (150) =

0.167 atm

SUMMARY The first law of thermodynamics or the principle of conservation of energy implies that energy can neither be created nor be destroyed. In other words, "during an interaction between a system and its surroundings, the amount of energy gained by the system is exactly equal to the amount of energy lost by the surroundings". Heat is a form of energy that is transformed between two systems by virtue of a temperature difference between them, Energy is a property, but heat is not a property. Energy is associated with a state, but heat is associated with a process. A process during which there is no heat transfer is called an adiabatic process. Work is an energy interaction between a system and its surroundings. Like heat, work is also a form of energy. There are several ways of doing work, each in some way related to a force acting through a distance. In many thermodynamic problems, mechanical work is the only form of work involved. It is associated with the movement of the boundary of a system, or with the movement of the entire system as a whole.

50 Fundamentals of Engineering Thermodynamics The moving boundary work or simply the boundary work is a form of mechanical work frequently encountered in practice and is associated with the compression or expansion of a gas in a piston-cylinder device. The gravitational work can be defined as the work done by or against the gravitational force field. The acceleration work may be defined as the work associated with a change in velocity of a system. From our discussion on heat and work, we conclude that work and heat are the only two mechanisms by which the energy of a closed system can be changed. For a broader interpretation, we can define work as follows: Work is energy transferred. without transfer of mass, across the boundary of a system, and work is said to be done by a system on its surroundings if the sole effect external to the system could be the raising of a weight. Work as defined above is energy in transit. Once this form of energy crosses the boundary of a system, it "disappears" and becomes part of the energy content of the system or the surroundings, as the case may be. Therefore, we cannot speak of a system having a certain amount of work at a given state. In other words, the quantity of work is not a thermodynamic property. Work interaction is possible only when a change of state occurs. The first law of thermodynamics, also known as the principle of conservation of energy, provides a firm basis for studying the relationships among the various forms of energy and energy transformations. For a closed system or a fixed mass, the first law may be expressed as Q — W = AE (kJ) or Q — W = + AKE + APE For stationary closed systems, the first law reduces to Q — W = AU (kJ) For a cyclic process, the first law simplifies to Q — W = 0 (1(.0 That is, the net heat transfer and the net work done during a cyclic process are equal. The specific heat of a substance is defined as the energy required to raise the temperature of a unit mass of the substance by one degree. In thermodynamics, we are interested in two kinds of specific heats, namely the specific heat at constant volume CE and the specific heat at constant pressure Cp. They are defined as

C,

au

— =( 67' 1.

cp - (ah

Energy and the First Law of Thermodynamics 51

The ideal-gas state equation is Pv = RT For an ideal gas, the internal energy and enthalpy are functions of temperature only. At low pressures, all real gases approach ideal-gas behaviour. The specific heats of real gases at low pressure are called ideal-gas specific heats, or zero-pressure specific heats. The relation between specific heats for ideal gases is Cp = C, + R [kJ/(kg K)] The gas constant R is determined from R=

[kJ/(kg K) or kPa m3/(kg K)]

where M is the molar mass (also called the molecular weight). The universal gas constant same for all substances and its value is 8.314 kJ/(kmol K). The specific heats ratio y is defined as Y=

is

Cp CV

For incompressible substances such as solids and liquids, it can be mathematically shown that Cy and Cp are identical. The specific heats for solids and liquids depend on temperature only.

PROBLEMS 2.1 A system is moved through a distance of 100 m by the application of a 2 kN force. Determine the work done by the force. [Ans. 200 kJ] 2.2

An elevator, having a mass of 9000 kg is to be raised a distance of 100 m at a location where the gravitational acceleration is 9.81 m/s2. What is the minimum work required for this? [Ans. 8829 kJ]

2.3

A hydraulic turbine in run by water from a reservoir at an elevation of 100 m above it. If the turbine generates 50 MW, determine the mass flow rate of water released from the reservoir. [Ans. 50968.4 kg/s]

2.4 The mass of an automobile and its occupants is 4000 kg. What is the minimum energy required to accelarate it from rest to a speed of 100 kmph? [Ans. 1543.2 kJ] 2.5 A linear spring has a spring constant of 150 kN/m. How much work is required to stretch it from its free length of 1.2 m to a length of 2.5 m? [Ans. 360.75 kJ] 2.6 A 12 volt battery delivers current at the rate of 40 Ah. In 30 min, the heat lost from the battery is 90 Id. What is the change in the internal energy of the battery for the discharging period? [Ans. —954 kJ]

52

Fundamentals of Engineering Thermodynamics

2.7 A spherical balloon contains 5 kg of air at 200 kPa and 500 K. The pressure inside the balloon is always proportional to the square of the diameter. Determine the work done when the volume of the balloon doubles as a result of heat trransfer. Also, determine the final temperature. [Ans. 901.5 kJ, 1489 K] 2.8

The stored energy of a closed system decreases by 5 Id while 50 Id of work is done on to it. Determine the amount of heat transfer associated with this process. [Ans. —55 kJ]

2.9 One kilogram of a fluid undergoes a quasi-static process inside a cylinder at a constant pressure of 250 kPa. If 7 Id of heat is removed while its volume changes from 0.5 m3 to 0.2 m3, determine the change in (a) the internal energy and (b) the enthalpy of the fluid. [Ans. (a) 0, (b) —75 kJ] 2.10 The normal stress a acting on an elastic bar when it is stretched is given by a=cT(L-1) L2 where c = 1.5 x 10-2 W/K is required to stretch the bar from its initial length of 1 m to 3 m at a contant temperature of 300 K. [Ans. 15 J/rn2] 2.11 A closed system undergoes a cycle consisting of two processes. During the first process, 40 Id of heat is transferred to the system while the system does 60 kJ of work. During the second process, 40 Id of work is done on the system. Determine (a) the heat transfer during the second process and (b) the net work and heat transfer for the cycle. [Ans. (a) —20 Id, (b) 20 Id, 20 kJ] 2.12 During the charging of a storage battery, the current is 20 ampere and the voltage is 12 volts. The rate of heat transfer from the battery is 12 W. Determine the net rate of increase in the internal energy of the battery. [Ans. 228 W] 2.13 Ten kilograms of saturated steam at 150 kPa is heated at constant pressure until the temperature reaches 400°C. Calculate the work done by the steam during this process. [Ans. 1.75 MJ] 2.14 A piston-cylinder device contains 0.1 m3 of a gas initially at 500 kPa. A spring exerts a force on the piston, which is proportional to the displacement of the piston. Heat is transferred to the gas, causing the piston to rise and compress the spring until the volume inside the cylinder doubles and pressure becomes 1 MPa. If the cross-sectional area of the piston is 0.2 m2, determine the spring constant. [Ans. 200 kN/m] 2.15 An ideal gas obeying PV = MRT is heated at constant volume until its temperature becomes twice its initial value. It is then cooled at constant pressure to its initial temperature. Draw the P—V diagram for the process and find an expression for W. [Ans. MRTI]

Energy and the First Law of Thermodynamics 53

2.16 Compressed air from a high pressure cylinder is released slowly into a large balloon. Emptying the cylinder fills the balloon to a volume of 25 m3. What is the work done by the compressed air if the atmospheric pressure is 101 kPa? [Ans. 2.525 MJ] 2.17 A frictionless piston-cylinder system has 150 kPa initial pressure of air in the cylinder. The piston is held in position by an external force F1 acting on the piston. The initial volume of 0.05 m3 is doubled by slowly reducing the external force, allowing the gas to expand. The expansion is found to obey the rule that the pressure is inversely proportional to volume. Find the work done by the expansion of air. [Ans. 7.5 In (2) kJ] 2.18 A spherical balloon of 0.5 m diameter contains air at a pressure of 500 kPa. The diameter increases to 0.55 m in a reversible process during which P is proportional to diameter. Determine the work done by the air in the balloon during this process. [Ans. 11.4 kJ] 2.19 In a frictionless piston-cylinder device, 5 kg of air at 200 kPa and 30°C is compressed to 400 kPa. During the process heat is transferred from the air such that the temperature inside the cylinder is kept constant. Determine the work done during the process and also sketch the P—V diagram for this process. [Ans. —301.53 kJ] 2.20 Nitrogen gas at 300 K, 101 kPa, and 0.1 m3 is compressed slowly in an isothermal process to 500 kPa. Calculate the work done during this process. [Ans. —16.15 kJ] 2.21 A gas expands in a cylinder according to the relation PV1-3 = constant, from an initial state of 0.3 m3 and 1000 kPa to a final state of 101 kPa. Calculate the work done on the piston by the gas pressure. [Ans. 410.83 kJ] 2.22 A piston-cylinder device shown below, contains a gas at 200 kPa and 0.0015 m3. At this initial state, the spring does not exert any force on the piston. Now the gas is heated to double its volume. The pressure at this state is 600 kPa. Draw the P—V diagram for the process. Also, calculate the work done by the gas. [Ans. 593.3 J]

2.23 Calculate the work that a crane has to do to lift a hollow metallic sphere of mass M and volume V through a distance s in a fluid of density p. Assume the motion to be slow and also neglect the frictional forces. [Ans. (M — pV)gs] 2.24 A cylinder of volume 0.1 m3 contains nitrogen gas at 101 kPa and 20°C. If 0.5 kg of nitrogen is now pumped into the cylinder, calculate the new pressure when the cylinder has returned to its initial temperature. The molar mass of nitrogen is 28 kg/kmol. Assume nitrogen to be a perfect gas. [Ans. 536.14 kPa]

54 Fundamentals of Engineering Thermodynamics 2.25 A perfect gas of mass 0.01 kg occupies a volume of 0.002 m3 at pressure and temperature of 286.4 kPa and 30°C, respectively. The gas is allowed to expand until the pressure is 101 kPa and the final volume is 4 times the initial volume. Calculate (a) the molar mass of the gas and (b) the final temperature. [Ans. (a) 44 kg/kmol, (b) 154.48°C] 2.26 A perfect gas has Cp = 1.039 kJ/(kg K) and C, = 0.742 kJ/(kg K). Calculate the gas constant and the molecular weight of the gas. [Ans. 297 J/(kg K), 28 kg/lanol] 2.27 A perfect gas of unit mass contained in a rigid vessel at 3 atm and 300°C is cooled until the pressure falls to 1 atm If the molar mass of the gas is 32 kg/lanol and y = 1.4, calculate the heat rejected. [Ans. —248.17 kJ/kg] 2.28 Calculate the internal energy and enthalpy of 1 kg of air occupying 0.03 m3 at 3 MPa. [Ans. 224.94 kJ/kg, 314.9 kJ/kg] 2.29 In an air compressor the pressures at the inlet and the exit are 1 atm and 4 atm, respectively. The inlet temperature is 20°C and the volume at the beginning of compression is 3 times that of at the end of compression. Calculate the temperature of air at the exit and the increase in internal energy per kg of air. [Ans. 118°C, 70.315 kJ/kg] 2.30 An insulated rigid tank contains 0.5 kg of argon gas at 30°C and 3 atm A paddle wheel with a power rating of 0.03 hp is operated within the tank for 45 minutes. Calculate the final pressure and temperature of the argon gas. [Ans. 694.2 kPa, 692.3 K] 2.31 A 10 kg copper block at 90°C is immersed into an insulated rigid tank containing 0.3 m3 of liquid water at 25°C. Calculate the equilibrium temperature of the system. [Ans. 25.2°C] 2.32 A piston-cylinder device contains unit mass of a fluid at 20 atm and 0.04 m3. The fluid is expanded reversibly according to the law PV I-5 = constant until the volume becomes two times that of its initial value. The fluid is then cooled reversibly at constant pressure until the piston comes back to its original position. Now, heat is supplied reversibly with the piston locked in position until the pressure rises to the original value of 20 atm Calculate the net work done by the fluid. [Ans. 18,842.06 J] 233 A steam turbine in a power plant develops 2000 kW. The heat supplied to the steam in the boiler is 3500 kJ/kg, the heat rejected by the steam to the cooling water in the condenser is 1800 kJ/kg. The feed-pump work required to pump the condensate back into the boiler is 7.5 kW. Calculate the mass flow rate of the steam. [Ans. 1.172 kg/s] 234 During the compression stroke, an internal combustion engine rejects 25 kJ/kg of heat to the cooling water. The work input to the stroke is 75 kJ/kg. Calculate the change in the specific internal energy of the working fluid. [Ans. 50 kJ/kg]

Energy and the First Law of Thermodynamics 55

2.35 An air tank of volume 3 m3 contains air at 10 atm and 35°C. A valve is opened and some air is allowed to escape from the tank to atmosphere. The tank pressure dropped sharply to 7 atm when the valve was closed. Calculate the mass of the air that escaped from the tank, assuming that the air mass in the tank undergoes a reversible adiabatic process. [Ans. 7.72 kg] 2.36 The internal energy of compressed air in a cylinder is 300 kJ/kg at the beginning of the expansion and 120 kJ/kg at the end of the expansion. If 90 kJ/kg of work is done during the expansion process, calculate the heat transfer associated with the process. [Ans. — 90 kJ/kg] 2.37 In an insulated vessel containing 5 kg of liquid water at 38°C, 1 kg of dry ice at 0°C is placed. Calculate the temperature of the water when the ice is completely melted. Take the specific latent heat of fusion of ice as 335 kJ/kg and the specific heat of water as 4.19 kJ/(kg °C). [Ans. 18.34°C] 2.38 Calculate the heat required to convert 2 kg of liquid water at 60°C into steam at 100°C at standard atmospheric pressure. Take the specific heat of water in the temperature range from 60 to 100°C to be 4.19 kJ/(kg °C) and the specific latent heat of vaporization to be 2.257 MJ/kg. [Ans. 4.8492 MJ]

CHAPTER

3 Thermodynamic Analysis of Control Volume

3.1 INTRODUCTION In Chapter 2, the first law of thermodynamics was applied to control mass systems. Now we extend the thermodynamic analysis, based on the first law, to control volume systems. Control volume analysis is suitable for numerous problems of practical interest. For instance, a water heater, a car radiator, a turbine, a compressor all involve mass flow and should be analysed as control volumes. In general, any arbitrary region, as that shown in Fig. 3.1, in space can be selected as a control volume (CV). The boundaries of a control volume are called control surfaces. The control , L_

—40. — Mass leaving _I Control volume (CV) Mass —ripentering —1 Figure 3.1 A control volume. surfaces can be real or imaginary. A control volume can be fixed in size and shape, or it may involve a moving boundary as illustrated in Fig. 3.2. In this chapter, we will be using the terms steady and uniform frequently. The term steady implies no change with time. The opposite of steady is unsteady or transient. The term uniform implies no change with location over a specified region. 56

Thermodynamic Analysis of Control Volume

57

Moving boundary CV

Figure 3.2 Control volume with a moving boundary.

3.2 CONSERVATION OF ENERGY PRINCIPLE The principle of conservation of energy for a control volume undergoing a process can be expressed as [ Total energy Total energy crossing the + of the mass CV boundary entering the CV as heat and work

—

Total energy Net change of the mass = in energy of leaving the CV the CV

or Q — W + ZEif, — EEout = AEcv

(3.1)

A control volume, like the system shown in Fig. 3.3, may involve one or more forms of work at the same time. The energy required to push fluid mass into and out of a control volume is called the flow work or flow energy. It is considered to be part of the energy transported during the process under consideration.

Moving boundary

.kkMkt.Wki,kkkkikkkkti r

Mass in

Mass out

Wsh

Figure 3.3 A control volume involving more than one form of work.

58

Fundamentals of Engineering Thermodynamics

3.2.1 Flow Work Examine the fluid element of volume V, as shown in Fig. 3.4. The force F owing to the fluid

Imaginary piston Figure 3.4 A fluid element entering a control volume. element pressure P, acting at the face of the imaginary piston is given by F = PA where A is the cross-sectional area of the piston. To push the entire mass of the fluid element into the control volume, the force F must act through a distance L. Thus, the work done in pushing the fluid element across the boundary, that is, the flow work is W

= FL = PAL = PV (kJ)

(32)

where AL = V, the volume of the fluid element. The flow work per unit mass is obtained by dividing both sides of Eq. (3.2) by the mass of the fluid element. wow, = Pv (kJ/kg)

(3.3)

It is important to note that, unlike other work quantities, flow work is expressed in terms of the properties. In fact, it is the product of two properties of the fluid, as seen from Eq. (3.3). For this reason it is viewed as a combination property (like enthalpy) and referred to as flow energy or convected energy or transport energy, instead of flow work.

3.2.2 Total Energy of a Flowing Fluid The total energy of a simple compressible system consists of three parts, namely the internal energy, the kinetic energy, and the potential energy. On a unit-mass basis, the total energy of a simple compressible system can be expressed as v2 e = u + ke + pe = u + — + gz (ki/kg) (3A) 2 where V and z are the velocity and elevation of the system, respectively. The fluid entering or leaving a control volume possesses an additional form of energy, namely the flow energy Pv. Then the total energy of a flowing fluid per unit mass (denoted by 9) becomes 0 = Pv+ e = Pv + u + ke + pe Or

0 = (u + Pv) + ke + pe

Thermodynamic Analysis of Control Volume 59 But u + Pv = h, the specific enthalpy. Therefore, 9

=

V2 h + ke + pe = h + — + gz (10/kg) 2

(3.5)

Equation (3.5) represents the energy of a fluid stream flowing into or out of a control volume. From now onwards, Eq. (3.5) will be used for the energy of the stream and no reference will be made to flow work or flow energy, in this book. Thus, the work term W in the control volume energy Eq. (3.1) will represent all forms of work: boundary work, shaft work, electrical work, etc. except flow work.

3.3 THE STEADY-FLOW PROCESS A large number of engineering devices such as nozzles, compressors, and turbines operate for long periods of time under same conditions, and are classified as steady-flow devices. Processes involving steady-flow devices can be represented reasonably well by a somewhat idealized process, called the steady-flow process. A steady-flow process may be defined as a process during which a fluid flows through a control volume steadily. That is, the fluid properties can change from point to point within a control volume, but at any fixed location they remain the same during the entire process. A steady-flow process is characterized by the following features: • The properties (intensive or extensive) within the control volume do not change with time. • The properties at the boundaries of the control volume do not change with time. • The heat and work interactions between the system and its surroundings do not change with time. Steady-flow conditions can be closely approximated by devices that are not intended for continuous operation, such as pumps, boilers, condensers, heat exchangers of steam power plants, turbines, etc.

Conservation of mass The principle of conservation of mass for a general steady-flow system with multiple inlets and outlets may be expressed in the rate form as ( Total mass entering the 1 = (Total mass leaving the control control volume per unit time) l volume per unit time ) or =

(3.6)

where the subscripts i and e refer to inlet and outlet, respectively. For a single-stream steadyflow system, Eq. (3.6) reduces to riti = ?he, or Pi AiVi = P2A2V2

(3.7)

60 Fundamentals of Engineering Thermodynamics where pi, A1 , VI , respectively are the density, area and velocity at the inlet and p,, A2, V2, respectively are the density, area and velocity at the exit.

Conservation of energy The principle of conservation of energy or the first law of thermodynamics for a general steadyflow system with multiple inlets and exits can be expressed as Total energy crossing Total energy transported the boundary as heat and = out of control volume with mass per unit time [ work per unit time

—

Total energy transported into control volume with mass per unit time

Or —

= Eke, — IAA

(3.8)

where 6 is the total energy of the flowing fluid, including the flow work, per unit mass. Equation (3.8) on combination with Eq. (3.5) results in

0 —iil =1 lime

he +

2. + gze )—E riti ( h. +-L V + gz.) (kW) 2 ' 2 '

(3.9)

This is the general form of the first law for steady-flow processes.

Special cases Consider a single-stream steady-flow system. Let the mass flow rate through the entire control volume remain constant (i.e. ?hi = rii2 ). For this system the energy conservation Eq. (3.9) becomes Vi2 + hi + V; — [ h2 — g(z2 — zi )] (kJ/kg) (3.10) 2 — W = th where the subscripts 1 and 2 stand for inlet and outlet conditions. Dividing Eq. (3.10) by m, we obtain q — w = h2 —

+

v — v2 22 1

2

g(z2 — zi ) (kJ/kg)

(3.11)

If , the fluid experiences negligible change in its kinetic and potential energies as it flows through the control volume, then the energy equation for a single-stream steady-flow system, given by Eq. (3.11), further gets reduced to q — w = Ah

(3.12)

Equation (3.12) is the simplest form of the first-law relation for control volumes. This form resembles the first-law relation for closed systems, Eq. (2.20), except that Au is replaced by Oh in this case.

3.4 SOME STEADY-FLOW ENGINEERING DEVICES Devices which operate under the same conditions for long periods of time are called steady-flow

Thermodynamic Analysis of Control Volume

61

devices. Some of the popular steady-flow devices used in engineering applications are nozzles, diffusers, turbines, compressors, throttling valves, mixing chambers, heat exchangers and so on. We will investigate the flow processes through these devices with the help of the first-law relation for steady-flow processes.

3.4.1 Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecrafts, and even in garden hoses. A nozzle is a device which increases the velocity of a fluid at the expense of pressure. A diffuser is a device that increases the pressure of a fluid stream by decelerating it. A typical nozzle and a typical diffuser used for fluid flows at subsonic speeds (the flow speed less than the speed of sound is called the subsonic speed) are shown in Fig. 3.5. We will now examine, for nozzle and diffusers, the relative importance of the terms appearing in the energy Eq. (3.10).

-sb.• Nozzle —Ir.

Diffuser

Figure 3.5 Nozzle and diffuser. • The term Q = 0. This is because in nozzles and diffusers, the rate of heat transfer between the fluid flowing through them and the surroundings is usually insignificant, even when they are not insulated from one another. This is mainly because of the fact that the fluid has high velocity and thus passes through a nozzle or diffuser quickly without any significant heat transfer taking place. Therefore, the flow through nozzles and diffusers may be assumed to be adiabatic. • The term W = 0. This is because nozzles and diffusers are properly shaped ducts to achieve flow acceleration and deceleration, without involving any work like shaft work and so on. • The term AKE # 0, since the devices involve very high velocity and the fluid experiences large changes in its velocity. • The term APE = 0. That is, the fluid usually experiences little or no change in its elevation as it flows through a nozzle or diffuser. Moreover, if the fluid is in gaseous state, gravitational effects can be neglected even if there is a significant variation in elevation. Thus for nozzles and diffusers, the steady-flow energy Eq. (3.11) becomes q - w = Ah + Ake + Ape

(3.13)

But for nozzles and diffusers, q = 0, w = 0, Ape = 0, thus, we get h, + V2- = h, + VI2 2 2

(3.14)

where h i and h2 are the static entha py at the inlet and exit of the device, respectively.

I

II

II :

62 Fundamentals of Engineering Thermodynamics At this stage it will be useful to note the following. When we analyse control volumes, we find it convenient to combine the internal energy and flow energy of a fluid into a single term, namely the enthalpy, defined as h = u + Pv. Whenever the kinetic and potential energies of the fluid are negligible, the enthalpy represents the total energy of a fluid. For flow through nozzles and diffusers, the potential energy of the fluid is negligible, but the kinetic energy is quite large. In such cases, it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called stagnation (or total) enthalpy ho, defined as ho = h +

v2

(kJ/kg)

(3.15)

With the definition of stagnation enthalpy ho, Eq. (3.14) becomes (3.16)

hot = hoz

Thus, for flow through nozzles and diffusers, the stagnation enthalpy of the fluid is conserved. That is, in the absence of any heat and work interaction and any changes in potential energy, the stagnation enthalpy of a fluid remains constant during a steady-flow process. The stagnation enthalpy represents the enthalpy of a fluid when it is brought to rest adiabatically.

3.4.2 Turbines and Compressors In gas, hydroelectric, or steam power plants, the turbine is the device used to drive the electric generators. As the fluid passes through the turbine, work is done against the turbine blades which are attached to its shaft. As a result, the shaft rotates and the turbine produces work. Compressors, pumps, fans, etc. are devices used to increase the pressure of a fluid. A compressor is capable of compressing a gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. Schematic diagrams of a compressor and a turbine are shown in Fig. 3.6. P2 T2

Compressor

P2 T2

Figure 3.6 Schematics of compressor and turbine. We now apply the principle of energy conservation to flow through compressors and turbines, by studying the terms involved in the energy equation one by one.

Thermodynamic Analysis of Control Volume

63

• The heat transfer term Q = 0. This is because the heat transfer in respect of these devices is generally small compared to the shaft work unless there is internal cooling (as in the case of some compressors). • The term W # 0, since there is shaft work involved. • The change in potential energy Ape = 0. • The change in kinetic energy Ake = 0. The velocities involved in devices, with the exception of the turbine, are usually too low to cause any significant change in the kinetic energy. For a compressor (involving some heat loss), the conservation of energy Eq. (3.11), therefore, becomes q — w = Ah + 4A40+ ae0 or q — w = Ah (3.17)

3.4.3 Throttling Valves Any kind of a flow-restricting device that causes a significant pressure drop in the fluid is called a throttling valve. For example, the adjustable valve, capillary tube, and porous plug are all throttling devices. A typical throttling valve is shown in Fig. 3.7. Unlike turbines, throttling valves produce a pressure drop without involving any work. The pressure drop in the fluid is often accompanied by a large drop in temperature, and for this reason the throttling devices are commonly used in refrigeration and air-conditioning applications. The magnitude of temperature Throttling valve

Figure 3.7 Throttling valve. drop (or, sometimes, the temperature rise) during a throttling process is governed by a property called the Joule-Thomson coefficient, which is a measure of the change in temperature with pressure during a constant-enthalpy process (see Section 13.6). Throttling valves are usually small devices, and the flow through them may be assumed to be adiabatic (q = 0) since there is neither appreciable time nor large enough area available for any significant heat transfer to take place. Further, w = 0 and Ape = 0. In many cases, even the increase in kinetic energy is insignificant (Ake = 0) even though the exit velocity is often considerably higher than the inlet velocity. Therefore, the energy conservation equation for throttling devices becomes h2 = h1 (kJ/kg)

(3.18)

That is, the enthalpy values at the inlet and exit of a throttling valve are the same. For this reason, they are also called isenthalpic devices.

64

Fundamentals of-Engineering Thermodynamics Equation (3.18) can also be written as tri + Po) = u2 +

The above relation implies that Internal energy + flow energy = constant If P2v2 > Ply!, then the internal energy has to decrease. As a result, there would be a drop in temperature (i.e. T2 < T1 ). In the case of an ideal gas h = h(T), and thus the temperature T has to remain constant during a throttling process as per Eq. (3.18).

3.4.4 Mixing Chambers In a flow system, the section where a mixing process takes place is commonly referred to as the mixing chamber. It need not have to be a distinct "chamber". An ordinary T-elbow or a Y-elbow in a shower, for example, serves as a mixing chamber for the cold and hot water streams. A T-elbow is shown schematically in Fig. 3.8. Mixing chambers are usually well insulated and therefore, q = 0. Also, they do not involve any kind of work, hence w = 0. Further, the kinetic

Hot water Figure 3.8 T-elbow for water mixing. and potential energies of the fluid streams are usually negligible; ke = 0, pe = 0. The principle of conservation of energy requires that the total incoming energy must be equal to the total outgoing energy of the control volume. Therefore, for mixing chambers, the conservation of energy equation becomes analogous to the conservation of mass equation. No heat or work is crossing the boundaries (Q 0, W = 0) and the kinetic and potential energies are negligible (KE = 0, PE = 0). Therefore, by conservation of energy Eq. (3.9), we have 0 HO v2 V2 W I the he + + gz. + gz, — Ent; h• + 2 2 Or

Erikh; = Ithehe Equation (3.19) is simply the mass conservation equation.

(3.19)

Thermodynamic Analysis of Control Volume

65

3.4.5 Heat Exchangers Heat exchangers are devices where two moving fluid streams exchange heat without mixing. The simplest form of a heat exchanger is a double-tube heat exchanger, shown in Fig. 3.9. Under Fluid B 80°C

•

Fluid A 20°C

60°C

30°C Figure 3.9 Double-tube heat exchanger. steady operation, the mass flow rate of each fluid stream flowing through the heat exchanger remains constant. Heat exchangers involve no work interaction, and also for each fluid stream the kinetic and potential energy changes are negligible. The heat transfer associated with a heat exchanger depends on how the control volume is chosen. The outer shell of a heat exchanger is usually well insulated to prevent heat loss to the surrounding medium.

3.4.6 Pipe and Duct Flow Fluid flow through pipes and ducts finds applications in numerous engineering devices. The flow through a pipe or a duct usually satisfies the steady-flow conditions. For pipe or duct flow: • Under normal operating conditions, the amount of heat gained or lost by the fluid may be significant, particularly when the pipe or duct is so long. Thus, Q # 0. • If the control volume involves a heating section (electrical wires), a fan, or a pump (shaft), then the work interaction should be considered. Therefore, pi/ # 0. • In case of liquids, the velocities involved are relatively low and hence AKE = 0. However, for gas flow in ducts with variable cross-sectional areas the change in.kinetic energy (AKE) may be significant. • When there is considerable elevation change, the change in potential energy can be significant, that is APE # 0.

66

Fundamentals of Engineering Thermodynamics

3.5 THE UNSTEADY-FLOW PROCESS During a steady-flow process, no changes occur within the control volume. However, many processes of practical importance involve changes within the control volume with time. Such processes are called unsteady-flow or transient-flow processes. For instance, charging of rigid vessels from a supply line, discharging of fluid from a pressurized vessel, driving a gas turbine with pressurized air stored in large containers, inflating tubes or balloons, cooking with an ordinary or pressure cooker, etc. are all processes, involving transient-flow. Unlike steady-flow processes, the unsteady-flow processes start and stop over some fmite time instead of continuing indefinitely. Therefore, for transient-flow processes we have to deal with changes that occur over some time interval At instead of with the rate of changes (i.e. changes per unit time). An unsteady system is similar, in some respects, to a closed system, except that the mass within the system boundary does not remain constant during the process.

Conservation of mass For a control volume undergoing an unsteady-flow process for a time interval At, the principle of conservation of mass can be expressed as Total mass entering Total mass leaving Net change in mass the control volume — the control volume = within the control during At during At volume during At That is, (320) Emi - Erne = Amcv where the subscripts i and e refer to inlet and exit of control volume (CV), respectively. If m1 and m2 are the values of mass in the control volume at the beginning and end of the time interval At, then Eq. (3.20) can be written as E

— I me = (m2 — mi)cv (kg)

(321)

This is the mass conservation equation for an unsteady-flow process. In the rate form, the mass conservation equation (in the time interval At and taking the limit as At 0) becomes E

— z the — dmcv (kg/s) dt

(322)

For a steady-flow process (dmcvldt = 0), Eq. (3.22) takes the form of Eq. (3.6), which as we know is the mass conservation equation for steady-flow processes.

Conservation of energy For a control volume undergoing any unsteady-flow process for a time interval At, the principle of conservation of energy can be expressed as Total energy \ crossing the boundary as heat and work during At j

Total energy Total energy transported by transported by mass into the — mass out of the control volume control volume during & during At ,

Net change in energy of the control volume during At

Thermodynamic Analysis of Control Volume 67

That is, Q — W + E 19; — E 19, = Ecv (kJ)

(323)

where 0 represents the total energy transported by mass into or out of a control volume. Now, V. 0; = 1 0;(5m; = L(h; + t + gz; )45m; .1 or

r

t

.,

v

h;

2

+ gdmidt

Therefore, Eq. (3.23) becomes v2 Q—W=EL (he + + gze Sme — E 2

+

In,

V4

+ gzi 8m4

AEcv (1d)

(324)

This is the energy conservation equation for unsteady-flow processes. To perform the integrations in Eq. (3.24), we should know how the properties of the mass at the inlets and the exits change during a process. In the rate form, the energy conservation equation becomes i 2 Ii7 = Eris, he +— + gz,. 2

v2 dEcv Ent; hi + -- + gz; + (kJ/s) 2 dt

(3.25)

3.6 THE UNIFORM-FLOW PROCESS The uniform-flow process is a simplified model which represents some of the unsteady-flow processes reasonably well. The unsteady-flow processes are generally difficult to analyse because the integrations in Eq. (3.24) are difficult to perform. Therefore, the uniform-flow processes are made use of for analysing the unsteady-flow processes, whenever possible. The simplifying idealizations associated with the uniform-flow process, which reduce the complications of the unsteady-flow process, are the following: • The state of the control volume is uniform at all instants during the process. • The state of the control volume may change with time, but it will do so uniformly. • The fluid flow at an inlet or exit is uniform and steady even though it may differ from one inlet or exit to another. With these idealizations, the integrations in Eq. (3.24) can be performed to get the energy conservation equation for a uniform flow as v.2 Ve2 Q — W = I me ( he + — + gze — Z m; h; + + gz; + (m2e2 — miei)cv (kJ)

2

2

(3.26)

68 Fundamentals of Engineering Thermodynamics When the kinetic and potential energies associated with control volume and fluid stream are negligible, the energy Eq. (3.26) reduces to Q — W = E mehe — E mihi + (m2u2 — miudcv (kJ)

(3.27)

Note that when mi = me = 0, this equation reduces to the first-law relation for control systems, i.e. Eq. (2.20). EXAMPLE 3.1 An insulated rigid tank having 5 kg of air at 3 atm and 30°C is connected to an air supply line at 8 atm and 50°C through a valve. The valve is now slowly opened to allow the air from the supply line to flow into the tank until the tank pressure reaches 8 atm, and then the valve is closed. Determine the final temperature of the air in the tank. Also, find the amount of air added to the tank. Solution The boundary of the control volume is indicated by dashed lines in the figure below. This is not a steady-flow process since the state of the control volume changes during the process.

Pi = 8 atm = 50°C

Air

r

Ti= 30°C P1 = 3 atm P2 = 8 atm T2 = ?

The inlet conditions of the air during the process remain constant, and the state within the control volume may be assumed to be changing uniformly since the flow into the tank is let in slowly. Therefore, the flow process can be treated as a uniform-flow process. Let the subscripts 1 and 2 refer to the initial and final states of the process in the tank and the subscript i to the state of air at the inlet of the tank. For the present control volume with a single inlet, the mass conservation equation gives mi = (m, — mi)cv The tank is insulated and no work is involved. Therefore, Q = 0 and W = 0. Thus, the conservation of energy Eq. (3.27) for this case reduces to MA =

(m2u2

- MIUI

V

Thermodynamic Analysis of Control Volume

69

For air, being an ideal gas, u1 = C, T1, u2 = GT2 and hi = CPT,. Therefore, miCpTi = Cv(m2 T2 But in, = m, - m l . Thus, (m2 - mi )CpTi = CE(ni2 T2 - M

)

By state equation we have PV = mRT. Therefore, =

m1

PV RT

,„

9 "

_ P2V2 RT,

Or m, = 8 T1 mi Pi T2

(... v, = v2)

_ 8 T, 3 T, Substituting this in the energy equation, we get 7 — ( 3 T, 8 -

- mi

CPT-

8T = C (- T2 m, 3 T,

- m1 1

Or

3

) CPT T2 -1

Cr(-8 3

-T)

or =8 -1)T, 3

Y

3 T2

Y = CP

Or

8 303.15 1 _ (8 3 T, 3

i ) 303.15 1 ) 323.15 1.4 '

(... y = 1.4

= 1.117 Therefore, Now,

T, = 381.86 K

MI

8 303.15 - 2.117 3 381.86

Therefore, 111 2. = 2.117 x 5 = 10.585 kg The mass of the air added Am =

ni 2

- mi = 10.585 - 5 = 5.585 kg

Note that the final temperature of the air in the tank is more than the supply line temperature. This is due to the fact that the enthalpy h = u + Pv contains the flow energy Pv, and this flow energy is converted to sensible internal energy once the flow comes to rest in the control volume and this is reflected as increase in temperature.

70

Fundamentals of Engineering Thermodynamics Imaginary piston

Alternative Solution This problem can also be solved by assuming the region within the tank and the mass of air that is entering the tank as a closed system, as shown in the figure above. The imaginary piston is assumed to push the enclosed mass of air in the supply line into the tank at constant pressure of 8 atm. The boundary work done during the process is

Wb = Pa

= PAV2 — VI) = Pi[Vtank — (Vtank Vi)] = — PiVi

where Vi is the volume occupied by the air before entering the tank and Pi is the pressure at the moving boundary, which is the face of the imaginary piston. By first law, for the closed system under consideration, we have

Q — Wb = U2 — or

Q = 0)

— Wb = U2

Now,

Wb = PiVi and U2 = M2C,,T2 U1 = initial IE in the tank + initial IE of the mass In; in the supply line. That is, = mi C,T1 + (m2 — mi )C,T; Therefore, the first-law relation becomes

PiVi = m2CvT2 — miC,T1 — (m2 — mi )C,Ti By state equation, we have PV = mRT. Therefore, we get (m2

P2V2

MI )Rri = M2Cv D

MICA — (1112 MI)Cv

Thermodynamic Analysis of Control Volume

71

or (A ni)(Y — 1 ) C,

= C,

P2V2

vmiTi —

R C

Amc,T; (... C,,

or

r-I

Amyl: = P2 V 2 „,, T '"I I R

Also V2

= =

5 x 287 x 303.15 — 1.431 m3 3 x 101325

mi RTI

Therefore, 8 x 101325x 1.431 287

Am x 1.4 x 323.15 =

(5 x 303.15)

or Am = 5583 kg That is, m2 — m1 = 5.583 or m2 = 5 + 5.583 = 10.583 kg

Therefore, T2 =

P2V2 m2 R

_8x

101325 x 1. 431 10.583 x 287

381.9 K

EXAMPLE 3.2

A high-velocity water jet nozzle has 0.02 m and 0.0001 m as its inlet and outlet diameters, respectively. Water discharges from the nozzle at a rate of 0.0060 kg/s. If the outlet pressure is 0.2 MPa, determine the pressure at the inlet, assuming the flow to be isothermal. Solution Since water is the fluid used, the density can be assumed to be constant at 103 kg/m3. Treating the flow to be steady, we have by Eq. (3.14) h2

v2 14,2 = + 2 2

where the subscripts 1 and 2 refer to inlet and outlet of the nozzle. The enthalpy h = u + Pv or dh = du + vdP + Pdv

For the isothermal case here,

du =

0 and for an incompressible fluid, dv = 0. Thus, dh = vdP

72 Fundamentals of Engineering Thermodynamics or

h2 - hi = v(P2

PO= — (P2 —Pi )

Therefore, the energy Eq. (3.14) becomes 2 P2 +I PV2 =

1 2 Pi+ — PV1 2

This is the well known incompressible Bernoulli's equation. The velocity of water is given by V = Therefore, m

and V2

m

=

PA2

and

A — IrD2 4 Substituting these values in the energy equation, we get 6,12 ( 1 ;72--

= P2 +

2

) 2

Or p p,

8 rh2

1 1

p

= 200 +

8m2

Di

D4 i

1

1

p[ (0 .0001)4 (0 . 02)4

= (200 + 291,800) kPa = 292 MPa Note that the pressure at the nozzle inlet Pi = 292 MPa and this is approximately 2882 atm. Such a high pressure is required for cutting metal, concrete, and plastic with water jets. Further, because of a very high pressure involved, the incompressible assumption is expected to introduce some error. EXAMPLE 33 An insulated rigid tank initially at zero pressure is connected through a valve to an infinite supply line of steam at 600 kPa and 250°C. The valve is slowly opened to allow the steam to flow into the tank until the pressure reaches 600 kPa, and at which point the valve is closed. Calculate the fmal temperature of the steam in the tank.

Thermodynamic Analysis of Control Volume 73

Solution

This is a non-steady flow and a non-steady-state process, since both the flow rate into the tank and the mass of the steam in the tank vary with time. Choose the control volume as shown in the figure below.

Steam

Control volume

The inlet conditions of the steam during this process remain constant. Also, the state within the control volume may be assumed to be changing uniformly, since the flow process is taking place slowly. Therefore, the filling process can be analysed as a uniform process. By Eq. (3.21), we have Emi — Eine = (m2 — mi)cv But mi =0 and me = O. Thus, mi = m2 That is, the final mass in the tank is equal to the amount of mass that entered the tank. By energy conservation Eq. (3.25), for the present case, we have

Q — w = " idle —

mil;

(m2u2 — miul)cv

But Q = 0, W = 0,

mehe = 0 and moil =0.

Thus, we get 0 = - m li, + M2 U2 But mi = m2. Therefore, u2 = hi. The enthalpy of the steam at the inlet, hi, is obtained as hi = 2957.2 kJ/kg (from Table 5 of the Appendix) = u2

Therefore, now for P2 = 600 kPa and u2 = 2957.2 kJ/kg, using again Table 5, we get T2 = 397 °C

74 Fundamentals of Engineering Thermodynamics Note that the temperature of the steam in the tank has increased by 147°C. This is due to the fact that the flow energy part Pv of enthalpy is converted to sensible internal energy once the steam comes to rest in the control volume. EXAMPLE 3.4 A fluid flows through a steady-flow open system at the rate of 3 kg/s. At the system inlet, the pressure, velocity, and internal energy are 5 atm, 150 m/s, and 2000 kJ/kg, respectively, and the specific volume is 0.4 m3/kg. The fluid leaves the system with 1.2 atm, 80 m/s, an internal energy of 1300 kJ/kg and specific volume of 1.1 m3/kg. The fluid loses 25 kJ/kg through heat transfer during the process. Determine the power output of the system, neglecting the change in potential energy. Solution The steady-flow energy relation for the system, given by Eq. (3.11) is q

w

=

+ 112 4., — VI. 2 2

h 2

or u2 1

v2

q — w = (u2 — ui ) + (P2v2 — Therefore,

+ 2

2

v2 v2 W = (U1 - U2 ) + (Pv 1 1— Pv 22 )+

'2

2

+q

= (2000 — 1300) 103 + [(5 x 101325 x 0.4) — (1.2 x 101325 x 1.1)] +

1502 — 802

2

+ (— 25) x 103 = 700 x 103 + 68901 + 8050 — 25000 = 751.951 kJ/kg Power output of the system = mw = 3 x 751.951 = 2255.85 kW

SUMMARY A control volume (CV) may be defined as any arbitrary region chosen in space. The boundaries of a control volume are called control surfaces. The control surfaces can be real or imaginary. The energy conservation equation for a control volume undergoing a process can be written as Q- W ZEin 1E0w = AEcv The energy required to push fluid mass into and out of a control volume is called the flow work or flow energy and is given by wflo,„ = Pv (kJ/kg) Unlike other work quantities, flow work is expressed in terms of the properties. Therefore, it is viewed as a combination property (like enthalpy) and referred to as flow energy, convected energy or transport energy, instead of flow work.

Thermodynamic Analysis of Control Volume

75

The total energy of a simple compressible system consists of three parts, namely IE, KE, and PE. The energy (6) of a fluid stream flowing into or out of a control volume is given by 17 2 8 = h + ke + pe = h + — + gz (kJ/kg) 2 The principle of conservation of mass for a general steady-flow system with multiple inlets and outlets may be expressed in the rate form as Emi = IMe The principle of conservation of energy or the first law of thermodynamics for a general steady-flow system with multiple inlets and outlets can be expressed as 0—gl =E nie

+

Ve2 +gze 2

K2

• h; +

2

+ gzi)(kW)

A nozzle is a device which increases the velocity of a fluid stream at the expense of pressure. A diffuser is a device which increases the pressure of a fluid stream by decelerating it. The steady-flow energy equation for nozzles and diffusers is h2

V2 V,2 = h 2 1 2

For flow through nozzles and diffusers, the stagnation enthalpy of the fluid is conserved. The stagnation enthalpy represents the enthalpy of a fluid stream when it is brought to rest adiabatically. Compressors, pumps, fans, etc. are devices used to increase the pressure of a fluid. A compressor is capable of compressing a gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. Any kind of a flow-restricting device that causes a significant pressure drop in the fluid is called a throttling valve. Unlike turbines, throttling valves produce a pressure drop without involving any work. The magnitude of temperature drop (or, sometimes, the temperature rise) during a throttling process is governed by a property called the Joule-Thomson coefficient, which is a measure of the change in temperature with pressure during a constant-enthalpy process. A throttling process is a constant enthalpy process. In a flow system, the section where a mixing process takes place is commonly referred to as the mixing chamber. Heat exchangers are devices where two moving fluid streams exchange heat without mixing. The mass conservation equation for an unsteady-flow process can be written as — E me = (m2 — mdcv (kg) The energy conservation equation for an unsteady-flow process can be written as Q — W = me he +

V 22

2

+ gze — E ?it; (hi + t + gzi + (m2e2 — iv! )cv (kJ)

The uniform flow process is a simplified model which represents some of the unsteadyflow processes reasonably well.

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Fundamentals of Engineering Thermodynamics

PROBLEMS 3.1

Crude oil of density 700 kg/m3 flows into a storage tank at a rate of 5 m3/min. At the same time, the oil is pumped out from the tank at a steady rate of 250,000 kg/h. After two hours of operation, what will be the change in weight of crude oil, in kilogram, in the storage tank? [Ans. —79992 kg]

3.2

Atmospheric air at 101 kPa and 30°C is allowed to get in slowly into an evacuated cylinder, through a valve. If no heat is transferred to or from the air in the bottle, what will be the final temperature in the cylinder when the cylinder pressure reaches 101 kPa? [Ans. 144.26°C]

33 A small turbine working under the following conditions runs an aircraft refrigeration unit. Air at 4 bar and 40°C flows steadily into the turbine at 40 m/s. At the turbine exit, the condition of the air is 1 bar, 2.5°C and 200 m/s. The shaft work delivered by the turbine is 52 kJ/kg of air. Determine the heat transfer per kg of air flowing through the turbine. [Ans. 33.5 kJ/kg] 3.4

A reciprocating air compressor takes 0.05 m3/s of air at 101 kPa and 25°C and delivers it at 150 kPa and 120°C to an aftercooler where the air is cooled at constant pressure to 30°C. The power supplied to the compressor is 5 kW. Determine the heat transfer in the compressor, and in the aftercooler. [Ans. 0.6302 kJ/s, 5.33 kJ/s]

3.5 An automobile tube has a constant volume of 0.08 m3. While driving on a hot road, the tube picks up heat energy at the rate of 0.6 kJ/km. Calculate the number of kilometres to be driven so that the pressure inside the tube rises to 200 kPa, if the initial pressure and temperature of the air inside the tube are 165 kPa and 310 K, respectively. [Ans. 11.69 km] 3.6 Oxygen passes through an adiabatic, steady-flow compressor at the rate of 1000 kg/h, entering as a saturated vapour at 2.5 atm and emerging at 17.5 atm and 175 K. Find the shaft power required per kg of oxygen and the required power of the motor to run the compressor. Treat oxygen to be an ideal gas. [Ans. —18.87 kW, — 67.92 kJ/kg] 3.7 Air at 8 atm and 80°C from a high-pressure tank of volume 8 m3 is allowed to escape slowly through a valve to the atmosphere. Determine the maximum work that can be obtained from the escaping air, assuming the process to be adiabatic. [Ans. 3368 kJ] 3.8 At the end of the exhaust stroke of an internal combustion engine, the combustion chamber is filled with 0.10 g of exhaust gases at 700 K and 101 kPa. The piston then travels out on the intake stroke, inducting 2 g of air at 293 K and 101 kPa. Neglecting any heat transfer associated with this process, determine the pressure and temperature of the mixture at the end of the intake stroke. [Ans. 101 kPa, 312.4 K] 3.9

Steam at 800 kPa and 200°C enters an insulated nozzle with a velocity of 50 m/s. It leaves the nozzle at 100 kPa with a velocity of 550 m/s. Calculate the enthalpy of steam at the nozzle outlet. [Ans. 2689.3 kJ/kg]

Thermodynamic Analysis of Control Volume 77

3.10 A turbine, operating under steady-flow conditions, receives 6000 kg of steam per hour. At the entry, the steam velocity and enthalpy are 50 m/s and 2839 kJ/kg, respectively. At the turbine exit, the steam leaves at 120 m/s with an enthalpy of 2100 kJ/h. Determine the power output of the turbine. Neglect the changes in potential energy. [Ans. 1220.36 kW] 3.11 A gas turbine unit developing 15 MW power has at its inlet, hi = 1100 kJ/kg and Vi = 50 m/s. At the exit, 112 = 300 kJ/kg and V2 = 200 m/s. If the mass flow rate through the turbine is 20 kg/s, calculate the rate of rejection of heat from the turbine. [Ans. 625 kW] 3.12 A gas turbine receives gases from the combustion chamber at 7 atm and 700°C, with a velocity of 10 m/s. The gases leave the turbine at 1 atm with a velocity of 50 m/s. Assuming the adiabatic and reversible processes of an ideal gas, calculate the power output of the turbine per unit mass flow rate. Take y= 1.3 and Cp = 1100 J/(kg K). [Ans. 386.165 kW] 3.13 Water vapour at 100 kPa and 150°C enters a subsonic diffuser with a velocity of 150 m/s. The inlet diameter is 10 cm. The fluid leaves the diffuser at 200 kPa with a velocity of 55 m/s, and 1.6 kJ/kg of heat is transferred to the surroundings. Determine (a) the final temperature, (b) the mass flow rate, and (c) the exit diameter. Treat the flow to be steady. [Ans. (a) 150°C, (b) 0.985 kg/s, (c) 14.79 cm] 3.14 Water flowing through a 25 mm diameter pipe with a velocity of 0.8 m/s falls into an empty tank of volume 0.1 m3. How long will it take to fill the tank? Assume uniform velocity across the pipe and take density of water to be 1000 kg/m3. [Ans. 4.245 min] 3.15 Air at 9 bar and 200°C is throttled to 5 bar before being expanded through a nozzle to a pressure of 1.1 bar. Assuming that flow through the nozzle is a reversible steady-flow process, and that no heat is rejected, calculate the velocity of the air at the nozzle outlet when the inlet velocity is 80 m/s. [Ans. 583.29 m/s] 3.16 A water-pump pumps water at the rate of 0.0167 m3/s from an open well to a tank. If the tank pressure is constant at 200 kPa, determine the work that the pump must do upon the water in order to keep pumping water into the tank against the tank pressure. Also, determine the hp required to run the pump. [Ans. 1.65 kJ, 2.21 hp] 3.17 Air flows steadily through a rotary compressor. At the entry, the air is at 20°C and 101 kPa and at the exit at 200°C and 600 kPa. Assume the flow to be adiabatic. (a) Evaluate the work done per unit mass of air assuming the velocities at the entry and exit to be negligible.(b) Estimate the increase in work input required if the velocities at the inlet and exit are50 m/s and 110. m/s, respectively. [Ans. (a) —180.81 kJ/kg, (b) 4.8 kJ/kg] 3.18 A gas contained in a piston-cylinder device is expanded at a constant pressure of 202 kPa by adding 5 kJ of heat to the gas. The volume change associated with this process is 0.008 m3. The friction between the piston and cylinder wall amounts to 0.3 U. Calculate the increase in the internal energy of the system consisting of the device and the gas. [Ans. 3.68 kJ]

78

Fundamentals of Engineering Thermodynamics

3.19 Air at 105 m/s and 1.25 kg/m3 enters a gas turbine of inlet area 0.05 m2. The air stream exits from the gas turbine at 135 m/s and 0.67 kg/m3. During the flow process, the air loses 27 kJ/kg of heat and its specific enthalpy comes down by 145 kJ/kg. Determine (a) the mass flow rate of the air through the turbine, (b) the turbine exit area, and (c) the power developed by the turbine. [Ans. (a) 6.5625 kg/s, (b) 0.0726 m2, (c) 774.375 kW] 3.20 Air from a compressed air supply line at 6 atm and 100°C is connected to an insulated piston-cylinder device through a valve. The cylinder has an initial volume of 0.02 m3 with air at 2 atm and 30°C. The valve is opened and the high pressure air is allowed to flow into the cylinder till its volume becomes 3 times its initial volume, maintaining constant pressure. Determine the mass of the air that entered the cylinder and the final temperature of the air in the cylinder after attaining equilibrium. [Ans. 0.1189 kg, 256 K] 3.21 An adiabatic compressor is run by directly coupling it to an adiabatic steam turbine. Steam enters the turbine at 10 MPa and 500°C at a rate of 20 kg/s and exits at 10 kPa and a quality of 0.9. Air entrs the compressor at 100 kPa and 300 K at a rate of 8 kg/s and exits at 1 MPa and 500 K. Determine the net power delivered by the turbine. [Ans. 18926.36 kW] 3.22 An insulated vertical piston-cylinder device initially contains 0.3 m3 of air at 300 kPa and 20°C. At this state, a linear spring just touches the piston without exerting any force on it. The cylinder is connected by a valve to a line that supplies air at 600 kPa and 20°C. The valve is opened, and air from the high-pressure line is allowed to enter the cylinder. The valve is closed when the presure inside the cylinder reaches 500 kPa. If the enclosed volume inside the cylinder doubles during this process, determine (a) the fmal temperature of air inside the cylinder, and (b) the mass of air entered the cylinder. [Ans. (a) 933 K, (b) 0.05 kg] 3.23 A gas of mass 0.23 kg at 1.4 MPa and 360°C is expanded adiabatically to a pressure of 100 kPa. The gas is heated at constant volume to again reach 360°C. The pressure after this heating is found to be 220 kPa. Finally it is compressed isothermally to reach the original pressure of 1.4 MPa. Determine (a) the adiabatic index of the expansion process, and (b) the change in internal energy during the adiabatic expansion. Assume CP for the gas to be 1005 J/(kg K). [Ans. (a) 1.426, (b) -55.967 kJ] 3.24 An oxygen cylinder of volume 300 litres has oxygen at 3.1 MPa and 18°C. The stop valve is opened and some gas is used. (a) If the pressure and temperature of the oxygen left in the cylinder fall to 1.7 MPa and 15°C, respectively, determine the mass of oxygen remaining in the cylinder. (b) If after the valve is closed, the oxygen in the cylinder gradually attains its intial temperature of 18°C, determine the amount of heat transferred from atmosphere to the gas. Assume the denisty of and specific heats ratio for oxygen to constants at 1.429 kg/m3 and 1.4, respectively. [Ans. (a) 5.48 kg, (b) 13.29 kJ] 3.25 A gas has a density of 1.9 kg/m3 at 1 bar and 15°C. To raise the temperature of 1 kg of the gas from 15°C to 250°C, at constant pressure, a heat transfer of 200 Id is required. Detrmine (a) the gas constant, (b) the Cp and C,, of the gas, (c) the change of internal energy, and (d) the work transfer. [Ans. (a) 182.7 J/(kg K), (b) Cp = 0.851 kJ/(kg K),

C,, =

0.668 kJ/(kg K), (c) 157 Id, (d) 43 kJ]

Thermodynamic Analysis of Control Volume

79

3.26 A pressure vessel is connected to a gas main maintained at 1.4 MPa and 85°C, thorugh a valve. The valve is opened and 2.7 kg of gas is allowed into the vessel which was at vacuum intially. When the valve is closed, the gas in the vessel stabilizes at 700 kPa and 60°C. Determine the heat transfer associated with this filling process. Also, determine the volume of the pressure vessel and the initial volume of the gas allowed into the vessel. Assume Cp = 0.88 kJ/(kg K), C,, = 0.67 kJ/(kg K), and neglect the velocity of the gas in the main. [Ans. —248.2 kJ, Initial volume of the 2.7 kg gas allowed into the vessel = 0.145 m3, Vessel volume = 0.27 m3] 3.27 An air reservoir of volume 4.25 m3 contains air at 650 kPa and 120°C. The air is cooled to 40°C. Determine (a) the final pressure of the air and (b) change of internal energy of the air. [Ans. (a) 517.7 kPa, (b) —1405.8 kJ] 3.28 Steam at 2 MPa and 350°C enters a turbine at 50 in/s and leaves with 100% quality at 0.1 MPa and 200 in/s. The elevation of inlet and exit above the reference level are 6 m and 3 m, respectively. Taking g = 9.8 m2/s, determine the power generated by the turbine if the heat transfer from the turbine is 8.5 kW. [Ans. 655.7 kW] 3.29 Steam at 800 kPa and 300°C is throttled to 200 kPa. Changes in kinetic energy are negligible for this process. Determine the final temperature of the steam. [Ans. 292.41°C] 3.30 One kilogram of a perfect gas is compressed in a piston-cylinder device according to the law pv1-3 = constant, until the pressure becomes 660 kPa. Calculate the heat flow to or from the cylinder walls, if (a) the gas is ethane of molar mass 30 kg/kmol and Cp = 2100 J/(kg K) and (b) if the gas is argon with molar mass 40 kg/kinol and Cp = 520 J/(kg K). [Ans. (a) 138.2 kJ/kg, (b) —58.58 kJ/kg]

CHAPTER

4 The Second Law of Thermodynamics 4.1 INTRODUCTION The first law of thermodynamics which is the law of conservation of energy has been applied to processes involving closed and open systems, in Chapters 2 and 3. From our discussion in these chapters it is clear that energy is a conserved property, and all known processes are found to obey the first law of thermodynamics. In other words, we can say that no process is known to have taken place in violation of the first law of thermodynamics. However, there is no way in the first law to ensure that all processes satisfying it will actually take place. For instance, a hot cake kept on a plate eventually cools down (Fig. 4.1). This process satisfies the first law of thermodynamics since the amount of energy lost by the cake is equal to the amount of energy gained by the surrounding atmospheric air. Now let us consider the reverse process—can a hot cake get even hotter in a cool surrounding as a result of heat transfer from the surrounding air? We all know that this process can never take place, even though the occurrence of this phenomenon would not violate the first law.

Figure 4.1 A hot cake kept in atmosphere. Likewise, we can enlist a number of processes which can take place only in a particular direction but not in the reverse, even though in both the directions the processes will satisfy the first law. From these observations, it is clear that some processes proceed in a certain direction only but not in the reverse direction. The first law stipulates no restriction on the direction of a process, but satisfying the first law does not ensure that the process will actually take place. This inadequacy of the first law to identify whether a process can actually take place is 80

The Second Law of Thermodynamics 81 compensated by introducing another general principle, namely the second law of thermodynamics. It will be shown in the course of our discussion in this chapter that the reverse process discussed above violates the second law of thermodynamics. This violation is easily detected by using a property called entropy (defined in Chapter 5). From the above discussion, it is clear that a process will not take place unless it satisfies both the first law and the second law of thermodynamics. There are many valid statements of the second law of thermodynamics. In this chapter we will examine some of the popular statements in relation to some engineering devices that operate on cycles. In addition to ascertaining the direction of a process, the second law also asserts that energy has quality as well as quantity. Besides, the second law of thermodynamics is used to determine the theoretical limits for the performance of many popular engineering systems, such as heat engines and refrigerators. This law is even used for predicting the degree of completion of chemical reactions.

4.2 THERMAL ENERGY RESERVOIRS A thermal-energy reservoir is a hypothetical body with a relatively large capacity for thermal energy (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in its temperature. The atmospheric air, the large expanse of water such as oceans, lakes, and rivers can be modelled as thermal energy reservoirs since they have large thermal energy storage capabilities, that is, large thermal masses. For example, megajoules of waste energy dumped in an ocean or a large river, by power plants, does not cause any significant change in the temperature of the water. The industrial furnace is another well known example of a thermal reservoir. If the temperature of most furnaces is carefully controlled, they are capable of supplying large quantities of thermal energy as heat in an essentially isothermal manner. Any physical body whose thermal energy capacity is large, compared to the amount of energy it supplies or absorbs, can be modelled as a thermal reservoir. In other words, a body need not have to be very large to be considered as a reservoir. For instance, in the analysis of heat dissipation from a radio or a television set, the air in the room can be treated as a reservoir, since the amount of heat transfer from the electronic devices in these products is not large enough to cause any noticeable effect on the room temperature. A thermal reservoir that supplies energy in the form of heat is called the source, and one that absorbs energy in the form of heat is called the sink. TherMal-energy reservoirs are also referred to as heat reservoirs.

4.3 HEAT ENGINES It can be observed from the operation of practical devices that work can easily be converted to other forms of energy, but converting other forms of energy to work is not so easy. In fact, work can be converted to heat directly and completely, but converting heat to work requires the use of some special devices. These devices are called heat engines. That is, heat engines are devices that convert heat to work. They differ considerably from one another, but they all can be characterized by the following:

82

Fundamentals of Engineering Thermodynamics

• • • • •

Schematically, all heat engines can be represented as shown in Fig. 4.2. They receive heat from a source. They convert part of this heat to work. They reject the remaining heat to a sink. They operate on a cycle. High-temperature source Qin

HEAT ENGINE

Wnet.out

Low-temperature sink

Figure 4.2 Schematic diagram of a heat engine. All cyclic devices including heat engines usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid. At this stage, it will be advantageous to note that the term heat engine is often used in a broader sense to include all work-producing devices, that is, even those which do not operate on a thermodynamic cycle. Engines that involve internal combustion such as automobile engines and gas turbines fall into this category. These engines operate on a mechanical cycle but not on a thermodynamic cycle, since the working fluid does not undergo a complete cycle. A typical example of a heat engine is the steam power plant, shown schematically in Fig. 4.3. This is an Energy source (such as a furnace) °In

System boundary

Wan

Qout Energy sink (such as the atmosphere)

Figure 4.3 Schematic diagram of a steam power plant.

The Second Law of Thermodynamics

83

external-combustion engine, that is, the combustion process takes place outside the engine, and the thermal energy released during this process is transferred to steam as heat. In Fig. 4.3, is the heat supplied to the steam in a boiler from a high-temperature source, Q., is the heat rejected from the steam in a condenser to a low-temperature sink, Win is the work required to compress water to boiler pressure, Wout is the work delivered by the steam as it expands in the turbine. It is essential to realize that the heat and work interactions indicated by the subscripts "in" and "out" are all positive quantities.

an

4.3.1 Thermal Efficiency Thermal efficiency is defined as the fraction of heat input that is converted to net work output. It is a measure of the performance of a heat engine. In general, the efficiency can be expressed in terms of the desired output and the required input as Efficiency —

desired output required input

(4.1)

Following Eq. (4.1), the thermal efficiency of a heat engine can be expressed as Thermal efficiency —

net work output total heat input

or ilth =

Wnet out

(4.2)

Qin

But

Wnetout = Qin — Qout,

therefore, Eq. (4.2) can also be expressed as 77th = I -

Qout

(4.3)

Qin

Examine the heat engine shown in Fig. 4.4. In the cyclic process of the heat engine shown, QH High-temperature reservoir at TN L 7QH Wnet,out

n ss

QL Low-temperature reservoir at 7-,

Figure 4.4 Schematic diagram of a heat engine.

84

Fundamentals of Engineering Thermodynamics

is the magnitude of the heat transfer between the cyclic device and the high-temperature reservoir at temperature TH, and QL is the magnitude of the heat transfer between the cyclic device and the low-temperature reservoir at temperature TL. The net work output and the thermal efficiency relations for the heat engine can be expressed as Wnet,out

QH

(4.4)

QL

and gth

Wnet,out

(4.5)

QH

Or

Rth = 1 —

(46)

QH

From Eq. (4.6) it is seen that the thermal efficiency of a heat engine is always less than unity, since both QL and QH are defined as magnitudes and therefore are positive quantities. Thermal efficiency is a measure of the capacity of a heat engine to convert the heat (that it receives) to work. Even though engineers constantly strive to increase the efficiency of these devices, the thermal efficiencies of work-producing devices are very low. Usually, the sparkignition automobile engines have a thermal efficiency of about 20 per cent. The thermal efficiency of diesel engines is about 30 per cent and that of large gas-turbine plants is also about the same. Steam power plants operate at about 40 per cent efficiency. EXAMPLE 4.1 A heat engine working with a thermal efficiency of 35 per cent receives 2 kW of heat from a furnace. Determine the waste heat rejected from the engine. Solution The adjacent figure shows the schematic diagram of the heat engine.

QH = 2 kW

Given that, qui = 0.35 and QH = 2 kW The thermal efficiency given by Eq. (4.6) is qui = —

=11

Therefore,

Wnetput QL = (1 — 17th)QH = (1 — 0.35)2

= 1.3 kW

QL = ?

The waste heat rejected from the engine is therefore LI:3 kW I.

4.4 KELVIN-PLANCK STATEMENT OF THE SECOND LAW From the heat engine shown in Fig. 4.4, it is evident that a heat engine must reject some heat to a low-temperature reservoir in order to complete the cycle. In other words, no heat engine

The Second Law of Thermodynamics 85

can convert all the heat it receives to useful work. The Kelvin-Planck statement of the second law of thermodynamics is based on this limitation on the thermal efficiency of heat engines. The Kelvin-Planck statement of the second law is: It is impossible for any device that operates on a cycle to exchange heat with just a single reservoir and produce a net amount of work. That is, to keep operating, a heat engine must exchange heat with a low-temperature sink as well as a high-temperature source. The Kelvin-Planck statement can also be expressed as: For a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. Or

No heat engine can have a thermal efficiency of 100 per cent. It is essential to realize that the impossibility of having a 100 per cent efficient heat engine is not due to friction or other dissipative effects, but is a limitation that applies to both the idealized and the actual heat engines.

4.5 REFRIGERATORS, HEAT PUMPS AND AIR-CONDITIONERS A refrigerator is a special device meant for transfer of heat from a low-temperature medium to a high-temperature medium. Like heat engines, refrigerators are also cyclic devices. The working fluid used in the refrigeration cycle is called the refrigerant. The most popular refrigeration cycle is the vapour-compression cycle. A typical refrigeration system is shown in Fig. 4.5. As seen from this figure, the vapour-compression cycle involves four main components—a compressor, a condenser, an expansion valve, and an evaporator. Surrounding medium such as the room atmospheric air

aH Condenser 800 kPa 30°C Expansion valve

800 kPa 60°C Wnetin

Compressor

LI 120 kPa —20°C

120 kPa —25°C Evaporator al_ Refrigerated space

Figure 4.5 A refrigeration system.

86 Fundamentals of Engineering Thermodynamics The refrigeration cycle process is as follows. The refrigerant enters the compressor as a vapour and is compressed to the condenser pressure. It leaves the compressor at a high temperature and cools down and condenses as it flows through the coils of the condenser by rejecting heat to the surrounding medium. Then it enters a capillary tube where its pressure and temperature drop drastically because of the throttling effect. The low-temperature refrigerant then enters the evaporator, where it evaporates by absorbing heat from the refrigerated space. The cycle is completed as the refrigerant leaves the evaporator and re-enters the compressor. In domestic refrigerators, the freezer compartment where heat is picked up by the refrigerant serves as the evaporator. The coils behind the refrigerator where heat is dissipated to the room atmospheric air serve as the condenser. Using the QH , QL, TH and TL notations of the heat engine, a refrigerator can be shown schematically as in Fig. 4.6. It is essential to note that, here QL is the magnitude of heat removed from the refrigerated space at temperature TL, QH is the magnitude of heat rejected to the environment at temperature TH, and W„Lin, is the net work input to the refrigerator. Also, TH is always greater than TL. Warm environment at TH > TL Required input Wnet,in

Desired ) output

Cold refrigerated space at TL

Figure 4.6 Schematic diagram of a refrigerator.

Coefficient of performance for a refrigerator The efficiency of a refrigerator is expressed in terms of its coefficient of performance (COP). The refrigerator is meant for removing heat (QL) from the refrigerated space. To achieve this, it requires a work input of in. The COP of a refrigerator can be expressed as COPR

— net output _ QL net input Wnetin

(4.7)

where the subscript R designates refrigerator. By the principle of conservation of energy for a cyclic device, we have Wnetin =

QH

(kJ)

(4.8)

The Second Law of Thermodynamics 87 Combining Eqs. (4.7) and (4.8), we obtain 1

COPR =

Qtl — Qt. (QH/QL) —1

(4.9)

From Eq. (4.9) it is seen that COPR can be greater than unity. That is, the amount of heat removed from the refrigerated space can be greater than the amount of work input. We should note that this is in sharp contrast to the thermal efficiency which can never be greater than unity. Therefore, intentionally, the efficiency of a refrigerator is expressed by another term—the coefficient of performance—in order to avoid the odd situation of having efficiencies greater than unity.

4.5.1 Heat Pumps A heat pump is a device that transfers heat from a low-temperature medium to a high-temperature medium. Even though the heat pumps and refrigerators operate on the same cycle, they differ in their objectives. The refrigerator is meant for maintaining the refrigerated space at a low temperature by removing heat from it. Discharging this heat to a higher temperature environment is merely a necessary part of the operation, not the purpose. However, the objective of a heat pump is to maintain a heated space at a high temperature. This is achieved by absorbing heat from a low-temperature source, such as water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a house. A typical heat pump is shown in Fig. 4.7. The efficiency of a heat pump is also expressed in terms of the coefficient of performance COPHp, defined as Warm heated space at TH > TL Desired output Wnet,in

Required input Cold environment at TL Figure 4.7 Schematic diagram of a heat pump. COPHp

= net output

QH

(4.10)

net input

Or

QH

COPHp QH

-a

1-

(a/QH

(4.11)

88 Fundamentals of Engineering Thermodynamics Comparison of Eqs. (4.7) and (4.10) results in (4.12)

COPHp = COPR + 1

for given values of QL and QH. Equation (4.12) implies that the coefficient of performance of a heat pump is always greater than unity since COPR is a positive quantity. Most of the heat pumps of present days have COPs between 2 and 3. EXAMPLE 4.2 A household refrigerator maintains the refrigerated space at 3°C by removing heat from it at the rate of 4 kW. The power required to run the refrigerator is 1.8 kW. Determine the COP of the refrigerator. Solution The schematic of the refrigerator is shown in the adjacent figure. The COP defined by Eq. (4.7) is COPR =

L. Wnet,in

4

2.22

1• 8

Refrigerated space

EXAMPLE 43 The temperature of a room is being maintained at 25°C by a heat pump. The room loses heat at the rate of 1.5 kJ/s when the outside temperature is 0°C. Under the given conditions if the COP of the heat pump is 2.22, determine the power input to the heat pump.

1.5 kJ/s

Solution The schematic diagram of the heat pump cycle is shown in the adjacent figure. By Eq. (4.10), the COP COP — HP

is defined as Wnet,in

QH Wnet,in

QH is the heat supplied by the pump to the room. Therefore,

WneLni

1 '5 = 0.676 kJ/s QH COPHp 2.22

The power input required to the pump is therefore 0.676 kW

Outside air at 0°C

The Second Law of Thermodynamics

89

4.5.2 Air-conditioners Air-conditioners are basically refrigerators whose refrigerated space is very large, like a room or a building, instead of a small volume as in the household refrigerators. A window air-conditioner cools a room by absorbing heat from the room air and discharging it to the outside atmosphere. If the operation is reversed, the same air-conditioner can be used as a heat pump in winter. In this mode, the unit will be installed backward and it will pick up heat from the cold outside air and deliver it to the room.

Energy efficiency rating (EER) In modern terminology, the performance of refrigerators and air-conditioners is expressed in terms of energy efficiency rating (EER) which is the amount of heat removed, from the cooled space, in British thermal iviits (Btus) for 1 Wh (Watt-hour) of electricity consumed. Considering that 1 kWh = 3412 Btu and thus 1 Wh = 3.412 Btu, a unit which removes 1 kWh of heat from the cooled space for each kWh of electric power it consumes (COP = 1) will have an EER of 3.412. Therefore, the relation between EER and COP is EER = 3.412COPR

(4.13)

Thus, most air-conditioners with (O!'s between 2.3 and 3.5 will have EERs between 8 and 12.

4.6 CLAUSIUS STATEMENT OF THE SECOND LAW The Kelvin-Planck statement, as may be seen from Section 4.5, is related to heat engines. The Clausius statement is related to refrigerators as well heat pumps. The Clausius statement of the second law is: It is impossible to construct a device that operates on a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body.

It is a matter of general knowledge that heat does not, on its own, flow from a cold medium to a hot medium. On the surface, it looks that as per Clausius statement it is impossible to construct a device that transfers heat from a colder medium to a warmer medium. The Clausius statement implies that a refrigerator will not operate unless its compressor is driven by an external power source, such as an electric motor. Like any other physical law, the statements of the second law of thermodynamics by Kelvin-Planck and Clausius are also based on experimental observations. These statements cannot be proved. However, as yet, no experiment has been conducted that contradicts the second law, and this should be taken as the evidence of its validity. The Kelvin-Planck and Clausius statements are two equivalent expressions of the second law of thermodynamics. It can be shown that any device that violates the Kelvin-Planck statement would also violate the Clausius statement, and vice versa.

4.7 EQUIVALENCE OF KELVIN—PLANCK AND CLAUSIUS STATEMENTS We have studied the Kelvin-Planck be stated as follows:

Clausius statements of the second law, which can also

90 FunJamentals of Engineering Thermodynamics

Kelvin—Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of this heat into work

Clausius statement No process is possible whose sole result is the transfer of heat from a cooler !os a hotter body. At first sight, the Kelvin—Planck and the Clausius statements appear to be quite unconnected, but indeed they are in all respects equivalent to each other. Two propositions of statements are said to be equivalent when the truth of one implies the truth of the second and the truth of the second implies the truth of the first. Using the symbol to mean "implies" and the symbol u to denote equivalence, we have, by definition, KP s C

where KP is the Kelvin—Planck s'atement and C is the Clausius statement. The above equivalence is valid only when KP C and C KR It can be easily shown that KP C also, when —KP —C and —C = —KP, where —KP and —C stand for the falsity of the Kelvin— Planck and Clausius statements. Thus, in order to demonstrate the equivalence of KP and C, we have to show that a violation of one statement implies a violation of the second and vice versa. • To prove that —C —KP, consider a refrigerator, shown on the left side of the Fig. 4.8, which requires no work to transfer Q2 units of heat from a cold reservoir to a hot reservoir

Hot reservoir

W=

No work Refrigerator

p

Q2

Heat engine

Cold reservoir

Figure 4.8 Proof that —C = —KP. The refrigerator on the left is a violation of C, the refrigerator and engine together constitute a violation of KP.

The Second Law of Thermodynamics 91 and which, therefore, violates the Clausius statement. Now assume that a heat engine (on the right) also operates between the same two reservoirs in such a way that heat Q2 is delivered to the cold reservoir. The engine, of course, does not violate any law, but the refrigerator and engine together constitute a self-acting device whose sole effect is to convert all this heat into work. Therefore, the refrigerator and engine together constitute a violation of the Kelvin—Planck statement. • To prove that — KP = — C, consider a heat engine, shown in Fig. 4.9 which rejects no heat to the cold reservoir and which, therefore, violates the Kelvin—Planck statement.

Hot reservoir

02

Heat engine

Refrigerator

No heat Cold reservoir

Figure 4.9 Proof that —KP = —C. The heat engine on the left is a violation of KP, the heat engine and refrigerator together constitute a violation of C. Suppose that a refrigerator (shown in the same figure) also operates between the same two reservoirs and uses up all the work liberated by the engine. The refrigerator violates no law, but the engine and the refrigerator together constitute a violation of the Clausius statement. Thus, it can be concluded that both the above statements of the second law are equivalent to one another.

4.8 PERPETUAL-MOTION MACHINES Any device that violates either the first or the second law of thermodynamics is called a perpetual-motion machine. It is only a hypothetical one, and despite numerous attempts, no perpetual-motion machine is known to have worked. A device that violates the first law of thermodynamics (creates energy) is called a perpetual-motion machine of the first kind (PMM1) and a device that violates the second law of thermodynamics is called a perpetual-motion machine of the second kind (PMM2).

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4.9 REVERSIBLE AND IRREVERSIBLE PROCESSES A reversible process is that process which can be reversed without leaving any trace on the surroundings. That is, both the system and the surroundings are returned to their initial states at the end of the reversible process. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combination of original and reverse processes. Processes that are not reversible are termed irreversible processes. The reversible processes are merely idealizations of actual processes. They actually do not occur in nature. That is, all the processes occurring in nature are irreversible. At this stage, it is natural to question why such fictitious processes need to be considered in our study of thermodynamics? The answer to this question is the following: • They are easy to analyse since a system passes through a series of equilibrium states during a reversible process. • They serve as idealized models to which actual processes can be compared. The reversible processes can be viewed as theoretical limits for the corresponding irreversible ones. Further, the concept of reversible processes leads to the definition of secondlaw efficiency for actual processes, which is the degree of approximation to the corresponding reversible processes. The performance of different devices that are designed to do the same task can be compared on the basis of their second-law efficiencies. A better design will result in lower irreversibilities and a higher second-law efficiency.

Irreversibilities Irreversibilities are factors that cause a process to be irreversible. The well known irreversibilities are friction, electric resistance, inelastic deformation of solids, unrestrained expansion, heat transfer through a finite temperature difference, nonquasi-equilibrium changes, and so on.

Internally and externally reversible processes An internally reversible process is that for which no irreversibilities occur within the boundaries of the system during the process. That is, the system proceeds through a series of equilibrium states during an internally reversible process, and when the process is reversed, the process passes through exactly the same equilibrium states while returning to its initial state. In other words, the paths of the forward and reverse processes coincide for an internally reversible process. The quasi-equilibrium processes can be closely approximated to internally reversible processes. An externally reversible process is that for which no irreversibilities occur outside the system boundaries during the process. For example, heat transfer between a reservoir and a system is an externally reversible process if the surface of contact between the system and the reservoir is at the temperature of the reservoir. A process for which no irreversibilities occur either within the system or outside the system (i.e. its surroundings) is called a totally reversible process or simply a reversible process. That is, a totally reversible process involves no heat transfer across a finite temperature difference, no friction or other dissipative effects, and no nonquasi-equilibrium changes. Let us examine the two systems shown in Fig. 4.10. The systems are identical and the transfer of heat to these systems takes place at constant pressure. That is, the transfer of heat

The Second Law of Thermodynamics 93

Boundary at 20°C 20°C

Heat Thermal energy reservoir at 20.0...0.1°C (a) Totally reversible

Thermal energy reservoir at 30°C (b) Internally reversible

Figure 4.10 Heat transfer processes. takes place at constant temperature. Under the stipulated conditions, both processes are internally reversible, since they are take place isothermally, and both pass through the same equilibrium states. Further, the first process is externally reversible as well, since heat transfer for this process takes place across an infinitesimal temperature difference. However, the second process is externally irreversible, since it involves heat transfer across a finite temperature difference.

4.10 THE CARNOT CYCLE The Carnot cycle is a reversible cycle. Probably, it is the best known reversible cycle in thermodynamics. It was proposed by Sadi Camot, a French engineer, in 1824. We can infer from the second-law statements that no actual cycle can be completely reversible. This implies that Camot cycle is only a fictitious theoretical cycle. The theoretical heat engine that operates on the Camot cycle is called the Carnot heat engine. The Camot cycle is composed of four reversible processes—two isothermal and two adiabatic—and if can be executed either in a closed system or in a steady-flow (open) system. The four reversible processes of a Camot cycle in a closed system are: • • • •

Reversible isothermal expansion Reversible adiabatic expansion Reversible isothermal compression Reversible adiabatic compression

The P—v diagram of a Camot cycle is shown in Fig. 4.11. Recalling that on a P—v diagram the area under the process curve represents the boundary work for quasi-equilibrium (internally reversible) processes, we see that the area under the curve 1-2-3 is the work done by the gas during the expansion part of the cycle, and the area under the curve 3-4-1 is the work done on the gas during the the compression part of the cycle. The area enclosed by the path of the cycle (area 1-2-3-4-1) is the difference between these two areas and represents the net work done during the cycle.

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P

.V Figure 4.11 The P—v diagram of a Carrot cycle. In our discussion on power cycles in Chapter 9, we will see that the Carrot cycle can be executed in a steady-flow system as well. Being a reversible cycle, the Carnot cycle is the most efficient cycle operating between two specified temperature limits. At this stage, it is essential to realize that: • Carnot cycle is only a fictitious theoretical cycle and not an actual cycle. • Even though the Carnot cycle cannot be achieved in reality, the efficiency of an actual cycle can be improved by attempting to approximate it more closely to the Carrot cycle.

4.10.1 The Reversed Carnot Cycle The Carrot heat-engine cycle given in Fig. 4.11 is a totally reversible cycle. When all the processes of the cycle are reversed it becomes the Carnot refrigeration cycle. For the reversed case, the cycle remains exactly the same, except that the directions of both heat and work interactions are reversed. That is, QL is absorbed from the low-temperature reservoir and QH is rejected to a high-temperature reservoir, and a work input Wneon is required to accomplish this. The P—v diagram of the reversed Carnot cycle is shown in Fig. 4.12. This cycle is the same as that for the Carnot cycle, shown in Fig. 4.11, except that the directions of the processes are reversed.

4.11 THE CARNOT PRINCIPLES The Carrot principles are the two conclusions pertaining to the thermal efficiency of reversible (ideal) and irreversible (actual) heat engines, drawn from the Kelvin-Planck and Clausius statements of the second law of thermodynamics. The Carrot principles are: 1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two thermal reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two thermal reservoirs are the same.

The Second Law of Thermodynamics 95

P

•v Figure 4.12 The P—v diagram of the reversed Carnot cycle. The first principle is called the Carnot theorem and the second the corollary of the Carnot theorem. These principles can be proved by demonstrating that the violation of either one, results in a violation of the second law of thermodynamics.

Proof of the first principle Consider the two heat engines, shown in Fig. 4.13, operating between the same two thermal reservoirs. Let one engine be reversible, and the other irreversible. High-temperature reservoir at TH Irreversible HE

°H. Reversible HE

°Liffey < 0L, rev

(assumed)

QL.rev

Low-temperature reservoir at TL.

Figure 4.13 Two heat engines operating between the same two reservoirs. For argument sake, we assume that the thermal efficiency of the irreversible engine, nutiffev is more than that of the reversible engine, rid,rey. This assumption is in violation of the first principle of Carnot. That is, lith,irrev >

nth,rev

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With Eq. (4.6), the above inequality results in QH

QH

QH But by Eq. (4.4), QH —

QH

Therefore, the above relation becomes Ket,out

QH

Woet,out

irrev

QH

But (QH)i,„v = (QH)„.v (Fig. 4.13). Thus, (WnetmOirrev > (W.LoOrev

Now let the reversible heat engine be reversed and operated as a refrigerator. This refrigerator will receive a work input of W„„ and reject heat to the high-temperature reservoir. Since the refrigerator is rejecting heat QH to the high-temperature reservoir and the irreversible heat engine is receiving the same amount of heat from the same high-temperature reservoir, the net heat exchange for this reservoir is zero. Thus the reservoir can be eliminated by having the refrigerator discharge QH directly into the irreversible heat engine. This combination of the refrigerator and the irreversible heat engine, without the high-temperature reservoir, is schematically shown in Fig. 4.14. The combination of the refrigerator and the irreversible engine Wirrev Wrev

Combined HE + R °Linen,- — QL,rev

Low-temperature reservoir at T1

Figure 4.14 Refrigerator-heat engine combination. produces a net work of (Win, — Wrev) while exchanging heat with a single reservoir—a violation of the Kelvin-Planck statement of the second law. Therefore, our initial assumption that ritkirrev > qtkrev is not correct. Hence we conclude that no heat engine can be more efficient than a reversible heat engine operating between the same thermal reservoirs. The second principle of Carnot can also be proved in a simple manner and shown that the efficiency of a reversible engine is independent of the nature or amount of the working substance undergoing the cycle, as long as the engine is operating between the same thermal reservoirs. The second principle of Carnot is called the corollary of Carnot theorem.

The Second Law of Thermodynamics 97

4.12 THE THERMODYNAMIC TEMPERATURE SCALE A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Such a temperature scale proves to be very useful in thermodynamic calculations. The thermodynamic temperature scale can be defined with the help of reversible heat engines. As we know, the thermal efficiency of reversible heat engines is a function of the reservoir temperatures only. That is, ihkrev = 11TH1 TL) Using Eq. (4.6), this relation can be expressed as QH =f( TH , TL)

a

(4.14)

where TH and TL are the temperatures of the high- and low-temperature reservoirs, respectively. If some functional relationship is assigned between TH, TL and QH, QL, Eq. (4.14) then becomes the defmition of a temperature scale. The functional form off(TH, TL) can be developed by considering the three reversible beat engines shown in Fig. 4.15. Engines A and C are supplied with the same amount of heat Qi from the high-temperature reservoir at temperature Ti . Engine B receives heat Q2 rejected by Engine A at temperature T2 and rejects heat G to the low-temperature reservoir at T3. Engine C rejects heat Q3 to the reservoir at T3. Thermal-energy reservoir at T1

Thermal-energy reservoir at T3

Figure 4.15 Three reversible heat engines.

98 Fundamentals of Engineering Thermodynamics Engines A and B can be combined into one reversible engine operating between the same reservoirs as Engine C and thus this combined engine will have the same efficiency as Engine C. Since the heat input to Engine C is the same as the heat input to the combined Engines A and B, both systems must reject the same amount of heat. Using Eq. (4.14), we can write for Engines A, B and C, respectively Qi =1*(7' T ) 1, 2 Q2

-F711 = Q3 Qi

Q3

T3)

=f(7i, T3)

Also, Q1 /Q3 can be expressed as Q2 Q3 Q2 Q3

Qi = Qi

That is, f(TI , T3) = f(TI , T2) x f( T2, T3) It can be inferred from this equation that, since the left-hand side is a function of Ti and T3, the right-hand side, too, must be a function of Ti and T3 only, and not T2. That is, the value of the product on the right-hand side of this equation is independent of T2. To satisfy this condition, the function f must have the following form: 1(7i, 7;)

11)(T)

T2)

and (T2, 7;)

OUP 4;47.3)

so that

= f ( 7;) ATI) (4.15) Q3 AID For a reversible heat engine operating between two reservoirs at TH and TL, Eq. (4.15) can be written as QH

TH )

QL

ATI)

(4.16)

This is the only condition that the second law stipulates on the ratio of heat flows to and from the reversible heat engines. Since the function 0(7) is completely art itrary, several values of it will satisfy Eq. (4.16). Lord Kelvin first proposed taking 0(T) = T to define the thermodynamic temperature scale as QH Qt.

= TH (4.17)

The Second Law of Thermodynamics

99

This temperature scale is called the Kelvin scale, and the temperatures on this scale are called absolute temperatures. On the Kelvin scale: • The temperature ratios depend on the ratios of heat transfer between a reversible heat engine and the reservoirs. • The temperature ratios are independent of the physical properties of any substance. • Temperatures vary between zero and infinity. From Eq. (4.17) it is obvious that the thermodynamic temperature scale is not completely defined by it, since it gives only a ratio of the absolute temperatures. At the International Conference on Weights and Measurements held in 1954, the triple-point of water (the state at which all three phases of water coexist in equilibrium) was assigned the value 273.15 K. The magnitude of a kelvin is defined as 1/273.15 of the temperature interval between absolute zero and the triplepoint temperature of water. The magnitudes of temperature units on the Kelvin and Celsius scales are identical (1K = 1°C). The temperatures on the two scales are related by a constant 273.15 (the temperature of the ice point) as T(°C) = T(K) —27115

(4.18)

It is essential to note that even though the thermodynamic temperature scale is defined with the help of reversible heat engines, it is not possible, in reality, to operate such an engine to determine numerical values on the absolute temperatures. However, absolute temperatures can be measured accurately by devices such as ideal-gas thermometers together with the usage of extrapolation techniques.

4.13 THE CARNOT HEAT ENGINE The Carrot heat engine is the hypothetical engine that operates on the reversible Carnot cycle. For any heat engine, reversible or irreversible, the thermal efficiency is given by nth = 1

QL

QH

For reversible heat engines, by Eq. (4.17), we have

QH

TH

QL m TL

Therefore, the efficiency of a Carnot engine, or any reversible engine, can be expressed as ilth,rev

TL H

(4.19)

This relation is popularly known as the Carnot efficiency. • This is the highest possible efficiency for a heat engine operating between the two thermal-energy reservoirs at temperatures TH and TL.

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Fundamentals of Engineering Thermodynamics

• All irreversible heat engines operating between these two temperature limits, TH and TL, will have efficiencies less than that of the Carnot engine. This is due to the fact that an actual heat engine cannot be free from all irreversibilities associated with its cycle. It is important to note that the temperatures TH and 7'1 in Eq. (4.19) are absolute temperatures. The thermal efficiencies of the actual and reversible heat engines operating between the same two temperature limits compare as follows: < ilth,rev rith = rith.rev qth,rev

irreversible engine reversible engine

(4.20)

impossible engine

For assessment of the performance of actual heat engines, their efficiencies should be compared with the efficiency of a reversible heat engine operating between the same two temperature limits, since this is the true theoretical upper limit for the thermal efficiency, not the 100 per cent. Most work-producing devices of today have efficiencies less than 40 per cent. Further, it is seen from Eq. (4.19) that the efficiency of a Camot engine can be increased by increasing TH or decreasing TL. In fact, as TI approaches zero, the Carnot efficiency approaches unity. This is also true in respect of actual heat engines. The thermal efficiency of an actual heat engine can be maximized by supplying heat to the engine at the highest possible temperature and rejecting heat from the engine at the lowest possible temperature. Note that the maximum and minimum temperatures for a work-producing device are limited by the material strength of the devices and the temperature of the sink, respectively.

4.14 THE QUALITY OF ENERGY The quality of energy may be defined as its potential capacity to do work, in a broad sense. Let us examine a Carnot heat engine working between the temperature limits 1000 K and 300 K. By Eq. (4.19), the efficiency of this engine is 70 per cent. We now examine how 77th varies with TH when TL is held constant at 300 K. The variation of qd, with different values of TH, with TL fixed at 300 K, calculated from Eq. (4.19), is listed in Fig. 4.16. High-temperature reservoir at TH TH(K) 1000 800 600 400

1(%) 70.0 62.5 50.0 25.0

7

Low-temperature reservoir at Ti= 300 K

Figure 4.16 Efficiency of a Carnot engine at different values of T11.

The Second Law of Thermodynamics 101 It is clear from the thermal efficiency values in Fig. 4.16 that energy has both quality and quantity. Further it can be inferred that more of the high-temperature thermal energy can be converted to work. Thus, the higher the temperature, the higher the quality of the energy. We all know that the quantity of energy is conserved. But still we keep hearing speeches and seeing articles on how to "conserve" energy. Then what do we mean by "conservation" of energy? The answer to this question is that even though the quantity of energy is conserved, the quality or the work potential of energy is not conserved. Thus, our energy conservation slogans imply the conservation of the quality of energy. One unit of high-quality energy can be more valuable than many units of lower-quality energy. For example, a finite amount of heat energy at high temperature is more valuable for power generation than a vast amount of heat energy at low temperature, such as the energy stored in the upper layers of the oceans.

4.15 THE CARNOT REFRIGERATOR AND HEAT PUMP A Carnot refrigerator, or a Carnot heat pump, is a refrigerator or a heat pump, that operates on the reversed Camot cycle. The general relations for the coefficient of performance of refrigerators and heat pumps, given by Eqs. (4.9) and (4.11) are COPR —

1

(QH/a) - 1

and 1

COPHP = 1—

(a/Qin)

These relations are valid for both reversible and irreversible devices. But for reversible refrigerators or heat pumps, the QH/QL ratio is given by Eq. (4.17) as ( QH ) = TH

QL rev

TL Using this, the COPs for reversible refrigerators and heat pumps can be written as 1 COPR,,ev = (TH / TL ) -1

(421)

and

1

COPHp ,„ ' = 1— (TL/TH)

(4.22)

These are the highest possible coefficients of performance that a refrigerator or a heat pump operating between the temperature limits TH and Ti.. can have. Since these are the theoretical maximum limits, all actual refrigerators and heat pumps operating between the temperature limits of TH and TL will have COPs less than those of the reversible ones. Thus, we have < COPR.,„ irreversible refrigerator COPR = COPR.„„ reversible refrigerator > COP,, impossible refrigerator Replacement of COPR in Eq. (4.23) by COPE results in similar relations for heat pumps.

(4.23)

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Fundamentals of Engineering Thermodynamics

From Eqs. (4.21) and (4.22), it is seen that a decrease in the value of TL results in a decrease in COPs of both the refrigerator and the heat pump. That is, more work is required to absorb heat from lower temperature media. In the limiting case, as the temperature of the refrigerated space goes down to zero, the amount of work required to produce a finite amount of refrigeration goes up to infinity and COPR tends to approach zero. EXAMPLE 4A A heat engine operates between a source at 600°C and a sink at 20°C. Determine the least rate of heat rejection per kW net output of the engine. Solution For the least rate of heat rejection, the engine must operate on the reversible Carrot cycle. The efficiency of reversible engines given by Eq. (4.19) is 7i.

ritimev =1

TH

293.15 — 0.664 873.15

—1

Now, 7/th,rev =

Wnet QH

Therefore, QH =

0.664

— 1.5 kW per kW net output of the engine

The amount of heat rejected QL can be obtained from the relation = TH

QH ( QL

L,

TL

or = Qt"'`"

TH

QH

,rev

(293.15)(1.5)= 0.5 kW 873.15

The least rate of heat rejection per kW of net output is therefore 0.5 kW EXAMPLE 4.5 The minimum power required to drive a heat pump which maintains a house at 20°C is 3 kW. If the outside temperature is 3°C, estimate the amount of heat which the house loses per minute. Solution For minimum power requirement, the heat pump must operate on the reversible cycle. By the definition of COP (Eq. (4.10)), we have

The Second Law of Thermodynamics

103

COPHpm QH Wriet.in

Here, W,Kin = 3 kW Also by Eq. (4.22), =

1 1 — (TL/TH )

1 1 276.15

— 17.24

293.15 Therefore, QH = (COPHRrevXWnet.0= 17.24 x 3 = 51.72 kW = 51.72 kJ/s The amount of heat which the house loses per minute is therefore 51.72 x 60 = 3103.2 kJ

SUMMARY The second law of thermodynamics asserts that a process will not take place unless it satisfies both the first law and the second law of thermodynamics. In addition to ascertaining the direction of a process, the second law also asserts that energy has quality as well as quantity. The second law of thermodynamics is also used to determine the theoretical limits for the performance of many popular engineering systems, such as heat engines and refrigerators. It is even used for predicting the degree of completion of chemical reactions. A thermal-energy reservoir is a hypothetical body with a relatively large capacity for thermal energy that can supply or absorb fmite amounts of heat without undergoing any change in its temperature. Heat engines are devices to convert heat to work. They receive heat from a source, convert part of this heat to work, reject the remaining heat to a sink, while operating on a cycle. It is important to note that the term heat engine is often used in a broader sense to include all work-producing devices that do not operate on a thermodynamic cycle. Thermal efficiency is defined as the fraction of heat input that is converted to net work output. It is a measure of the performance of a heat engine. Thermal efficiency 7711, of a heat engine can be expressed as Thermal efficiency —

net work output total heat input

or tltn

— W> Qin

Or

qth = -

104 Fundamentals of Engineering Thermodynamics Thermal efficiency is a measure of the capacity of a heat engine to convert the heat (that it receives) to work. Usually, the spark-ignition automobile engines have a thermal efficiency of about 20 per cent. The thermal efficiency of diesel engines is about 30 per cent and that of large gas-turbine plants is also about the same. Steam power plants operate at about 40 per cent efficiency. The Kelvin-Planck statement of the second law is: It is impossible for any device that operates on a cycle to exchange heat with just a single reservoir and produce a net amount of work. The Kelvin-Planck statement can also be expressed as: For a power plant to operate. the working fluid must exchange heat with the environment as well as the furnace Or

No heat engine can have a thermal efficiency of 100 per cent. A refrigerator is a special device meant for transfer of heat from a low-temperature medium to a high-temperature medium. Like heat engines, refrigerators are also cyclic devices. The efficiency of a refrigerator in terms of its coefficient of performance (COP) can be expressed as 1

QL

COPR = QH

QL (QH/QL ) -1

From which it is seen that COPR can be greater than unity. A heat pump is a device that transfers heat from a low-temperature medium to a hightemperature medium. Unlike a refrigerator which is meant for maintaining the refrigerated space at a low temperature by removing heat from it, the objective of a heat pump is to maintain a heated space at a high temperature. The coefficient of performance of a heat pump is given by COPS =

QH QH

— QL

1 1 - /QH

Most of the heat pumps of present days have COPs between 2 and 3. Air-conditioners are basically refrigerators whose refrigerated space is very large. In modern terminology, the performance of refrigerators and air-conditioners is expressed in terms of energy efficiency rating which is the amount of heat removed, from the cooled space, in British thermal units (Btus) for I Wh (Watt-hour) of electricity consumed. The Clausius statement of the second law is: It is impossible to construct a device that operates on a cycle and produces no effect other than the transfer of heat from a lowtemperature body to a high-temperature body. Any device that violates either the first or the second law of thermodynamics is called a perpetual-motion machine. A reversible process is that process which can be reversed without leaving any trace on the surroundings. Processes that are not reversible are termed irreversible processes. An internally reversible process is that for which no irreversibilities occur within the boundaries of the system during the process. An externally reversible process is that for which no irreversibilities occur outside the system boundaries during the process. A process for which no irreversibilities occur either within the system or outside the system (i.e. its surroundings) is called a totally reversible process or simply a reversible process.

The Second Law of Thermodynamics 105 The Carnot cycle is a reversible cycle. Probably, it is the best known reversible cycle in thermodynamics. It is only an imaginary cycle. The theoretical heat engine that operates on the Carrot cycle is called the Carnot heat engine. The Carnot principles are the two conclusions pertaining to the thermal efficiency of reversible (ideal) and irreversible (actual) heat engines. They are: 1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two thermal reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two thermal reservoirs are the same. The first principle is called the Carnot theorem and the second the corollary of the Carnot theorem. A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Lord Kelvin defined thermodynamic temperature as

QUI = TH a rev TL This temperature scale is called the Kelvin scale, and the temperatures on this scale are called absolute temperatures. The magnitudes of temperature units on the Kelvin and Celsius scales are identical (1K = 1°C). The temperatures on the two scales are related by a constant 273.15 as T(°C) = T(K) — 273.15 The Carnot heat engine is the hypothetical engine that operates on the reversible Carnot cycle. The efficiency of a Carnot engine can be expressed as

, TL =— H This relation is popularly known as the Carnot efficiency. The thermal efficiencies of the actual and reversible heat engines operating between the same two temperature limits compare as follows:

nth

< nthsev

irreversible engine

= nth,rev

reversible engine

nthsev

impossible engine

The quality of energy may be defined as its potential capacity to do work. A Carnot refrigerator, or a Carnot heat pump. is a refrigerator or a heat pump that operates on the reversed Carnot cycle. The thermal efficiencies of the actual and reversible refrigerators operating between the same temperature limits compare as follows: 1 < COPR,,,v irreversible refrigerator COPR = COPR,„„ reversible refrigerator > COPR.,„ impossible refrigerator

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Fundamentals of Engineering Thermodynamics

PROBLEMS 4.1

A refrigerator removes heat at an average rate of 760 MJ/h from the cold compartment. If the coefficient of performance of the refrigerator is 4.0, determine (a) the power input to the refrigerator and (b) the amount of heat rejected by the refrigerator. [Ans. (a) 52.78 kW, (b) 263.89 kW]

4.2 A heat pump delivers heat at the rate of 200 MJ/h to a room. If the coefficient of performance of the pump is 4.0, determine the power required to run the pump. [Ans. 13.89 kW] 4.3 A refrigerator, operating in a cycle, removes heat from a low temperature reservoir at TL = 250 K and rejects heat to a high temperature reservoir at TH = 300 K. Determine whether this machine is reversible, irreversible, or impossible if (a) QL = 1 kJ, = 250 J, (b) QL = 2 kJ, QH = 2.4 kJ, and (c) QH = 3 kJ, Wne, = 500 J. [Ans. (a) COP < COPR,hence irreversible, (b) COP = COPR,„„, hence reversible, (c) COP = COPR,,,v, hence reversible] 4.4 A heat pump delivers heat at the rate of 10 kJ/s to heat a room maintained at 25°C, by receiving heat from a reservoir at —10°C. If the COP of the heat pump is 50% of the COP of an ideal heat pump operating between the same temperature limits, determine the power required to run the heat pump. [Ans. 2.353 kW] 4.5 A reversible heat engine receives heat from a high-temperature reservoir at TH and rejects heat to a reservoir at 1000 K. A second reversible heat engine receives the heat rejected by the first engine at 1000 K and rejects heat to a reservoir at 300 K. If the thermal efficiency of both the engines have to be the same, what should be TH? [Ans. 3333.3 K] 4.6 Two reversible refrigerators are connected in series. The fast one removes heat from a cold reservoir at TL and discharges heat to an environment at 10°C. The second refrigerator absorbs the heat discharged by the first at 10°C and discharges heat to an environment at 25°C. If the COPs of both the refrigerators are equal, determine TL. [Ans. 268.76 K] 4.7 A house is to be maintained at 20°C with a heat pump at all times in the winter. When the ambient temperature outside drops to —10°C, the rate at which heat is lost from the house is estimated to be 25 kW. What is the minimum electrical power required to drive the heat pump? [Ans. 2.56 kW] 4.8

A Carrot engine receives 620 kJ of heat from a source and rejects 270 kJ of it to a sink at 20°C. Determine (a) the thermal efficiency of the heat engine and (b) the temperature of the source. [Ans. (a) 0.565, (b) 673.6 K]

4.9

An inventor claims to have developed a heat engine which receives 950 kJ of heat from a source at 400 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 293 K. Is this a reasonable claim? Why? [Ans. Not reasonable, since the rhi,,,c,(0.316) > 77,1,,,,„(0.268), which is impossible]

The Second Law of Thermodynamics

107

4.10 An air conditioning system maintains a house at 20°C when the outside temperature is 35°C. If the air conditioning system draws 5 kW of power for its operation, determine the maximum rate of heat removal that it can provide from the house. [Ans. 5859 kJ/min] 4.11 A domestic refrigerator which runs one-third of the time removes on an average 1800 kJ/11 of heat from the food compartment. If the power consumed is 0.50 kW, determine its COP. [Ans. 3] 4.12 A household refrigerator has to freeze 10 kg of water to 0°C. If the water is initially at 20°C, how long it will take for the water to freeze. The COP of the refrigerator is 2.5 and the power input required to run it is 400 W. The specific heat of water is 4.2 kJ/(kg °C). [Ans. 14 min] 4.13 Two reversible heat engines are arranged in series, as shown in the diagram below. The work output of the first engine is twice that of the second. Determine 7'2, rh , rh and Q3. [Ans. 166.77°C, 43.1 per cent, 37.9 per cent, 106 Id]

4.14 A reversible heat engine receives 300 kJ from a heat source, delivers 100 kJ of work and rejects the balance heat to a heat sink. If the sink is at 30°C, determine the temperature of the heat source. [Ans. 181.55°C] 4.15 The efficiency of a Carnot engine rejecting 1200 kJ/min to a heat sink at 20°C is 33 per cent. Determine the temperature of the heat source and the power of the engine. [Ans. 164.4°C, 9.85 kW] 4.16 Determine the power required to run a refrigeration plant for keeping a cold storage at — 40°C by removing 7400 kJ/min and discharging it to the atmospheric air at 30°C. Assume the refrigeration cycle to be the reversed Carrot cycle. [Ans. 49.64 hp]

108 Fundamentals of Engineering Thermodynamics 4.17 A heat pump working on the Carnot cycle maintains the inside temperature of a house at

22°C by supplying 45,000 kJ/h. If the siutside temperature is 0°C, determine the heat taken from the outside air and the power required to run the pump. [Ans. 41,645.8 kJ/h, 1.25 hp] 4.18 A refrigerator having a COP of 4.5 is run by an engine of 28 per cent thermal efficiency. Determine the heat input into the engine for each kJ of heat removed from the refrigerated space. [Ans. 0.794 kJ] 4.19 An automobile engine with a power output of 75 kW has a thermal efficiency of 30 per cent. If the heating value of the fuel used is 40,000 kJ/kg, determine the fuel consumption rate. [Ans. 6.25 g/s] 4.20 A steam power plant with a thermal efficiency of 20 per cent consumes coal at a rate of 55 tonnes/h. If the heating value of the coal is 27,000 kJ/kg, determine the power output of the plant. [Ans. 82.5 kW] 4.21 Determine the Carnot efficiency of an engine working between thermal reservoirs at 950°C and 20°C. [Ans. 76 per cent] 4.22 A reversible heat engine operating between thermal reservoirs at 800°C and 30°C drives a reversible refrigerator which refrigerates a space at —15°C and delivers heat to a thermal reservoir at 30°C. The heat input to the heat engine is 1900 kJ and there is a net work output from the combined plant (heat engine and refrigerator) of 290 Id. Determine the heat transfer to the refrigerant and the total heat transfer to the 30°C thermal reservoir. [Ans. 5099 kJ, 6709.4 kJ] 4.23 A Carrot heat engine operates between two heat reservoirs at 300°C and —5°C. If the engine receives 120 kJ of heat from the source, find the net work done and the heat rejected to the sink. Also, calculate the thermal efficiency of the heat engine. [Ans. 63.84 kJ, 56.16 kJ, 53.2 per cent] 4.24 A reversible heat engine operates under two environments. In the first, it draws 12,000 kJ/s from a thermal source at 400°C and in the second environment, it draws 25,000 kJ/s from a thermal source at 100°C. In both the operations, the engine rejects heat to a thermal sink at 20°C. Determine the operation in which the engine delivers more power. [Ans. In the first operation] 4.25 A cold storage plant requires 30 tonnes of refrigeration. The freezing temperature is —20°C and the ambient temperature is 30°C. If the performance of the refrigeration plant is 25 per cent of the Carnot coefficient of performance of the plant working between the same temperature limits, determine the power supply required to run the plant. One tonne of refrigeration is equivalent to 211 kJ/min. [Ans. 83.07 kW] 4.26 A heat pump maintains the inside temperature of a house at 22°C. The outside temperature is — 3°C. The heating load is estimated to be 100,000 kJ/h. Calculate the theoretical minimum power required to run the heat pump. Neglect heat losses. [Ans. 3.15 hp]

The Second Law of Thermodynamics 109

4.27 During winter the inside temperature of a house is maintained at 21°C by a heat pump consuming a power of 6 kW. The heat loss rate from the house is 5000 kJ/h per °C temperature difference. Determine the lowest outside temperature for which the heat pump can maintain the house temperature at 21°C. [Ans. —14.68°C] 4.28 A heat pump is proposed to maintain a house at 25°C by extracting heat from the outside air on a day when the outside air temperature is 0°C. The house is estimated to lose heat at a rate of 100,000 kJ/h, and the heat pump consumes 6 kW of electric power. Is this heat pump powerful enough to meet the heating requirements of the house? [Ans. Yes] 4.29 A gas turbine has an efficiency of 17 per cent. If it develops a power output of 6000 kW, determine the fuel consumption rate of this gas turbine when a fuel of heating value 46,000 kJ/kg and density 0.8 is used. [Ans. 57.54 litres/min] 430 A Carnot heat engine is operating between a source at TH1 and a sink at TL. If the thermal efficiency of this engine has to be tripled by increasing the source temperature while the sink temperature TL is maintained at the same level, what should be the new source temperature Tm in terms of THI and TL? [Ans. ( THi TL)/(3TL — 271-a

CHAPTER

5 Entropy 5.1 INTRODUCTION The second law of thermodynamics introduced in Chapter 4 leads to the definition of a new property called entropy. Entropy is an abstract property, and it is difficult to give a precise physical definition to it. Entropy is best understood by studying its uses in commonly encountered work-producing practical processes. We will follow this method to understand entropy. Unlike energy, entropy is a non-conserved property. There is nothing like conservation of entropy. We will begin our discussion with the Clausius inequality, which forms the basis for entropy and continues with the principle of the increase-of-entropy.

5.2 THE CLAUSIUS INEQUALITY We have seen in Chapter 4 that the second law of thermodynamics often leads to expressions that involve inequalities. For example, an actual (i.e. irreversible) heat engine is less efficient than a reversible heat engine, operating between the same two thermal reservoirs. Another important inequality that has major consequences in thermodynamics is the Clausius inequality, stated by the German physicist R.J.E. Clausius (1822-1888), one of the founders of thermodynamics, which is expressed as QL,rev

This may also be expressed as QL,irrev = QL,rev + aliff

where Qdjff is a positive quantity. Performing the cyclic integration of 5QIT for this irreversible heat engine, we obtain QH TH

(1 8 ley

QL,irrev TL

=QH — QL.rev Qdiff

TH TL TL =

Qairr

0 irreversible process = 0 reversible process . < 0 impossible process

This is the criterion for determining whether a process is reversible, irreversible, or impossible.

5.3.6 The Increase-of-Entropy Principle for Closed Systems The change in entropy AS for a closed system is simply the difference between the initial and final entropies of the system, since a closed system involves no mass flow across its boundary. The entropy of the surroundings of a closed system will be affected by a process, only if the system exchanges heat with the surroundings during the process. The surroundings can usually be treated as the thermal energy reservoir at Ts„,T, and the change in its entropy AS is related to heat transfer by the relation (kJ/K)

AS = o

The increase-of-entropy principle for closed systems is by Eq. (5.15) expressed as Sgen = A.Stot = ASsys + ASsu, ?. 0 (kJ/K) where ASsy, = S2 — = M(S2 — S ) and ASsun.

ant. Tun

122 Fundamentals of Engineering Thermodynamics The subscripts 1 and 2 designate the initial and final states of the system and m is the mass of the system. For an adiabatic process, Q0,,, = — Q,,, = 0 and the increase-of-entropy principle reduces to r„ = ASsot = m(s2 — si ) 0

(5.16)

This implies that the entropy of a closed system can never decrease during an adiabatic process. EXAMPLE 5A One kilogram of water at 273 K is brought into contact with a heat reservoir at 373 K. (a) When the water has reached 373 K, find the change in entropy of the water, of the heat reservoir, and of the universe. (b) If the water had been heated from 273 K to 373 K by first bringing it in contact with a reservoir at 323 K and then with a reservoir at 373 K, what would have been the change in entropy of the universe? Solution (a) Let us take the water and the heat source as our system and the surrounding atmosphere as the surroundings, as shown in the adjacent figure. The heating of water can be treated as a constant 7 Heat /, source volume process since water is incompressible. Thus, for As Atmosphere // water, the change in entropy is given by

0

Water

S = 8Q = mC In —1 1. For water at 273 K, C, = 4.217 kJ/(kg K) from standard tables. Thus,

= 4217 (In -373 ) = 1316 J/K 273 The heat loss by the source 8Q = heat gained by the water = mC,AT = 4217 x 100 Therefore, the change in entropy of the heat source is A Sheet source =

—8Q T

since (5Q is negative.

—421,700 —1130 J/K 373 The total change in entropy of the system = 1316 — 1130 = 186 J/K. This much entropy is added to the entropy of the universe, since the system considered is part of the universe. Thus, the entropy change of the universe is 186 J/K (b) Here the heat addition to water is done in two stages. In the first stage, the water temperature increases from 273 K to 323 K and in the second stage the temperature goes up from 323 K to 373 K. For stage one at 273 K, C,, = 4.217 kJ/(kg K) for water and for stage two at 323 K, C, = 4.180 kJ/(kg K).

Entropy 123

Therefore, for water A Si =

In

T,

= 4217 In

323 = 709.2 J/K 273

and 373 A S2 = 4180 In — = 601.6 J/K 323 Therefore, the net change in entropy of water is AS,„„„, = AS1 + AS2 = 709.2 + 601.6 = 1310.8 J/K For the heat source, the change in entropy for stage one is A =

-5Q -mC„(AT) -4217 x 50 = 652.8 Jac = = T T 323

And for stage two, the change in entropy is AS2 -

-4180 x 50 - 560.3 J/K 373

Thus, the net change in entropy of the heat source is ASh,m

= - 652.8 - 560.3 = -1213.1 J/K

The change in entropy of the universe is, therefore = 1310.8 - 1213.1 = 97.7 J/K

5.3.7 The Increase-of-Entropy Principle for Control Volumes The second-law analysis of control volumes (open systems) is similar to that for closed systems, but for this case we have to consider the mass flow across the boundary of the control volume along with other terms. The mass flow that enters or leaves a control volume carries some entropy (as well as energy) with it, and therefore the mass transfer will affect the entropy content of both the control volume and the surroundings. With this consideration, the change in entropy of a control volume and its surroundings can be expressed, respectively, as AScv = (S2 - SI)cv

(5.17)

ASSUIT = 9 -11-r + Se - S.

(5.18)

and

Tsun.

where the subscripts 1 and 2 designate the initial and final states, respectively, in the control volume and subscripts i and e refer to the states at the inlet and exit of the control volume, respectively. Si is the total entropy transported from the surroundings into the control volume, with the mass entering the control volume and Se is the total entropy transported out of the control volume into the surroundings, with the mass leaving the control volume. Note that in Eq. (5.17), the influence of Si and Se on AScv is reflected in 52. Substituting

124 Fundamentals of Engineering Thermodynamics Eqs. (5.17) and (5.18) into Eq. (5.15), we get the equation for the increase-of-entropy principle for control volumes as Sgen =

AStotai = (S2 — Si)cv

Se —

+

QUIT > 0 (kJ/K) — T.

(5.19)

In general, the fluid properties at an inlet or an exit may vary during a process. For such cases, the transported entropy terms should be determined by integration, that is, S = EL

.5',5 nii

and

S, = E J S,8 me With these relations for Si and Se, the general form of the increase-of-entropy principle for a control volume becomes Sgen

= (S2

)CV +

f

0,c

.3,8 me — EL Sig

+ QUIT 0 (kJ/K) su,

(5.20)

5.3.8 The Uniform Flow Process For this case, the state of the control volume changes uniformly, and the fluid properties at any inlet or exit are assumed to remain constant. Therefore, Eq. (5.20) simplifies to Sgen = ( M252

Si )0/ + E MeSe

E m,s, + QUIT Ts=

0 (kJ/K)

(5.21)

5.3.9 The Steady Flow Process For steady flow processes the properties anywhere within the control volume, including the inlets and the exits, remain constant with respect to time during the process. Therefore, AScv = 0 and the total entropy change for a steady-flow process becomes the entropy change of the surroundings only. Therefore, the increase-of-entropy principle for steady-flow processes becomes the total entropy change of the surroundings only. Thus, we have Sgen = ASsurr =

(Se — Si) +

QUIT -

IS=

(k-IIK)

(522)

that is, ;en=

L1SsuR = mese - MoSi

0

The rate of change of total entropy generation associated with a control volume undergoing a steady-flow process can be written as Sgen

= Z tite Se

— E ihr S +

0 (kW/K) TSUIT

(5.23)

Entropy

125

For a single-system steady-flow device, this relation reduces to =

suff > 0 (kW/K) TSUIT

(5.24)

%tin = Se — Si + — 0 [kJ/(kg K)]

(5.25)

On unit mass basis, Eq. (5.22) becomes s

m Ts=

From the relations of the increase-of-entropy principle for flow processes through a control volume, it is evident that the entropy of a fluid will increase because of irreversibilities as it flows through an adiabatic steady-flow device. It is essential to note that •

is not a property. Its value depends on how the process is executed, and thus it is a path function, like heat and work.

Sgen

• ..Sze„ is the rate of change of total entropy, or the rate of entropy generation. • The equality of the relations of the increase-of-entropy principle or that of the relations of entropy balance applies to reversible processes, and the inequality to irreversible processes.

5.3.10 The Concept of Lost Work We have seen that the amount of entropy generated is a measure of thermodynamic irreversibility. The irreversibility associated with a thermodynamic process can also be quantified with the concept known as lost work. By definition, lost work Win, is Wlost = TRSgen

where TR is a reference temperature. Since Sgen is always positive, lost work is also always positive for all real processes. It is essential to note that Sge„ depends solely on the degree of thermodynamic irreversibility of the process involved. On the other hand, the amount of lost work is a relative quantity that depends on the choice of the reference temperature TR. If the reference temperature is taken as the absolute temperature of the environment To, then the lost work is called the thermodynamic lost work, to distinguish it from the lost work using any other reference temperature. That is, the thermodynamic lost work Wcon„, by definition, is = To Sgen It can be shown that the thermodynamic lost work is identical to the concept of lost availability or exergy destruction. In Chapter 6, it will be seen that Wo.ins, is nothing but irreversibility. EXAMPLE 53 From a thermal reservoir at 1200 K, 2500 Id of heat is transferred to another reservoir at 600 K. Determine the amount of entropy generated for the process, and the amount of lost work with the environment at 30°C.

126

Fundamentals of Engineering Thermodynamics

Solution Let us consider the two reservoirs as our system. The entropy generated is Sgen = AS1 + AS2 where AS1 is the change in entropy of reservoir 1 and AS2 that of reservoir 2. AS1

(8Q) T )1

1200

2.083 kJ/K

AS2 = (5Q) = 25: — 4.167 kJ/K T 2 6 Thus, Sim = —2.083 + 4.167 = 2.084 kJ/K The amount of lost work is Wo,k, t = To Sgen = (30 + 273.15)(2.084) = 631.76 kJ

5.3.11 Causes of Entropy Change From our discussion on entropy so far, it is clear that heat transfer, mass flow, and irreversibilities are the three mechanisms which cause the entropy of a system to change. Heat transfer to a system increases the entropy of that system, and heat transfer from a system decreases the entropy of that system. This effect is illustrated schematically in Fig. 5.9. In fact, this is the only way to decrease the entropy of a closed system. Entropy change as a result of reversible heat transfer is called entropy flow or entropy transfer, and it does not involve any entropy generation. Therefore, the entropy of the universe does not increase as a result of entropy flow. A

Heat

11r ASB > 0 ASA < 0 B

Figure 5.9 Entropy change with heat transfer. Mass flow transports both energy and entropy into or out of a control volume. Entropy transfer via mass flow is called entropy transport. For closed systems the entropy transport is zero, since there is no mass flow involved. Irreversibilities such as friction, heat transfer through a finite temperature difference, and fast expansion or compression always cause the entropy to increase. Thus, the entropy of a system cannot decrease during an adiabatic process. The entropy generation Sgen = 0 for a reversible process. The entropy of a system, which undergoes a process involving no heat transfer (adiabatic) and no irreversibilities within the system (internally reversible), must remain constant during that process. Such a process is called an

Entropy 127

isentropic process or internally reversible adiabatic process. It is obvious from this fact that it is impossible to have a reversible process. An isentropic process is also an idealization, like the quasi-equilibrium process discussed in Section 1.9. However, the isentropic process serves as a model for actual processes. In a nutshell, it can be stated that: • A process can occur in a certain direction only such that AStotai 0 and any process that violates this principle is impossible. • Entropy is a non-conserved property, and there is no such thing as the principle of the conservation of entropy. • The performance of engineering systems is degraded by the presence of irreversibilities, and the amount of entropy generated is a measure of the magnitudes of the irreversibilities present during that process.

5.3.12 What is Entropy? We have so far seen that entropy is a useful property and serves as a valuable tool in the second-law analysis of engineering devices. This understanding, however, does not mean that we have understood entropy well. In fact, with our discussions so far, it may not be possible to give an answer to the question—what is entropy? A close look at the microscopic nature of matter will give us a better idea about this useful property. We, therefore, now look at entropy from microscopic point of view. Entropy can be viewed as a measure of disorder or molecular disorder or molecular randomness. As the system becomes more disordered, the positions of the molecules become less predictable and the entropy increases. Thus, the entropy of a substance is the lowest in the solid phase and the highest in the gas phase. The gas molecules possess considerable amount of kinetic energy. But we know that gas molecules will not rotate a paddle inserted into a gas tank, as shown in Fig. 5.10, and produce work, however large may be their kinetic energy. This is due to the fact that the energy of the gas molecules, is disorganized. Probably, the number of molecules trying to rotate the wheel and the number of molecules trying to prevent the wheel from rotating match all the time, thus not enabling the wheel to rotate. This demonstrates that we cannot extract any useful work directly from disorganized energy. In other words, disorganized energy does not create much useful effect, no matter how large it may be.

Figure 5.10 A paddle wheel in a gas tank.

128 Fundamentals of Engineering Thermodynamics As another example, we consider a rotating shaft as shown in Fig. 5.11. The energy of the molecules is completely organized since the molecules of the shaft are rotating together in the same direction. The organized energy can be used to perform useful tasks such as lifting a weight, generating electricity, and so on. The work, being an organized form of energy, is free of disorder or randomness and thus free of entropy. There is no entropy transfer associated with energy transfer as work. Assuming that there is no friction present in the rotating shaft process, we can say that there is no entropy associated with this process. Hence, this process can be reversed and used for lowering the weight, since any process that does not produce a net entropy is reversible. Therefore, during this process the energy is not degraded and the potential to do work is not lost. W$h

Weight Figure 5.11 A rotating shaft used for lifting a weight. Let us now consider the operation of the paddle wheel of Fig. 5.10 with a shaft work Wsh. The paddle wheel work will be converted to the internal energy of the gas, resulting in a higher level of disorder or molecular chaos in the container. In this case, only a portion of the supplied energy can be converted to work by partially reorganizing it through the use of a heat engine. Therefore, energy is degraded during this process and the ability to do work is reduced. The molecular disorder produced results in increase of entropy of the system. From this example: • It is evident that even though the quantity of energy is always preserved during an actual process, its quality is bound to be degraded. • This degradation in quality is always accompanied by an increase in entropy. Heat is a form of disorganized energy, and the flow of heat always carries some entropy with it. This causes the entropy, or the level of molecular disorder, of a hot body to decrease and that of a cold body to increase. The second law requires that the increase in entropy of the cold body be greater than the decrease in entropy of the hot body, thereby allowing the net entropy of the combined system to increase. Thus, processes can take place only in the direction of increase of overall entropy or molecular disorder. Thus, from the microscopic or statistical point of view, entropy is a measure of molecular randomness, i.e. the uncertainty about the positions of molecules at any instant. Even in solid phase, the molecules of a substance are in a state of random motion. However, these molecular oscillations subside as the temperature is decreased and eventually the molecules freeze at

Entropy

129

absolute zero temperature. This represents a state of ultimate molecular order or minimum entropy. Therefore, the entropy of a pure crystalline substance at absolute zero temperature is zero, since there is no uncertainty about the state of the molecules at that instant. A pure substance is one which has a fixed chemical composition throughout. The above statement about the zero entropy state is known as the third law of thermodynamics. It provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called the absolute entropy. The absolute entropy plays a dominant role in the thermodynamic analysis of chemical reactions. EXAMPLE 5.6 Two identical bodies of constant heat capacity C at temperatures Ti and T2, respectively, are used as reservoirs for a heat engine. If the bodies remain at constant pressure and undergo no change of phase, show that the amount of work obtainable is W = C(7'1 + T2 — 2 7j) where Tf is the fmal temperature attained by both bodies. Also, show that for maximum work output T1 = friT; Solution Let T1 be the temperature of the source and T2 that of the sink. Heat supplied by the source per unit mass of working substance is = — C(Tf — T1) since the heat going out of the body is negative. Heat received by the sink is qL = C(7'f — T2) Assuming no heat loss from the engine, the net work delivered is W= 9H — 91 =—

Tf —

—

Tf — T2)

For maximum work output, the heat engine cycle should be reversible. That is, the heat transfer process should be reversible. In other words, Ass„, = eSSO1RCe + ASsink = 0 Now,

essource = and ASsink = C In

T,T2

130 Fundamentals of Engineering Thermodynamics Therefore, Tr 7', AS,,,, = C ln —1— + C In —/— T1 T2 = C In 12— = 0 7i7; That is, T2

f =1

TT

Or

rf = 7 7 ; For maximum work output we, therefore, obtain rf = 1172 7 EXAMPLE 5.7 A 10 kg metal piece with constant specific heat of 0.9 kJ/(kg K) at 200°C is dropped into an insulated tank which contains 100 kg of water at 20°C. Determine the final equilibrium temperature and the total change in entropy for this process. Solution We take the metal piece and the water as our system. Applying the principle of conservation of energy for this process, we get Q— W = AU + AKE + APE where Q= 0, W = 0, AKE = 0 and APE= 0. Therefore, AUnr i +AUK =0 For liquid water, specific heat = 4.184 kJ/(kg K). Thus, 10x 0.9(T2 — 200) + 100 x 4.184(T2 — 20) = 0 where T2 is the final temperature. Solving this, we get T 10168 2 = 427.4

23.79°C

The entropy generation is given by AS

. 94) = mC„ In a =10 x 0.9 x In 1 473.15 i = — 4.192 kJ/IC

AS

=100 x 4.184x In (

296.94) 293.15)

= 5.375 kJ/K Thus,

Astoud = As..1 +

=

1.183 kJ/K

Entropy 131

5.4 PROPERTY DIAGRAMS We will use the property diagrams extensively in Chapter 7, namely the P-v and T-v diagrams in conjunction with the first law of thermodynamics. They are helpful visual aids in the thermodynamic analysis of processes. Similarly, the property diagrams with entropy as one of the coordinates, namely the T-S and h-S diagrams are used extensively in the second-law analysis of thermodynamic processes.

5.4.1 The

T-5 Diagram

Equation (5.5) which defines entropy can be expressed as Vi„tre,„ = TdS (kJ)

(5.26)

The total heat transfer during an internally reversible process can be determined by integrating the above equation, which gives

= Jl TdS (kJ)

(5.27)

This corresponds to the area under the process curve on a T-S diagram, as shown in Fig. 5.12. T

Internally reversible process

dA = TdS = 80

Area

r

TdS =Q

S Figure 5.12 The T-S diagram for an internally reversible process. Therefore, it can be concluded that the area under the process curve on a T-S diagram represents the internally reversible heat transfer. It is important to note that this has no meaning for irreversible processes. On unit mass basis, Eqs. (5.26) and (5.27) can be expressed as = Tds (kJ/kg)

(5.28)

= Tds (kJ/kg)

(5.29)

or Sint rcv

The relation between T and s should be known in order to evaluate the above integral. The

132 Fundamentals of Engineering Thermodynamics internally reversible process is a special case for which these integrations can be performed to result in QinLrev

= To AS (kJ)

9i„,,,„ = To As (kJ/kg) In the above two relations, To is the constant temperature and AS is the change in entropy of the system during the process. In Eqs. (5.26) to (5.29), T is the absolute temperature, which is always positive. Therefore, during an internally reversible process, the heat transfer is positive when entropy increases and negative when entropy decreases.

5.4.2 The Isentropic Process A process during which the change in entropy is zero is called an isentropic process. That is, for an isentropic process, the entropy of the system remains constant. The process diagram of an isentropic process on a T—S diagram is essentially a vertical line segment as shown in Fig. 5.13. T

1 Isentropic process 2

S2 = Figure 5.13 The

S

T—S diagram of an isentropic process.

The T—S diagram for the liquid and vapour regions of a pure substance is as shown in Fig. 5.14. It is seen from these process curves that: • At any point in the single phase region, the constant-volume lines are steeper than the constant-pressure lines. • The constant-pressure lines are parallel to the constant-temperature lines, in the saturated liquid-vapour mixture region. • The constant-pressure lines coincide with the saturated liquid line, in the compressed liquid region.

Entropy 133 T, °C

Critical state

Saturated liquid line

Saturated vapour line 0

4

2

6

8

s, kJ/(kg K)

Figure 5.14 The T—s diagram for water.

5.4.3 The

h—s Diagram

The h—s or enthalpy-entropy diagram is quite useful in the analysis of steady-flow devices, such as compressors, turbines, and nozzles. A typical h—s diagram for an adiabatic process through a steady-flow device is shown schematically in Fig. 5.15. In the h—s diagram, Air is a measure of work and As is a measure of irreversibilities. The process diagrams on h—s coordinates are also h 1 •

Figure 5.15 The h—s diagram for a steady adiabatic process. called Mollier diagrams. The general features of an h—s diagram are illustrated in Fig. 5.16. Note that:

134

Fundamentals of Engineering Thermodynamics h, kJ/kg

3200

3000

2800

2600

2400

Figure 5.16 The h—s diagram for water. • The Tom, lines are straight in the saturated liquid-vapour mixture region. They become almost horizontal in the superheated vapour region, particularly at low pressures. This is not surprising since steam approaches ideal-gas behaviour as it moves away from the saturation region, and for ideal gases h is a function of T only. The details about the liquid-vapour mixture region, and liquid line, etc. are given in Chapter 7.

Note:

5.5 THE

TdS RELATIONS

The conservation of energy equation for an internally reversible process of a closed system consisting of a simple compressible substance can be expressed, using Eq (2.19), as But by Eq. (5.26),

tin*rev — OWint.rev = dU eQint,rev = TdS

and by Eq. (2.6), PdV

Therefore, the energy conservation equation becomes TdS = dU + PdV

Entropy 135 On unit mass basis, we have Tds = du + Pdv

(530)

This equation is known as the first Tds, or Gibbs equation. Note that the boundary work is the only type of work interaction that a simple compressible substance may involve as it undergoes an internally reversible process. Eliminating du from Eq. (5.30) and the enthalpy relation h = u + Pv, we get the second Tds equation, namely Tds = dh — vdP

•

(531)

Since both Eqs. (5.30) and (5.31) are property relations, they are independent of the type of the processes. The Tds relations are developed for internally reversible processes, even though entropy is a property and the change in a property between two states is independent of the type of process the system undergoes. The Tds equations are relations between the properties of a unit mass of a simple compressible system as it undergoes a change of state. They are applicable to both closed and open systems. By solving Eqs. (5.30) and (5.31), the explicit relation for differential changes in entropy can be obtained as du Pdv ds = + (532) T T and dh vdP ds = — (533) T The integration of these equations between the initial and final states will give the change in entropy during a process. For systems involving more than one mode of quasi-equilibrium work interactions, namely non-simple systems, the Tds relations can be obtained in a similar manner by including all the relevant quasi-equilibrium work modes. EXAMPLE 5.8 A simple compressible substance inside a cylinder undergoes a change of state quasi-statically, isothermally, and at constant pressure. If the enthalpy change for the system is 3344 kJ at 140°C, determine the corresponding change in its entropy. Solution By Eq. (5.33), we have the entropy change for a simple compressible substance as dh vdP ds = — — T T For a constant pressure process dP = 0, therefore, dh T

ds = — Thus, the entropy change for the system is dS =

3344 dH = T 140 + 273.15

8.094 kJ/K

136 Fundamentals of Engineering Thermodynamics

5.6 ENTROPY CHANGE FOR PURE SUBSTANCES The Tds relations developed are valid for all pure substances at any phase or combination of phases. However, the use of these relations depends on the availability of the property relations between T and du or dh and the P-v-T behaviour of the substances. Usually, for a pure substance, these relations are too complicated. This makes it impossible to obtain simple relations for entropy changes. The values of entropy, therefore, are determined from measurable property data following some complicated procedures and are tabulated in the same manner as the properties u, v, and h. The value of entropy it a specified state is calculated just as any other property. In the compressed liquid and superheated vapour regions, it can be determined directly from the tables. In the saturated mixture region, it is obtained from the relation s = sf + xsfg where x is the quality (refer Chapter 7) and sr and sfg values are given in the saturation tables. In the absence of compressed liquid data, the entropy of a compressed liquid can be approximated by the entropy of the saturated liquid at the given temperature. That is, S®p, The entropy change of a pure substance during a process is simply the difference between the entropy values at the final and initial states, that is AS = m(s2 - si )(kJ/K) On unit mass basis, we have As = (s2 - si )[ld/(kg K)]

5.7 ENTROPY CHANGE FOR SOLIDS AND LIQUIDS Solids and liquids can be idealized as incompressible substances and thus, dv = 0. Therefore, Eq. (5.32) becomes ds = du Also, du = CdT, since Cp = C, = C for solids and liquids. Then, we have s2 - s1 =

C(T) f-. [kJ/(kg K)]

(5.34)

The specific heat C for liquids and solids, in general, depends on T. However, in many cases, C may be treated as a constant at some average value over the given range of temperature. Therefore, we have s2 — si = C.,„ In Z [kJ/(kg K)] T

(535)

Entropy

137

EXAMPLE 5.9 Determine the specific entropy of water at 100°C. Solution By Eq. (5.35), we have the entropy change for liquids from state 1 to 2 as 7' In -I

s, - =

where subscripts 1 and 2 refer to initial and final states. Let the final temperature be the given temperature of 100°C. Now we need to fix the temperature of the initial state. If we choose a temperature 7'1 such that s1 = 0, then it becomes easy to calculate the entropy change with the above equation. In the case of steam, this arbitrary zero is chosen as the triple point, whose temperature is 273.15 K. Hence, in the above equation, si = 0 when Ti = 273.15 K Therefore, s2-0=

Ca° In

7' 2 273.15

For water C., is assumed to be 4.187 kJ/(kg K). Therefore,

s2 = 4.187 In

T

273.15

Note that Ca, for water actually varies with temperature, but 4.187 kJ/(kg K) is an average value at normal low range temperatures. At T2 = 100°C = 100 + 273.15 = 373.15 K s2 = 4.187 In

373.15 273.15

1306 J/(kg K)

From the saturated water--temperature table (Table 3 of the Appendix), we find that the entropy of water at 100°C is listed as 1306.9 J/(kg K)

5.7.1 Isentropic Processes for Solids and Liquids For this case, from Eq. (5.35), we have T2 =0 Ca, In — This gives T2 =

T

That is, the isentropic process of an incompressible substance is also isothermal.

1 138 Fundamentals of Engineering Thermodynamics

5.8 ENTROPY CHANGE FOR IDEAL GASES For ideal gases, du = CvdT and Pv = RT. Therefore, Eq. (5.32) becomes dT ds = C,— + Rdv T

v

This results in r2 v2 dT s2 - st = j Cv(T)— + R 111 — T v,

(536)

Similarly, Eq. (5.33) with dh = CpdT results in 2 dT P2 32 —s, = 1 Cp(T) — - R ln

T

(537)

g

For all ideal gases, except monatomic gases, the specific heats depend on temperature. Therefore, C, and Cp must be known as functions of temperature for evaluating the integrals in Eqs. (5.36) and (5.37). But performing the integrals every time, even when C„(T) and Cp(T) functions are available, is a difficult task. To overcome this difficulty, either the integrations are performed by assuming C, and Cp as constants, or the integrals are evaluated and kept as printed tables.

5.8.1 Constant Specific Heats Even though it is an approximation to assume Cv and Cp as constants, the error introduced by this assumption is very small for temperatures up to 800 K. For most of the practical processes, the temperatures involved are less than 800 K and thus this approximation proves to be useful in many cases. With C, and Cp as constants, Eqs. (5.36) and (5.37) result in TZ

(5.38)

T P s2 - si = Cp In -2 - R In 1 [kJ/(kg K)] Ti g

(539)

7;

+ R In

v2

[kJ/(kg K)]

s2 — S1 = Ci, In

v,

and

EXAMPLE 5.10 Air is compressed from 1 atm at 15°C to 6 atm at 110°C, in a steady-flow device. (a) Determine the entropy change for the air. (b) If there is heat transfer between the air and the environment at 15°C, and the actual shaft work input is 200 kJ/kg, determine the entropy change for the overall process. Solution (a) The entropy change for air, being an ideal gas, is given by Eq. (5.39) as s2 -s, = As„fr = Cp In

a - R ln P2 li

li

Entropy 139

For air, R = 287 J/(kg K) and Cp =

r R - x 287 =1004.5 J/(kg K) y —1 0.4

Therefore, Asair .1004.5 in 11 15 0+ 273.15

287 In 6)

T)

— 228 J/(kg K) Note that this entropy change is negative. This does not contradict the second law, because the process is not adiabatic. Further, this As is only for the system and if we consider the entropy change for the system and surroundings together, it will be seen that the entropy change for the combination is positive. (b) Now let us consider the entropy change for the system and the surroundings. Our system is the control volume of the steady-flow device in which air is being compressed and the surroundings is the environment at 15°C. The environment is a thermal reservoir and, therefore, its temperature will remain at 15°C in spite of the heat transfer from the system to the surroundings. Thus, &ammo =(!) — (=q 7L T e„,fron )rnviron The energy conservation equation for the system, neglecting the changes in potential and kinetic energies, is q—w=Ah=CP AT Thus, q = 1004.5(110 — 15) — (200 x 103) = — 104.57 kJ/kg That is, heat is transferred from the system to the surroundings. Now, A saiviron

= Tinvinm

_ 104.57 x 103 288.15 — 362.9 J/(kg K)

Note that the positive value of q is used in Ase„ • . Since the environment receives heat, and heat input is taken positive as per our convention, therefore, 46Stotal = fair '6Scnviron

= — 228

+ 362.9

134.9 J/(kg K)

5.8.2 Variable Specific Heats For these cases, the variation of C,. and Cp with temperature should be properly accounted for by utilizing proper relations for the specific heats as functions of temperature.

140 Fundamentals of Engineering Thermodynamics For tabulating the integrals in Eqs. (5.36) and (5.37), the following method is generally adopted: Choose a reference temperature of zero degree absolute and define a function s° as dT s° = Cp(T)— T

(5.40)

As per the definition, s° is a function of T alone, and its value is zero at absolute zero temperature. The integration of Eq. (5.40) between states 1 and 2 becomes

Ji

,,rx dT Cpki )— =s2 —

where 4 and .5.° are the values of s° at T2 and T,, respectively. Therefore, the entropy change during a process from state 1 to 2 can be expressed as s2 — s, = s2°—s°1 — R In

[1c1/(kg

(5.41)

The entropy changes given by Eqs. (5.38), (5.39) and (5.41) can also be expressed on a unit-mole basis by multiplying them by molar mass. Note that, unlike internal energy and enthalpy, the entropy of an ideal gas varies with specific volume v or pressure P as well as with the temperature. Therefore, the entropy cannot be tabulated as a function of T alone. The s° values in the tables account for temperature dependence of the entropy. The variation of s with P is accounted for by the last term in Eq. (5.41). EXAMPLE 5.11 Oxygen gas is heated at constant pressure from 420 K to 600 K. Determine the entropy change for the heating process. Solution For the constant pressure process, by Eq. (5.41), we have the entropy change as As =

4-

where subscripts 1 and 2 refer to initial and final states and s° is the entropy at any temperature T, and its value is zero at absolute zero temperature. From Table 10 of the Appendix, we can find s° and

.4

as = 215.241 kJ/(kmol K) s2° = 226.346 kJ/(kmol K)

Therefore, As = 226.346 — 215.241 = 11 .1 kJ/(kmol K)

Entropy 141

5.8.3 Isentropic Processes of Ideal Gases (a) Ideal gases with constant specific heats For this case, by Eq. (5.38), we obtain in

2 ..,

R in vx vi

Or ( VI

( T2

(5.42) v2

s=const

since for ideal gases, R = Cp — C, y= Cp/C,,, and thus R/C, = y — 1. Equation (5.42) is called the first isentropic relation for ideal gases. Similarly, with Eq. (5.39), we can obtain fr-1)/r ( r2

)s-const =

(5.43) I

This is called the second isentropic relation for ideal gases. The third isentropic relation for ideal gases is obtained, by combining Eqs. (5.42) and (5.43), as P2

PI

s=const

= V2

(5.44)

Note that in Eqs. (5.42) and (5.44) the specific volume can be replaced with density p(p = 1/v) of the gas to obtain the isentropic relations in terms of p. The isentropic relations with constant specific heats play a dominant role in gas dynamic processes since most of them can be closely approximated to isentropic processes with constant Cp and C„. (b) Ideal gases with variable specific heats For this case, the equation which accounts for the variation of specific heats with temperature, namely Eq (5.41) should be used. By Eq (5.41), we get 0= 4 — — R In Or

sZ

P = .4) + R hl —L

(5.45)

where .s° and s2 are the values of s° at the beginning and end of the isentropic process, respectively.

142 Fundamentals of Engineering Thermodynamics

5.8.4 Entropy Postulations From our discussion so far, we may stunmarize that: • The entropy content of an isolated system can never decrease. • The entropy of any substance approaches a constant value as its temperature approaches absolute zero. For a perfect crystalline substance, its entropy is zero at the absolute zero of temperature.

5.8.5 The Nature of Entropy • Entropy is a primitive concept. • Every system has entropy. That is, with the exception of a work reservoir, entropy change is always possible for any system. • Entropy is an extensive property. • One interpretation of entropy is that it is an index of that portion of the energy content in a system that is not available to do work. • Entropy content of an isolated system is not conserved. In fact, the entropy of an isolated system is a monotonically increasing function of time. In this respect, entropy differs fundamentally from energy; energy is a conservative property, entropy is not. EXAMPLE 5.12 Air is compressed adiabatically in a piston-cylinder device from 15°C and 101 kPa to 300°C in a reversible manner. Determine the compression ratio of the piston-cylinder device. Solution Let the subscripts 1 and 2 refer to the initial and final states, respectively. The compression ratio is given by the ratio of initial volume to final volume (VINO. The process is a reversible adiabatic. Also, for a closed system V1

= mvi =

Vl

V2 MV2 V2

By Eq. (5.42), we have (

y = ( T2 ) V2

Therefore, tio, -I) = T2 ) T1 v2

(300 + 273.15 15 + 273.15

1/0.4

( y =1.4 for air)

= 5.58 The compression ratio of the piston-cylinder device is therefore 5.58

Entropy 143

5.9 REVERSIBLE STEADY-FLOW WORK The heat transfer and work done during a process depend on the path followed as well as on the properties at the end states. We know that the reversible (quasi-equilibrium) moving boundary work associated with closed systems and expressed in terms of fluid properties, given by Eq. (2.6), is Wb = f PdV I Note: It is extremely important to note that the quasi-equilibrium work interactions lead to the maximum work output for work-producing devices and the minimum work input for workconsuming devices. It would also be very much useful to express the work associated with steady-flow devices in terms of fluid properties. By energy conservation law, we have Eq,„ — &m„= dh + dke + dpe But Ofq,„= dh — vdP Therefore, dh — vdP — env„„= dh + dke + dpe This yields, wtv = — 1.2 j 1 vdP — Ake— Ape

(5.46)

When Ake and Ape are negligible, Eq. (5.46) reduces to Wrev

= — fi vdP (kJ/kg)

(5.47)

If the working fluid is incompressible, v remains constant and therefore, Eq. (5.46) results in wre,, = v(P1 — P2) — Ake — Ape (kJ/kg)

(5.48)

For a steady-flow of a fluid through a device which involves no work interactions, such as a nozzle, the work term is zero, and the Eq. (5.48) reduces to v(p

_ p) + 2 i

V2 2

_ v2 I

2

+ g(z2 — zi) = 0

(5.49)

This is the well known Bernoulli equation which plays a dominant role in fluid mechanics. EXAMPLE 5.13 Determine the work required to compress steam isentropically from 150 kPa to 1.2 MPa assuming that steam exists as saturated vapour at the initial state. Neglect the changes in potential and kinetic energies. Solution The process is isentropic and therefore, ds = 0.

144

Fundamentals of Engineering Thermodynamics For the process with AKE = 0 and APE = 0, the reversible work is given by Eq. (5.47) as wrev = — v dP

By Eq. (5.31), we have, for ds = 0 vdP = dh Therefore, r2 = — j dh= i

— h2

From Table 4 of the Appendix, for state 1 (saturated vapour), h1 = 2693.6 kJ/kg, s1 = 7.2233 kJ/(kg K) From Table 5 of the Appendix, for state 2, at s2 = si and 1.2 MPa, we have h2 = 3160.8 kJ/kg Thus, wrev = 2693.6 — 3160.8 = —467.2 kJ/kg

5.9.1 Advantage of Reversible Processes One of the major advantages of reversible processes is that the steady-flow devices deliver the most and consume the least work when the process is reversible. Consider two steady-flow devices, one reversible and the other irreversible, operating between the same inlet and outlet states. By energy conservation law, we have egi„e„ — Swir,„ = dh + dke + dpe and etqr, — Avre„ = dh + dke + dpe Thus, ?irrev OWirrev = &hey &'rev

or Scare „ — Owirrev — ds —

&lirrev

>0

(..• dq„„, = Tds)

Since ds > Sq "' , and also as T is the absolute temperature which is always positive, we get, AVrev

5Wirrev

Or Wrev -?- Wirrev

(5.50)

This implies that the work-producing devices such as turbines deliver more work, and workconsuming devices such as compressors require less work when they operate on reversible , cycles.

Entropy 145

5.10 ADIABATIC EFFICIENCY OF COMPRESSORS AND PUMPS The adiabatic efficiency of a compressor is defined as the ratio of the work input necessary to raise the pressure of a gas to a specified value in an isentropic manner to the actual work input. /lc —

isentropic compressor work _ ws Wa actual compressor work

(5.51)

It is important to note that ws is always smaller than wa. When the changes in kinetic and potential energies of the gas being compressed are negligible, the i7 becomes /lc — h2s h2a

(5.52)

where h2s and h2a are the enthalpy values at the exit state for isentropic and actual compression processes, respectively, and h i is the enthalpy at the inlet. For a pump, when Ake and Ape of a liquid are negligible, its adiabatic efficiency is defined similarly as 17p

P1) v(P2 ws — h2. — hl

(5.53)

The reversible isothermal process is a realistic model process for a compressor that is intentionally cooled during the compression process. For such cases, we can define isothermal efficiency, by comparing the actual process to a reversible isothermal process, as (5.54) where WT is the isothermal work input and wa is the actual work input to the compressor.

5.10.1 Adiabatic Efficiency of Turbines The adiabatic efficiency of a turbine is defined as the ratio of the actual work output of the turbine to the work output that would be achieved if the process between the initial and the exit pressures were isentropic. /kith —

actual turbine work isentropic turbine work

Usually the changes in kinetic and potential energies associated with a fluid stream flowing through a turbine are small compared to enthalpy changes associated with the flow, and can be neglected. Thus, the adiabatic efficiency can be expressed as hl — h2a qturb

hl his

where h2a and h2, are enthalpy at the exit state for the actual and isentropic states, respectively, and h1 is the enthalpy at the inlet. For a turbine under steady operation, the inlet state of the working fluid and the exhaust pressure are fixed. Therefore, the ideal process for an adiabatic turbine is an isentropic process between the inlet state and the exhaust pressure.

146 Fundamentals of Engineering Thermodynamics The value of q mainly depends on the design of the individual components that make up the turbine. Properly designed large turbines have efficiencies above 90 per cent. For small turbines, however, it may be even less than 70 per cent. EXAMPLE 5.14 Air is compressed adiabatically from 101 kPa and 20°C to 750 kPa in a compressor. If the mass flow rate is steady at 0.18 kg/s and the adiabatic efficiency of the compressor is 85 per cent, determine the required power input to the compressor and the exit temperature of the air. Treat air to be an ideal gas. Solution Let subscripts 1 and 2 refer to inlet and outlet conditions of the compressor. For steady-flow devices, the conservation of energy equation is

Q—

= ph(Ah + Ake + ape)

where Q = 0, Ake = 0, and 4e = 0. Therefore, = th(hi — h2s) hi = CpT1 = 1004.5 x 293.15 x

= 294.47 kJ/kg

h2s = CpT2s By isentropic relation, r,

T2 s =

F

= 293.15 (7500.4/1.4) 101)

= 520.14 K

Therefore, = 1004.5 x 520.14 x 10--3 = 522.48 kJ/kg By Eq. (5.52), qc _

h2, —

_ 522.48 — 294.47 — 0.85 h2, — 294.47

or h2s = 562.72 kJ/kg For this 112,„ from Table 8 of the Appendix, Tea = 557.65 K The power input = 0.18(294.47 — 562.72) = —48.285 kW Note that for power input, hz, is used and not h2s. This is because h2, is the actual enthalpy and h25 is only an imaginary value.

Entropy 147

5.11 ADIABATIC EFFICIENCY OF NOZZLES Nozzles are flow accelerating devices and the flow pro.cesses through them are essentially adiabatic. The adiabatic efficiency of a nozzle is defined as the ratio of the actual kinetic energy of the fluid at the nozzle exit to the kinetic energy value at the exit of an isentropic nozzle for the same inlet state and exit pressure. 71N

actual KE at the nozzle exit — V2 .isentropic KE at the nozzle exit Ves2

(5.55)

In terms of enthalpies, rhq can be expressed as hi — he 71ri

(5.56)

— hes

where h, is the enthalpy at the nozzle inlet and he. and he, are the enthalpies at the nozzle exit for the actual and isentropic processes, respectively.

SUMMARY Entropy is an abstract property. It serves as a valuable tool in the second-law analysis of engineering devices. Entropy can be viewed as a measure of disorder or molecular disorder or molecular randomness. Unlike energy, entropy is a non-conserved property. There is nothing like conservation of entropy. The entropy S is defined as 8Q dS = (T )int,rev Entropy is an extensive property of a system and it is also referred to as total entropy. Entropy per unit mass, s, is an intensive property. The change in S during a process can be expressed as AS S2 —

r2 (8Q)

= .11

T

OcRIO

int,rev

The statistical definition of entropy of a system is S=kln11 where k is the Boltzmann's constant and f2 is the thermodynamic probability of the system's macrostate, namely the number of microstates belonging to it. The entropy of an adiabatic process of a closed (fixed mass) system always increases or, in the limiting case, remains constant during a process. In other words, for an adiabatic process of a closed system, the entropy never decreases. This is called the increase-of-entropy principle. The increase-of-entropy principle for any process is expressed as Sge„ = A Stow = A Ss). + A S,

0 (kJ/K)

148 Fundamentals of Engineering Thermodynamics This is a general expression for increase-of-entropy principle. It is applicable to both closed and open systems. From the increase-of-entropy principle, it can be summarized that ggen = Astow

> 0 irreversible process = 0 reversible process < { 0 impossible process

It is essential to note that • Sion is not a property. Its value depends on how the process is executed, and thus it is a path function, like heat and work. • Sr„ is the rate of change of total entropy, or the rate of entropy generation. • The equality of the relations of the increase-of-entropy principle or that of the relations of entropy balance applies to reversible processes, and the inequality to irreversible processes. A process during which the change in entropy is zero is called an isentropic process. The process diagrams on h—s coordinates are also called Mollier diagrams. The isentropic process of an incompressible substance is also isothermal. The first, second and third isentropic relations for ideal gases, respectively are = VI

T2 ( 7i s.const

T-1

1,2)

( )(7-017 P2 ( 7i)s=const (P2

7

=

( V2 )

- s=const

The Bernoulli's equation for incompressible flows is given by v(p ps ) + 2 1

v2 v2 1

2

g(z2

zi) = 0

2

The adiabatic efficiency of a compressor is defined as /lc —

isentropic compressor work _ ws actual compressor work wa

When the changes in kinetic and potential energies of a gas being compressed are negligible, the ric becomes h2s — hl h2, — hi

Entropy 149 For a pump, when Ake and Ape of a liquid are negligible, its adiabatic efficiency is defined as

_

ws _ v(P2 — P1) wa

The adiabatic efficiency of a nozzle is defined as the ratio of the actual kinetic energy of the fluid at the nozzle outlet to the kinetic energy value at the outlet of an isentropic nozzle for the same inlet state and exit pressure. 71N —

actual KE at the nozzle exit _ Ve isentropic KE at the nozzle exit V 2 es

In terms of enthalpies, ritq —

—

The adiabatic efficiency of a turbine is defined as the ratio of the actual work output of the turbine to the work output that would be achieved if the process between the initial and the exit pressure were isentropic. Must

actual turbine work isentropic turbine work

Usually the changes in kinetic and potential energies associated with a fluid stream flowing through a turbine are small compared to enthalpy changes associated with the flow, and can be neglected. Thus, the adiabatic efficiency can be expressed as curb

L nt

— h2. L u2s

PROBLEMS 5.1 Water is evaporated in a reversible isothermal process at 300 kPa. Determine the heat transfer of this process per unit mass of water. [Ans. 2163.95 la/kg] 5.2

From a saturated liquid state at 200 kPa, 3 kg of steam is heated to a temperature of 250°C at constant pressure. Determine the entropy change for this process. [Ans. 18.54 kJ/K]

5.3 Air is throttled through a throttling valve from 2 bar and 30°C to 1 bar, adiabatically. Assuming the kinetic energy to be negligible, calculate the change in entropy of the air. [Ans. 0.1989 kJ/(kg 5.4 Air is compressed reversibly in a polytropic process from 1 bar and 40°C to 3 bar. The polytropic index is 1.3. Determine the change in entropy per unit mass of the air. [Ans. — 0.0607 kJ/(kg

150 Fundamentals of Engineering Thermodynamics 5.5

From a thermal reservoir at 600 K to a thermal sink at 300 K, 800 kJ of heat is transferred. Calculate the change in entropy of the universe resulting from this heat exchange process. [Ans. 1.34 kJ/K]

5.6

Steam is expanded reversibly and adiabatically in a turbine from 25 bar and 400°C to 3 bar. Calculate the work done per unit mass of steam in a steady-flow process. Assume the kinetic energy change to be negligible. [Ans. 478.3 kJ/kg]

5.7

A gas undergoes a change of state reversibly and isothermally in a piston-cylinder device. Work done by the gas is 120 J. Determine the change in entropy of the gas if its temperature is 125°C. [Ans. 0.304 J]

5.8 A heat engine is designed to operate between heat reservoirs at 1800 K and 330 K. The engine delivers 1.5 MW of power when heat is added at the rate of 3 MJ/s. Determine the rate of lost work production owing to irreversibility. [Ans. 950 kW] 5.9

A steel tank of mass 25 kg and volume 1.5 m3 is full of liquid water. Initially, both the steel tank and the water are at 40°C at noon. By midnight the entire system cools to the surrounding temperature of 28°C. Determine the total entropy change for this process. [Ans. 4.862 kJ/K]

5.10 A 1 kW electric heating element, 10 cm long and 10 mm in diameter, is immersed in 25 kg of water initially at 12°C, in an insulated container. Determine the time required for the heater to raise the water temperature to 65°C. Also, fmd the entropy generated during this process. [Ans. 1.54 h, 17.83 kJ/K] 5.11 Two kilograms of water at 20°C is converted to steam at 300°C at constant atmospheric pressure. Assuming the heat capacity of liquid water to be 4.2 kJ/(kg K) and the heat of vaporization at 100°C to be 2260 kJ/kg, and using the relation, Cp

— = 3.634 + 1.195T + 0.13572 find the change in entropy of the steam. [Ans. 41.617 MJ/K] 5.12 The working fluid of a Carnot engine experienced an entropy change of 1.35 kJ/K during the isothermal heat rejection process. If the temperature of the thermal energy sink is 20°C, determine (a) the amount of heat rejected to the sink, (b) the change in entropy of the sink, and (c) the total energy change for this process. [Ans. —395.75 kJ, 1.35 kJ/K, 0] 5.13 During the isothermal heat addition process of a Carnot cycle, 750 kJ of heat is added to the working fluid from a source at 475°C. Determine (a) the change in entropy of the working fluid, (b) the change in entropy of the source, and (c) the total entropy change for the process. [Ans. 1.0025 kJ/IC, —1.0025 kJ/IC, 0]

Entropy

151

5.14 Saturated liquid water at 150 kPa in a frictionless piston-cylinder device receives 300 Id of heat from a source at 450°C, and a part of the liquid water vaporizes at constant pressure. Determine the total entropy change for this process. Is this process reversible? [Ans. 0.365 kJ/K, No] 5.15 A rigid tank contains air at 35°C which is being stirred by a paddle wheel. The paddle wheel does 500 Id of work on the air. During the stirring process, the air temperature remains constant owing to transfer of heat between the system and the surrounding medium which is at 15°C. Calculate (a) the change in entropy of the air in the tank and (b) the change in entropy of the surroundings. Does this process satisfy the increase-ofentropy principle? [Ans. 0, 1.735 kJ/K, Yes] 5.16 Air in a compressor is compressed from Pi to P2 at a constant temperature of 30°C. The temperature during compression is maintained constant as a result of heat transfer to the surrounding medium which is at 22°C. If the power input to the compressor is 12 kW, determine the rate of change in entropy of (a) the air and (b) the surroundings. Check whether the process satisfies the second law of thermodynamics. [Ans. —0.0396 kW/K, 0.0406 kW/K, satisfied] 5.17 One kilogram of water at 20°C is converted to ice at —10°C at constant atmospheric pressure. Assuming that heat capacity per gram of the liquid water remains practically constant at 4.2 J/(g K) and that of the ice at one-half of this value, and taking the heat of fusion of ice at 0°C to be 335 J/g, calculate the total entropy change for the system. [Ans. —1.59 kJ/K] 5.18 Calculate the change in entropy of the universe owing to each of the following processes: (a) A 0.5 kg copper block at 100°C is placed in a lake of water at 10°C. (b) The same block at 10°C is dropped from a height of 100 m into the lake. (c) Two such blocks at 100°C and 0°C, respectively are joined together. [Ans. 8.3 J/K, 1.73 J/K, 4.7 J/K] 5.19 A body at temperature Ti and of constant heat capacity Cp is put in contact with a thermal reservoir at temperature T.f . which is higher than T,. The pressure remains constant while the body comes to equilibrium with the reservoir. Show that the change in entropy of the universe is equal to Cp[.x — In (1 + x)], where x = Ti)/Tf. 5.20 A fuel-air mixture enters a duct combustion chamber at V1 = 80 m/s and T1 = 325 K. Estimate the amount of heat required to be added to the combustion chamber for this flow to exit as a choked flow. Assume the fuel-air mixture to be equivalent to air. [Ans. 1272.7 kJ/kg] 5.21 Air flowing through a frictionless diffuser has at one section 0°C, 140 kPa, and 900 m/s. At a downstream station, the velocity has decreased to 300 m/s. Assuming the flow to be adiabatic, calculate the increase in pressure and temperature of the flow. [Ans. 2.631 MPa, 358.39 K]

152 Fundamentals of Engineering Thermodynamics 5.22 Hydrogen gas in a cylinder at 7 atm and 300 K is expanded isentropically, through a nozzle to a fmal pressure of 1 atm Assuming hydrogen to be a perfect gas with y= 1.4, determine the velocity and Mach number corresponding to the fmal pressure. Also, fmd the mass flow through the nozzle for an exit area of 10 cm2. [Ans. 1923 m/s, 1.93, 0.275 kg/s] 5.23 Prove the following relation for an ideal gas. ds = Cp

dv

dP

Using this result, show that for an ideal gas undergoing an isentropic change of state with constant specific heat, PO' = constant. 5.24 Air undergoes a change of state isentropically from 300 K and 110 kPa to a final pressure of 550 kPa. Assuming ideal gas behaviour, determine the change in enthalpy of the air. [Ans. 175.94 kJ/kg] 5.25 Nitrogen gas is compressed reversibly and isothermally from 100 kPa and 25°C to a final pressure of 300 kPa. Calculate the change in entropy of the gas. [Ans. — 0.3263 kJ/(kg K)] 5.26 Air enters a compressor at 100 kPa and 1.175 kg/m3 and exits at 500 kPa and 5.875 kg/m3. Determine the difference in enthalpy of the air between the outlet and inlet states. [Ans. 0] 5.27 Air at low pressure inside a rigid tank is heated from 50°C to 125°C. Find the change in entropy of the air associated with this heating process. [Ans. 149.75 J/(kg K)] 5.28 Oxygen gas is heated from 25°C to 125°C. Determine the increase in its internal energy and enthalpy. Take y= 1.4. [Ans. 64,950 J/(kg K), 90,930 J/(1cg K)] 5.29 A rigid tank contains a certain amount of nitrogen gas, initially at 120 kPa and 30°C. Two kilograms of nitrogen gas is then added to the tank. If the fmal pressure and the final temperature of the gas in the tank are 240 kPa and 30°C, respectively, what is the volume of the tank? [Ans. 1.5 m3] 5.30 A quantity of air has a pressure of 0.7 MPa and it occupies a volume of 0.014 m3 at a temperature of 150°C. The gas is then expanded isothermally to a volume of 0.084 m3. Calculate the change in entropy of the air. [Ans. 513.4 J/(kg K)] 5.31 A quantity of air weighing 0.3 kg and at 350 kPa and 35°C receives heat energy at constant volume until its pressure becomes 700 kPa. It then receives heat energy at constant pressure until its volume becomes 0.2289 m3. Calculate the change in entropy caused by each process. [Ans. 149.2 J/K, 333 J/K]

Entropy 153 5.32 Determine the work required to compress steam isentropically from 300 kPa to 1.2 MPa, assuming that the steam exits as (a) saturated liquid and (b) saturated vapour at the initial state. Neglect the changes in kinetic and potential energies. [Ans. (a) — 0.9567 kJ/kg, (b) —298.7 kJ/kg] 5.33 Combustion gases enter an adiabatic gas turbine at 1200 K and 1 MPa, and leave at 300 kPa. Treating the combustion gases as air and assuming an adiabatic efficiency of 88 per cent, determine the work output of the turbine. [Ans. 309.4 kJ/kg] 5.34 Steam expands in a turbine at a steady rate of 32,000 kg/h, entering at 10 MPa, 400°C and leaving at 75 kPa as saturated vapour. If the power generated is 3.7 MW, determine the rate of entropy generation of the process, assuming that the surrounding medium is at 30°C and the kinetic and potential energies are negligible. [Ans. 11.57 kW/K] 5.35 Liquid water at 200 kPa and 25°C enters a mixing chamber at the rate of 2 kg/s and is mixed with superheated steam at 200 kPa and 400°C entering the chamber. The chamber looses heat at the rate of 520 kJ/min to the surroundings at 30°C. If the mixture leaves the mixing chamber at 200 kPa and 50°C, determine (a) the mass flow rate of the superheated steam and (b) the rate of entropy generation during this mixing process, assuming the potential and kinetic energies to be negligible. [Ans. (a) 0.0713 kg/s, (b) 0.1656 kW/K] 5.36 An iron piece of mass 30 kg and a copper piece of mass 40 kg, both initially at 110°C, are dropped into a lake at 20°C. As a result of heat transfer, thermal equilibrium is established in a short while. Determine the total entropy change of this process, taking the average specific heats of iron and copper to be 0.45 kJ/(kg K) and 0.386 kJ/(kg K), respectively. [Ans. 1.137 kJ/K] 5.37 A three-stage isentropic compressor has two inter-coolers which cool the gas to the initial temperature between the stages. Determine the two intermediate pressures (P. and Pb) in terms of the inlet and exit pressures (P, and Pe) that will minimize the work input to the compressor. [Ans. P° = (P Per Pb = (P2e Ps)15 ] 5.38 A turbine is supplied with steam at 0.4 MPa and 400°C. The steam expands through the turbine steadily to an exit pressure of 20 kPa, and a dryness fraction of 0.93. The mass flow rate is 3 kg/s, and the mechanical efficiency is 90 per cent. Neglecting the kinetic and potential energies, calculate the power output of the turbine. Show that the process is irreversible and calculate the change of specific entropy. Heat losses from the turbine are negligible. [Ans. 2237.76 W, 0.4853 kJ/(kg K)] 5.39 In a centrifugal compressor, air is compressed adiabatically from 100 kPa, 300 K to 400 kPa. If the entropy change associated with the compression is 0.105 kJ/(kg K), determine the final temperature. [Ans. 495 K] 5.40 Steam at 1.5 MPa is throttled to 0.1 MPa and to a temperature of 150°C. Calculate the initial dryness fraction and the change of specific entropy. [Ans. 0.992, 1.202 kJ/(kg K)]

CHAPTER

6 Availability, Irreversibility and Availability Analysis of Engineering Processes

6.1 INTRODUCTION From our discussion on first and second laws of thermodynamics, it is clear that the first law deals with the quantity of energy and asserts that energy can neither be created nor be destroyed. The second law deals with the quality of energy. It is concerned with the degradation of energy during a process, the entropy generation, and so on. In this chapter we will examine the performance of engineering devices in the light of the second law of thermodynamics.

6.2 AVAILABILITY The availability of a system is defined as the property that enables us to determine the useful work potential of a given amount of energy contained in the system at some specified state. The work potential of the energy contained in a system at a specified state may be viewed as the maximum useful work that can be obtained from the system. We know that work done during a process depends on the initial and fmal states, and the process paths. That is, W = f(initial state, fmal state, process path) In an availability analysis, the initial state is specified and thus it is not a variable. Further, for maximizing the work output, the process between the two specified states has to be executed in a reversible manner. That is, all the irreversibilities are disregarded in determining the work potential. Finally, the system must be in the dead state at the end of the process to maximize the work output. The dead state is that state at which a system is in thermodynamic equilibrium with its 154

Availability, Irreversibility and Availability Analysis of Engineering Processes

155

surroundings. That is, in the dead state the system is at the same T and P as those of its surroundings, it has no kinetic or potential energy relative to its surroundings, and it does not react with the surroundings. The properties of a system in the dead state are denoted by the subscript zero. Unless specified otherwise, the dead state temperature and pressure are assumed to be To = 25°C and P0 = 1 atm A system has zero availability in the dead state. Therefore, it can be concluded that a system will deliver the maximum possible work only when it undergoes a reversible process from the specified initial state to the state of its surroundings, i.e. the dead state. The maximum work represents the useful work potential of the system at the specified state and is termed availability. It is important to note that the availability only represents the "upper limit on the amount of work a device can deliver without violating any thermodynamic law", and it does not represent the amount of work that a work-producing device will actually deliver. Another term meaning the same thing as availability and popularly used today is energy. We will use both these terms interchangeably in this text. EXAMPLE 6.1 A thermal reservoir at 1000 K can supply heat at a steady rate of 2000 kJ/s. Determine the availability of this energy if the environment is at 20°C. Solution Availability is the useful work potential of a system, that is, the maximum possible amount of work that can be extracted from it. This is nothing but the amount of work that a reversible heat engine operating between the source and the environment can produce. Therefore, T qth.max

i ith,rev

I

7,13 H

1 - 293.15 — 0.707 1000

It means that a heat engine can convert a maximum of 70.7 per cent of the heat received from the furnace to work. In other words, only 70.7 per cent of heat energy of the reservoir is available for producing work. Therefore, the availability of the reservoir, in rate form, is equivalent to the power produced by the reversible engine. That is, *max = Wm = The availability is therefore

rith,rev Oin

= 0.707 x 2000 = 1414 kJ/s

1414 kW

Note that 29.3 per cent of the energy transferred from the furnace as heat is not available for doing work. The portion of energy which cannot be converted to work is called the unavailable energy.

6.3 REVERSIBLE WORK AND IRREVERSIBILITY The adiabatic efficiencies discussed in Chapter 5 are of limited use because of the idealized model (isentropic) process used in their definitions. The exit state of the isentropic process is not the same as the actual exit state. In this section, two new quantities that are related to the exact initial and final states, namely reversible work and irreversibility and which serve as valuable tools in the optimization studies of complex thermodynamic systems, are described.

156 Fundamentals of Engineering Thermodynamics The work done by or against the surroundings during a process is called the surroundings work. The work done by work-producing devices is not always entirely in a usable form. For example, when a gas in a piston-cylinder device is expanding, part of the work done by the gas is used to push the atmospheric air out of the way of the piston. This work, which cannot be recovered and utilized for any useful purpose, is equal to the atmospheric pressure P. times the change in volume (V2 — V1 ) of the system. That is, Wsun = Pa(V 2 — V1 ) = mPa(v2 — v1 ) (kJ)

(6.1)

where V1 and V2, respectively, are the initial and final volumes of the gas and m is its mass. The difference between the actual work W and the surrounding work W. is called the actual useful work or useful work Wu. Wu

W Want (ki)

(6.2)

Note that W has significance only for systems that involve moving boundary work. For cyclic devices and systems with fixed boundaries, Ku„ has no significance. Reversible work, W , is defined as the maximum amount of useful work that can be obtained as a system undergoes a process between the specified initial and final states. When the final state is the dead state, the W equals availability. For processes that require work, reversible work represents the minimum amount of work necessary to carry out the process. The irreversibility I of any process is the difference between IV, and Wu which is due to the presence of irreversibilities during the process. Examine the process illustrated in Fig. 6.1. It is seen from this figure that 1 = Wm/ — Wu (k)

(63)

i = wre„ — wu Oa/kg)

(6.4)

On unit niass basis, we have

The irreversibility rate, namely the irreversibility generated during a process per unit time is given by

I =141 — Wu (kW)

(6.5)

For a totally reversible process, I = 0, as expected, since a totally reversible process generates no entropy, which is a measure of irreversibility occurring during a process. Initial state Actual process

ssik

Wu < Wrev

Reversible process Wre, Final state I= Wrev

Figure 6.1 An irreversible process.

Availability, Irreversibility and Availability Analysis of Engineering Processes 157

6.3.1 Availability Transfer with Heat and Work Interactions We have seen that the availability or the available work potential of the energy in a heat source at temperature T is the maximum work that can be obtained from that source in an environment at temperature To, and it is equal to the work produced by a Carnot heat engine operating between the source and the environment. Therefore, the Camot efficiency rk = 1 — To/T represents the fraction of the available energy in a heat source at temperature T > To. Also, we have seen in Chapter 5, that heat transfer Q at a location at temperature T is always accompanied by entropy transfer in the amount of QIT. Hence we infer that heat transfer Q is always accompanied by availability transfer, and the amount of availability transfer with heat, AQ, is

AQ

=(' --

71 °)Q

In this relation, when T < To, the heat transfer Q is the heat rejected to the cold medium as waste heat. Also, the availability transferred with heat becomes zero when T = To. When T > To, both the availability and the energy content of the medium to which heat is rejected increase. But when T < To, the availability and the heat transfer are in opposite directions. That is, the entropy of the cold medium increases as a result of heat transfer, but its availability decreases. The availability of the cold medium becomes zero when its temperature reaches To. The relation AQ = (1 — T°17) Q can also be viewed as the availability of thermal energy Q at temperature T. When the temperature T at a location where heat transfer takes place is not constant, we have AQ

= -+7 1°Q

The availability transfer accompanying work, A w, is simply the useful work. That is,

4 =W„ = - Pc,(v2 -

) for boundary woric

Aw = Wu = W for other forms of work It is interesting to note that the availability of either kinetic or potential energy is equal to itself, since each is fully available to work. That is, I

AKE =KE= — 2m V

2

and APE = PE = mgz EXAMPLE 6.2 A heat engine receives heat at a rate of 320 kJ/s from a source at 700 K and rejects the waste heat to a medium at 305 K. If the power output of the engine is 98 kW, determine the reversible power and the irreversibility rate for this process.

158 Fundamentals of Engineering Thermodynamics Solution The irreversibility rate generated during a process per unit time is given by = ikre„ — Wn where Fice,„ is the amount of power which a reversible heat engine, such as a Carnot heat engine would produce when operating between the same temperature limits, and giu is the actual or useful power delivered by the engine. Now, Wmax = th,rev Oin = (1

•H

=1—

700

Qin

)320 = 180.6 kW

This is the maximum power which a heat engine operating between the given TH and TL and receiving heat at the rate of 320 kJ/s can produce. This would also represent the available power if 305 K were the lowest temperature available for heat rejection. The irreversibility rate, which is the difference between the reversible power and the useful power is given by = 180.6 — 98 = 82.6 kW That is, 82.6 kW of power potential is wasted during this process owing to irreversibilities present in the cycle. EXAMPLE 6.3 A metal piece of 1 kg mass with constant specific heat of 0.4 kJ/(kg K) is cooled from 200°C to 100°C by transferring heat to the surrounding air at 27°C. Determine the reversible work and the irreversibility for this process. Solution The heat transferred from the metal piece to the environment is Q = mC(AT) = 1 x 0.4 (200 — 100) = 40 kJ That is, 40 kJ of heat is removed from the metal. Now, the reversible work that can be done by a heat engine supplied with 40 kJ of heat and rejecting waste heat to the surroundings at To = 27°C is Wre, = Q — T0dS The change in entropy associated with the cooling of the metal piece is S2 — St = mC InT2 --= 0.41n

373.15 = 0.095 kJ/K 473.15

Therefore, wrev =Q —T0(51— 52) = 40 — (300.15X+ 0.095) 11.49 kJ The irreversibility I = Wre, — Wu W.

Availability, Irreversibility and Availability Analysis of Engineering Processes

159

In this case, since the system is rigid, the actual useful work is zero. Therefore, the irreversibility of the process is simply W. That is, 1 = 11.49 kJ Note that reversible work and irreversibility are identical for processes that involve no actual useful work. Alternative Method The irreversibility I = ToSgeT, where Sgen = ASsys + LSsur =

MC 111

T

+Qsurr

Ti

=1 x 0.4 In

To 373.15 0 + — 0.0383 kJ/K 473.15 300.15

Therefore, / = 300.15 x 0.0383 = 11.49 kJ EXAMPLE 6.4 Calculate the increase in availability of 1 kg of air when heated reversibly from 40°C to 250°C at constant pressure of 2 atm. Assume the environment to be at 20°C. Solution The availability transfer associated with a heating process is

tr: where SQ = Cp dT and, therefore,

AQ = 1r2

ll

(1—+ 3, Cp dr

=Cp(T2 — T) CpTo

(

To is a constant)

=1004.5[(250 — 40) — 293.15 In 523.15] 313.15 59.83 kJ/kg

6.4 SECOND-LAW EFFICIENCY mi The thermal efficiency and coefficient of performance were defined as the measures of performance, in Chapter 4. They were defined on the basis of the first law only, and they are sometimes referred to as the first-law efficiency or conversion efficiency.

160 Fundamentals of Engineering Thermodynamics The first-law efficiency makes no reference to the best possible performance, and thus it may be misleading at times. For example, consider two heat engines A and B having the same of 30 per cent. Let Engine A work between reservoirs with TH = 600 K and TL = 300 K and B between TH = 1000 K and Ti, = 300 K. The first-law efficiency tin of 30 per cent simply implies that both are performing equally well. Now, we will calculate their efficiencies assuming that the best possible efficiency is achieved when they perform as reversible (Carrot) engines. We have nKvn =I

1_

TH A

= (. lrevs

TH B

ll

= _ 300 _ — 50 per cent 600 =1

300 _ _ iu per cent

1000

It is seen that Engine B has a greater work potential available to it. Therefore, it can be said that B is performing poorly compared to A even though both have the same thh. Thus, the first-law efficiency is not a realistic measure of the performance of engineering devices. To overcome this deficiency, the second-law efficiency rh1, also called the effectiveness, is defined as the ratio of the thermal efficiency to the maximum possible (reversible) thermal efficiency under the same conditions. That is, Ott — lib (heat engines) 'lth.rev

(6.6)

Therefore, 0.30 = /7/LA = — 0.60 and on 0.50

0.30 = = — 0.43 0.70

That is, Engine A converts 60 per cent of the available work potential into useful work, whereas Engine B converts only 43 per cent of the available work potential into useful work. The second-law efficiency can also be expressed as /hi = — Wu (work-producing devices) Wrev

(6.7)

This defmition is more general, since it can be applied to processes as well as cycles. Note that whatever be the definition, thi cannot exceed, 100 per cent. For work-consuming devices, such as compressors, fit may be defined as tlu =

(work-consuming devices) Wu

(6.8)

For cyclic devices, such as refrigerators, we can define 7711

COP (refrigerators, heat pumps) COPre,

(6.9)

In all the above relations, W should be determined using the same initial and final states as in the actual process.

Availability, Irreversibility and Availability Analysis of Engineering Processes 161

6.5 SECOND-LAW ANALYSIS OF CLOSED SYSTEMS Consider a closed system undergoing a process from state 1 to state 2 in an environment which is at P0 and T0, as shown in Fig. 6.2. The first and second laws of thermodynamics for this Surroundings To, Po r

I

Closed system Heat

Figure 6.2 A closed system exchanging heat with its surroundings. system can be expressed, respectively, as Q — W = U2 — and Sgen = (S2 — )sys

QUIT

TS=

But, Ts„„ = T0,Qs~ur = —Q, and Ste„ is the amount of entropy generated (i.e. the total entropy change) for the process. Therefore, we can write

w = (u1 - u2 ) - ;(51 - 52) - Tosg.

(6.10)

This is the total actual work done during the process. The useful work is given by

=w-

= W - Po(v2 - vi)

Or Wu

=(ui - u2) - To(5, - s2)+Po(v, - v2) - rosge„

(6.11)

Now, consider the case where the process is executed reversibly between the same initial and fmal states. For reversible processes, Sge„ = 0 and W = W. Thus,

w =(U, - U2 ) - To(Si - 52) ± Po(vi - v2) (d)

(6.12)

This is the maximum useful work possible for the system under consideration. The availability of a closed system 4:0 is the maximum useful work potential of the system at a given state. It is obtained from Eq. (6.12) as

_(U

— U0 ) — To(S — S0) + P0(V — Vo) (ci)

(6.13)

162 Fundamentals of Engineering Thermodynamics On unit mass basis, we have = (u

uo )

To (s — so ) + Po (v — vo ) (kJ/kg)

(6.14)

where no subscript refers to the existing state and the subscript 0 refers to the dead state. It is seen from Eqs. (6.13) and (6.14) that for a system in the ddad state, the closed-system availability is zero. The reversible work in terms of availabilities can be expressed as Wrev = 4)i — 4)2 (kJ)

(6.15)

On unit mass basis, we have = 01 — 02 (kJ/kg)

(6.16)

where 0 = elWm. The irreversibility of a closed system, for any process, is given by I =

WK„ — Wu = ToSgen (kJ)

(6.17)

i = wrev — wu = ToSgen (kJ/kg)

(6.18)

Or

Therefore, the irreversibility or the energy that becomes unavailable for work during an irreversible process is To times the entropy generation (entropy change of the universe) that is brought about by the irreversible process. Since no energy becomes unavailable during a reversible process, it follows that the maximum amount of work is obtained when a process takes place reversibly. Since irreversible processes are continuously going on in nature, energy is continually becoming unavailable for work. This conclusion, known as the principle of the degradation of energy and first developed by Kelvin, provides an important physical interpretation of the entropy change of the universe. It is essential to realize that the energy that becomes unavailable for work is not the energy which is lost. The first law is obeyed at all times. Energy is merely transformed from one form to another. EXAMPLE 65 A piston-cylinder device contains 2 kg of air at 1 MPa and 400°C. The air expands to a final state of 300 kPa and 150°C, doing work. Heat losses from the system to the surroundings are estimated to be 75 Id during the process. If the surrounding environment is at 100 kPa and 300 K, determine (a) the availability of the air at the initial and fmal states, (b) the reversible work, (c) the irreversibility, and (d) the second-law efficiency for the process. Treat the air to be a perfect gas. Solution Let us take the air contained in the piston-cylinder device as our system. It is a closed system. At state 1: P1 = 1 MPa and T1 = 400°C = 673.15 K = C,T1 = 0.7175 x 673.15 = 482.985 kJ/kg R

=

= 0.193 m3/kg

Availability, Irreversibility and Availability Analysis of Engineering Processes

163

At state 2: P2 = 300 kPa, T2 = 150°C = 423.15 K Therefore, u2

= CvT2 = 303.61 kJ/kg

v2 = 0.405 m3/kg At the surroundings (a dead state): P0 = 100 kPa and To = 300 K Therefore, u0 = 215.25 kJ/kg vo = 0.861 m3/kg (a) The availability of the air at state 1 is =m

— uo) — To (si — so) + Po (vi — vo)1

= 2 [(482.985 — 215.25) — 300 (1.0045 In

67 3.15 300

1 0.287 In" — )+ 100 (0.193 — 0.8601 1 00

= 2(267.735 — 45.294 — 66.88) = 311.12 kJ The availability of the air at state 2 is = 2(88.36 — 9.057 — 45.6) = 67.4 Id (b) The reversible work Wre„, is Wrev =

— 4112 = 311.12 — 67.4 = 243.72 kJ

Alternative Method The reversible work is also given by Wre, = m[(ul — 112) —

— s2)

Po(vi — v2)]

= 2 [(482.985 — 303.61) — 300 (1.0045 In

423'15 673.15

0.287 In

00 ) +100 (0.193 — 0.405)] 1 000

= 2 (179.375 — 36.237 — 21.2) = 243.876 la (c) The irreversibility of a process is simply the difference between the reversible work and the useful work for that process. That is, I = Wre, — Wu The useful work is the difference between the actual work and the surrounding work. That is, Wu = Wact Ws = Wact

P0(V2 —

v1)

The actual work can be determined from the energy conservation equation Q — Wad= m(u2 — u1) Or — 75 — W,ct = 2(-179.375)

164 Fundamentals of Engineering Thermodynamics or Wee, = 283.75 kJ Therefore, Wu = Wad — Po(v2 — vi) = 283.75 — 100(0.405 — 0.193X2) = 241.35 Id Thus, I=

w,- Wu

= 243.876 — 241.35 = 2.526 kJ The irreversibility is also given by I

= ToSgen

where Sgen = (S2 — SI) + auff — 0.00842 To Therefore, / = 300 x 0.00842 = 2.526 kJ (d) The second-law efficiency mi for the process is _ W,, _ 241.35 _ 0.99 = 99 per cent W 243.876

6.5.1 Heat Transfer with Other Systems (or Bodies) So far we have analysed processes involving heat transfer with the surrounding medium only. We will now consider heat transfer of a system with other bodies at different temperatures as well as with its surroundings, as shown in Fig. 6.3. These kinds of processes can be handled easily by treating the main system and the individual bodies as separate systems and then superimposing the results. That is, Ototal = Osys °body Wrev,total = Wrev,sys

Wrevbody

Surroundings

Figure 6.3 Heat transfer of a system with the other body as well as with the surroundings.

Availability, Irreversibility and Availability Analysis of Engineering Processes 165

A simple relation for lice, can be obtained when the temperatures of the bodies with which the system exchanges heat remain constant during the process. Consider a closed system receiving heat QR from a reservoir at TR. Let the system be located in an environment at To and let it change state from 1 to 2 owing to the heat transfer process. Using the first- and secondlaw relations for this process, the reversible work can be expressed as — 02)sys — QR

W rev =

(1-°) (a)

(6.19)

The irreversibility relation remains the same. But the Sr „ becomes

Sgen =

— S !Lys +

Qs= + a (kJ/K) To Tit

(6.20)

Note that the entropy change of the reservoir is included in the entropy generation. It is important to note that QR in the above two equations is to be assigned the proper sign. That is, it is -QR for the hot reservoir (source) and +QR for the cold reservoir (sink). EXAMPLE 6.6 A rigid tank of volume 2.5 m3 contains air at 200 kPa and 300 K. The air is heated by supplying heat from a reservoir at 600 K until the temperature reaches 500 K. The surrounding atmosphere is at 100 kPa and 300 K. Determine the maximum useful work and the irreversibility associated with the process. Solution The mass of the air is m—

PV RT

—

200 x 103 x 2.5 — 5 . 8 kg 287 x 300

The initial specific volume is 2.5 v1 = = — = 0.431 m3 /kg m 5.8 The final specific volume is v2 = vi , since the tank is rigid. The final pressure is P2 =

f

= 200 ( 51 = 333.33 kPa

300

The change in entropy is

S2

-

s1 = Cp In

— R In p 2 =1004.5In

500 300

—287In

= 0.3665 kJ/(kg K) Heat transferred to the air in the tank is q = Au = Cv(T2 —

= 0.7175 x 200 = 143.5 kJ/kg

333.33 200

1 166 Fundamentals of Engineering Thermodynamics This process involves heat transfer with a reservoir other than the surrounding medium, and thus, the reversible work can be obtained from Wrev = ((D1 — (D2)sys — QR (1 — 4 7 )

1

= m[(u1 — u2) —;(s, — s 2) — qR (1 — ° 7 )1 1 =5.8 [-143.5 — 300( —0.3665) — (-143.5) (1— ÷-)] = 5.8 (-143.5 + 109.95 + 71.75) = 221.56 kJ The useful work done Wu = 0, since the system is with rigid boundaries. Therefore, / = W — W. = W That is, Irreversibility = 221.56 kJ Checking I = ToSge„ where Sge„ = (M)sys + AS., , = (S2 — Si isys _,i_ Qs,,,,T _,, QR To TR = 5.8 [0.3665 + 0 +

(-143.5)]

600

= 0.739 kJ/K Therefore, / = 300 x 0.739 = 221.7 k.1

6.6 SECOND-LAW ANALYSIS OF STEADY-FLOW SYSTEMS Consider a steady-flow process through a system, as shown in Fig. 6.4. The system may have multiple inlets and exits and may exchange heat with its surroundings at P0 and To. For the system under consideration, the first and second laws, can be expressed as v2 ) V — IV = Enre he + s—2 + gze )—Enti lhi + t + gzi 2 and ; ±gen = E !hese — Irks. + Alwr respectively.

I

Availability, Irreversibility and Availability Analysis of Engineering Processes

167

ine

Figure 6.4 A steady-flow device with heat transfer. But

a.. =

Elimination of the Q term between the above two relations results in v2 i gzi —Tosi — gl := Ent; ( Is; +—+

v2

e gze —Tose — ..5' +—+

e

2

(6.21)

This is the actual work during the process, which is also the useful work since there is no boundary work associated with the rigid boundaries of the system. The reversible work gl„„ is obtained by setting 4,.„ = 0, as Filmy = E Ai i hi +

v.2 2

v2

+ gzi —Tosi

e

e gze —Tose ) (kW) +—+ 2

(622)

For a single fluid stream entering and leaving the steady-flow device, we have [v z_v2 lice,,, = fit

(hi — he )— To

— se )

F

i

'e

2

+ g(zi — ze ) (kW)

(6.23)

g(Zi — 4) (kJ/kg)

(624)

Or wlev =

(hi — he) — To(si — se) +

— V2

2

e

That is,

wrev = Toils — Ah — Ake — Ape (kJ/kg)

(6.25)

6.6.1 Stream Availability The availability of a fluid stream is called stream availability iv, and is obtained by replacing the subscript i with no subscript and the exit state by the dead state in Eq. (6.24). That is,

v2 y =(h — ho )—To(s — so)+ — + gz (kJ/kg) 2

(6.26)

since for the dead state, 1/0 = 0 and zo = 0. The reversible work in terms of stream availability becomes

168 Fundamentals of Engineering Thermodynamics ' rev =

— I they/e (kW)

(6.27)

For a single stream of fluid through a steady-flow device, we have (6.28)

141rev = Oz(Vi We) (kW) EXAMPLE 6.7

Evaluate the steady flow availability per unit mass of water at 100°C and 0.5 MPa. Estimate the availability if the temperature is 200°C. Solution The availability of a fluid per unit mass, given by Eq. (6.26) is =

— ho) — Tds — o)

where the subscript 0 refers to the reference state at 25°C and 1 atm Using the saturated water properties table (Table 3 of the Appendix), we obtain yi = (419.04 — 104.89)— 293.15(1.3069 — 0.3674)

38.74 kJ/kg When T = 200°C, the water would be superheated since at 0.5 MPa, Tcd = 151.86°C. Therefore, using the superheated vapour table, we obtain yi = (2855.6 — 104.89) — 293.15(7.0592 — 0.3674)

789 kJ/kg

6.6.2 Irreversibility Rate The irreversibility rate i associated with a steady-flow process is 1 = gire, — 41„ = To, g,„ (kW)

(6.29)

On unit mass basis, we have i = wrev — wu = Tosgen (kJ/kg)

6.6.3 Heat Transfer of a Steady-Flow Device with Other System (or Bodies) Heat transfer between a steady-flow device and bodies other than the surroundings can be handled in the same manner as that discussed in relation to a closed system. For example, when a steady-flow device exchanges heat with a reservoir at TR at a rate of Q it, the IV, relation is obtained by eliminating the heat transfer term (with the surroundings) from the first- and secondlaw relations for this process, resulting in *rev = E m1 ii — E the We — 62

(1 — 7'1 ) (kW)

TR

(6.30)

Availability, Irreversibility and Availability Analysis of Engineering Processes

169

The second-law efficiency of the steady-flow devices can be determined from the general relation, namely —

availability recovered availability supplied

6.7 SECOND-LAW ANALYSIS OF UNSTEADY-FLOW SYSTEMS Processes like filling or draining of storage tanks, starting-up or shutdown of equipment, and so on involve variations of properties with time. Therefore, they should be analysed as unsteadyflow processes.

6.7.1 Uniform-Flow Processes These are flow processes with constant flow properties at any inlet or exit. Consider a uniform flow through a stationary control volume (CV) from state 1 to state 2, as shown in Fig. 6.5. Surroundings r

Po, To CV )

m1

Figure 6.5 A uniform-flow device with heat exchange. The control volume may involve multiple inlets and exits and may exchange heat with the surroundings at Po and To. Some mass may enter the CV at a uniform state i and some mass may exit at another uniform state e. The first- and second-law relations for this process are Q—W = I me (he +

V2 + gze 2

Sgen = (S2 — S1)cv +

V2 ++ gzi j+ (U 2 — 2 I mese — I misi +

)cv

QSUff

To

where the subscripts 1 and 2 refer to initial and final states and Q5,, = — Q. Eliminating between these two relations, we obtain v W=E

+

22

+ gzi — Tosi j— me (he, +

+ [(U1 — U2) — To(Si — S2)]cv — ToSgen

V2

Q

+ gze — Tose ) (6.31)

This is the actual work done during the process. The uniform-flow systems may involve moving boundaries and thus, W,„, = Po(V2 — V1).

170 Fundamentals of Engineering Thermodynamics

6.7.2 Reversible Work This is the maximum possible work and is obtained from Eq. (6.31) by setting Sgen = 0, and adding Ws,,,„ as 2 2 Ve = Emi (hi + --- + gz; — Tosi — Eme he + — + gze — Tose 2

+ [(U1 — U2) — TaSi — S2) + WTI — V2)]cv (kJ)

(6.32)

wre, = Emivri - Eno', + (01 - 02)Cv (a)

(6.33)

or

The availability is given by Availability = Key,„„, = E no/ + (1:01,cv (kJ) The irreversibility is given by

= Toss,. (kJ)

I = Ky.—

(634) (635)

6.7.3 General Unsteady-Flow Processes When an unsteady flow through a control volume has significant variation in the properties of the stream over the inlet and exit areas, it cannot be treated as a uniform flow. This variation of flow properties during a process should be accounted for. This is done by replacing the finite v2 quantity m(h + — + gz —7'os 2

by the sum of differential quantities as V2 f 2 (h + — + gz — Tos Sm 2

where Sm = pVdAdt and p is density, V is the velocity normal to the area, dA is the differential cross-sectional area of an inlet or exit, and dt is the differential time interval. The reversible work for a CV exchanging heat only with the surroundings can be expressed as W = E 12 k

gz; —

2

— E 12 ( +

+ RU1 - U2) — TaSi— 52)

V2 2

gze —T s)Sm Oee

PaVi — V2)Jcv (d)

(6.36)

Or Wr y

2

= E 1 V/,Sint

E

f

2 Ve5nie

+ (01

)CV (d)

(637)

The variation of fluid properties with time as well as position must be known to perform the integrals in these equations. When such variation relations are not available, the integrals can be evaluated only numerically.

Availability, Irreversibility and Availability Analysis of Engineering Processes 171 In the limit that the time interval At —> °°, we have =E

4

2 (k + -a-- + gzi — Tosi )

v2

— E f (he + :-!-- + gze — Tose ) Sthe A. 2

d(U + PoV — ToS)cv (kW) dt

(6.38)

Equations (6.36) and (6.38) can be modified to handle heat transfer with more than one body (say N bodies) by subtracting N

..1 Q„ (1 - n1))

and

1 1 a,

,

(i_a)

respectively, from Eqs. (6.36) and (6.38) to result in =

r2

j

vi2

— + gzi — Tosi ) Smi — I f 2

2

(

he +

V2 + gze Thse) Sme 2 N

-I- [(U1 —

2 ) — To (S1 — S2 ) A- Po (VI — V2 Bcv — E Q.(1— — 4)

11.1

2

(6.39)

v2

=E f (hi +-a-- + gzi — Tos &hi — E f (he + + gze — ;se ) 8iiie A 2 A. d(U + P0V — ToS)cv dt

7i) n=1

rn

(6.40)

The above reversible work relations [Eqs. (6.39) and (6.40)] can be applied to any thermodynamic system.

SUMMARY The availability of a system is defined as the property that enables us to determine the useful work potential of a given amount of energy contained in the system at some specified state. The dead state is that state at which a system is in thermodynamic equilibrium with its surroundings. A system has zero availability in the dead state. Therefore, it can be concluded that a system will deliver the maximum possible work only when it undergoes a reversible process from the specified initial state to the state of its surroundings, i.e. the dead state. It is important to note that the availability only represents the "upper limit on the amount of work a device can deliver without violating any thermodynamic law", and it does not represent the amount of work that a work-producing device will actually deliver. Reversible work, Wre,„ is defined as the maximum amount of useful work that can be obtained as a system undergoes a process between the specified initial and final states. When the final state is the dead state, the We,„ equals availability.

172

Fundamentals of Engineering Thermodynamics

The irreversibility I of any process is the difference between Wev and Wu, which is due to the presence of irreversibilities during the process. The thermal efficiency and coefficient of performance were defined as the measures of performance, in Chapter 4. They are sometimes referred to as the first-law efficiency or conversion efficiency. The first-law efficiency makes no reference to the best possible performance and thus it may be misleading at times. To overcome this deficiency, the second-law efficiency thi, also called the effectiveness, is defined as the ratio of the thermal efficiency to the maximum possible (reversible) thermal efficiency under the same conditions. That is, ihi —

rid,

(heat engines)

qth,rev

The availability of a closed system 1 is the maximum useful work potential of the closed system at a given state. It is given by

el) =(U — U0 ) — T(S — So ) + Po(V — Vo ) (kJ) The irreversibility of a closed system, for any process, is given by 1 = Wm — Wu = ToSgeu (kJ) The availability of a fluid stream is called stream availability. The second-law efficiency of the steady-flow devices can be determined from the general relation, namely tin —

availability recovered availability supplied

PROBLEMS 6.1 A heat engine receives heat from a source at 1000 K at the rate of 500 kJ/s, and rejects waste heat to a sink at 350 K. The surrounding environment is at 300 K. If the measured power output of the heat engine is 200 kW, determine (a) the availability, (b) the rate of irreversibility, and (c) the second-law efficiency of the heat engine. [Ans. 325 kJ/s, 125 kW, 0.615] 6.2 A heat engine that receives heat from a furnace at 630°C and rejects waste heat to a river has a thermal efficiency of 30 per cent and the second-law efficiency of 45 per cent. Determine the temperature of the river water. [Ans. 24.85°C] 6.3 A heat engine operating between thermal reservoirs at 775 K and 310 K has a thermal efficiency of 36 per cent. Determine the second-law efficiency of the engine. [Ans. 60 per cent] 6.4 How much of the 1200 kJ of thermal energy at 750 K can be converted to useful work when the environment is at 300 K? [Ans. 720 kJ]

I

Availability, Irreversibility and Availability Analysis of Engineering Processes 173 6.5 A thermal energy reservoir at 1200 K can supply heat at a rate of 100,000 Id/h. If the environment is at 20°C, determine the availability of this supply heat. [Ans. 21 kW] 6.6

Determine the power generation capacity of a windmill having 7 m diameter rotors, installed in a place where the wind is blowing steadily with an average velocity of 15 m/s. Take the density of air to be 1.225 kg/m3. [Ans. 79.55 kW]

6.7 A turbine expands air adiabatically from 6 bar, 400 K, 200 m/s to I bar, 290 K, 50 m/s. Determine the actual work output and the optimum work output possible for the same operating conditions. Also, find the irreversibility of the process. Take the surrounding atmosphere to be at 1 bar and 280 K. [Ans. 129.24 kJ/kg, 182.78 kJ/kg, 53.54 kJ/kg] 6.8

Determine the closed system availability of nitrogen gas at 10 MPa and 320 K, when the environment is at 101 kPa and 25°C. [Ans. 313.55 kJ/kg]

6.9

Determine the stream availability of a nitrogen gas stream at 10 MPa, 320 K and 10 m/s, if the environment is at 101 kPa and 298 K. [Ans. 457.26 kJ/kg]

6.10 Carbon dioxide gas is contained in a rigid tank of volume 1 m3. The initial state is 1.2 bar and 300 K. The temperature of the gas is increased to 400 K (a) either by stirring with a paddle wheel, (b) or by supplying heat from a reservoir at 500 K. The surrounding atmosphere is at 1 bar and 300 K. Of the two methods which is better from the thermodynamic point of view? Treat carbon dioxide to be an ideal gas. [Ans. (a) I = 248.42 Id, (b) 1= 35.08 kJ; thus, direct heating is better] 6.11 Three kilograms of air contained in a piston-cylinder device at 102 kPa and 300 K is compressed quasi-statically at 300 K to 25 times its initial pressure. During the compression process, heat from the air in the cylinder is transferred to the surrounding atmosphere at 101 kPa and 300 K. Determine (a) the work done on the air in the cylinder, (b) the amount of heat transferred to the environment, (c) the irreversibility associated with the process. Assume air to be an ideal gas. [Ans. (a) - 831.42 Id, (b) - 831.42 Id, (c) 0] 6.12 A steady-flow compressor compresses air from 101 kPa and.30°C to 800 kPa in an adiabatic process. The first-law efficiency for the process is 87 per cent. Calculate the irreversibility for this process and the second-law efficiency. The atmosphere is at 20°C. [Ans. 19.2 kJ/kg, 0.932] 6.13 Two kilograms of air in a rigid tank at 4 bar and 650 K is cooled at constant pressure to 350 K. Calculate the maximum useful work possible with the cooling process. If the air from the given initial state attains a final dead state which is 101 kPa and 20°C instead of 350 K, how much will be the increase in useful work? [Ans. 170.07 U, 30.59 kJ] 6.14 An insulated container of volume 0.02 m3 contains air at 1 bar and 40°C. Work is added to the air through a paddle wheel until the temperature rises to 150°C. Calculate the change in availability of the air if the surrounding environment is at 20°C. [Ans. 3.173 kJ]

174

Fundamentals of Engineering Thermodynamics

6.15 A thermal reservoir at 1000 K can supply heat at the rate of 120,000 kJ/h. Determine the availability of this supplied energy, assuming the surrounding environment to be at 300 K. [Ans. 23.33 kW] 6.16 Air-fuel mixture is burnt at atmospheric pressure in a furnace. If the furnace temperature is constant at 1800 K, find the availability of the system when the surrounding atmosphere is at 22°C. Take the C, of the burnt gas to be 1005 J/(kg K). [Ans. 1264.34 kW] 6.17 Water enters a heat exchanger at the rate of 30 kg/s as compressed liquid at 0.4 MPa and 100°C and leaves at 0.4 MPa and 130°C. Heat is supplied to the water stream by a hot air stream which enters at 600 K and 0.5 MPa and leaves at 200°C and 0.5 MPa. The environmental temperature is 300 K. Determine (a) the change in the stream availability of the water, (b) the change in the stream availability of the air, and (c) the overall irreversibility of the heat-exchange process. Neglect the changes in KE and PE. [Ans. (a) 870.6 kJ/s, (b) 1557 kJ/s, (c) 686.4 Id/s1 6.18 A heat engine with thermal efficiency of 30 per cent and a second-law efficiency of 55 per cent draws heat from a source at 700 K. Determine the temperature of the sink to which the engine rejects waste heat. [Ans. 318.5 K] 6.19 Nitrogen gas at 5 bar and 30°C is throttled through a well-insulated throttle valve to a pressure of 1.3 bar. The surrounding atmosphere is at 20°C. Determine the optimum work and the irreversibility for this process. [Ans. 117.25 kJ/kg, 117.25 kJ/kg] 6.20 Dry saturated steam at 100°C is condensed in a heat exchanger to saturated liquid at the same temperature of 100°C. Determine the maximum useful work obtainable from this process. Assume the environment is at 101 kPa and 25 °C, and neglect the changes in kinetic and potential energies. [Ans. 450.76 kJ/kg] 6.21 A refrigerator has a second-law efficiency of 65 per cent, and heat is removed from the refrigerated space at a rate of 300 kJ/min. If the space is maintained at 4°C while the surrounding air temperature is 30°C, determine the power input to the refrigerator. [Ans. 0.72 kW] 6.22 Steam enters an adiabatic turbine at 3 MPa and 450°C and exits at 300 kPa and 200°C. If the second-law efficiency of the turbine is 0.88, determine the maximum output that the turbine can deliver, the irreversibility, and the environment temperature to which the steam is ejected by the turbine. Assume that the kinetic and the potential energies are negligible. [Ans. 543.64 kJ/kg, 65.24 kJ/kg, 286 K] 6.23 Helium gas of volume 0.5 m3 at 150 kPa and 20°C is compressed in a piston-cylinder device in a polytropic process Pr' = constant, to 400 kPa and 140°C. If the surrounding environment is at 100 kPa and 20°C, determine (a) the actual work done for the compression and (b) the minimum work with which the compression can be achieved. [Ans. (a) -33.5 Id, (b) -31.59 kJ]

Availability, Irreversibility and Availability Analysis of Engineering Processes 175

6.24 An insulated rigid tank of volume 0.8 m3 contains 1.2 kg of ethane gas at 100 kPa. Paddlewheel work is done on the system until the pressure in the tank rises to 150 kPa. Assuming the specific heats to be constant, determine the entropy change of ethane during the process. [Ans. 0.726 kJ/K] 6.25 Air is compressed steadily by a compressor from 101 kPa and 22°C to 1100 kPa at a rate of 0.05 kg/s. Determine the minimum power input required if the process is (a) adiabatic and (b) isothermal. Assume air as an ideal gas with variable specific heats, and neglect the changes in kinetic and potential energies. [Ans. (a) —14.48 kW, (b) —10.1 kW]

CHAPTER

7 Properties of Pure Substances 7.1 INTRODUCTION A pure substance is a substance with fixed chemical composition throughout its mass. For example, water, helium, nitrogen, and carbon dioxide are all pure substances. A pure substance does not have to be a single chemical element or compound. For example, air which is a mixture of several gases is considered to be a pure substance, since it has a uniform chemical composition. A mixture of two or more phases of a pure substance is also a pure substance as long as the chemical composition of all the phases is the same. A mixture of liquid water and ice is a pure substance.

7.2 PHASES OF A PURE SUBSTANCE The three principal phases of a substance are solid, liquid, and gas. However, a substance may have several phases within a principal phase, each with a different molecular structure. It will be useful to gain some understanding of the molecular phenomena involved in each phase, and phase transformation, even though we need not be concerned with the molecular structure and behaviour of different phases, while studying phases or phase changes in thermodynamics. In a solid the molecules are arranged in a three-dimensional pattern which is repeated throughout its mass. The small distance between the molecules makes the attractive forces of molecules on each other very large and keeps the molecules at fixed positions within the solid. In the liquid phase, the molecular spacing is not much different from that of the solid phase, except that the molecules are no longer at fixed positions relative to each other. In the gas phase, the molecules are far apart from each other, and a molecular order is nonexistent. Compared to solid and liquid phases, in the gas phase the molecules are at considerably higher energy levels. Therefore, a gas must release a large amount of heat before it can condense or freeze. 176

Properties of Pure Substances 177

7.3 PHASE-CHANGE PROCESSES Two phases of a pure substance coexist in equilibrium in many practical situations. For example, water exists as a mixture of liquid and vapour in the boiler and the condenser of a steam power plant. A liquid at a state where it is not about to vaporize is called a compressed liquid or subcooled liquid. For example, water, say, at 1 atm and 20°C is a supercooled liquid. A liquid which is about to vaporize is called a saturated liquid (e.g. water at 1 atm and 100°C). A vapour which is about to condense is called a saturated vapour. A substance at states between the saturated liquid and saturated vapour states is referred to as a saturated liquid-vapour mixture, since the liquid and vapour phases coexist in equilibrium at these states. A vapour which is not about to condense is called a superheated vapour. All the phases from subcooled liquid to superheated vapour can be achieved by heating water from (say) 20°C to (say) 300°C at a constant pressure of 1 atm. This constant-pressure phase-change process can be represented as a T—v diagram, as shown in Fig. 7.1. T, °C

P= 1 atm • 5 300 tr

2 b•

100 —

To

3 •

co • 4

e,te0 0 20 — • 1 Figure 7.1 Heating process of water at constant pressure.

Saturation temperature Tsat it is the temperature at which a pure substance starts boiling, at a given pressure. Saturation pressure P.m It is the pressure at which a pure substance starts boiling, at a given temperature. A plot of Tut vs. Put, called a liquid-vapour saturation curve, for water, is shown in Fig. 7.2. A curve of this kind is characteristic of all pure substances. From this figure it is evident that a substance at high pressures will boil at higher temperatures.

178 Fundamentals of Engineering Thermodynamics Put, kPa

600 400 200 O

o

100

I 200

I, Tut °C

Figure 7.2 Liquid-vapour saturation curve for water.

7.4 PROPERTY DIAGRAMS The property diagrams are extremely useful in the study of variation of properties during phasechange processes. We will discuss the T—v, P—v, and P—T diagrams for pure substances, in this section.

7.4.1 The

T—v Diagram

The T—v diagram in Fig. 7.1 describes the phase-change process of water at 1 atm pressure. We now repeat this process at different pressures to develop the T—v diagram for water, as shown in Fig. 7.3. As the pressure increases, the specific volume of water decreases and the boiling point increases. For example, the boiling point of water at 1 MPa is 179.9°C. T, °C A

Critical point co 374.14

Saturated liquid

Saturated vapour v m3/kg

6.003155

Figure 73 The

T—v

diagram of constant-pressure phase-change process of water.

Properties of Pure Substances 179 Further, the specific volume v of the saturated liquid is larger and the v of the saturated vapour is smaller than their corresponding values at 1 atm pressure. That is, the horizontal line that connects the saturated liquid and saturated vapour states, on the T—v plot becomes shorter and shorter with increase in pressure. Ultimately, this line becomes a point when the pressure reaches 22.09 MPa, for water. This point is called the critical point. In other words, the critical point is that point at which the saturated liquid and saturated vapour states are identical. The T, P, and v of a substance at the critical point are referred to as the critical temperature T critical pressure Pcii, and critical specific volume vai, respectively. At pressures above Pai, there will not be a distinct phase-change process. The specific volume of the substance will continually increase, and at all times there will be only one phase present. A substance is referred to as superheated vapour at temperatures above Tcri and as compressed liquid at temperatures below Tai. The saturated liquid line is the locus of all saturated liquid points. The saturated vapour line is the locus of all saturated vapour points. These lines on the T—v diagram will look as shown in Fig. 7.4. T 4

Compressedliquid region

Critical point

, , I ,,

Saturated liquid-vapour region

Saturated liquid line

Superheated vapour region

Saturated vapour line

Figure 7.4 The T—v diagram of a pure substance. All the states that involve both liquid and vapour phases in equilibrium are located under the dome (Fig. 7.4) which is called the saturated liquid-vapour region, or the wet region.

7.4.2 The P—v Diagram The P—v diagram of a pure substance will look like the T—v diagram, but the T = constant lines on this diagram have a downward slope, as shown in Fig. 7.5.

180

Fundamentals of Engineering Thermodynamics P A Critical point

Superheated vapour region

Saturated liquid-vapour region

Saturated liquid line

Saturated vapour line .v

Figure 7.5 The P—v diagram of a pure substance.

7.4.3 Solid Phase on Process Diagram The basic principles discussed with respect to the liquid—vapour phase-change process apply to solid—liquid and solid—vapour phase-change processes as well. Most substances contract during their solidification (freezing) process. However some substances, like water, expand as they freeze. The P—v diagrams for these two groups of substances are shown in Figs. 7.6 and 7.7, respectively. P

Solid + Vapour

Figure 7.6 The P—v diagram for a substance that contracts on freezing.

Properties of Pure Substances

181

p A

Solid

v Figure 7.7 The P—v diagram for a substance that expands on freezing. ►

As seen from Figs. 7.6 and 7.7, the P—v diagrams for the above two groups of substances differ only in the solid—liquid saturation region. Just as two phases can coexist in equilibrium, under some conditions all three phases of a pure substance can also coexist in equilibrium. On P—v or T—v diagrams, these triple-phase states form a line called the triple line. The states on the triple line of a substance have the same P and T but a different v. On a P—T diagram, the triple line appears as a point and, therefore, it is often called the triple point. For water, the triple point T and P are 0.01°C and 0.6113 kPa, respectively. No substance can exist in the liquid-phase at pressures below the triple.point pressure. In the same manner, no substance can exist in the liquid-phase at temperatures below the triple point temperature for substances that contract on freezing.

7.4.4 The

P—T Diagram

The P—T diagram or the phase diagram of a pure substance in P—T coordinates is shown in Fig. 7.8. As seen from this figure, the sublimation line separates the solid and vapour regions, the vaporization line separates the liquid and vapour regions, and the melting line separates the solid and liquid regions. All the above three lines meet at the triple point.

7.5 THE

P—v—T SURFACE

We know that the state of a simple compressible substance is fixed by any two independent, intensive properties. Once the two appropriate properties are fixed, all the other properties become dependent properties. Realizing that any equation with two independent variables of the form Z = Z(x, y) represents a surface in space, we can represent the P—v—T behaviour of a substance as a surface in space, as shown in Figs. 7.9 and 7.10. Here T and v may be treated as independent variables and P as the dependent variable.

182 Fundamentals of Engineering Thermodynamics P

Substances that contract on freezing

Substances that expand on freezing

Critical point

ie (Ps" ..9 Triple point

Vapour

T Figure 7.8 The P—T diagram of a pure substance. Solid Solid-liquid Liquid

Gas Critical point Liquid—vapour Triple line Vapour

Figure 7.9 The P—v--T surface of a substance that contracts on freezing. All points on the P—v--T surface represent equilibrium states. All states along the path of a quasi-equilibrium process lie on the P—v—T surface, since such a process must pass through equilibrium states.

Properties of Pure Substances 183

Figure 7.10 The P—v—T surface of a substance that expands on freezing.

7.6 PROPERTY TABLES Generally, the relationships among thermodynamic properties are too complex to be expressed by simple equations. Therefore, the thermodynamic properties are often represented in the form of tables. These tables are extremely useful for quantitative calculations. The intensive properties which are of primary interest in thermodynamic analysis of systems are pressure P, specific volume v, temperature T, specific internal energy u, specific enthalpy h, and specific entropy s. In a single-phase region, such as the superheat region, two intensive properties are required to designate or specify the equilibrium state. Variables such as v, u, h, and s are usually tabulated in superheat tables as functions of P and T, since P and T are easily measurable properties. To get an insight into the superheat table, let us consider the P—T diagram shown in Fig. 7.11. In this figure, the superheat vapour region is divided into grids. The lines of these grids represent P

liquid region

T

Figure 7.11 The P—T diagram to illustrate superheated vapour table.

184

Fundamentals

of Engineering Thermodynamics

the integral values of the pressure and the temperature. A superheated vapour table then represents the values of u. v, h, and s at the grid points. The values of u, v, h and s are completely arbitrary, each being an assigned value at some reference state. Tables of superheated vapour data for few substances are given in the Appendix. If the data required for a particular T or P is not given in the table, the required values can be obtained by interpolation. In fact, in many calculations the interpolation of data becomes inevitable. The saturation tables list specific property values such as v, u, h, and s for the saturatedliquid and saturated-vapour states. The properties of these two phases are denoted by the subscripts f and g, respectively.

7.6.1 Superheated Tables We know that the superheat region is a single-phase region (only vapour phase), T and P are no longer the dependent properties and they can be used as the two independent properties in the table.

7.6.2 Compressed Liquid Tables Compressed liquids are also called subcooled liquids. There is not much data available on compressed liquids. However, since water is used as the fluid in fossil-fuel power plants, considerable data is available on it in the liquid region. This data is given in Table 6 in the Appendix. Compressed-liquid data in most cases can be approximated closely by using the property values of the saturated-liquid state at the given temperature. That is, the compressed-liquid data is more temperature dependent than it is pressure dependent. However, when the pressure difference between the saturated-liquid and the compressed-liquid is very large, the enthalpy value should be corrected for the pressure difference. In the absence of compressed-liquid data, the compressed liquid may be treated as saturated liquid at the given temperature.

7.6.3 Reference State and Reference Values The quantities u, h, and s cannot be measured directly. They are calculated from measurable properties such as P and 7', using the relations among the thermodynamic properties. But, these relations give only the changes in properties and not their absolute values. Thus, the tabulated values of u. h, and s in all the tables are not absolute values. Each is the difference between the value at any state and the value of the respective property at a reference state. Therefore, we need to choose a convenient reference state and assign the value of zero to a convenient property or properties at that state. For water, the state of saturated liquid at 0.01°C is taken as the reference state and the internal energy and entropy are assigned zero values at that state.

Enthalpy Enthalpy was defined with the properties U, p, and V as H = U + PV (kJ)

Properties of Pure Substances 185 On unit mass basis, we have h = u + Pv (kJ/kg) Mollier referred to the quantity (u + Pv) as heat content or total heat. But in the modern thermodynamic terminology, this quantity is termed enthalpy (from the Greek word enthalpien which means heat). We know that H is the total enthalpy or simply enthalpy and h is the specific enthalpy. The property enthalpy may be viewed as a combination property.

7.6.4 Saturated Liquid-Vapour Mixture A substance exists as a mixture of saturated liquid and saturated vapour during a vaporization process. For a proper analysis of this mixture, the properties of the liquid and vapour phases in the mixture should be known. This is done by defining a new property called the quality x as the ratio of the mass of vapour to the total mass of mixture. x

- '° "hot

(7.1)

where mud = mass of liquid + mass of vapour = ml + ms. Also, we can write the total volume V of the mixture as the sum of the volume of saturated liquid Vf and the volume of saturated vapour That is, Vg.

V = Vf

Vg

But V = my, where m is the mass of the mixture and v its specific volume. Therefore, we can write av =

" tav

Mfif M gVg

= (NO

Mg) Vf MgVg

This results in vas = (1— x)vf + xvg (m3/kg)

(7.2)

where v.,. is the average specific volume of the mixture, x is the quality and of and vg are the specific volumes of saturated liquid and saturated vapour, respectively, in the mixture. The above analysis can be done alternatively as follows: Equation (7.2) can also be expressed as (73) vav = of XVfg (m3/kg) where vig = vg — vp is the volume change per unit mass during the vaporization process. This yields x=

Vav

Yfg

(7.4)

Similarly, it can be shown that u.„ = ul + xufg (kJ/kg)

(7.5)

hay = hi + xhfg (kJ/kg)

(7.6)

186

Fundamentals of Engineering Thermodynamics

The quantity hik is termed the enthalpy of vaporization or latent heat of vaporization. It simply means the amount of energy required to vaporize a unit mass of saturated liquid at a given temperature or pressure. Its value decreases as the temperature or pressure increases, and becomes zero at the critical point. EXAMPLE 7.1 Determine whether water at each of the following states is a compressed or subcooled liquid, an equilibrium mixture of liquid and vapour, or a superheated vapour. (a) T = 303 K; P = 50 kPa (b) P = 1.0 MPa; v = 1.5 m3/kg Solution (a) From the saturated water-temperature table (Table 3 of the Appendix), at 303 K (30°C) P.,( = 4.246 kPa. But the given pressure is 50 kPa, therefore, the water will exist as a subcooled or compressed liquid. (b) At

P=

1.0 MPa from the saturated water-pressure table (Table 4 of the Appendix), we get vi= 0.001127 m3/kg vg = 0.19444 m3/kg

The given specific volume is much more than of and vg. Therefore, the state is superheated vapour state. EXAMPLE 7.2 A rigid tank contains 200 kg of saturated water at 65°C. Determine the pressure in the tank and the volume of the tank. Solution The water in the tank exists as saturated liquid, therefore, the pressure must be the saturation pressure at 65°C. From the saturated water-temperature table (Table 3 of the Appendix) at 65°C, we get Put =

25.03 kPa

The specific volume of saturated water at 65°C is ,

vi- = 0.001020 m3/kg

The total volume of the tank becomes V = mv = 200 x 0.001020 = 0.204 m3 EXAMPLE 73 One hundred grams of saturated liquid water is vaporized completely at a constant pressure of

Properties of Pure Substances

187

75 kPa. Determine the change in volume. What is the amount of energy required for the vaporization process? Solution The change in volume per unit mass during a vaporization process is given by vfg = vg — VI. From the saturated water-pressure table (Table 4 of the Appendix), at 75 kPa vg = 2.217 m3/kg vf = 0.001037 m3/kg. Thus, vfg = vg — vf = 2.217 — 0.001037 = 2.215963 m3/kg Therefore, the change in volume of 100 g of saturated water during the vaporization process is AV = nivfg = 2.215963 x 0.1

0.2216 m3

The energy required to vaporize a unit mass of saturated liquid at a given T or P is the enthalpy of vaporization hfg At 75 kPa, hfg from Table 4 is hfg = 2278.6 kJ/kg Therefore, the amount of energy required for the vaporization process is mhik = 0.1 x 2278.6 = 227.86 kJ EXAMPLE 7.4 A rigid tank contains 3 kg of water as saturated liquid-vapour mixture at 80°C. If 2 kg of water is in the liquid form and the rest in the vapour form, determine the partial pressure of the mixture and the volume of the tank. Solution The mixture is given as saturated mixture. Therefore, the two phases coexist in equilibrium. Hence the pressure must be the saturation pressure at the given temperature. From the saturated water-temperature table (Table 3 of the Appendix), at 80°C Psat

= 4739 kPa

Also from Table 3, at 80°C, we get of = 0.001029 m3/kg, vg = 3.407 m3/kg Now the volume can be obtained in two ways: 1. Find the volume occupied by each phase and add them. 2. Find the average volume and get the final volume from it. Method 1 V= = 2 x 0.001029 + 1 x 3.407 Vg = MfVf MgV g

3.409 m3

0 •

ti

188 Fundamentals of Engineering Thermodynamics Method 2 = mvap — mat 3 1 v = of + xvfg = 0.001029 + i(3.407 — 0.001029) = 1.13635 m3/kg Thus, V = mv = 3 x 1.13635 = 3.409 m3 EXAMPLE 7.5 A rigid tank of 0.03 m3 volume contains a mixture of liquid water and water vapour at 80 kPa. The mass of the mixture in the tank is 12 kg. Calculate the heat added and the quality of the mixture when the pressure inside the tank is raised to 7 MPa. Solution The tank is rigid and, therefore, the volume is constant and the work done is zero. Thus, the first law becomes Q = AU = ndu The specific volume of the mixture is .03 v=V —= 0 — 0 . 0025 m3/kg m 12 The specific volume can also be written as v = of + xvfg Let the subscripts 1 and 2 refer to initial and final states, respectively. At state 1, water exists as a saturated liquid, therefore, from the saturated water-pressure Table 4, at 80 kPa, we have vfi = 0.001038 m3/kg, vgi = 2.087 m3/kg Therefore, 4. •

ipt • •

0.0025 = v = 0.001038 + x(2.087 — 0.001038)

or x= 0.0007 The quality of the mixture is 0.07 per cent. Again from Table 4, uI = uf + ufgi = 391.58 + 0.0007(2498.8 — 391.58) = 393.05 Id/kg The specific volume of the mixture remains the same, since the tank is rigid. Therefore, at state 2, v = 0.0025 =

1112 XVig2

From Table 4, at 7 MPa, we have vf2 = 0.001351 m3/kg, vg, = 0.02737 m3/kg

Properties of Pure Substances

189

Thus, 0.0025 = 0.001351 + x(0.02737 — 0.001351) or x = 0.0441 Now, the internal energy is given by (see Table 4) u2 = un + x2ufg2 = 1257.55 + (0.0441)(1323.0) = 1.316 MJ/kg The heat transferred to the mixture is given by Q = m(u2 — u1)= 12(1.316 — 0.39305)

= 11.075 MJ

7.7 THE IDEAL-GAS EQUATION\ OF STATE A state equation is an equation relating the pressure, temperature, and specific volume of a substance. Among the several equations of state, the simplest and the best known equation for substances in the gas phase is the ideal-gas equation of state. It was experimentally determined by J. Charles and J. Gay-Lussac, that at low pressures the volume of a gas is proportional to its temperature. That is, T) P=R (— v Or

Pv = RT

(7.7)

where the constant of proportionality R is called the gas constant and Eq. (7.7) is called the ideal-gas equation of state. A gas which obeys this equation is called an ideal gas. The gas constant R is determined from the relation R = RE [kJ/(kg K)l

(7.8)

where R. is the universal gas constant and M is the molecular weight of the gas. The gas constant R is different for each gas, whereas R„ is the same for all substances and its value is R.= 8.314 [kJ/(kg K)) The molecular weight or molar mass M is defined as the mass of one mole of a substance in kilograms (or grams). When we say that the molecular mass of oxygen is 32, it means that the mass of I kilomole of oxygen is 32 kg, or 1 grammole of oxygen is 32 grams. The mass of a system m is equal to the product of its molecular mass M and the mole number N. That is, m = MN (kg) The values of the gas constant R and molar mass M for many substances are given in Table 1 of the Appendix.

190 Fundamentals of Engineering Thermodynamics The ideal-gas state equation can be written in different forms as follows: (1) PV = mRT, since V = mv. (2) PV = NR„ T, since mR = (MN)R = NR,, (3) Pi,- = R.T, since V = where v is the volume per unit mole, called the molar specific volume (in m3/kmol). Many familiar gases such as air, nitrogen, oxygen, helium, argon, neon, krypton and even heavier gases such as CO2 can be treated as ideal gases with negligible error (less than I per cent). However, dense gases such as water vapour in steam power plants and the refrigerant in refrigerators should not be treated as ideal gases.

7.8 COMPRESSIBILITY FACTOR Basically, the compressibility factor Z may be viewed as a measure of the deviation from idealgas behaviour. It is defined as Pv Z= (7.9) RT It can be expressed as Z

Vactual videal

(7.10)

where vid,„,1 = RT/P. That is, Z = 1 for ideal gases. For real gases, Z can be greater than or less than unity. The ideal-gas equation is simple and thus convenient to use. But gases deviate from idealgas behaviour significantly at states near the saturation region and critical point. This deviation from ideal-gas behaviour at a given temperature and pressure can be accurately accounted for by the introduction of the correction factor called the compressibility factor Z. The farther Z is away from unity, the more the deviation of the gas from ideal-gas behaviour.

7.8.1 Principle of the Corresponding States At a given pressure and temperature, different gases behave differently. But they behave very much the same as T and P are normalized with respect to their Tc,i and F. The normalized pressure PR(= P/Pcia and normalized temperature TR(= T/Tcri) are called the reduced pressure and reduced temperature, respectively. At the same PR and TR, the Z for all gases is approximately the same. This is called the principle of the corresponding states.

7.8.2 Generalized Compressibility Chart A generalized compressibility chart is a set of curves fitted through a set of experimentally determined Z plotted versus PR and TR for several gases. This chart can be used for all gases. A typical compressibility chart is shown in Fig. 7.12. In Fig. 7.12, the experimentally determined Z values are plotted versus PR and TR for several gases. These gases were found to obey the principle of the corresponding states reasonably well. By curve fitting these data, we obtain the generalized compressibility chart

4.0

.0.11611 ►1

Compressibility factor, Z = Pv/RT

3.0

g5./...-.0.14.

iso 2.0

3.0 OO1115" co° 8.00 10.00 vR = 0.60

1.0

0.0

0

10

20 Reduced pressure, PR Figure 7.12 Compressibility chart.

30

40

Propertiesof Pure Substances191

TR = 1.00

192 Fundamentals of Engineering Thermodynamics which can be used for all gases. This chart is given in the Appendix for three ranges for reduced pressure, namely, 0 < PR < 1, 0 < PR < 7, and 0 < PR < 40. From the generalized compressibility chart, it can be observed that: • At very low pressures (PR 2), the ideal gas behaviour can be assumed with good accuracy regardless of pressures (except when PR >> 1). • The deviation of a gas from the ideal-gas behaviour is the greatest in the vicinity of the critical point.

7.9 OTHER EQUATIONS OF STATE The ideal-gas equation, even though simple, is applicable only over a limited range. Therefore, continued attempts were made to develop equations of state that could represent the P—v—T behaviour of substances over a larger region without any limitation. Several equations have been proposed; some of them are the van der Waals equation of state, Beattie-Bridgeman equation of state, Bertholet equation of state, and Dieterici equation of state.

7.9.1 The van der Waals Equation of State It has two constants which are determined from the behaviour of a substance at the critical point. The equation is given by (P+ — a2 )(v — b)= RT

(7.11)

This equation accounts for the following effects: 1. The intermolecular attraction forces 2. The volume occupied by the molecules themselves. These effects are not considered in the ideal-gas model. The term a/v2 accounts for the intermolecular attraction force and the constant b accounts for the volume occupied by the gas molecules. The constants a and b are given by a—

27R2 Tc2„. 64/16

b— RTcri 8Pcfi For any substance, a and b can be determined from the above critical point data.

7.9.2 The Beattie-Bridgeman Equation of State This equation is developed with five experimentally determined constants. It is given by P

=2 V

1

_ c Ti T3

A I

V

2

(7.12)

Properties of Pure Substances 193

a where A = A0(1— — ) and B = B0(1— — b ). The five constants a, b, c, A0, and B0 have to be determined experimentally.

7.9.3 The Bertholet Equation of State This equation is given by P=

RT —b

a T

(7.13)

The constants a and b are given by a—

27R27 64Pcd

b_ 9RT • 28Pcri

7.9.4 The Dieterici Equation of State This equation is given by P — RT —b

e-

a/ RTV

(7.14)

where a— b

4R2 Tgi 13,02 RTeri Pc02

2 •— 0.271 = — RTcd e2

Pcrii7en

This equation predicts properties very well in the neighbourhood of the critical point and on the critical isotherm, but results in large errors in other regions.

7.9.5 Virial Equations of State The equation of state of a substance can also be expressed in a series form as a(T) b(T) + P=R + 2 3 v

V

c(T) + d(T) V4 VS

(7.15)

This and similar equations are called virial equations of state. The coefficients a(7'), b(7'), etc. which are functions of T alone are called the virial coefficients. These coefficients can be determined experimentally or theoretically from statistical mechanics. All equations of state given above are applicable to the gas phase of the substances only.

194 Fundamentals of Engineering Thermodynamics EXAMPLE 7.6 A small capsule of volume 0.0002 m3, containing liquid water at 700 kPa and 40°C, is placed inside a large evacuated container of volume 1.30 m3. The capsule is broken by an appropriate device and the water is allowed to evaporate and fill the large container. The large container exchanges heat with its surroundings and reaches a fmal equilibrium temperature of 40°C. Determine the final quality of the water vapour mixture in the container. Also, determine the amount of heat exchanged with the surroundings. Solution Let us take the large container and the capsule within it as our system, as shown in the adjacent figure. The inside of the container forms the boundary of the system. The work done is zero, since the system has a rigid boundary. By the first law, we have Q — W = AU or Q = AU = m(u2 — u1 )

(... W = 0)

From the saturated water-pressure table (Table 4 of the Appendix), at 700 kPa, we have = 0.001108 m3/kg, u1 = 696.44 kJ/kg The mass of the water is V1 0.0002 m= — = — 0.18 kg v1 0.001108 The final specific volume is v2

V2 = 13 = 7.222 m3/kg = m 0.18

The final state is to be determined from

T2 =

40°C and v2 = 7.222 m3/kg.

From the saturated water-temperature table (Table 3 of the Appendix), at vg2 = 19.52 m3/kg, vf2 = 0.001008 m3/kg Since vg2 > v2, the mixture exists as a saturated mixture. Therefore, V2 = V12 + X2Vig2 Or

7.222 = 0.001008 + x2(19.52 — 0.001008) Or x2 =

0.37

Using Table 3 again, the internal energy at the final state is U2 = Uf2

X 2Ufg2

= 167.56 + 0.37(2262.6) = 1004.72 kJ/kg

T2 =

40°C, we have

Properties of Pure Substances 195 Therefore, the heat transfer between the system and the surroundings is Q = m(12 - u1 ) = 0.18(1004.72 - 696.44) 55.5 Id That is, 55.5 kJ of heat is supplied by the surroundings to the container to maintain the vapour at 40°C during the expansion-evaporation process. EXAMPLE 7.7 Determine the specific volume of methane gas at 10 MPa and 250 K using (a) the ideal-gas equation of state and (b) the compressibility chart. Solution (a) From the ideal-gas equation of state, we have v=

RT

0.5182 x 250

P

10 x 103

0.01296 m3 /kg

(b) For methane gas, the critical pressure and temperature are 4.64 MPa and 191.1 K, respectively. Therefore, the reduced pressure and temperature are, respectively 10

PR -

4.64

-

2.15

pen

and T 250 , = — = — - 1.31 7;,.; 191.1 For these values of PR and TR, from the compressibility chart (compressibility chart (b) 0 < PR < 7 of the Appendix), we have Z = 0.68 Therefore, v=

ZRT P

0.00881 m3/kg

Note that the specific volume obtained by the ideal-gas equation of state differs from the value obtained by the compressibility chart by 47 per cent. EXAMPLE 7.8 Determine the approximate slope of a constant-pressure line in the h-s diagram for steam at 1.0 MPa and 300°C. Solution The required slope is given by (dhr dA)p. Since only approximate slope is required, (dhl as)p can be taken as (AhlAs)p. The given state of the steam is superheated, since 300°C > Tcd for water at 1 MPa (i.e. 179.91°C).

196 Fundamentals of Engineering Thermodynamics From the superheated water table (Table 5 of the Appendix), we have T(°C)

h(kJ/kg)

s[1(.1/(kg K)]

250 300 350

2942.6 3051.2 3157.7

6.9247 7.1229 7.3011

Therefore,

(

Ali) _ 3157.7 — 2942.6 _ 571.5 K 7.3011— 6.9247 )p

Note: We can write dh = (Sq + vdp = Tds + vdp At constant pressure, dh = Tds, and thus,

( dh ) rAh) _ ds )p-- As ) P

=T

Thus, the slope (AhlAs)p is nothing but the thermodynamic temperature at the given state. Therefore, the correct answer for the problem is (AhlAs)p= 300 + 273.15 = 573.15 K EXAMPLE 7.9 A rigid tank of volume 0.5 m3 contains 80 per cent by volume of saturated liquid water and 20 per cent by volume of saturated steam at 200°C. Now 140 kg of liquid water is pumped into the tank. If the final temperature of the fluid in the tank is 80°C, determine the final pressure. Solution The initial volumes of the liquid water and steam in the tank are, respectively Vp = 0.8(0.5) = 0.4 m3 and Vg1 = 0.2(0.5) = 0.1 m3 At 200°C, from the saturated water-temperature table (Table 3 of the Appendix), we have vf = 0.001157 m3/kg and vg = 0.12736 m3/kg Therefore, the masses of the liquid water and steam are, respectively mf =

vf

—

0.4 — 345.72 kg 0.001157

and mg =

Vg1 5

0. 1 — 0.12736 — 0.785 kg

Properties of Pure Substances 197

Thus, the initial mass of the fluid in the tank is m1 = mi + mg = 346.5 kg The final mass is m2 = m i + Am = 346.5 + 140 = 486.5 kg The average specific volume of the fluid is V Vav —

0.5 —

In,

486.5

—0.

00102775 m3 /kg

Thus, the final state is given by v = 0.00102775 m3/kg and T = 80°C. For this v and T, from the compressed water table (Table 6 of the Appendix), we get (by extrapolation) P = 3 MPa = 30001cPa Note: From Table 6 of the Appendix, at T = 80°C, for P = 5 MPa and 10 MPa, we have v = 0.00102068 m3/kg and 0.0010245 m3/kg, respectively. Therefore, we may say that the change in v for every 1 MPa change in pressure is 4.6 x 10-7 m3/kg. Thus, at 80°C, v = 0.00102775 m3/kg corresponds to P = 3 MPa. EXAMPLE 7.10 A rigid tank of volume 10 m3 is filled with nitrogen at 3.0 MPa and 300 K. Determine the mass of nitrogen in the tank. Solution At 3.0 MPa, the critical temperature Tcri = 123.5 K. Therefore, the nitrogen is at superheated state. From the superheated nitrogen table (from Engineering Thermodynamics by Francis F. Huang), we have P(MPa)

v(m3/kg)

2 3 4

0.044396 ? 0.022179

The specific volume at 3 MPa has to be obtained by interpolation. The available data interval is two wide for linear interpolation. Using the inverse interpolation will yield a more accurate result. Let us solve the problem by both the methods. (1) Using linear interpolation, we obtain v — 0.044396 0.022179 — 0.044396 or v = 0.0332875 m3/kg

3—2 4—2

198 Fundamentals of Engineering Thermodynamics Therefore, V 10 mass m = — v 0.0332875

300 . 4 kg

(2) Using inverse interpolation, we obtain 1 1 v — 0.044396 _3— 0.022179 — 0.044396 1_ 1 4 2 Or

v = 0.02956 m3/kg Therefore, m=

10— 338.30 kg 0.02956

Note: The inverse interpolation will give a more accurate result, since at a given T v is inversely proportional to P. EXAMPLE 7.11 The average atmospheric pressure in Kanpur is 97.3 kPa. Determine the temperature at which water in an uncovered pan will boil in Kanpur. Solution The boiling temperature of water in Kanpur is the saturation temperature corresponding to the atmospheric pressure in Kanpur which is 97.3 kPa. From the saturated water-pressure table (Table 4 of the Appendix), we have P(kPa)

T(°C)

75 973 100

91.78 99.63

Since P is directly proportional to T, using linear interpolation, we obtain T — 91.78 99.63 — 91.78

97.3 — 75 100 — 75

Or

T= 98.78 °C Water will boil at 98.78°C in Kanpur. EXAMPLE 7.12 Plot the following processes on P—v and P—T diagrams. (a) A superheated vapour is cooled at constant pressure until the liquid just begins to condense.

Properties of Pure Substances 199 (b) A liquid vapour mixture with a quality 50 per cent is heated at constant volume until its quality becomes 100 per cent. Solution (a) In this process, the superheated vapour is brought to the state where it just condenses. The process can be represented on P—v and P—T diagrams as below: Saturated vapour line

1 Superheated region Saturated region

Vapour I T

(b) In this process, the liquid vapour mixture of quality 50 per cent is heated to reach 100 per cent quality. Therefore, the mixture state will change from saturated mixture state to superheated state. The process can be represented on P—v and P—T diagrams as below: P

P

T

►v

EXAMPLE 7.13 Determine the specific volume of water vapour at 20 MPa and 520°C: (a) by the ideal-gas equation of state, (b) by the principle of the corresponding states, and (c) by using the steam table. For water vapour, R = 0.4615 kJ/(kg K). Solution (a) By the ideal-gas equation of state, we have RT

0.4615 x (520 + 273.15) 20 x 103 0.0183 m3/kg

(b) The specific volume v3 = Z vid„i where Z is the compressibility factor which depends on the reduced pressure PR and reduced temperature TR. Here, P

20

PR = P • = 22.09 —

0.905

200

Fundamentals of Engineering Thermodynamics

and TR —

793.15 T — — 1.225 Tcr, 647.3

From the compressibility chart (compressibility chart (a) 0 < PR < 1 of the Appendix), for these values of PR and TR, Z = 0.83. Thus, v., = 0.83 x 0.0183 = 0.01519 m3 /kg (c) From the superheated water table (Table 5 of the Appendix), at 20 MPa and 520°C, we get (by linear interpolation) v= 0.01548 m3 / kg Note that v obtained from Table 5 differs from vac, obtained by the principle of the corresponding states by 1.87 per cent, whereas it differs from the vki,,,,I by 18.2 per cent. EXAMPLE 7.14 Find the change in u and h of water for a change of state from 20°C, 20 MPa, to 60°C, 30 MPa by using (a) the compressed liquid table, and (b) the saturated water table: Solution (a) From the compressed water table (Table 6 of the Appendix), we get u2 — u1 = 246.06 — 82.77 = 163.29 kJ/kg h2 — h1 = 276.19 — 102.62 = 173.57 kJ/kg (b) The u and h for compressed water may be taken as the uf and hp respectively, of saturated water at the given temperature. Thus, from the saturated water temperature table (Table 3 of the Appendix), we get u2 — u1 = 251.11 — 83.95 = 167.16 kJ/kg h2 — h1 = 251.13 — 83.96 = 167.17 kJ/kg Note that the error in u because of using the saturated liquid data in place of the compressed liquid data is 2.3 per cent, whereas the error in h is 3.7 per cent.

SUMMARY A pure substance is a substance with fixed chemical composition throughout its mass. The three principal phases of a substance are solid, liquid, and gas. A liquid at a state where it is not about to vaporize is called a compressed liquid or subcooled liquid.

Properties of Pure Substances 201 A liquid which is about to vaporize is called a saturated liquid. A vapour which is about to condense is called a saturated vapour. A vapour which is not about to condense is called a superheated vapour. Saturation temperature is the temperature at which a pure substance starts boiling at a given pressure. Saturation pressure is the pressure at which a pure substance starts boiling at a given temperature. The critical point is that point at which the saturated liquid and saturated vapour states are identical. A substance is referred to as superheated vapour at temperatures above Tai and as compressed liquid at temperatures below T. The saturated liquid line is the locus of all saturated liquid points. The saturated vapour line is the locus of all saturated vapour points. The triple line is the line formed by the triple-phase states on a P—v or T—v diagram. The triple point is that point on a P—T diagram at which all the three phases of a substance coexist. The quality x of a mixture is a property, defined as the ratio of the mass of vapour to the total mass of mixture. That is, lIkap X

"hot A state equation is an equation relating the pressure, temperature, and specific volume of a substance. For ideal gases, the state equation is Pv = RT The compressibility factor Z may be viewed as a measure of the deviation from ideal-gas behaviour. It is defined as Le =

Pv — RT

The normalized pressure PR (= P/Prn) and the normalized temperature TR (= TIT„;) are called the reduced pressure and reduced temperature, respectively. At the same PR and TR, the Z for all gases is approximately the same. This is called the principle of the corresponding states.

PROBLEMS 7.1

Determine the states of Freon-12 at the following states: (a) T = —15°C, P = 25 kPa, (b) T = 30°C, s = 0.5 kJ/(kg K), (c) P = 1 MPa, h = 220 kJ/kg. [Ans. (a) superheated vapour, (b) saturated mixture, (c) superheated vapour]

202 7.2

Fundamentals of Engineering Thermodynamics Determine the h, v and s of water at 150 kPa, and 10 per cent quality. [Ans. 689.76 ki/kg, 0.11688 m3/kg, 2.0126 kJ/(kg K)]

7.3 A piston-cylinder device contains steam at 500 kPa and 50 per cent quality. The initial volume is 0.04 m3. Determine the amount of heat addition required to increase the volume to 0.1554 m3 at constant pressure. [Ans. 387.27 kJ] 7.4 A rigid container is filled with steam at 600 kPa and 200°C. At what temperature will the steam begin to condense when it is cooled? Determine the corresponding pressure. [Ans. 154.4°C, 535.5 kPa] 7.5 Determine the specific volume and entropy for superheated steam at 250 kPa and 500°C. [Ans. 1.484 m3/kg, 8.4192 kJ/(kg K)] 7.6

A storage tank of volume 120 m3 contains oxygen at 200 K and 101 kPa. Determine the mass of oxygen in the tank. [Ans. 234.7 kg]

7.7 A storage tank of volume 120 m3 contains oxygen as a saturated liquid at 101 kPa. Determine the mass of oxygen in the tank. [Ans. 136,986.3 kg] 7.8

The state of nitrogen is changed from 180 K and 1 MPa to 280 K and 1 MPa. Determine the change in the property Pv because of the change of state. [Ans. 31.08 kJ/kg]

7.9 Saturated water at 30°C is changed to a final state given by 200 kPa. The entropy at the initial and final states is the same. Determine the change in enthalpy of the water. Assume water to be incompressible. [Ans. 0.197 kJ/kg] 7.10 Nitrogen at an initial state of 30 MPa and 160 K is changed to a final state of 101 kPa. The enthalpy at both the states is the same. Determine the fraction of the mass that is saturated liquid at the final state. [Ans. 0.122] 7.11 Oxygen at 200 kPa and 300 K is changed to a final state of 101.325 kPa, isentropically. Determine the final temperature of the fluid. [Ans. 247.4 K] 7.12 A rigid tank of volume 1 m3 contains 60 per cent of saturated liquid water by volume and 40 per cent of saturated vapour by volume at 75°C. Determine the quality of the mixture and the total mass of the fluid in the tank. [Ans. 0.0001655, 584.897 kg] 7.13 One kilogram of oxygen occupies a volume of 0.036 m3 at a temperature of 110 K. Find the pressure of the fluid. [Ans. 543.5 kPa] 7.14 Ten kilograms of water with a quality of 40 per cent is contained in a tank at 1 MPa. What is the volume of the tank? [Ans. 0.7845 m3]

Properties of Pure Substances 203 7.15 The state of 1 kg of oxygen is changed isothermally at 220 K from 4 MPa to 0.2 MPa. Determine the changes in enthalpy and entropy associated with this process. [Ans. 17.113 kJ, 0.8344 kJ/K] 7.16 Find the enthalpy and entropy of the steam when P = 3 MPa and the specific volume is 0.055 m3/kg. [Ans. 2484.55 kJ/kg, 5.5566 kJ/(kg K)] 7.17 Determine the enthalpy, entropy, and specific volume of the steam at 2.5 MPa and 320°C. [Ans. 3055.8 kJ/kg, 6.7224 kJ/(kg K), 0.103244 m3/kg] 7.18 A piston-cylinder device initially contains 120 litres of liquid water at 30°C and 200 kPa. Heat is transferred to the water at constant pressure until the entire liquid is vaporized. Determine (a) the mass of the water, (b) the final temperature, (c) the total enthalpy change, and (d) the total entropy change. [Ans. (a) 119.52 kg, (b) 120.23°C, (c) 308.47 MJ, (d) 799.62 kJ/K] 7.19 A rigid tank of volume 0.3 m3 contains saturated liquid-vapour mixture of water at 80°C. The water is now heated until it reaches the critical state. Determine the mass of the liquid water and the volume occupied by the liquid water at the initial state. [Ans. 95.03 kg, 0.098 nr3] 7.20 Water in a 300 mm deep container is found to boil at 99.8°C. Determine the temperature at which water in a 30 mm deep container will boil. Assume that both containers are full of water. [Ans. 99.01°C] 7.21 Enumerate the condition under which the real gases can be treated as ideal gases. [Ans. When T is high or P is low relative to its Teri and Pa;, respectively.] 7.22 A spherical balloon of 4 m diameter is filled with nitrogen at 22°C and 202 kPa. Determine the mole number and the mass of the nitrogen in the balloon. [Ans. 2.76 mol, 77.2 kg] 7.23 Steam at 0.5 MPa and 200°C is cooled at constant volume. Find the temperature at which the steam will become saturated vapour. Calculate the heat transferred per kg of steam in cooling from 200°C to saturation temperature. [Ans. 147°C, -86.16 kJ/kg] 7.24 Steam at 2 MPa and 250°C enters a turbine. If the steam is expanded reversibly and adiabatically in the turbine to a temperature of 50°C at the turbine exit, determine the work output per kg of steam. [Ans. 806.1 U/kg] 7.25 Steam flowing through a pipe with a mass flow rate of 0.8 kg/s is at 0.5 MPa and 300°C. Steam flowing through another pipe is at 0.5 MPa with a quality of 0.8. These two streams of steam are mixed together, as shown in the figure below. If the flow rate of the mixture is 2.2 kg/s and temperature 60°C, determine the quality of the steam mixture. Assume the mixing process to be adiabatic. [Ans. 0.994]

204 Fundamentals of Engineering Thermodynamics Steam 1

Mixture

Steam 2 7.26 A steam boiler contains 3.5 in3 of steam and 2 m3 of liquid water at 0.8 MPa. Steam is taken out at constant pressure until 1 m3 of water is left in the boiler. Determine the heat transfer associated with this process. Assume that the steam taken out is with quality 1.0. [Ans. 1836.76 MJ] 7.27 A large insulated rigid container is divided into two compartments. In compartment 1, there is 2 kg of dry steam at 0.5 MPa. In compartment 2, there is 5 kg of steam with quality 0.7 at 0.8 MPa. The partition is removed and the steam in the two compartments is allowed to mix and attain an equilibrium state. Determine the final pressure, quality, and the change in entropy associated with the mixing process. [Ans. 658.9 kPa, 0.78, 0.5684 kJ/K] 7.28 One kilogram of steam at 1.4 MPa and 400°C in a piston-cylinder device is cooled at constant pressure until two-third of the mass condenses. Determine (a) the final temperature, and (b) the change in volume. [Ans. (a) 195.07°C, (b) — 0.17 m3] 7.29 An automobile tube of volume 0.014 m3 has air at 20°C and 130 kPa (gauge). If the pressure inside the tube has to be raised to 220 kPa (gauge), how much air needs to be pumped in? Assume the temperature and volume to remain constant and take atmospheric pressure to be 101 kPa. [Ans. 0.01497 kg] 7.30 An air tank of volume 1.8 m3 at 30°C and 700 kPa is connected through a valve to another air tank having 11 kg of air at 40°C and 220 kPa. The valve is opened, and the entire system is allowed to come to an equilibrium state at 25°C. Determine the final pressure at the equilibrium state. [Ans. 346.44 kPa] 7.31 Determine the percentage error involved in treating carbon monoxide as an ideal gas at 2.5 MPa and 8°C. [Ans. —3.1 %] 7.32 Six kilograms of a perfect gas of molar mass 26 kg/kmol and y= 1.3 is contained in a rigid vessel at 300 kPa and 600 K. Calculate the heat to be rejected from the gas if it has to be cooled to a state where the pressure is 150 kPa. [Ans. —1918.8 kJ] 733 In an experiment the value of y for carbon dioxide was found to be 1.3. Assuming that carbon dioxide is a perfect gas, calculate R, C,,, and C,,. [Ans. 0.189 kJ/(kg K), 0.819 kJ/(kg K), 0.630 kJ/(kg K)]

CHAPTER

8 Nonreacting Gas Mixtures 8.1 INTRODUCTION So far we have discussed thermodynamic systems which involve a single pure substance, such as water. However, many thermodynamic applications of practical importance involve mixtures of several pure substances. Any nonreacting gas mixture can be treated as a pure substance, since it is usually a homogeneous mixture of different gases.

8.2 COMPOSITION OF A GAS MIXTURE The composition of a mixture and the properties of its individual components should be known for determining the properties of the mixture. The composition of a mixture is described in the following two ways: 1. By specifying the number of moles of each component, called molar analysis. 2. By specifying the mass of each component, called gravimetric analysis.

8.2.1 Mass Fraction mf It is the ratio of the mass of a component ma to the mass of the mixture mm. mi —1, — — m„,

m

k

The mass of the mixture m„, = l mi . i l

205

(8.1)

206 Fundamentals of Engineering Thermodynamics

8.2.2 Mole Fraction Y Ni to the mole number of the mixture N„,.

This is the ratio of the mole number of a component Y=

Ni N„,

(8.2)

From Eqs. (8.1) and (8.2), it is obvious that k

k

Mi; =

1 and I Y=1 i=1

(8.3)

The mass m of a substance can be expressed as m=

NM

(kg)

where N is the mole number and M is the molar mass. In the same way, the average molar mass M. of a mixture can be expressed as MM

The

= mm = Lmi = ENi M i

N,,,

N,,,

N.

— YM. a=i

(kg/kmol)

(8.4)

average gas constant R„, of a mixture can be determined from . R. =— [kJ/(kg

In thermodynamic language, the above average quantities are called

(8.5)

apparent

quantities.

EXAMPLE 8.1 A gas mixture consists of 0.5 kg of carbon monoxide and 1 kg of CO2. Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass and the gas constant of the mixture. Solution (a) The mass of the mixture m„,= mco mco2 =- 1.5 kg The mass fractions are _ mco = 0-5 = 0.3333 ink° — in,,, 1.5 and

mco2

Ink°, = nt.

1

=

0.6667

(b) For getting the mole fractions, the mole number of the components should be known. Thus, "co —

me° Mco

—

0.5 kg = 0.0174 lanol 28 kg/lcmol

Nonreacting Gas Mixtures 207 and ?Pico, Nco2

Mco,

1.0 kg 0.0227 kmol 44 kg/kmol

Thus, Nm = N co + Nco2 = 0.0179 + 0.0227 = 0.04061=01 Now, NCO - 0. 0179 N„, 0.0406

y CO

0.441

and 'CO2

en Nco2

—

0.0227 0.0406

N.

0.559

(c) The average molar mass M. is

M Mm =

m 1.5 = m = 0.0406

= YcoMco

Mco, =

36.95 kg/kmol ( 0.5 ) + 0.559 (— LL) 0.0227 0.441 0.0179 36.95 kg/kmol

The gas constant of the mixture is, therefore, R„ Mm

8.314 kJ/(kmol K) 36.95 kg/kmol

0.225 kJ/(kg K)

8.3 P-v-T BEHAVIOUR OF GAS MIXTURES An ideal gas is that gas whose molecules are spaced sufficiently far away so that the behaviour of any one molecule is not influenced by the presence of other molecules. This is exactly the situation encountered at low densities. Also, we know that even real gases at low pressures or high temperatures relative to their critical temperatures approximate this behaviour closely. For ideal gases, the P—v—T behaviour is given by the ideal-gas equation of state, i.e. Pv = RT. For real gases, the P—v—T behaviour is expressed as Pv = ZRT, where Z is the compressibility factor. In a mixture of two or more ideal gases, the behaviour of any one molecule is not normally influenced by the presence of other similar molecules. Therefore, a nonreacting mixture of ideal gases also behaves as an ideal gas. For example, air is treated as an ideal gas in the range of P and T where nitrogen and oxygen behave as ideal gases. The P—v—T behaviours of gas mixtures are, predictably, generally based on the following two models: I. Dalton's law of additive pressures. 2. Amagat's law of additive volumes.

208 Fundamentals of Engineering Thermodynamics

8.3.1 Dalton's Law of Additive Pressures Dalton's law of additive pressures states: The pressure of a gas mixture is equal to the sum of the pressures which each gas would exert if it existed alone at the temperature and volume of the mixture. That is, Pm = E Pi(T„„ Vm) i=t

(8.6)

where P. is the pressure of the mixture and Pi is called the component pressure.

8.3.2 Amagat's Law of Additive Volumes Amagat's law of additive volumes states: The volume of a gas mixture is equal to the sum of the volumes which each gas would occupy if it existed alone at the temperature and pressure of the mixture. That is, VAT., Pm) Vm = iZ =i

(8.7)

where V„, is the volume of the mixture and Vi is called the component volume. The ratios m and Vi N., respectively, are called the pressure fraction and volume fraction of the component i. The above two laws hold accurately for ideal-gas mixtures, but only approximately for real-gas mixtures.

8.4 IDEAL-GAS MIXTURES For ideal gases, the component pressure P, and the component volume V; can be related to mole number, by using the relations for both the components and the gas mixture: Pi(l n,Vm) _ IsliRuTmNin _ —Y Pm N„,RuT„,N„, N. Vi (T.,P„,) _ Ni RuT,,,IF,'„ N. V„, Nm RuTm/Pm N„, Therefore, = = = P. V„, N„,

y

(8.8)

The quantities YiP„, and YiV„, are called the partial pressure and partial volume, respectively. It is interesting to note from Eq. (8.8) that, for an ideal gas mixture, the pressure fraction, the volume fraction, and the mole fraction of a component are identical. The composition of an ideal-gas mixture, such as the exhaust gas leaving a combustion chamber, is often determined by volumetric analysis and Eq. (8.8). A sample gas of known V, P, T is passed into a vessel containing a reagent that absorbs one of the gases. The volume of the remaining gas is then measured at the original P and T. The ratio of the reduction in volume to the critical volume represents the mole fraction of that particular gas which was absorbed by the reagent.

Nonreacting Gas Mixtures 209

8.5 REAL-GAS MIXTURES Dalton's law and Amagat's law can also be used for real gases. For this case, the component pressures or component volumes should be evaluated from relations which account for real gas effects. This is done either by using more exact equations of state like van der Waals equation or by using the compressibility factor as PV = ZNRuT

(8.9)

It can be shown that, Zrn =

-1

YiZi

(8.10)

where Z„, is the compressibility factor for the mixture and zi is that for a component. Further, by experience it is established that Dalton's law is more appropriate for gas mixtures at low pressures and Amagat's law at high pressures.

8.5.1 Kay's Rule This is yet another approach for predicting the P-v-T behaviour of gas mixtures. This method was proposed by W.B. Kay in 1936. In this case, the gas mixture is treated as a pseudopure substance. A pseudocritical pressure and pseudocritical temperature for the mixture are defined in terms of and Tcfi,,,, as =

YiPc,, i and Teri m =

The compressibility factor of the mixture Z„, is then determined from these pseudocritical properties. EXAMPLE 82 A rigid vessel contains 0.5 kg of carbon monoxide and 2 kg of nitrogen, at 300 K and 8 MPa. Calculate the volume of the tank using (a) the ideal-gas equation of state, (b) the Kay's rule, and (c) the compressibility factors and Amagat's law. Solution (a) By the ideal-gas equation of state, we have the mixture volume V„, expressed as vm

NH,R„T„,

where N„, = Nco

NN 2

mco MCO

MN:

0.5 2 = — + — = 0.0893 kmol 28 28

210

Fundamentals of Engineering Thermodynamics

Therefore,

v 011

N„,R„T,„ _ 0.0893 x 8.314 x 300 8 x 103

0.02784 m3

(b) In Kay's rule, the mixture is treated as a pseudopure substance. Therefore, to use the rule, the pseudocritical temperature and pressure of the mixture have to be determined by using the critical point properties of CO and N2. The mole fraction of the components are Nco _ 0.01786 — 0.200 N„, 0.08930

co = and

NN2 YN =

0.07143 _ 0.800 0.08930

Ars,

2

The pseudocritical pressure and temperature are, respectively Pcsri,in

=

YiPcrij =

1'COPcri.00

YN2 PCIVV2

= 0.200 x 3.5 + 0.800 x 3.39 = 3.412 MPa

and

=

= YCOrcri,C0

YN2 Tcri•N2

= 0.2 x 133 + 0.8 x 126.2 = 127.56 K IR

=

T„, 300 = — 2.35 Tc'd,„, 127.56 8

P„,

PR " 3.412=2.35 Pc;;,„,

From the compressibility chart (compressibility chart (b) 0 < PR < 7 of the Appendix), z,„ = 0.97. Therefore, V „, —

ZNRT P„,

Z,„V = 0.97 x 0.02784 = 0.027 m3

(c) For using the Amagat's law, we need the compressibility factor Z of each component. On the basis of the Amagat's law: For CO, T.

TR,C0 and PR.00

—

300

= 2.26

Tai.co 133 =

Pcri,C0

= 2.29

3.5

Nonreacting Gas Mixtures 211 For N,, TR.N2 =

T„,

=

Tcri.N2

and PR.N2

300 = 2.38 126.2

8 P. = = 2.36 Pcn•.N2 3• 39

From the compressibility chart, Zco = 0.96, ZN2 = 0.98 Therefore, Z„, = E

= YcoZco YN242 = 0.2 x 0.96 + 0.8 x 0.98 = 0.976

and V. = z.Videm = 0.976 x 0.02784 = 0.0272 m3

8.6 PROPERTIES OF GAS MIXTURES The internal energy, enthalpy, and entropy of a gas mixture can be expressed, respectively, as U

= I U. = E mini = E NA (kJ) ' i= i= 1

m i=1 k

H=

a=1 k

Sm

=

(8.11)

k

k

= E m.h. = I N.11; (kJ) a-1 " a=1 k

(8.12)

k

S. = Z nvs. = E i.17; (kJ/K) ; 1N a1 ==1 ; "=

(8.13)

In a similar manner, changes in internal energy, enthalpy, and entropy during a process can be expressed as k

k

k

iAt7; (kJ) DUm = I AU; = miAtt; ai N a a== 1

(8.14)

and so on. The above properties of a gas mixture per unit mass or per unit mole of the mixture can be determined by dividing the respective equations by the mass or mole fraction of the mixture. The resulting expressions are k

U

k

= I m fiu; and U = I Y1 17; (kJ/kg or kJ/kmol) i= 1.1

(8.15)

and so on. Similarly, the specific heats of gas mixtures can be expressed as k

= E mi C„ a=1 ' Cp m

_

and C„,„, = E Y; Cv ; [kJ/(kg°C) or kJ/(Icinol°C)] i=1

= E m fi Cpj and a=1

=E a=1

epj [kJ(kg°C) or kJ)(kmol°C)]

(8.16) (8.17)

212 Fundamentals of Engineering Thermodynamics All the property relations given above are valid, in general, for both ideal- and real-gas mixtures. The major difficulty associated with these relations is the determination of properties of each individual gas in the mixture.

8.6.1 Ideal-Gas Mixtures The gas mixture and its components can be treated as ideal gases when the mixture is at a high temperature and low pressure relative to the critical point values of the individual gases. Under this condition, each gas component in the mixture behaves as if it exists alone at the mixture temperature T„, and mixture volume V„,. This principle is known as the Gibbs-Dalton law. The h, u, C„, and C, of an ideal-gas depend only on temperature in the ideal-gas mixture. The partial pressure of a component i in an ideal-gas mixture is given by P, = YiP„„ where P,„ is the pressure of the mixture. Evaluation of Au or Ah of an individual gas in the mixture, during a process, requires only a knowledge of the initial and final temperatures. But As of a component depends on P or V of the component as well as on its temperature. The change in entropy of individual gases in an ideal-gas mixture during a process can be determined, respectively, from A

P.2

= .4).2 — 4.1 — R C,,j

In

(8.18)

Pi.1

= T2

Pi 2

R

(8.19)

11,1

and P2

i = S42 Si.1 In

Ku Ti 2 Ti,1

— R„ ln

(8.20)

P.2 11,1

(8.21)

Note that in the evaluation of the change in entropy only the partial pressure of each component is used, and not the pressure of the mixture.

8.6.2 Real-Gas Mixtures The analysis of real-gas mixtures becomes more complex when the components of the mixture do not behave as ideal gases. The u, h, and Cp depend on pressure or specific volume as well as on the temperature. Also, the effects of deviation from the ideal-gas behaviour on the mixture properties must be accounted for. EXAMPLE 83 A gas mixture of 2.2 kg mass, which consists of 75 per cent nitrogen, 22 per cent oxygen and 3 per cent carbon dioxide by mass, is contained in a piston-cylinder device. The mixture is initially at 101 kPa and 310 K. It is then compressed to 500 kPa in a reversible polytropic process

Nonreactitig Gas Mixtures 213 with an index of 1.3. Determine the work done, heat transfer, and change in entropy associated with the compression process. Take y for N2, 02, CO2, as 1.4, 1.4, and 1.3 respectively. Solution The gas constants for N2, 02, and CO2 are 8314 — 296.9 J/(kg K), RN = 2 28

8314 R02= — = 259.8 J/(kg K) 32

314 R cot ==189 J/(kg K) 44 The gas constant for the mixture can be obtained from R,,, = I til R, = 0.75(296.9) + 0.22(259.8) + 0.03(189) m = 285.5 J/(kg K) The gas constant for the mixture is also given by R. — Ru M. where M„, is the molar mass of the mixture, given by MIN —

m„, — mass of the mixture N„, mole number of the mixture

or M.= I Yilili where Y, is the mole fraction of the component i and M1 is its molar mass. The mole numbers of the constituents are 2.2 x 0.75 MN2 NN2 = — Ai -

28

- 0.059 1=01

= mo, _ 2.2 x 0.22 _ 0.01511anol 02 2, 32 ivi o,

N

Nco2 =

moo, _ 2.2 x 0.03 — 0.0015 kmol AA. 44 ''' CO2

The mole number of the mixture N. =I Ni = 0.059 + 0.0151 + 0.0015 = 0.0756 lanol

244

Fundamentals of Engineering Thermodynamics

Therefore, the mole fractions are YN2

0.0590 0.0756

Yo -2 -

0.0151 - 0.1997 0.0756

-

0.7804

0.0015 - 0.01984 0.0756

Yco2 Thus,

M = m = 2.2 = 29.1 kg/kmol m N„, 0.0756 Also,

Mm = E Yikli = 0.7804 x (28) + 0.1997 x (32) + 0.01984 x (44) = 29.1 kJ/kmol Therefore, R -

8314

m 29.1

=

285.7 J/(kg K)

The Cp for the mixture is Cp = The

Cps

for the constituents are CD

=

N2

sv:s1 y

1

2

-eM

0.4 (296.9) =1.039 kJ/(kg K)

1.4 C,,0 = -R0 = - (259.8) = 0.909 kJ/(kg K) Y-1 2 0.4 Cpc02

co2 = 1 (189) = 0.819 kJ/(kg K) = y -1 R 0. 3

Therefore, for the mixture Cp = (0.75 x 1.039) + (0.22 x 0.909) + (0.03 x 0.819) = 1.0038 kJ/(kg K) and

C, = Cp- R = 1.0038 - 0.2857 = 0.718 kJ/(kg K)

The final temperature of the mixture is In-1)41

T, =7;H

( 500 11.3-1)/1 3

= 310

= 448.4 K 101

The work done becomes

W

V _ P2 V2 - P I I _ mR(T2 n-1 n-1 (2.2 x 0.2857)(448.4 - 310) 0.3

290 kJ

Nonreacting Gas Mixtures The heat transfer

215

Q is Q = mc(T2 — T,)— W = 2.2 x 0.718(448A — 310) — 290 —71.4 Id

That is, 71.4 Id of heat is rejected from the system. The change in entropy is

S2

I

— miCp

— R1n)

= 2.21(1.0038) In

448.4 310

(0.2857) In500 — 101

—0.19 kJ/K EXAMPLE 8.4 A gas mixture of 5 kg mass at 135°C consists of 20 per cent 02, 70 per cent N2 and 10 per cent CO2 by volume. If the pressure of the mixture is 300 kPa, determine the gravimetric analysis and the partial pressures of the gases in the mixture. If the mixture is cooled to 20°C at constant pressure, find the final volume of the mixture. Solution The gravimetric analysis specifies the mass of each component. Given that, Vo

VCO2 •.2 = 0.2, — = 0.7,

V„,

v,:,

V„,

— 0.1

The molar masses of the constituent gases are Moe = 32 kg/lanol,

MN2 = 28 kg/lanol, and Mc02 = 44 kglkmol

We know that the mole fraction, Yi = The molar mass of the mixture

V. N. V.=N.

M. is

M. =1 YiMi = (0.2 x 32) + (0.7 x 28) + (0.1 x44) = 6.4 + 19.6 + 4.4 = 30.4 lcg/lanol The mass fraction is

ml

In; _ in; _ (ilV)Mi _ mm In; E (VilV)M i

(VilV)M Mm

216 Fundamentals of Engineering Thermodynamics Thus, MI,o2 =

0.2 x 32 30.4

0.2105

0.7 x 28 _ 0.6447 411'N2 - 30.4 0.1 x 44 _ 0.1447 Alf'c°2 - 30.4 The partial pressures are PO2 = Yo2 X P„, = 02 x 300 = 60 kPa PN2 = 0.7 x 300 = 210 kPa Pc02 = 0.1 x 300 = 30 kPa The mole number of the mixture is N = "1" =

m M„,

5 - 0.1645 kmol 30.4

The final volume of the mixture when it is cooled to 20°C is 0.1645 x 8.314 x 293.15 300

N„,R.T V

p

1.336 m3

EXAMPLE 8.5 A rigid tank of volume 3 m3 is divided into two parts by a partition. One part contains nitrogen gas at 600 kPa and 400 K, and the other carbon dioxide at 200 kPa and 290 K. The partition is removed and the gases are allowed to mix adiabatically. Find the equilibrium temperature and pressure of the mixture, and the change in entropy associated with the mixing process. Treat N2 and CO2 as ideal gases at the initial state. Solution The mole numbers of N2 and CO, are, respectively -2

PV 600 x 1.5 - 0.2706 lcmol RuT 8.314 x 400

and N

-

200 x 1.5 - 0.1244 lcmol 8.314 x 290

For the mixture, the mole number is N„,= 0.2706 + 0.1244 = 0.395 kmol

Nonreactiag Gas Mixtures 217 For the system under consideration, W = 0 and Q = 0. Therefore, by the principle of energy conservation, we have AU = AUN2 + AUcc,2 = 0 Or [mC,(T„, — 73 + [mC,,(T„, — 7;) =0 42 02 For N2, y= 1.4 and for CO2, y= 1.3. Using the ideal-gas relations, we have Cv

y-1

Therefore, (CON,

=.314 1 x — = 0.742 kJ/(kg K) 0.4 28

8.314 1 (C„)co = x — = 0.630 kJ/(kg K) 2 44 0.3 The mass of N2 and CO2 are, respectively mN2 = NN2MN2 = 0.2706 x 28 = 7.577 kg and mc02 = Nco2Mco2 = 0.1244 x 44 = 5.474 kg Thus, the equilibrium temperature of the mixture is (7.577 x 0.742X T„, — 400) + (5.474 x 0.630X T„, — 290) = 0 or 9.07 T„, = 3248.95 Therefore, T„, = 358.2 K The equilibrium pressure of the mixture is P — N„,R„T 0.395 x 8.314 x 358.2 — . — 392.1 kPa V 3 The change in entropy associated with the mixing process is the sum of the changes in entropy of each constituent. For N2 T Pm ASN2 = mN2 ( Cp In — R In T = 7.577[1.4 x 0.7421n

358 2 . 400

8.314 92.11 In 3 28 600

= 0.0884 kJ/K

J

For CO2 ASCO2 = 5.474 [(1.3 x 0.63) In 3 5 8:2 290 = 0.2506 kJ/K

8.314 in 392.1 1 44 20 j

218 Fundamentals of Engineering Thermodynamics Thus, AS„, = 0.0884 + 0.2506 = 0.339 kJ /K EXAMPLE 8.6 A rigid tank contains 0.35 kg of steam of quality 0.2 and 0.1 kg of nitrogen gas. If the temperature of the mixture is 90°C, determine the volume of the tank and the pressure of the mixture. Solution The mixture consists of steam and nitrogen in gas phase and water in liquid phase. The quality of the steam is x=

m g =0.2 miot

Therefore, the mass of the dry steam is

mg = 0.2 x 0.35 = 0.07 kg Thus, the mass of the liquid water mf = 0.35 — mg = 0.28 kg From the saturated water-temperature table (Table 3 of the Appendix), at 90°C, we get Put = Pg = 70.14 kPa, vf = 0.001036 m3/kg and vg = 2.361 m3/kg The volume of the gas mixture, namely the mixture of the dry steam and nitrogen gas is equal to either the volume of the dry steam at the partial pressure of dry steam or the volume of nitrogen at the partial pressure of nitrogen. Therefore, the volume of the tank is V = mf vf + mgvg = (0.28 x 0.001036) + (0.07 x 2.361) 0.166 m3 The partial pressure of the nitrogen gas is

(0. i ) ( 8.31 4 ) PN -

mN 2RT

2

VN2

-

28

(36315)

0.07 x 2.361

- 65.24

kPa, since VN2 =

Vg =

0.07 x 2.361

Thus, the pressure of the mixture

P., = Pg + PN2 = 70.14 + 65.24 =

135.38 kPa

EXAMPLE 8.7 Steam of quality 93 per cent and at 40°C flows through a condenser at the rate of 3500 kg/h. If the leakage of air from the condenser is at the rate of 3.9 kg/h, determine the amount of water vapour carried away by the air for condensate temperature of 30°C.

Nonreacting Gas Mixtures

219

Solution The pressure in the condenser is given by the inlet conditions of steam. From the saturated water-temperature table (Table 3 of the Appendix), at 40°C, we get Pam = Pg = 7.384 kPa and vg = 19.52 m3/kg The amount of dry steam (water vapour) flowing through the condenser is = 3500 x 0.93 = 3255 kg/h There is 3.9 kg of air for every 3255 kg of vapour. Therefore, the mass of air in a mixture having volume of 19.52 m3/kg is ma =

3. 9

— 1.198 x 10-3 kg

The partial pressure of the air is given by a

ma g,T _ (1.198 x 10-3)(0.287 x 313.15) — 5.516 Pa vg 19.52

The condenser pressure P = Pa = 7.384 — 0.005516 = 7.378 kPa. At 30°C, from the saturated water-temperature table (Table 3 of the Appendix), we get Pg = Psat = 4.246 kPa, vg = 32.89 m3/kg The partial pressure of the air for 30°C is Pa = P — Pg = 7.378 — 4.246 = 3.132 kPa The mass flow of the air per kg of vapour is ma —

Pa y. _ 3.132 x 32.89 _ 1.184 kg RaT 0.287 x 303.15

Thus, the mass of water vapour carried away per kg of air is 1 1 n7 . = .184 = 0'845 kg Therefore, the mass of water vapour carried away by the air per hour is 0.845 x 3.9 = 3.3 kg Note: In the calculations of partial pressure, the mass flow of air per kg of vapour and the mass of water vapour carried away by the air, it appears that the units do not match in the expressions used for their calculations. It should be realized that there is no violation of unit matching in these steps since the calculations are on 1 kg basis and, therefore, V becomes v. EXAMPLE 8.8 A steady stream of equimolar N2 and CO2 mixture at 100 kPa and 27°C is to be separated into N2 and CO2 gases at the same pressure and temperature. Determine the minimum work required per unit mass of the mixture to accomplish the separation process. Assume the environment to be at the mixture temperature.

220 Fundamentals of Engineering Thermodynamics Solution The work required will be minimum if the process is reversible. Thus, the minimum work required to separate the mixture into its components is equal to the reversible work associated with the mixing process, i.e. the irreversibility associated with the mixing process. Therefore, i = wre,„ Or = i = Tesgen The entropy change associated with the mixing process is AS„, = E AS; = ZNiAsi

Ta

= Ni(Cpi

In Pi,2

But Tu = T. Thus, YP Sion = AS„, = —R„ ln'Z —

R„ IN; In Yi

Pa = (-8.314X0.5 In 0.5 + 0.5 In 0.5) = 5.763 kJ/(kmol K) Mm = E YiMi = 0.5 x 28 + 0.5 x 44 = 36 kg/lanol Sgen

=5363 0—160 . 1d/(kgK) Sgen Min 36

i = Tosion = 300 x 0.160 48 kJ/kg The minimum work required to accomplish the separation process is therefore 48 kJ/kg.

SUM MARY Any nonreacting gas mixture can be treated as a pure substance. The composition of a mixture is described in the following two ways. 1. By specifying the number of moles of each component, called molar analysis. 2. By specifying the mass of each component, called gravimetric analysis. The mass fraction mfi is the ratio of the mass of a component mi to the mass of the mixture mm. Mi

m

Nonreacting Gas Mixtures 221 The mole fraction Yi is the ratio of the mole number of a component number of the mixture N.

Ni to the mole

Y = Ni N„, The

average molar mass M. of a mixture can be expressed as M m

m Ens. I Ni M• Y.M. (kg/kmol) = N. N., N., i=t

The average gas constant

R„, of a mixture can be determined from R„, =— R'' [kJ/(kg M.

An ideal gas is that gas whose molecules are spaced sufficiently far away so that the behaviour of any one molecule is not influenced by the presence of other molecules. For ideal gases, the P—v—T behaviour is given by the ideal-gas equation of state; i.e.

Pv = RT. For real gases, the P—v—T behaviour is expressed as Pv = ZRT. For gas mixtures, the P—v—T behaviour is, predictably, generally based on the following two models: 1. Dalton's law of additive pressures. 2. Amagat's law of additive volumes. Dalton's law of additive pressures states that the pressure of a gas mixture is equal to the sum of the pressures which each gas would exert if it existed alone at the temperature and volume of the mixture. That is,

Pm = I Pi (T„„V„,) Amagat's law of additive volumes states that the volume of a gas mixture is equal to the sum of the volumes which each gas would occupy if it existed alone at the temperature and pressure of the mixture. That is, k

Vrn=

i=i Vi (T„„ P.)

The ratios PilP„, and VW Vm, respectively, are called the pressure fraction and volume fraction of the component i. For ideal gases, the component pressure Pi and the component volume Vi can be related to mole number as

= Vi = =y P., V. Nm

222 Fundamentals of Engineering Thermodynamics The quantities Y,P„, and YiV„, are called the partial pressure and partial volume, respectively. It is interesting to note from the above equation that, for an ideal-gas mixture. the

pressure fraction, the volume fraction, and the mole fraction of a component are identical. Kay's rule is yet another approach for predicting the P-v-T behaviour of gas mixtures. In this case, the gas mixture is treated as a pseudopure substance. The U, H, and S of a gas mixture can be expressed as k

k

Um = I Ui = I I=I a=1

k

= I NA (Id)

a-1

k k k _ H=L I = I nigh; = I NA (id) a I

S = I = I misi = E N is; (k.J/K) i=1

i=1

i=1

The gas mixture and its components can be treated as ideal gases when the mixture is at a high temperature and low pressure relative to the critical point values of the individual gases. Under this condition, each gas component in the mixture behaves as if it exists alone at the mixture temperature T„, and mixture volume Vm. This principle is known as the Gibbs-Dalton

law. The change in entropy of individual gases in an ideal-gas mixture during a process can be determined from 2 Asa = s10-,2 - s-/0.1 - R- In P --t-

For real-gas mixtures, u, h, C,, and C,, depend on pressure or specific volume as well as on the temperature. Also, the effects of deviation from the ideal-gas behaviour on the mixture properties must be accounted for.

PROBLEMS 8.1 Determine the (a) mole fraction of the components, (b) the partial pressure of the components, and (c) the gas constant, the molar mass, the volume, and the density of a mixture consisting of 2 kg of oxygen and 7 kg of nitrogen at 400 kPa and 300 K. [Ans. (a) ro2 = 0.2, YN2 = 0.8, (b) P02 = 80 kPa, PN2 = 320 kPa, (c) R„,= 0.289 kJ/(kg K), Mm = 28.8 Icg/lanol, V„, = 1.95 m3, p„, = 4.615 kg/m3] 8.2

A gas mixture consists of 65 per cent N2 and 35 per cent CO2 by mole basis. Determine the gravimetric analysis of the mixture. Also, determine the molar mass and the average gas constant for the mixture. [Ans. mf N2 = 0.542, mfco2 = 0.458, Mm = 33.6 kg/kmol, R„,= 0.2474 kJ/(kg K)]

8.3 A gas mixture consists of 50 per cent CH4 and 50 per cent CO2 by mass. Determine the mole fractions of the constituents and the average gas constant for the mixture. [Ans. You = 0.7334, Yc02 = 0.2666, R,,, = 0.3543 kJ/(kg K)]

Nonreading Gas Mixtures

8.4

223

A gas mixture consists of 6 kmol of H2 and 4 kmol of N2. Determine the mass of each gas and the gas constant of the mixture. [Ans. mH2 = 12 kg, mN2 = 112 kg, R„, = 0.6705 kJ/(kg K)]

8.5

A mixture of 1.5 kg of hydrogen and 4.5 kg of nitrogen is compressed in a piston-cylinder device until its initial temperature of 20°C is raised to 100°C in a reversible polytropic process with index 1.2. The Cps of the gases at 300 K are taken at their ideal-gas state. Determine the heat transfer and the entropy change for the process. [Ans. -1822.684 kJ, - 4.536 kJ/K]

8.6 Air is considered to be a mixture of 78 per cent nitrogen and 22 per cent oxygen by volume if the small amounts of argon and carbon dioxide present in the mixture are neglected. Treating nitrogen and oxygen in air to be ideal gases, find the gas constant, the molar mass and C4 and C, for air at 25°C. [Ans. 0.2879 kJ/(kg K), 28.88 kJ/lanol, 1.0106 kJ/(kg K), 0.722 kJ/(kg K)] 8.7

The exhaust of a fuel has the following composition by volume: N2(0.77), CO2(0.13), and 02(0.1). Determine the heat transfer per kg of the mixture required to cool it from 820°C to 100°C assuming that the exhaust flows steadily at constant velocity. Assume the average molar heat capacities at constant pressure of the constituents, over the temperature range of the problem, to be 31.08 for N,, 50.51 for CO2, and 33.38 for 02, all in kJ/(kmol K). [Ans. -799.3 kJ]

8.8

A rigid tank of volume 0.33 m3 contains 0.25 kg of carbon monoxide and 1.25 kg of air, at 30°C. Calculate the partial pressures of the constituents and the pressure of the mixture in the vessel. Treat air to be a mixture of 77 per cent nitrogen and 23 per cent oxygen. [Ans. PN2 = 262.54 kPa, Poe = 68.618 kPa, Pao = 68.19 kPa, Pm = 399.35 kPa]

8.9

A mixture of 2 kmol of CO2 and 3 kmol of air is contained in a tank at 100 kPa and 20°C. Treating air to be a mixture of 79 per cent nitrogen and 21 per cent 02 by volume, calculate (a) the individual mass of CO2, N2 and 02 by volume, (b) the percentage content of carbon by mass in the mixture, and (c) the molar mass, gas constant, and the specific volume of the mixture. [Ans. (a) 88 kg, 66.36 kg, 20.16 kg, (b) 13.75 per cent by mass, (c) 34.904 kg/kmol, 0.2382 kJ/(kg K), 0.6983 m3/kg]

8.10 A mixture of hydrogen and oxygen is in theeratio 2 to 1 by volume. Determine the mass of hydrogen required and the volume of the container, per kilogram of oxygen, if the temperature and pressure of the mixture are to be 100 kPa and 300 K, respectively. [Ans. 1/8 kg, 2.34 m3] 8.11 A gas mixture in an engine cylinder has 12 per cent CO2, 11.5 per cent 02 and 76.5 per cent N2, by volume. The mixture at 1000°C expands reversibly, according to Pv1.25 = constant, to 7 times its initial volume. Determine the work done and heat transfer per unit mass of the mixture. The average Cp values for CO2, 02, and N2 are 1.271 kJ/(kg K), 1.110 kJ/(kg K), and 1.196 kJ/(kg K), respectively. [Ans. 536.94 Id/kg, 83.32 kJ/kg]

224 Fundamentals of Engineering Thermodynamics 8.12 A tank of volume 2 m3 containing 02 at 600 kPa and 310 K is connected through a pipe line to another tank of volume 4 m3 containing CO2 at 100 kPa and 290 K. The gases mix adiabatically and come to an equilibrium state. Determine the final temperature and pressure of the mixture and the entropy change for the system. [Ans. 303.56 K, 265.8 kPa, 4.8228 kJ/K] 8.13 An air tank of volume 0.5 m3 is at 70 kPa and 95°C. Now, water is injected into the tank, keeping the temperature constant at 80°C. Determine the mass of water required to be injected so that the tank is just filled with saturated vapour. If the water injection continues up to 20 per cent more than what is required for saturated vapour, calculate the total pressure in the tank. [Ans. 0.2523 kg, 70.014 kPa] 8.14 A tank of volume 0.3 m3 contains a mixture of steam of quality 0.75 and air. If the mixture is at 700 kPa and 120°C, calculate the mass of the liquid water, the mass of the water vapour, and the mass of the air in the mixture. [Ans. 0.3364 kg, 0.1121 kg, 1.861 kg] 8.15 A rigid vessel contains a mixture of air and dry saturated steam at 36°C. If the ratio of the mass of air to that of steam is 0.1, determine the pressure in the vessel. [Ans. 6.2935 kPa] 8.16 A rigid tank contains 8 kmol of 02 and 4 kmol of CO2 at 200 kPa and 300 K. Determine the volume of the tank. [Ans. 149.65 m3] 8.17 A rigid tank contains 2 kmol of argon and 3 kmol of nitrogen at 202 kPa and 320 K. The mixture is now heated to 500 K. Determine the volume of the tank and the final pressure of the mixture, treating the constituents and the mixture as ideal gases. [Ans. 65.85 m3, 315.64 kPa] 8.18 A rigid tank of volume 1.2 m3 is divided into two equal parts by a partition. One compartment contains neon gas at 22°C and 110 kPa, and the other contains argon gas at 47°C and 225 kPa. Now the partition is removed, and the two gases are allowed to mix. If 10 Id of heat is lost to the surroundings at 20°C during the mixing process, determine the final temperature and pressure of the mixture. [Ans. 27.97°C, 161.94 kPa] 8.19 Ethane (C2H6) at 30°C and 101 kPa arid methane (CH4) at 40°C and 101 kPa enter an adiabatic mixing chamber. The mass flow rate of ethane is 6 kg/s, which is half the mass flow rate of methane. Determine the temperature of the mixture and entropy generation during the mixing process. Assume the gases and their mixture to be ideal gases. [Ans. 37.05°C, 4.0428 kJ/K] 8.20 An ideal-gas mixture of CO (33.3 per cent), CO2 (50 per cent) and 0, (16.7 per cent), by volume enters a steady-state compressor at 30°C and 60 m/s and leaves at 120°C and 90 m/s. Determine the shaft-work input required to run the compressor if a heat loss of 9.5 kJ/kg occurs during the process. If the volume flow rate at the entry is 12 m3/min and the pressure is 101 kPa, determine the power input to the compressor in kilowatts. [Ans. -94.82 kJ, 27.88 kW]

Nonreacting Gas Mixtures 225

8.21 A mixture of nitrogen and water vapour is contained in a tank of volume 0.5 m3. The partial pressure of water vapour is 5 kPa. Calculate the heat transfer when the mixture in the tank is cooled to 10°C. [Ans. —338.33 kJ] 8.22 0.5 kg of saturated air is contained in a rigid tank along with water vapour at 160°C and 1 MPa. Calculate the amount of heat required to be transferred to cool the mixture to 80°C. How much of moisture will be condensed in this process? Assume that the mixture is nonreacting. [Ans. —1108.95 kJ, 0.4818 kg]

CHAPTER

9 Vapour Power Cycles 9.1 INTRODUCTION A cycle which continuously converts heat into work is called the power cycle. In a power cycle the working fluid repeatedly performs a succession of processes. If the working fluid is alternately vaporized and condensed, then the cycle is called a vapour power cycle. Steam is the most commonly used working fluid in vapour power cycles because of its many desirable characteristics, such as low cost, easy availability, and high enthalpy of vaporization. For high-temperature applications, sodium, potassium, and mercury are used as working fluids. In this chapter, we would be mainly concerned with steam power plants. Steam power plants are referred to as coal plants, nuclear plants, natural gas or geothermal plants, depending on the type of fuel used to supply heat to the steam.

9.2 THE CARNOT VAPOUR CYCLE The Carrot vapour cycle serves as a perspective ideal cycle for vapour power plants, even though it is not a practical cycle. Consider a steady-flow Carrot cycle executed within the saturation dome of a pure substance such as water, as shown in Fig. 9.1. In the cycle at Fig. 9.1(a), water is heated reversibly and isothermally in a boiler (process 1-2), expanded isentropically in a turbine (2-3), condensed reversibly and isothermally in a condenser (3-4), and compressed isentropically by a compressor to the initial state (4-1). This is purely an imaginary cycle with several impracticalities. Some of the problems associated with it can be avoided by executing the cycle as shown in Fig. 9.1(b). However, isentropic compression to extremely high pressures and isothermal heat transfer at variable pressures pose problems. Thus, the Carnot cycle can at best serve as an ideal model and cannot be approximated to actual devices.

9.3 THE RANKINE CYCLE The Rankine cycle is an ideal cycle for vapour power cycles. Many of the impracticalities 226

Vapour Power Cycles 227 T

T

(a) Figure 9.1

(b) T—s diagram of two Carnot vapour cycles.

associated with the Carnot cycle can be eliminated by superheating the steam in the boiler and condensing it completely in the condenser, as shown in Fig. 9.2. The processes involved are: Stn

Figure 9.2 The Rankine cycle. isentropic compression in a pump (1-2), constant pressure heat addition in a boiler (2-3), isentropic expansion in a turbine (3-4), and constant pressure heat rejection in a condenser (4-1). The cycle that results in these processes is the Rankine cycle. The pump, boiler, and condenser associated with a Rankine cycle are steady-flow devices, and thus all the processes of this cycle can be analysed as steady-flow processes. The Me and

228 Fundamentals of Engineering Thermodynamics Ape of the steam are usually small compared with the work and heat transfer terms and are,

therefore, neglected. Thus, the steady-flow energy equation per unit mass of steam is q — w = he — hi (kJ/kg)

(9.1)

Assuming the pump and turbine to be isentropic and noting that there is no work associated with the boiler and the condenser, the energy conservation relation for each device becomes wptunp,in = h2

h1 = v(P2 — P1)

(9.2)

9boi,in = h3 h2

(93)

wturb,out = h3 h4

(9.4)

hi

(93)

qcond,out = h4

The thermal efficiency of the Rankine cycle is given by nth

= Writ = 1 — gout gin

(9.6)

gin

where w„,., = qth — gout = Wwthmut — Wpump,in The qth can also be interpreted as the ratio of the area enclosed by the cycle on a T—s diagram to the area under the heat addition process. EXAMPLE 9.1

A steam power plant operates between a boiler pressure of 4 MPa and 300°C and a condenser pressure of 50 kPa. Determine the thermal efficiency of the cycle, the work ratio, and the specific steam flow rate, assuming (a) the cycle to be a Carnot cycle, and (b) a simple ideal Rankine cycle.

T

A

Solution (a) The T—s diagram of a Carnot cycle is shown in the

adjacent figure. Process 1-2 is reversible and isothermal heating of water in the boiler. Process 2-3 is isentropic expansion of steam at state 2 in the turbine. Process 3-4 is reversible and isothermal condensation of steam in the condenser. Process 4-1 is isentropic compression of steam to initial state. At state 1: P1 = 4 MPa, T1 = 300°C At state 2: P2 = 50 kPa, the steam is in a saturated state. From the saturated water-pressure table (Table 4 of the Appendix), at 50 kPa, we get T2 = T = T.t = 81.33°C

Vapour Power Cycles 229 Therefore, the thermal efficiency for the given Carrot cycle is nit,cantot

Ti

=

_

T

I

81.33 + 273.15 _ 0 3815 300 + 273.15

38.15 per cent

The work ratio = net work output gross work output

wnet out w rossout

Heat supplied = h2 — hi = ha@ 4mpa = 1714.1 kJ/kg (From Table 4 of the Appendix) Wnet,in

Wttet,out qth,carnot =

gross heat supplied

= 0.3815

Therefore, wnei.old — wnei, = 0.3815 x 1714.1 = 653.9 kJ/kg That is, the net work output = 653.9 kJ/kg. To find the expansion work for the process 2-3, h3 is required. From Table 4, h2 = 2801.4 kJ/kg and s2 = 53 = 6.0701 kJ/(kg K) But or

s3 = 6.0701 = Sp + x3sfg3 = 1.0910 + x3(7.5939 — 1.0910) x3 = 0.766

Now, h3 = hp + x3h153 = 340.49 + 0.766(2645.9 — 340.49) = 2106.4 kJ/kg Therefore, w32 = h2 — h3 = 2801.4 — 2106.4 = 695 kJ/kg That is, the gross work output, wipossma = 695 kJ/kg Therefore, Work ratio =

Wnet,out

wgross,oui

—

653.9 _ 0.94 695

The specific steam flow rate (ssfr) is the steam flow required to develop unit power output. That is, 1 ssfr = _steam — Wnet, out mswout _ 1 — 653.9

0.00153 kg/kW

(b) The T—s diagram of the ideal Rankine cycle is shown below. The turbine and the pump are assumed to be isentropic, since the cycle is ideal; there are no pressure drops in the boiler and

230 Fundamentals of Engineering Thermodynamics T

the condenser. Also, it is assumed that the steam leaves the condenser and enters the pump as saturated liquid at the condenser pressure. At state 1: Pi = 50 kPa and water is saturated liquid. From the saturated water-pressure table (Table 4 of the Appendix), we get h 1 = hi® 50 kpa = 340.49 kJ/kg vi = Vf® 50 kPa = 0.001030 m3/kg si = 1.0910 kJ/(kg K) At state 2: P2 = 4 MPa, s2 = Si wp

= h2 - hi = v1(P2 -

= 0.001030(4000 - 50) = 4.0685 kJ/kg

Therefore, h2 = hi + w

= 340.49 + 4.0685 = 344.56 kJ/kg

At state 3: P3 = 4 MPa, 7'3 = 300°C From the superheated water table (Table 5 of the Appendix), we get h3 = 2960.5 kJ/kg, s3 = 6.3615 kJ/(kg K) At state 4: P4 = 50 kPa, 54 = s3 (saturated mixture) Therefore, s4 _ s3 _ 6.3615 - 1.0910 _ x4 0.81 sfg sfg 7.5939 - 1.0910 Now, h4 = hi + x4hfg = 340.49 + 0.81(2645.9 - 340.49) = 2207.87 kJ/kg qi„ = h3 - h2 = 2960.5 - 344.56 = 2615.94 kJ/kg 4014 = h4 - hi = (2207.87 - 340.49) = 1867.38 kJ/kg Therefore, rith - 1

qout =1- 1867.38 - 0.286 = 28.6 per cent 2615.94 gin

The thermal efficiency may also be determined as follows: wiurb,oui = wg o1e = h3 - h4 = 2960.5 - 2207.87 = 752.63 kJ/kg wnei,oui = wu.b.0ut - wp ,i„ = 752.63 - 4.0685 = 748.56 kJ/kg

l'apour Power Cycles

231

Also, Wnet,out = gin — gout

= 2615.94 — 1867.38 = 748.56 kJ/kg

Thus, nth

Wnetout

748.56

gin

2615.94

— 0.286 = 28.6 per cent 748.56

The work ratio = wnet,out 9gross,out The ssfr =

1

1

w„ei.c,„,

748 .56

752.63

0.995

0.001336 kg/kW

9.4 ACTUAL VAPOUR POWER CYCLES The irreversibilities associated with the various components make the actual vapour power cycles differ from the ideal Rankine cycle. The ideal and actual vapour power cycles are shown in Fig. 9.3. The deviation of the actual pumps and turbines from the ideal isentropic ones can be T Ideal cycle Pressure drop in the boiler

Irreversibility in the pump

3 Irreversibility in the turbine

Pressure drop in the condenser ►

Figure 9.3

s

Comparison of ideal and actual vapour power cycles.

properly accounted for by using adiabatic efficiencies, defined as ws

=

h2s — ht

wa h2. — ht

SIT —

ws

h3 — h4. = L L .13 — n4s

(9.7)

(9.8)

where the subscripts 2a and 4a refer to actual exit states of the pump and the turbine, respectively, and 2s and 4s refer to the corresponding isentropic states. The effect of the irreversibilities associated with pumps and turbines on the ideal Rankine cycle is shown in Fig. 9.4.

232 Fundamentals of Engineering Thermodynamics T

Figure 9.4 Effect of pump and turbine irreversibilities on the ideal Rankine cycle. EXAMPLE 9.2 A steam power plant operates on the cycle shown below with 3 MPa and 400°C at the turbine inlet and 10 kPa at the turbine exhaust. The adiabatic efficiency of the turbine is 85 per cent and that of the pump is 80 per cent. Determine (a) the thermal efficiency of the cycle, and (b) the mass flow rate of the steam if the power output is 20 MW. T

Solution All the components are treated as steady-flow devices. The changes, if any, in the kinetic and potential energies are assumed to be negligible. Losses other than those in the turbine and pump are neglected. vi (P2 — 13) (a)

Wptimp,in

=

rip

0.001010 (3000 —10) = 3.77 kJ/kg 0.80

Turbine work output is Wthth,out = 7rWtrkin =

(h3 —

h4s)

= 0.85(3230.90 — 2192.21) = 882.89 kJ/kg

Vapour Power Cycles 233

Boiler heat input is gin

= 3230.9 — 195.59 = 3035.31 kJ/kg

= h3

Thus, Wnet.out = wturb.out Wpump,in =

71in

=

%clout

gin

882.89 — 3.77 = 879.12 kJ/kg

879.12 = 0.2896 = 28.96 per cent 3035.31

If there are no losses in the turbine and the pump, the thermal efficiency would be 28.99 per cent. (b) The power generated by the power plant is

Finet,out = thWnet,out = 20,000 kW Therefore, the mass flow rate, m = 20, 000= 22.75 kg/s 879.12

9.5 MEANS TO INCREASE THE EFFICIENCY OF THE RANKINE CYCLE Most of the electric power in the world is produced by steam power plants. Therefore, even a small increase in iith can result in large savings of fuel. Hence, continued efforts are being made to improve the nth of the cycle on which the steam power plants operate. The methods to improve the thermal efficiency of the vapour power cycles aim at • increasing the average temperature at which heat is transferred to the working fluid in the boiler, or • decreasing the average temperature at which heat is rejected from the working fluid in the condenser. The means to accomplish these aims are described in subsections 9.5.1 to 9.5.3.

9.5.1 Lowering the Condenser Pressure The lowering of the operating pressure of the condenser automatically lowers the temperature of the steam and thus the temperature at which heat is rejected, since steam exists as a saturated mixture in the condenser at the saturated temperature corresponding to the pressure inside. The effect is shown on the T—s diagram in Fig. 9.5. The shaded area represents the increase in the net work output. The heat input requirement also increases, but this increase is small. The condensers of steam power plants usually operate well below the atmospheric pressure in order to take advantage of the increased efficiency at low pressures. This does not present a major problem since the vapour power cycles operate in a closed loop. However, the condensation pressure cannot be lowered below the saturation pressure corresponding to the temperature of the cooling medium.

9.5.2 Superheating the Steam to High Temperatures By superheating the steam to high temperatures, the Ta, at which heat is added to the steam can

11

234 Fundamentals of Engineering Thermodynamics T

wJ (

a

Increase in %et

►s

Figure 9.5 Increase in wne owing to lowering of the condenser pressure. be increased without increasing the boiler pressure. Further, superheating the steam decreases the moisture content of the steam at the turbine exit, and this is highly desirable from the efficiency point of view. The effect of superheating the steam on the performance of vapour power cycles is shown in Fig. 9.6. Even though, in principle, it is advantageous to raise the steam temperature as high as possible by superheating, the maximum temperature is limited by metallurgical considerations. Presently, the maximum possible steam temperature is 620°C. T A

Increase in wnet

4 4'

$

Figure 9.6 Effect of superheating steam on Rankine cycle performance.

9.5.3 Increasing the Boiler Pressure An increase in the operating pressure of the boiler automatically raises the temperature at which boiling takes place. This in turn increases the average temperature at which heat is added to the steam, which results in an increase in the thermal efficiency of the cycle. The effect of increasing the boiler pressure on the cycle performance is shown in Fig. 9.7. It is seen that for a fixed turbine inlet temperature, the cycle shifts to the left and the moisture content of the steam at the turbine exit increases. However, this undesirable effect can be corrected by reheating the steam.

v apour Power Cycles

235

T Increase in Wnet

$

Figure 9.7 Effect of increase in boiler pressure on Rankine cycle performance. The boiler pressure of the present-day power plants is about 30 MPa. Many steam plants operate at supercritical pressure, P > 22.09 MPa, and have th of about 40 per cent for fossil-fuel plants and 34 per cent for nuclear plants. The process diagram of a supercritical Rankine cycle is shown in Fig. 9.8. T

Figure 9.8 A supercritical Rankine cycle.

9.6 THE IDEAL REHEAT RANKINE CYCLE The ideal reheat Rankine cycle aims at taking advantage of the increased efficiencies at high boiler pressures without facing the problem of excessive moisture at the final stages of the turbine. This is achieved by: • Superheating the steam to very high temperatures before it enters the turbine in order to achieve an increase in Tav at which heat is added to the steam. This will result in an increase in 77,h.

236 Fundamentals of Engineering Thermodynamics • Introducing a reheat process into the ideal Rankine cycle. This is done by expanding the steam in the turbine in two stages, and reheating it in between. This being a practical situation, it is used quite frequently in modern steam power plants. The schematic diagram of a power plant working on Reheat. Rankine cycle and its process diagram are shown in Figs. 9.9(a) and 9.9(b), respectively. This cycle differs from the simple ideal Rankine cycle in the fact that the expansion process takes place in two stages. In the first stage Low-pressure High-pressure turbine turbine

T High-pressure turbine

Low-pressure turbine

Condenser

(b)

(a)

Figure. 9.9 The ideal reheat Rankine cycle. with a high-pressure turbine, the steam is expanded isentropically to an intermediate pressure and sent back to the boiler where it is reheated at constant pressure, usually to the inlet temperature of the first turbine stage. In the second stage, the steam expands isentropically, through a low-pressure turbine, to the condenser pressure. Thus for a reheat cycle, we have gin = %Timmy + =

—

h4) (kJ/kg)

(9.9)

(h5 —

h6) (kJ/kg)

(9.10)

— h2)

Wturb,out = Wturb,1 =

preheat

—

Wturb,11

The T„ at which heat is added is not significantly influenced by the reheat process. Therefore, nth is also not influenced much by reheating. The Tay during the reheat process can be increased by increasing the number of expansion and reheating stages. As the number of stages increase, the expansion and reheat processes approach an isothermal process at the maximum temperature, as shown in Fig. 9.10. The optimum number of stages is determined by economical considerations. The use of more than one or two reheat stages, in general, is not economically viable even in large power plants because the savings resulting from increased Ah are more than offset by the enhanced costs associated with the increased number of reheat stages.

Vapour Power Cycles 237 T

Figure 9.10 Rankine cycle with a large number of reheat stages. EXAMPLE 93 In a steam power plant, operating on the ideal Rankine cycle, the steam enters the turbine at 3 MPa and 400°C and is condensed in the condenser at 10 kPa. Determine (a) the thermal efficiency of the power plant, (b) the thermal efficiency if the turbine inlet temperature is 500°C instead of 400°C, and (c) the thermal efficiency if the boiler pressure is raised to 10 MPa while the turbine inlet temperature is kept constant at 400°C. Solution (a) This steam power plant is the same as discussed in Example 9.2, except that the efficiencies of the turbine and pump are assumed to be both equal to 100 per cent. State 1: Pi = 10 kPa, water is saturated liquid. From the saturated water pressure table (Table 4 of the Appendix), we get h i = hi= 191.83 kJ/kg, vi = of = 0.001010 m3/kg State 2: P2 = 3 MPa, s2 = si wpurnpir, = vi(P2 — PO= 0.001010(3000 — 10) = 3.0199 kJ/kg h, = hi + wpunip.,„ = 191.83 + 3.0199 = 194.85 kJ/kg State 3: P3 = 3 MPa, T3 = 400°C, from the superheated water table (Table 5 of the Appendix), we get h3 = 3230.9 kJ/kg, s3 = 6.9212 kJ/(kg K) State 4: P4 = 10 kPa, s4 = s3, saturated mixture x4 =

s4 — sf

53 — Sf

sfg

sfg

=

6.9212 — 0.6493 = 0.836 8.1502 — 0.6493

238

Fundamentals of Engineering Thermodynamics h4 = hf + x4hfg = 191.83 + 0.836(2584.7 - 191.83) = 2192.27 kJ/kg qm = h3 - h2 = 3230.90 - 194.85 = 3036.05 kJ/kg gout = h4 - hi = 2192.27 - 191.83 = 2000.44 kJ/kg

Thus, -1

2000.44 - 0.3411 = 34.11 per cent 3036.05

qoui - 1 qi„

(b) Now the turbine inlet temperature is increased to 500°C. Therefore, h3 = 3456.5 kJ/kg, 53 = 7.2338 kJ/(kg K) Thus, x4 =

53 - sf sfg

7.2338 - 0.6493 = 0.878 8.1502 - 0.6493

and h4 = 191.83 + 0.878(2584.7 - 191.83) = 2292.77 kJ/kg Therefore, qin = h3 - h2 = 3456.5 - 194.85 = 3261.65 kJ/kg q,„,t = h4 - hi = 2292.77 - 191.83 = 2100.94 kJ/kg and gib _

1 /hut _ qin

2100.94 _ 0.3559 = 35.59 per cent 3261.65

The increase in the turbine inlet temperature thus results in an increase in the value of 77th as well as an increase in the value of steam quality x from 0.836 to 0.878. That is, the moisture content comes down. (c) In this case, the turbine inlet pressure is increased from 3 MPa to 10 MPa but 400°C only. Now, h2 = hi + w

T3

remains at

= hi + v1(10,000 - 10) = 191.83 + 0.001010(10,000 - 10) = 201.92 kJ/kg h3 = 3096.5 kJ/kg, s3 = 6.2120 kJ/(kg K) X4

6.2120 - 0.6493 8.1502 - 0.6493

0.742

h4 = 191.83 + 0.742(2584.7 - 191.83) = 1967.34 kJ/kg Thus, qi„ = h3 - h2 = 3096.5 - 201.92 = 2894.58 kJ/kg q0 = ha - hi = 1967.34 - 191.83 = 1775.51 kJ/kg and

= 1 - q" = 1

1775.51 - 0.3866 =

38.66 per cent 2894.58 gm It is seen that the increase in the boiler pressure from 3 MPa to 10 MPa results in an increase in the value of 77,h from 34.11 to 38.66 per cent.

Vapour Power Cycles 239 EXAMPLE 9A Consider a steam power plant operating on the ideal reheat Rankine cycle. The steam enters the high-pressure turbine at 3 MPa and 400°C. After expansion to 0.6 MPa, the steam is reheated to 400°C and then expanded in the low-pressure turbine to the condenser pressure of 10 kPa. Determine the thermal efficiency of the cycle and the quality of the steam at the outlet of the low-pressure turbine. Solution The schematic diagram of the power plant and the T—s diagram of the cycle are shown below. The cycle is the ideal reheat cycle, therefore, both the stages of the turbine as well as the pump are assumed to be isentropic; there are no pressure drops in the boiler and the condenser; and the steam at the outlet of the condenser and the inlet of the pump is saturated liquid at the condenser pressure. 3 MPa

T

Pump State 1: P1 = 10 kPa, water is saturated liquid, therefore, from Table 4 of the Appendix, we have

hi = hi = 191.83 kJ/kg, vi = vi= 0.001010 m3/kg, si = si= 0.6493 kJ/(kg K) State 2: P2 = 3 MPa, s2 = SI Wp

.in = vi(P2 — P1) = 0.001010(3000 — 10) = 3.0199 kJ/kg h2 = hi + wp„„,p,a, = 191.83 + 3.0199 = 194.85 kJ/kg

State 3: P3 = 3 MPa, 7'3 = 400°C

From Table 5 of the Appendix, we have h3 = 3230.9 kJ/kg, 53 = 6.9212 kJ/(kg K) State 4: P4 = 0.6 MPa, s4 = s3

240

Fundamentals of Engineering Thermodynamics

The entropy of saturated vapour at 0.6 MPa (from Table 4 of the Appendix) is sg @ 0.6 MPa = 6.7600 kJ/(kg K) That is, Sg (0 . 0.6 Mpa < 53.

Therefore, state 4 is still in the superheated region. From the superheated water table (Table 5 of the Appendix), at P4 = 0.6 MPa and s4 = 6.9212 kJ/ (kg K), we get T4 =

190.97°C, h4 = 2829.63 kJ/kg

State 5: P5 = 0.6 MPa and T5 = 400°C, therefore, from Table 5, we get h5 = 3270.3 kJ/kg, 55 = 7.7079 kJ/(kg K) State 6: P6 = 10 kPa, s6 = s5 Therefore, s6 - sf = s5 - sf = 7.7079 - 0.6493 mum X6 = 8.1502 - 0.6493 sfg Sig Now, h6 = hf + x6hfg = 191.83 + 0941(2584.7 - 191.83) = 2443.52 kJ/kg Thus, gin = (h3 - h2) + (h5 - h4) = (3230.9 - 194.85) + (3270.3 - 2829.63) = 3476.72 kJ/kg and flout = h6 - hi = 2443.52 - 191.83 = 2251.69 kJ/kg Therefore, yith

_

gout _ gin

2251.69 _ 0.3523 = 35.23 per cent 3476.72

The quality of the steam at the outlet of the low-pressure turbine is 94.1 per cent Note: This problem was worked out in Example 9.3 for the same pressure and temperature limits, but without the reheat process. From these examples, it is seen that reheating reduces the moisture content from 16.4 to 5.9 per cent and increases the thermal efficiency marginally from 34.11 to 35.23 per cent.

9.7 THE IDEAL REGENERATIVE RANKINE CYCLE Examine the T-s diagram of the Rankine cycle, redrawn in Fig. 9.11. It is seen that heat is added to the working fluid during the process 2-2' at a relatively low temperature. This lowers the Tav

Vapour Power Cycles 241 T

Low temp. heat addition

Stream exiting the boiler

Figure 9.11 The T—s diagram of the Rankine cycle. at which heat is added to the steam and consequently the cycle efficiency. To overcome this shortcoming, we look for ways to raise the temperature of the liquid leaving the pump (feedwater), before it enters the boiler. One way to accomplish this is to compress the feedwater isentropically to a high temperature, as in the Carrot cycle. However, because of the high pressures involved, this method is impractical. Another possibility is to heat the feedwater from the expanding steam in a counterflow heat exchanger built into the turbine, that is, to use regeneration. But this increases the moisture content of the steam at the final stages of the turbine, and also it is difficult to design such a heat exchanger, therefore, this method is also impractical. A practical regeneration process is achieved by extracting or bleeding steam from the turbine at various parts. This steam is used to heat the feedwater. The device where the feedwater is heated is called a regenerator. In addition to improving the //d„ the regenerator deaerates (removes the air that leaks in at the condenser) the feedwater which is necessary to prevent corrosion in the boiler. Because of these advantages, regeneration is used in all modem steam power plants. A regenerator is also called a feedwater heater. It is essentially a heat exchanger where heat is transferred from the steam to the feedwater either by mixing or without mixing the two fluid streams. The mixing type is called the open or direct-contact feedwater heater and the other is called the closed feedwater heater.

9.7.1 Open Feedwater Heater It is basically a mixing chamber, where the steam extracted from the turbine mixes with the feedwater exiting the pump. The mixture leaves the heater as a saturated liquid at the heater pressure. The ideal regenerative Rankine cycle with an open feedwater heater and its process cycle are shown in Figs. 9.12(a) and 9.12(b), respectively. In this cycle, the steam enters the turbine at the boiler pressure (stage 5) and expands partially in the turbine to an intermediate pressure. Some steam Y at this state is extracted and routed to the feedwater heater. The remaining (1 — Y) kg of steam expands completely to the condenser pressure. Therefore, the

242

Fundamentals of Engineering Thermodynamics

T

(a)

s

Pump

(b)

Figure 9.12 Regenerative Rankine cycle with open feedwater heater. mass flow rates of steam are different in different components. If the mass flow rate through the boiler is th, it will be (1 — Y)m through the condenser. The heat and work interactions for this cycle per unit mass of steam flowing through the boiler are as follows:

q,, = h5 — h4

(9.11)

gout = (1 — Y)(h2 — hi)

(9.12)

wturb,oui = (hs — hi)) + (1 — 11)(ho — h2) Wpump,in = ( 1

—

nWptunp I.in + Wpm') Ilin

(913) (9.14)

where Y=

616 m5

Wptutip Lin

= v1(P2 — Pi)

Wptmm II,in

= v3(P4 — P3)

The tith of the Rankine cycle increases as a result of regeneration. The cycle efficiency increases further with more number of feedwater heaters used as in present-day power plants. The optimum number is governed by economical considerations. EXAMPLE 9.5 Consider a steam power plant operating on the ideal regenerative Rankine cycle with one open feedwater heater. Steam enters the turbine at 3 MPa and 400°C and is condensed in the condenser at a pressure of 10 kPa. Some quantity of steam leaves the turbine at a pressure of 0.6 MPa and enters the open feedwater heater. Compute the fraction of the steam extracted from the turbine and the thermal efficiency of the cycle. Solution The power plant operates on the ideal regenerative Rankine cycle, hence both the turbine and the pump are isentropic; there are no pressure drops in the boiler, condenser, and the feedwater

Vapour Power Cycles 243 heater and the steam leaves the condenser and the feedwater heater as saturated liquid. The T-s diagram of the cycle is shown in the figure below: T

►s

State 1: P1 = 10 kPa, water is saturated liquid, therefore, from Table 4 of the Appendix, we have hi = hf = 191.83 kJ/kg, v1 = vf = 0.001010 m3/kg, si = sf = 0.6493 kJ/(kg K) State 2: P2 = 0.6 MPa, s2 = si Wp

Lin = V1(P2 - P1 ) = 0.001010(600 - 10) = 0.5959 kJ/kg h2 = hi + wp Li„ = 191.83 + 0.5959 = 192.43 kJ/kg

State 3: P3 = 0.6 MPa, water is saturated liquid, therefore, from Table 4 of the Appendix, we have h3 = hf@ 0.6 hen = 670.56 kJ/kg, v3 = of® 0.6 MPa = 0.001101 m3/kg State 4: P4 = 3 MPa, s4 = s3 (Lin = v3(P4 - P3) = 0.001101(3000 - 600) = 2.6424 kJ/kg

W

h4 = h3 + w1, tt m = 670.56 + 2.6424 = 673.20 kJ/kg State 5: P5 = 3 MPa,

T5 =

400°C, therefore, from Table 5 of the Appendix, we have h5 = 3230.9 kJ/kg, 55 = 6.9212 kJ/(kg K)

State 6: P6 = 0.6 MPa, s6 = S5 The entropy of the saturated vapour at 0.6 MPa is sg = 6.7600 kJ/(kg K) which is < 55. Therefore, state 6 is still in the superheated region. From the superheated water table (Table 5 of the Appendix), at P6 = 0.6 MPa and s6 = 6.9212 kJ/(kg K), we have T6 = 190.97°C, h6 = 2829.63 kJ/kg State 7: P7 = 10 kPa, = 55 Therefore, s7 - sfi . s5 X7 Sf$1

Sfgi

= 6.9212 - 0.6493 0.836 8.1502 - 0.6493

244

Fundamentals of Engineering Thermodynamics

and h7 = hi + x7hfg = 191.83 + 0.836 (2584.7 — 191.83) = 2192.22 kJ/kg Thus, qin = h5 — h4 = 3230.9 — 673.20 = 2557.7 kJ/kg and (but = (1 — Y)(h7 — hi) where Y is the fraction of the steam extracted from the turbine (Y = th6/ths ). The energy analysis of the open feedwater heater is identical to that of the mixing chamber discussed in Chapter 3. Thus, by Eq. (3.19), we have I ?hi

= I kho

or

Yh6 + (1 — Y)h2 = (1)(h3) Therefore, Y

- h3 — h2 _ h6 -

h2

670.56 — 192.43 2829.63 — 192.43

0.181

Thus, gout = (1 — 0.181X2192.27 — 191.83) = 1638.36 kJ/kg and 1638.36 0.359 = 35.9 per cent rich =1 _!l! = 1 _ 2557.7 gin Note: Comparing the present results with those of Example 9.3, it is seen that the thermal efficiency of the cycle has increased from 34.11 to 35.9 per cent as a result of regeneration. Also, note that the net work output has decreased by 362.08 kJ/kg and at the same time the net work input has decreased by 478.35 kJ/kg, which results in a net increase in the value of the thermal efficiency.

9.7.2 Closed Feedwater Heater The two fluid streams, namely the extracted steam and the feedwater can be at different pressures, since they do not mix. The schematic diagram of a steam power plant with one closed feedwater heater and its process diagram are shown in Figs. 9.13(a) and 9.13(b), respectively. In an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which leaves the heater below the exit temperature of the extracted steam because a temperature difference of at least a few degrees is required for any effective heat transfer to take place. The condensed steam is then pumped to the feedwater line or routed to another heater or to the condenser through a device called a trap, which allows the liquid to be throttled to a lower pressure region but traps the vapour. EXAMPLE 9.6 Consider a steam power plant operating on the ideal regenerative Rankine cycle with one closed feedwater heater. Steam enters the turbine at 3 MPa and 400°C and is condensed in the

Vapour Power Cycles 245 T

$

(b) Figure 9.13 Regenerative Rankine cycle with a closed feedwater heater. condenser at a pressure of 10 kPa. Some quantity of steam is extracted from the turbine at a pressure of 0.6 MPa and enters the open feedwater heater. Compute the fraction of the steam extracted from the turbine and the thermal efficiency of the cycle. Solution The T—s diagram of the cycle is given in the figure below: T

From Example 9.5, the following data for the present problem are valid: P1 = 10 kPa, saturated liquid, h1 = 191.83 kJ/kg, s6 = 6.9212 kJ/(kg K) P6 = 3 MPa,

7'6 = 400°C,

h6 = 3230.9 kJ/kg

P7 = 0.6 MPa,

s7 = s6,

h7 = 2829.63 kJ/kg

246

Fundamentals of Engineering Thermodynamics

P8 = 10 kPa, h2 = h1 + w h3 = hi +

s8 = = s6

h8 = 2192.27 kJ/kg

Lin = 191.83 + 0.001010(3000 - 10) = 194.85 kJ/kg

Wpm, 11,in

= hf®0.6 MPa Wpurnp 11,in = 670.56 + 0.001101 (3000 - 600) = 673.20 kJ/kg

The fluid at state 9 is assumed to be a compressed liquid at the same temperature as that at state 3. The saturation temperature at 0.6 MPa is 158.85°C. Thus, state 9 is at 3 MPa and 158.85°C. The enthalpy at state 9 is obtained from the compressed liquid water table (Table 6 of Appendix), by linear extrapolation, as h9 = 671.98 kJ/kg Note: From Table 6 of the Appendix, for P = 5 MPa and 10 MPa, at T = 158.85°C, we have h = 673.18 kJ/kg and 676.18 kJ/kg, respectively. Therefore, we may say that the change in h per 1 MPa pressure change is 0.6 kJ/kg. Thus, at 3 MPa and 158.85°C, we get h9 = 671.98 kJ/kg. To determine h5, we need to know the fraction of the steam Y that is extracted from the turbine at 0.6 MPa. The basic energy balance for the closed feedwater heater requires mitt = moho or

Y(h7 - h3) = (1 - n(h9 - h2) Therefore, 671.98 - 194.85 (h9 - h2) - 0181 (h7 - h3) + (h9 - h2 ) (2829.63 - 670.20) + (671.98 - 194.85) That is, 0.181 kg of steam is extracted at state 7 for every kilogram which enters the turbine at state 6. The enthalpy at state 5 is determined by applying the mass and energy conservation equations of the mixing chamber which is assumed to be insulated. Thus, we have y-

E ththe = E mihi or (1)h5 = (1 - Y)h9 + Yh3 or h5 = (1 - 0.181) 671.98 + (0.181 x 673.20) = 672.2 kJ/kg Thus, q„,= h6 - h5= 3230.9 - 672.2 = 2558.7 kJ/kg and gem= (1 - Y)(h8 - h1)= (1 - 0.181X2192.27 - 191.83) = 1638.36 kJ/kg Therefore, >)t6 = 1

0, _ - 9

1638.36 2558.7

qui The rite can also be determined as follows:

0 36 = 36 per cent •

Wnet,out = Wturb,out Wpurnp,in Wturb,out =

( h6 h7) + (1 - n(h7 - h8)

= (3230.9 - 2829.63) + (1 - 0.181X2829.63 - 2192.27) = 923.27 kJ/kg

Vapour Power Cycles 247 wpurni, i„ = p„nQ

+ Wr,ump ll,in

= 3.02 + 2.64 = 5.66 kJ/kg

Thus, w.=923.27-5.66917.61 kJ/kg wnei3O„i qth =

gin

917.61 2558.7

= 0.3586

= 35.86 per cent

Note that the rid, for the closed feedwater heater is the same as that for the open feedwater heater (Example 9.5). However, at higher pressures, it is a normal practice to use a closed feedwater heater rather than an open feedwater heater. EXAMPLE 9.7 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one open feedwater heater, one closed feedwater heater, and one reheater. The schematic of the power plant and its T—s diagram are shown below. Steam enters the turbine at 3 MPa and 400°C and exhausts at 10 kPa. Some steam is bled from the turbine at 1 MPa, for the closed feedwater heater, and the remaining steam is reheated to 400°C. The bled steam is completely condensed in the heater and is pumped to 3 MPa before it mixes with the feedwater at the same pressure. Steam for the open feedwater heater is bled from the low-pressure turbine at 0.6 MPa. Determine the fraction of the steam bled at 1 MPa and 0.6 MPa and also find the thermal efficiency of the plant. 1 kg

T

3 MPa 40TC Turbine

Mixing chamber

Pump

Solution Let the fractions of the steam bled for the closed and open feedwater heaters be Y and Z, respectively. The enthalpies at the various states and the pump work per unit mass of fluid are: h = 191.83 U/kg h3 = 670.56 kJ/kg h5 = 764.55 kJ/kg h7 = 765.06 kJ/kg h9 = 3230.9 kJ/kg hi = 3263.9 kJ/kg hi3 = 2366.24 kJ/kg

h2 = 192.43 U/kg h4 = 673.2 kJ/kg h6 = 762.81 kJ/kg hs = 764.57 kJ/kg h io = 2827.38 U/kg h 12 = 3117.06 kJ/kg

wpump "in = 0.596 kJ/kg = 2.64 kJ/kg . = 2.25 kJ/kg wpw,p Km WixomIt,in

248 Fundamentals of Engineering Thermodynamics The steam fractions Y and Z are obtained as follows: Closed feedwater heater: Ithihi = Ethehe or Y(h10 — h6) = (1 — Y)(h5 — h4) Therefore, h5 — its (h10 — h6) + (h5 — its)

764.55 — 673.2 — 0.0424 (2827.38 — 762.81) + (764.55 — 673.2)

Open feedwater heater:

Ethik = Eth.he or Zh12 + (1

-

Y — Z)h 2 = (1 — Y)h3

Therefore, Z

— (1 — Y)(h3 — h2 ) (1 —0.0424)(670.56 — 192.43) _ 0.1566 3117.06 — 192.43 hu — h2

The enthalpy h8 is obtained from the conservation of mass and energy equations of the mixing chambers. That is,

1 thek = Intik or (1Xh8) = (1 — Y)h5 + Yh7 Therefore, h8 = (1 — 0.0424)764.55 + (0.0424 x 765.06) = 764.57 kJ/kg Thus, (1/9 — h8) + (1 — YXh11 — h10) = (3230.9 — 764.57) + (1 — 0.0424) (3263.9 — 2827.38)

qin =

= 2884.34 kJ/kg go., = (1 — Y— ZXhi3 — h1 ) = (1— 0.0424 — 0.1566)(2366.24 — 191.83) = 1741.70 kJ/kg and _ 1 oth

q.,,, _ 1 gin

1741.70 —0 2884.34

.396= 39.6 per cent

Note: A similar problem has been solved in Example 9.4 with reheat. It may be seen that the thermal efficiency has gone up from 35.23 to 39.6 per cent as a result of regeneration.

9.7.3 Comparison of Open and Closed Feedwater Heaters Open systems are simple, inexpensive, and have good heat transfer characteristics. They also bring the feedwater to the saturation state. But a separate pump is required for each heater, to handle the feedwater.

Vapour Power Cycles 249 Closed systems are more complex because of the internal pumping network, and thus they are expensive. The heat transfer is also less effective since the two steams are not allowed to be in direct contact. To take the best of these systems, most steam power plants use a combination of open • and closed feedwater heaters.

9.8 SECOND—LAW ANALYSIS OF VAPOUR POWER CYCLES We know that the ideal Cannot cycle is totally reversible. However, the ideal Rankine cycles— simple, reheat or regenerative—are only internally reversible. They may involve irreversibilities external to the system, such as heat transfer through a finite temperature difference. The magnitude and the location of these irreversibilities can be determined by analysis using the second law. We know from Chapter 6, that the irreversibility for a steady-flow system which may be exchanging heat OR with a reservoir at TR can be expressed as i = To.tge„ = To im these — z ?nisi

+

TR

(kW)

(9.15)

or for a one-inlet, one-exit, steady-flow device, on unit mass basis, as i = Toss., = To (se — si + q13-) (kJ/kg) TR

(9.16)

where To is the ambient temperature. The irreversibility of a cycle depends on the heat transfer with reservoirs and their temperatures. Thus, we can express the specific irreversibility i as i = To E-21-(kJ/kg) ;

(9.17)

The specific irreversibility i for a cycle which involves only heat transfer with a source at TH and a sink at TL, becomes = T ( gout —) (kjikg) o

TL

TH

(9.18)

The availability of a fluid stream v at any state is given by v2

yr = (h — ho) — To(s — so) + — + 8z (kJ/kg)

2

(9.19)

where the subscript 0 refers to the surroundings. The change in yr of the working fluid during a process gives the W,„„ for that process. EXAMPLE 9.8 Calculate the irreversibilities of all the four processes of the Rankine cycle discussed in Example 9.1, assuming that heat is transferred to the steam in a furnace at 1400 K and rejected to

250

Fundamentals of Engineering Thermodynamics

a cooling medium at 300 K and 101 kPa. Also, calculate the availability of the steam exiting the turbine. Solution The T—s diagram of the ideal Rankine cycle is reproduced below. T

s Processes 1-2 and 3-4 are isentropic and, therefore, there are no irreversibilities (internal or external) associated with them. That is, iI2 = 0 and 134 = 0 Processes 2-3 and 4-1 are constant pressure heat addition and heat rejection processes, respectively. Also, they are internally reversible. However, the heat transfer between the working fluid and the source or sink takes place through a finite temperature difference, rendering both processes irreversible. From steam tables, we have s2 = si = 1.0910 kJ/(kg K) and 54 = s3 = 6.3615 kJ/(kg K) The irreversibility for the process 2-3 is

—) i23 = T0 (S3 — S2 _.i_qR23 TR Now, qR23 = h3 — h2 = 2960.50 — 344.56 = 2615.94 kJ/kg Thus, i23

= 300 [6.3615 — 1.0910 + (-2615.94)1 1400 j 1020.6 kJ/kg

Also, 9R I i41 = To (Si — S4 ' 7,7 . R- )

Vapour Power Cycles 251 Here, 91241

= ha — h1

= 2207.87 — 340.49 = 1867.38 kJ/kg

Thus, i41 =300[1.0910-6.3615+

] 1867.381 — 286.23 kJ/kg

L The irreversibility of the cycle is icyck = il2 i23 + 64 + 141 =

1020.6 + 286.23 = 1306.83 kJ/kg

The irreversibility of the cycle can also be obtained from icyck = To ( gout — gin ) TL TH =

300 (1867.38 300

2615.94 1400

1306.82 kJ/kg

The availability is given by W = (ha — ho) — To(sa — so) +

v2 2

gza

Neglecting the KE and PE, we get = (ha — ho) — rasa — so) where h0 = h® 300 Lim kPa = hf@ 300 K Thus,

=

50 = S(41 300 IC,101 kPa = sf@ 300 K =

112.62 id/kg

0.3931 kJ/(kg K)

}

(From Table 3 of the Appendix)

yr= (2207.87 — 112.62) — 300(6.3615 — 03931) 304.73 kJ/kg That is, 304.73 kJ/kg of work can be derived from the steam exiting the turbine if it is brought to the state of the surroundings in a reversible manner.

9.9 COGENERATION In our discussion on cycles so far, energy in the form of heat was transferred to the working fluid to derive work output. However, in many devices, heat, called the process heat, is required as the direct energy input. For example, chemical, textile, paper, oil production and refining, steel making, and food processing industries rely heavily on process heat. In these industries, process heat is generally supplied by steam at about 5 atm and 150-200°C. Let us consider a simple process heating plant shown in Fig. 9.14. If the heat losses in the piping are neglected, the process heating becomes a perfect operation with no wastage of energy. The quality of energy in the furnace is very high, since the temperature there is around

252 Fundamentals of Engineering Thermodynamics

Process heater

Figure 9.14 Process heating plant. 1370°C. This high-quality energy is transferred to water to produce steam at about 200°C (a highly irreversible process). It is thus obvious that a high-quality energy is transferred into a low-quality energy. This is not desirable from any point of view. Further, industries that use large amounts of process heat also consume large amounts of electric power. Therefore, a power plant that produced electricity while at the same time meeting the process heat requirements of certain individual processes was developed. Such a plant is called a cogeneration plant. Cogeneration is therefore the production of more than one useful form of energy from the same energy source. The power cycle in a cogeneration plant can use either a steam-turbine cycle or a gasturbine cycle (discussed later), or even a combined cycle (discussed later). An ideal cogeneration plant with steam-turbine is shown in Fig. 9.15. It is seen that there is no condenser involved in this plant. Thus, there is no waste heat that needs to be rejected. It is the usual practice to 3

20 kW Process heater 120 kW 100 kW

Figure 9.15 An ideal cogeneration plant with steam-turbine. study the performance of cogeneration plants by defining a parameter called the utilization factor rh, as 1lu

net work output + process heat delivered _ Yriet • total heat input 6n

(9.20)

Vapour Power Cycles 253 Or

nu

= —0out

(9.21)

On

where 0out is the heat rejected in the condenser. But here there is no condenser and hence Q = 0. Thus, the utilization factor of the ideal steam-turbine cogeneration plant is 100 per cent. But in actual plants, q, can go up to only 70 per cent, since there are heat losses in piping and other components. EXAMPLE 9.9 Steam enters the turbine of a cogeneration plant, shown in the figure below, at 12.5 MPa and 440°C. Some steam is extracted from the turbine at 700 kPa for process heating. The remaining steam continues to expand to 8 kPa. Steam is then condensed at constant pressure and pumped to the boiler pressure of 12.5 MPa. When considerable amount of process heat is required, some steam leaving the boiler is throttled to 700 kPa and routed to the process heater. The extraction fraction is so adjusted that the steam leaves the process heater as a saturated liquid at 700 kPa; it is then pumped to 12.5 MPa. The mass flow rate of the steam through the boiler is 75 kg/s. Neglecting the pressure drops and heat losses in the piping and assuming the pumps and the turbine to be isentropic, determine (a) the maximum rate at which process heat can be supplied, (b) the power produced and the utilization factor when no process heat is supplied. T

12.5 MPa 440°C

2— 3

Expansion V valve

Turbine

a —4 5

Boiler

Mixing chamber

700 kPa — 6 Process 8 kPa heater 7

10

Condenser

12.5 MPa 8 kPa

Solution The work inputs to the pumps are wp

= vg(P9 — P8) = 0.0010085(12500 — 8) = 12.6 kJ/kg

Wpump II,in = VAPIO — P7) = 0.001108(12500

= 13.07 kJ/kg

— 700)

254 Fundamentals of Engineering Thermodynamics The enthalpies at different states are: h, = h2 = h3 = = 3167.7 kJ/kg hs = 2552.74 kJ/kg and x5 = 0.898

h6 = 1947.37 kJ/kg and x6 = 0.738 h7 = 697.22 kJ/kg

h8 = 173.88 kJ/kg h9 = h8 + wp„„,i, = 186.48 kJ/kg h,0= h7 + tit'pump min = 710.29 kJ/kg

(a) The process heat achieved will be maximum when all the steam in the boiler is throttled and sent to the process heater and no steam is allowed to pass through the turbine. Thus, On. = rii(h4 — h7 ) = 75(3167.7 — 697.22) 185.286 MW In this case, the utilization factor is T/u

=1_

= 1 _ 0 =1 Qin

Qin

(b) When no process heat is supplied, all the steam leaving the boiler is expanded through the turbine to the condenser pressure of 8 kPa. Therefore, the turbine output is fkturb,out = m 3(h3 — h6) = 75(3167.7 — 1947.37) = 91.525 MW Power to run the pump is ikpunip,in

= Wpm" 1,iin

since the second pump is not involved here. Thus, 1.41

= 75 x 12.6 = 945 kW

Therefore, the net work output is CCI.OUt = Pkturb,out

ikp

= 91525 — 945 = 90.58 MW

Thus,

n„=

Pilnet + Op 90.58 + 0 185.286 — 0 A89 = 48.9 per cent Qin

Vapour Power Cycles 255

9.10 BINARY VAPOUR CYCLES Even though water is the best working fluid presently available, it is far from being the ideal one for vapour cycles. The binary cycle aims at achieving the desirable characteristics of the ideal working fluid by using two fluids. For best performance, the working fluid for vapour pressure cycles should have the following characteristics: • High critical temperature and safe maximum pressure • Low triple-point temperature • Condenser pressure which is not too low • High enthalpy of vaporization, hfg • Saturation dome that resembles an inverted U • Good heat transfer characteristics • Inert, easy availability at low cost By and large, water comes close to the expected nature of the working fluid. However, for water the Tcri is low (374°C) and the saturation pressure at high temperatures is very high (16.5 MPa at 350°C). To overcome these shortcomings of water, during the high temperature part of the cycle, water is replaced with any one of the more suitable fluids, such as mercury, sodium, potassium, and sodium-potassium mixtures. The result is a power cycle which is a combination of two cycles, one in the high-temperature region and the other in the low-temperature region, called the binary vapour cycle. In this cycle, the condenser of the high-temperature cycle called the topping cycle, serves as the boiler of the low-temperature cycle, termed the bottom cycle. The schematic diagram and the process diagram for a mercury-water binary vapour cycle are shown in Figs. 9.16(a) and 9.16(b), respectively. It is evident from this figure that the binary cycle approximates the Carnot cycle more closely than the steam cycle for the same temperature limits. Therefore, the binary cycle can be regarded as a means to increase the thermal efficiency of a power plant. Thermal efficiencies of the order of 50 per cent are possible with binary cycles, but the cycle is not attractive economically because of its high initial cost.

9.11 COMBINED GAS-VAPOUR POWER CYCLES The combined gas-vapour cycle is a modified cycle involving a gas power cycle topping a vapour power cycle. It takes advantage of the gas-turbine cycle at high temperatures and uses the high-temperature exhaust gases as the energy source for the bottom cycle such as a steam power cycle. A combined gas-steam power cycle is shown in Fig. 9.17. The energy is recovered from the exhaust gases by transferring it to the steam in a heat exchanger, which serves as the boiler. The combined cycle increases the efficiency without any appreciable increase in the initial cost. Thermal efficiencies of the order of 40 per cent are possible presently with these cycles.

256

Fundamentals of Engineering Thermodynamics

Boiler Mercury pump

MERCURY CYCLE

Mercury turbine

Heat exchanger

Superheater STEAM CYCLE Steam turbine

Steam pump

Saturation dome (mercury)

T

Saturation dome (steam)

s (b) Figure 9.16 Mercury-water binary vapour cycle.

Vapour Power Cycles

air, Combustion chamber

Compressor GAS CYCLE Gas turbine

Air in 5

9

Exhau;st gas

ANVV-

STEAM CYCLE Steam turbine

Pump

4

Condenser

6out

T

GAS CYCLE

Qin

STEAM CYCLE

60 u t I. i.-u

►

,S

()A 7 Combined gas-steam power cycle.

257

258

Fundamentals of Engineering Thermodynamics

SUMMARY A cycle which continuously converts heat into work is called the power cycle. A power cycle in which the working fluid is alternately vaporized and condensed is termed vapour power cycle. The Carnot vapour cycle serves as a perspective ideal cycle for vapour power plants. But it is an ideal model and cannot be approximated to actual devices. The Rankine cycle is an ideal cycle for vapour power cycles. It is the cycle obtained by eliminating many of the impracticalities associated with the Carnot cycle. The thermal efficiency of the Rankine cycle is given by rIth =

9m

=I—

flout 9

The methods to improve the thermal efficiency of the vapour power cycles aim at: • Increasing the average temperature at which heat is transferred to the working fluid in the boiler, or • Decreasing the average temperature at which heat is rejected from the working fluid in the condenser. The ideal reheat Rankine cycle aims at taking advantage of the increased efficiencies at high boiler pressures without facing the problem of excessive moisture at the final stages of the turbine. A regenerator is a device where the feedwater is heated. It is also called a feedwater heater. It is essentially a heat exchanger where heat is transferred from the steam to the feedwater either by mixing or without mixing the two fluid streams. The mixing type is called the open or direct-contact feedwater heater and the other is called the closed feedwater heater. The irreversibility for a steady-flow system which may be exchanging heat QR with a reservoir at TR can be expressed as

I = To&• = To (E these — E this; + gee

TR

(kW)

Cogeneration is the production of more than one useful form of energy from the same energy source. The utilization factor q,, is a measure of the performance of cogeneration plants. It is defined as %

= net work output + process heat delivered Wnet + Qp total heat input Qin

or

77. = 1 — Qou a,

t n

The binary vapour cycle aims at achieving the desirable characteristics of the ideal working fluid by using two fluids. The combined gas-vapour cycle is a modified cycle involving a gas power cycle topping a vapour power cycle. It takes advantage of the gas turbine cycle at high temperatures and uses the high-temperature exhaust gases as the energy source for the bottom cycle such as a steam power cycle.

Vapour Power cycles 259

PROBLEMS 9.1

Water enters the boiler of a steady-flow Camot engine as a saturated liquid at 1.3 MPa and leaves with a quality of 90 per cent. Steam leaves the turbine at 75 kPa. Show the Camot cycle on a T—s diagram, along with the saturation lines. Determine (a) the thermal efficiency of the cycle, (b) the quality at the end of the thermal heat rejection process, and (c) the net work output. [Ans. (a) 21.5 per cent, (b) 0.166, (c) 381.7 kJ/kg]

9.2 Water changes from saturated liquid to saturated vapour as heat is transferred to it from a source at 300°C, in a steady-flow Camot cycle. Heat rejection is at a constant pressure of 10 kPa. Draw the T—s diagram for the cycle relative to the saturation lines, and determine the amount of heat rejected, the thermal efficiency of the cycle, and the net work output. [Ans. 44.3 per cent, 781.89 kJ/kg, 622.4 kJ/kg] 93

Steam enters a turbine at 10 MPa and leaves at 20 kPa. If the quality of the steam has to be 90 per cent, what should be the temperature at the turbine inlet for an adiabatic turbine efficiency of 100 per cent? [Ans. 864°C]

9.4

Steam enters a turbine at 500°C and leaves at 60°C. If the quality of the steam has to be 90 per cent, what should be the pressure at the turbine inlet for an adiabatic turbine efficiency of 100 per cent? [Ans. 3.21 MPa]

9.5 A steam power plant operates between a boiler pressure of 40 bar and a condenser pressure of 1.0 bar. Determine the ideal Rankine cycle efficiency of the plant. Also, show the cycle on a T—s diagram with respect to the saturation lines. [Ans. 25.3 per cent] 9.6

For a steam power plant operating on an ideal Rankine cycle, the turbine inlet conditions are 4 MPa and 400°C and the condenser pressure is 20 kPa. Determine the thermal efficiency and the change in the quality of the steam leaving the turbine at the given temperature compared to saturation temperature at 4 MPa. How much is the increase in thermal efficiency owing to superheating? Also, sketch the T—s diagram of the cycle with respect to the saturation lines. [Ans. 33.1 per cent, 0.099, 1.7 per cent]

9.7

A steam power plant operates on a simple ideal Rankine cycle. The maximum and minimum pressures of the cycle are 15 MPa and 20 kPa. The mass flow rate through the cycle is 55 kg/s. The quality of the steam at the turbine exit should not go below 90 per cent. Draw the T—s diagram for the cycle and determine the minimum turbine inlet temperature and the thermal efficiency of the cycle. [Ans. 800°C, 44.52 per cent]

9.8

Steam enters the turbine of a 250 MW steam power plant operating on a simple ideal Rankine cycle at 15 MPa and 400°C and is cooled in the condenser at 20 kPa. Determine the moisture content of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam. [Ans. 29 per cent, 33.8 per cent, 272.72 kg/s]

260 Fundamentals of Engineering Thermodynamics 9.9 Consider a steam power plant operating on the ideal reheat Rankine cycle. Steam enters the fast-stage turbine at 6 MPa and 450°C and exits at 1 MPa. It is then reheated at 1 MPa to 370°C before expansion in the second-stage turbine to 0.2 MPa. The steam exiting the second turbine is again reheated to 320°C at constant pressure of 0.2 MPa. It is then expanded through the third-stage turbine to 20 kPa. Assuming the expansions in the turbine to be isentropic, determine (a) the power generated, (b) the cycle efficiency, and (c) the cycle efficiency if no reheat is involved. [Ans. (a) 1338.86 kW, (b) 36.2 per cent, (c) 35.5 per cent] 9.10 Consider a steam power plant operating on the ideal reheat Rankine cycle. Steam enters the turbine at 14 MPa and 500°C. After expansion to a dryness fraction of 0.95, the steam is reheated to 450°C before completing its expansion to the condenser pressure of 4 kPa. Find the reheat pressure and the cycle efficiency, assuming insentropic efficiency for the turbine and 87 per cent efficiency for the feed pump. [Ans. 900 kPa, 32.1 per cent] 9.11 A steam power plant operates on a simple ideal Rankine cycle. The steam enters the turbine at 3 MPa and 350°C and exits at 5 kPa. Determine the net work done per kg of steam and the thermal efficiency of the cycle. [Ans. 1057.34 kJ/kg, 35.5 per cent] 9.12 A steam power plant operating on a Rankine cycle has a pressure of 10 MPa and a temperature of 500°C at the boiler exit and a temperature of 35°C at the turbine exit. The isentropic efficiency of the turbine is 0.9 and the isentropic efficiency of the pump is 0.6. If the power plant generates 100 MW, determine the cycle efficiency and the steam flow rate. Also, show the T—s diagram of the cycle with reference to the saturation lines. [Ans. 37.4 per cent, 83.44 kg/s] 9.13 A steam power plant operating on a Rankine cycle with a superheater has a pressure of 0.8 MPa at the turbine inlet and the condenser temperature is 60°C. The superheater and the condenser operate in such a manner that the turbine and the pump handle water in single phase. Assuming isentropic operation for the turbine and the pump and neglecting the pump work, determine the degree of superheat required, the ideal efficiency of the cycle, and the specific steam consumption. Also, draw the T—s diagram for the cycle. [Ans. 345.47°C, 27.7 per cent, 3.97 kg/h] 9.14 A steam power plant generating 10 MW works on a Rankine cycle. Steam at 2.5 MPa and 350°C enters the high pressure turbine. After expansion to 0.4 MPa, the steam is reheated to 350°C at 0.4 MPa. The reheated steam is then expanded in a low-pressure turbine to the condenser pressure of 5 kPa. (a) Determine the thermal efficiency of the cycle and the steam flow rate. Also, draw the T—s diagram of the cycle. (b) If the power plant operates between the initial and final states on the Rankine cycle without reheat, fmd the corresponding thermal efficiency of the cycle. [Ans. (a) 35.5 per cent, 8.16 kg/s, (b) 34.7 per cent]

Vapour Power Cycles 261 9.15 Consider a steam power plant operating on the ideal regenerative Rankine cycle with one feedwater heater. Steam enters the turbine at 3 MPa and 400°C and is condensed in the condenser at a pressure of 5 kPa. Some quantity of steam is extracted from the turbine at a pressure of 0.5 MPa and fed into an open feedwater heater where it is directly mixed with the condensed water and then fed to the feed pump. Calculate the fraction of the steam extracted from the turbine and the thermal efficiency of the cycle, neglecting the work input to the pump. [Ans. 0.189, 38.2 per cent] 9.16 Consider a regenerative steam power plant operating on the ideal Rankine cycle. Steam enters the turbine at 4 MPa and 450°C and exits at 3 kPa. (a) In one case, steam at 0.3 MPa is extracted from the turbine and fed to an open feedwater heater before pumping into the boiler. (b), In the second case, steam is extracted at three bleeding points at pressures of 1.5 MPa, 0.3 MPa and 50 kPa. Calculate the fraction of steam extracted at each bleed point, and the ideal cycle efficiency neglecting the feed pump work. Compare the efficiency of the regenerative cycle with that of the ideal Rankine cycle. [Ans. (a) 0.176, 40.5 per cent, (b) Y1 = 0.114, Y2 = 0.094, Y3 = 0.102, 44.5 per cent] 9.17 Consider a steam power plant working between the limits of 2 MPa, 300°C, and 3 kPa. Assuming the isentropic efficiency of the turbine to be 0.71, determine the cycle efficiency, the work done, and the quality of the steam entering the condenser for the following cases: (a) a simple Rankine cycle, (b) a reheat Rankine cycle with the steam reheated to 300°C at the pressure where it becomes saturated vapour, (c) a regenerative Rankine cycle, with one bleed point (open feedwater heater) at the pressure where the steam becomes saturated vapour. [Ans. (a) 24.6 per cent, 718.75 kJ/kg, 0.9, (b) 25.5 per cent, 802.47 kJ/kg, 0.957, (c) 21.3 per cent, 489.52 kJ/kg, 0.9] 9.18 In a steam power plant, saturated steam at 3 MPa enters a high-pressure turbine and expands isentropically to a pressure at which its quality is 0.841. At this pressure, the liquid water (or moisture) is extracted and returned to the boiler through the feed pump. The remainder, assumed to be dry steam, is expanded isentropically to 4 kPa in a lowpressure turbine. Determine the ideal cycle efficiency of the plant. [Ans. 27.7 per cent] 9.19 Consider a steam power plant operating on a Rankine cycle. The steam enters the turbine at 2 MPa and 350°C and leaves at 5 kPa. The turbine and pump each have an efficiency of 80 per cent. Determine the net work output per kg of steam and the cycle efficiency. Also, find the percentage increase in net work output and the cycle efficiency if the cycle is assumed to be ideal. [Ans. 813.29 kJ/kg, 27 per cent, 25 per cent, 25.56 per cent] 9.20 A steam power plant operating on a simple ideal Rankine cycle has 12.5 MPa and 400°C as the steam state at the turbine inlet. The steam leaves the turbine at 10 kPa. Assuming a source temperature of 1200 K and a sink temperature of 300 K, determine the irreversibilities associated with each of the processes. [Ans. i12 = 0, i23 = 906.46 kJ/kg, i34 = 0, 141 = 105.07 kJ/kg]

262 Fundamentals of Engineering Thermodynamics 9.21 A steady-flow steam turbine operates between 3 MPa, 400°C and 10 kPa and quality x = 10. The work delivered by the turbine is 600 kJ/kg. If the surrounding environment is at 300 K, determine the irreversibility associated with the steam passing through the turbine. [Ans. 414.9 kJ/kg] 9.22 A food processing unit requires 3 kg/s of saturated or slightly superheated steam at 0.5 MPa, which is extracted from the turbine of a cogeneration plant. The boiler generates steam at 6 MPa and 450°C at the rate of 9 kg/s. The condenser pressure is 10 kPa. The steam leaves the process heater as saturated liquid. It is then mixed with the feedwater at the same pressure and this mixture is pumped to the boiler pressure. Assuming that both the pump and turbine have adiabatic efficiencies of 80 per cent, determine (a) the rate of heat transfer to the boiler, and (b) the power output of the cogeneration plant. [Ans. (a) 26562.69 kW, (b) 7069.58 kW] 9.23 A regenerative steam cycle operates between pressure limits 3.5 MPa and 15 kPa. The temperature of the steam entering the turbine is 400°C. Steam extraction occurs at 800 kPa and an open feedwater heater is employed. Calculate the thermal efficiency of the cycle assuming isentropic expansion in the turbine. [Ans. 35.28 per cent]

CHAPTER

10 Gas Power Cycles 10.1 INTRODUCTION From our studies in the previous chapters it is clear that power generation and refrigeration are the two important application areas for thermodynamics. Both these areas are generally accomplished through systems that operate on thermodynamic cycles. Further, the thermodynamic cycles can be divided into power cycles and refrigeration cycles. The devices that produce a net power output are called engines and the devices that produce refrigeration are called refrigerators, air-conditioners, or heat pumps. The cycles can also be categorized as gas cycles or vapour cycles depending on the phase of the working fluid. The thermodynamic cycles can also be classified as closed and open cycles. In closed cycles, the working fluid is returned to the initial state at the end of the cycle and is recirculated. In open cycles, the working fluid is renewed at the end of each cycle. Depending on how the heat is supplied to the working fluid, the heat engines are classified as internal combustion (IC) engines and external combustion engines.

10.2 ANALYSIS OF POWER CYCLES The cycles of actual devices are difficult to analyse because of the presence of friction and nonavailability of sufficient time for the establishment of equilibrium conditions during the cycle. To study these cycles analytically, we have to keep the complexities at a manageable level and utilize some idealizations. For example, when the actual cycle is stripped of all the internal irreversibilities and complexities, the resulting cycle still resembles the actual cycle closely but involves the internally reversible processes only. Such a cycle is called an ideal cycle, the process diagram of which is shown in Fig. 10.1. In this chapter, we will discuss the cycles of actual devices along with their idealized models.

263

264

Fundamentals of Engineering Thermodynamics

p

Actual cycle s Ideal cycle

•

Figure 10.1 Comparison of actual and ideal cycles. In general, heat engines are devices meant for converting other forms of energy (usually heat) to work. Their performance is expressed in terms of thermal efficiency thh. We know that W _w„et = Vin

net gin

(10.1)

We saw in Chapter 4 that qd, of the Carnot engine is the highest of all heat engines operating between the same temperature levels. In other words, the Camot cycle is the best possible cycle. Now, it is natural to ask the question: If Carnot cycle is the best cycle, why do we not use that cycle for heat engine operation instead of ideal cycles? The answer to this question is that the condition of total reversibility cannot be achieved for practical devices even at the idealized conditions and hence Carnot cycle is unsuitable as a realistic model. The efficiency of ideal cycles, even though much less than that of Carnot cycle, is still considerably higher than the qd, of an actual cycle because of the following idealizations made. • The ideal cycles involve no friction. • All expansion and compression processes take place in a quasi-equilibrium manner. • The heat loss through the piping system is made negligibly small by suitably insulating the pipes. • The changes in kinetic and potential energies of the working fluid are negligibly small, except in nozzles and diffusers where AKE is significant. The area under the process curves of a cycle represents the net work produced during a cycle, as shown in Fig. 10.2. This area is also equivalent to the net heat transfer for that cycle. The T—s diagram is particularly useful as a visual aid in the analysis of ideal power cycles.

10.3 THE CARNOT CYCLE The Carnot cycle consists of four totally reversible processes, as shown in Fig. 10.3. The cycle can be executed in a closed system, such as a piston-cylinder device or in a steady-flow system

Gas Power Cycles 265 P A

v Figure 10.2 Process diagram of a cycle. P

T

s

Figure 103 Process diagrams of a Carnot cycle. such as a combination of turbines and compressors, as shown in Fig. 10.4. Further, either a gas or a vapour can be utilized as the working fluid.

Isothermal compressor

Isentropic turbine

Figure 10.4 A steady-flow Carnot engine.

266 Fundamentals of Engineering Thermodynamics The thermal efficiency is given by TL rhh.c.arnot =1—

TH

(102)

Even though it is not practical to build an engine operating on a cycle which would closely approximate the Carrot cycle, it can be treated as a standard for comparison of the actual or other ideal cycles with it. EXAMPLE 10.1 Consider a Carrot heat engine using air as the working medium. At the beginning of the isothermal expansion, the P, V, and T are 600 kPa, 500 cm3, and 250°C, respectively. During the isothermal expansion process, 0.32 kJ of heat is added, and the maximum volume for the cycle is 5000 cm3. Determine (a) the volume after isothermal expansion, (b) the temperature of the lowtemperature reservoir (sink), (c) the volume after isothermal compression, and (d) the thermal efficiency of the cycle. Solution The T—s diagram of the Carnot cycle is shown below: TA

TH

T1

All the four processes which comprise the Carnot cycle are reversible. Therefore, (a) For process 1-2, we have Qin = TH AS12 = TH(S2 — Si) But the change in entropy of an ideal gas at constant temperature is S2 — S, = mR In (V2/V1 ) Therefore, Qin = mR TH In MN') = /WI In (V2/Vi) or 0.32 = 600(500 x

le) In (vivo

Gas Power Cycles 267 Or V2 = 2.906%71 = 2.906 x 500 = 1453 cm3 (b) Process 2-3 is isentropic expansion. Also, V3 = 5000 cm3. For ideal gases, we have T )1(7-1) = (T ry-" = V ) V2 C T3) or

5000 1.4- I

=r 500017 -1 _ TL

1453)11453 )

Or TL —

T 523. 15 —— 319.19 K= 46.04°C 1.639 1.639

(c) The volume after isothermal compression is V4. Therefore, for process 4-1, we have

v4 =( )14r- I) Vi

T) 4

But T= T 2 =T H 7'4 7; TL Thus, 1/0.4 V4 = (7. 523 .15) TL 14") (319 .19

=3.439

Or V4 = 3.439 x 500 = 1719.5 cm3 (d) The thermal efficiency is

qth -

TL. TH

-1

319. 19 523.15

- 0.39 =

39 per cent

10.4 AIR-STANDARD ASSUMPTIONS In devices, such as spark-ignition (SI) automobile engines, compression ignition (CI) or diesel engines, and the conventional gas turbine engines, which work on gas power cycles, the working fluid remains as gas throughout the entire cycle. These are called internal combustion (IC) engines. Working on an open cycle is the characteristic of all IC engines.

268

Fundamentals of Engineering Thermodynamics

To reduce the complexities associated with the analysis of actual gas power cycles to a manageable level, the following assumptions known as air-standard assumptions are utilized: • The working fluid is equivalent to air which behaves as an ideal gas and continuously circulates in a closed loop. • All the processes associated with the cycle are internally reversible. • The combustion process is replaced by a heat addition process from an external source. • The exhaust process is equivalent to a heat rejection process that restores the working fluid to its initial state. In addition to these assumptions, the air is assumed to have constant specific heats whose values are determined at 25°C. When this assumption is included, the air-standard assumptions are called the cold-air-standard assumptions. The resulting cycle with these assumptions is called an air standard cycle. Since the working fluid is an ideal gas with constant specific heats, we have from isentropic relations, -1 = 1,4 y T4 VI and -1 2= T3 v2 y Or V4 Vi

V3 V2

(*.*

=T2 and T4 = T3)

Denoting the isentropic compression or expansion ratio rt, as r = v4 =v 3 vt v2 the Carnot cycle efficiency can also be expressed as 1 -1 (103) Pv Equation (10.3) shows that the thermal efficiency of the Carnot cycle increases as the compression ratio r, increases. For an isentropic process, we have 1(r -1)/7 = 7.4 (P4 and T2 =rp2)(7-1)17 7/th,carnot —1

T3

8J

Or PI =

Pa

133

(•.= T2 and Ts = T3)

Gas Power Cycles 269 Denoting the isentropic pressure ratio as rp, we have _ P2 rp =- P4 In terms of rp, the thermal efficiency of the Carrot cycle can be expressed as 1 r(r -IYr

qth.carnot

(10.4)

It is evident from Eq. (10.4) that higher the pressure ratio, the higher is the Carrot cycle efficiency. That is, for high efficiency, the Carrot cycle must operate with high peak cycle pressure. EXAMPLE 10.2 An air-standard Carrot cycle operates between the temperature limits of 50°C and 500°C. The minimum and maximum pressure in the cycle are 100 kPa and 10 MPa. Draw the T—s and P—v diagrams of the cycle. Calculate the thermal efficiency and the net work output of the cycle. Solution The T—s and P—v diagrams of the cycle are shown below: A

P

A

TH = constant

500°C

50°C

$

rith - 1

TH

- 1

323.15 773.15

— 1

r(y-IYr '

- 0.582

= 58.2 per cent

Also, 1

where rp =

PZ

P3

Thus, (

ry-1)

rP (0.418j

p, = 21.18 = —*P3

270

Fundamentals of Engineering Thermodynamics

Or P2 = 21.18P3 = 21.18 x 100 = 2118 IcPa Now, gh, = q12 = TH(s2 — si) = TH (Cp In = —RTH in — P2 = — (0.287X773-

a_ R ln

(*: T2 = 2118 15) in 10,000 — 344.4 kJ/kg

Therefore, wdet = 7/dg;,, = 0.582 x 344.4 200.4 kJ/kg Alternative method The net work is also given by. w„et =

— qout

Now, gout = 34 = TL(S4 — 53) = 71.1Cp T3 = — RTL In

8

R In P4 P3

(..• T4 = T3)

The isentropic compression or expansion ratio is given by r = v4 = V3 v1 v2 Or V3 v2

T3 P2 T2 P3

323.15 x 21.18 = 8.85 = — v4 773.15

Therefore, v4 = 8.85v1 = 8.85

RT1

8.85 x 0.287 x 773.15 =

10 x103

— 0.1964 m3 /kg

Now, p = RT4 = 0•287 x 323.15 — 472.22 kPa 4 v4 0.1964 Thus, g34 = 0.287 x 323.15 In and wnet = qui —

472.22 — 143.96 kJ/kg 100

= 344.4 — 143.96 = 200.4 kJ/kg

Gas Power Cycles 271 Also, rIth = 1 —

cr

1 8.85(14

-

58.2 per cent

10.5 AN OVERVIEW OF RECIPROCATING ENGINES The reciprocating engine is basically a piston-cylinder device. It is one of the excellent inventions with a wide range of applications. It is the heart of the vast majority of automobiles, light aircraft, ships, electric power generators, and many other devices. The basic components of a reciprocating engine are shown, schematically, in Fig. 10.5. The piston reciprocates in the cylinder between two fixed positions called the top dead centre (TDC) and the bottom dead centre (BDC). At TDC and BDC the piston forms the smallest and the largest volumes in the cylinder, respectively. The distance between the TDC and BDC is called the stroke of the engine, which is the largest distance that, the piston can travel in one direction. The diameter of the piston is called the bore. Intake valve

U

Exhaust calve

------------Bore

TDC Stroke BDC

Figure 10.5 Basic components of a reciprocating engine. The valve through which the air or air-fuel mixture is drawn into the cylinder is called the intake valve, and the valve through which the combustion products are sent out of the cylinder is called the exhaust valve. The minimum volume in the cylinder with piston at TDC is called the clearance volume. The volume displaced by the piston movement between TDC and BDC is called the displacement volume. The ratio of the maximum volume formed in the cylinder to the minimum volume is called the compression ratio r of the engine. Thus, ;pc r = Vmvc

Vmin

VI-pc

(10.5)

The mean effective pressure (MEP) is another important term associated with reciprocating engines. It is a fictitious pressure and is defined as that pressure which when acting on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. The MEP is illustrated in Fig. 10.6.

272

Fundamentals of Engineering Thermodynamics

P

Wnet = MEP (0/max — limn) MEP

1

limn

TDC

BDC

Figure 10.6 Illustration of MEP. = (MEP x piston area x stroke) = MEP x displacement volume or MEP =

Wnet

V

— V„,;„

_

net

V

— vinin

(kPa)

(10.6)

If the combustion of the air-fuel mixture is initiated by a spark plug, the reciprocating engine is called a spark-ignition (SI) engine, and if the combustion is by self-ignition as a result of compression of the air-fuel mixture above its self-ignition temperature, the engine is called a compression-ignition (CI) engine.

10.6 THE OTTO CYCLE The Otto cycle is the ideal cycle for SI engines, named after Nikolaus A. Otto. In most SI engines, the piston executes four complete strokes within the cylinder, and the crank shaft completes two revolutions for each thermodynamic cycle. These engines are called four-stroke (IC) engines. The schematic diagrams of the strokes and the P—v diagrams for both actual and ideal four-stroke SI engines are shown in Figs. 10.7(a) and 10.7(b) respectively. In two-stroke engines, all the four functions described in Fig. 10.7 are executed in just two strokes, namely the power stroke and the compression stroke. The two-stroke engines are usually less efficient than their four-stroke counterparts because of the incomplete expulsion of the exhaust gases and the partial expulsion of the fresh air-fuel mixture along with the exhaust gases. However, they have the advantages of being simple and inexpensive, and have high power-to-weight and power-tovolume ratios. These features make them suitable for applications requiring small size and weight such as motor cycles.

Gas Power Cycles

Air-fuel mixture

Exhaust gases

Compression stroke

Power

Intake stroke

Exhaust stroke

stroke

End of combustion

Ignition

Exhaust valve opens E

Intake valve opens

BDC

TDC

(a)

2 —3 Air 4 —1 Isentropic

v=const

compression

heat addition

Isentropic expansion

v= const. heat rejection

P

►V

BDC

TDC

(b)

Figure 10.7

l

b

Comparison of actual and ideal four-stroke SI engines.

273

274

Fundamentals of Engineering Thermodynamics

The complexities associated with the actual four-stroke and two-stroke cycles can be significantly simplified with the air-standard assumptions. The resulting cycle which closely resembles the actual operating conditions is the ideal Otto cycle, consisting of four internally reversible processes (Fig. 10.7(b)): 1-2 Isentropic compression, 2-3 v = constant heat addition, 3-4 Isentropic expansion, and 4-1 v = constant heat rejection. The T—s diagram of an ideal Otto cycle is given in Fig. 10.8. The Otto cycle is executed in a closed system. Therefore, the T

$

Figure 10.8 Otto cycle T—s diagram. first-law relation for any of the processes becomes q — w = Au (kJ/kg)

(10.7)

The heat transfer to and from the working fluid, under the cold-air standard assumptions, can be expressed, respectively, as 9m= q23 = U3 — U2 = Cv(T3 — T2) (10.8) and (10.9) gout = — 941 = — (u1 — u4) = cgs — T1) The thermal efficiency of the Otto cycle becomes Rtkotto

wnet _

gout _

T4 —

qin

qin

T3 — T2

—1

TI[(T4/T1) — 1 ] T2RT3/ T2)

1]

Processes 1-2 and 3-4 are isentropic, and v2 = v3 and v4 = v1 . Thus, = V2 T2

VI

7

)

7 = ( V3 V4)

= 4 T3

(10.10)

Using these, the expression for gdotto reduces to hth Otto = .

1

1 r7 -I

(10.11)

where r = V„,,x1Vnin = V11112 = vi /v2 is the compression ratio and y= Cp/Cv. From Eq. (10.11) it is evident that the thermal efficiency of an ideal Otto cycle depends on r and y of the working fluid, under cold-air-standard assumptions. The Mikotto increases with increases in both r and y.

Gas Power Cycles 275 This is true for actual spark ignition IC engines too. A plot of ihh,o,„ versus r, for y = 1.4 is shown in Fig. 10.9. From Fig. 10.9 it is sect- that the increase in n -,th,Otto is not well pronounced at high compression ratios. When high compression takes place, the temperature of the air-fuel mixture rises above the autoignition t..mperature of the fuel during the combustion process, causing an early and rapid burning of the fuel. This premature ignition of the fuel, called autoignition, produces an audible noise, termed engine knock.

0.6 Typical compression ratios for gasoline engines

0

0 .E 0.4

0.2

0 1 l 0 2

1

I

I

14

10

6

Compression ratio, r

Figure 10.9

Variation of rith,0„,, versus r.

Further, for a given r, an ideal Otto cycle using a monatomic gas, such as argon or helium with y= 1.67, as the working fluid will have the highest 17th.0„,,. Thus, the 17th of the Otto cycle increases with an increase in y of the working fluid. The effect of yon n th.Otto is shown in Fig. 10.10. At room temperature y= 1.4 for air, 1.3 for carbon dioxide, and 1.2 for ethane. The working fluid in actual engines such as carbon dioxide contains larger molecules, and y decreases with temperature, which is one of the reasons why the actual cycles have lower efficiencies than the ideal Otto cycle. The thermal efficiencies of the actual SI engines range from about 25 to 30 per cent. A 0.8 2 0.6

0

_C

0.4

0.2 0

4

0

8

I . 12

Compression ratio, r

Figure 10.10

Effect of yon Thh,0„„.

276

Fundamentals of Engineering Thermodynamics

EXAMPLE 103 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression stroke, the air is at 101 kPa and 22°C, and 900 kJ/kg of heat is added to it during the constant-volume heat addition process. Considering the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure encountered during the cycle, (b) the net work output, (c) the thermal efficiency of the cycle, and (d) the mean effective pressure for the cycle. Solution The Otto cycle is shown below on a P—v diagram.

P

q

Note that the air contained in the cylinder forms a closed system. Also, note that we have to account for the variation of the specific heats of air with temperature, and therefore, any ideal-gas relation which uses the constant-specific heat assumption should not be used in the calculations. (a) As seen from the P—v diagram, the maximum temperature and pressure in an Otto cycle occur at state 3. Process 1-2 is isentropic compression of an ideal gas. Therefore, = v2 = 1 vd v, r or

„r2 =

- vrt

r 8 where yr is the relative specific volume. From the ideal gas properties table (Table 8 of the Appendix) for T1 = 295.15 K, we have vd = 647.9 and u1 = 210.49 kJ/kg. Therefore, 647.9 8 Again from Table 8, for va = 80.99, we have VT2 =

T2 =

- 80.99

662.74 K and u2 = 483.15 kJ/kg.

By state equation, we have P2 =

p, ( .2

71

v2

= 1011

74 ) 8 = 1814.31 1cPa 295.15

Gas Power Cycles 277 Process 2-3 is constant volume heat addition, therefore, by energy equation, we have q23 — W23 = u3 — u2 or 900 — 0 = u3 — 483.15 Or u3 = 1383.15 kJ/kg The corresponding temperature (from Table 8) is T3 = 1690.18 K and v13 = 4.853 Thus, P3 = /n2 T3

T2

II

V = 181431(1690.18 (1) = 4.627 MPa 662.74) 3

(b) Process 3-4 is isentropic expansion of the ideal gas. Thus, vr4 = v4 = r V3 V13 Or Vr4 =

rVr3 = 8 x 4.853 = 38.824

The corresponding T4 and u4 from Table 8 are T4 = 862.36 K, u4 = 643.35 kJ/kg Process 4-1 is constant volume heat rejection. Thus, 941 — w41 = Ul — U4 or q41 = 210.49 — 643.35 = — 432.86 kJ/kg That is, gout = 432.86 kJ/kg Thus, wnet = gnat = gin gout = 900 — 432.86 =

467.14 kJ/kg

wnet _ 467.14 — 0.519 = 51.9 per cent 900 gin

(c)

This is the thermal efficiency of the cycle. If the cold-air-standard assumptions are made and the specific heats are assumed to be constant volume at room temperature, the thermal efficiency given by Eq. (10.11) is 1 1 rith,Otto = 1 — r - — 1 — 80. 4 = 0.565 = 56.5 per cent which is different from the actual value. (d) The mean effective pressure given by Eq. (10.6) is MEP =

It

— w net _ V x — Vi.nn v1 — V2 %et

Wnet

VI C1

—

278

Fundamentals of Engineering Thermodynamics

Now, =

R7i Pi

—

287 x 295.15 101 x 103

— 0.839 m3/kg

Thus, 467.14 MEP — — 636.32 kPa ll 0.839 (1— ) 8

10.7 THE DIESEL CYCLE The Diesel cycle is the ideal cycle for CI reciprocating engines, named after its proposer Rudolph Diesel. In SI or petrol engines, the fuel-air mixture is compressed to a temperature which is below the autoignition temperature of the fuel, and the combustion is initiated by fixing a spark plug. But in CI or diesel engines the air is compressed to a temperature which is above the autoignition temperature of the fuel, and the combustion starts on contact as•the fuel is injected into the air. Therefore, the spark plug and the carburettor (the device in which air-fuel mixing takes place in SI engines) are replaced by a fuel injector in diesel engines, as shown in Fig. 10.11. The diesel engines can have higher compression ratios, typically between 12 and 24, compared to SI engines, since in diesel engines only air is compressed and thus there is no possibility of autoignition. Spark Fuel injector plug Fuel spray Air-fuel mixture

Gasoline engine

Diesel engine

Figure 10.11 Comparison of petrol and diesel engines. In diesel engines, the fuel injection process starts when the piston approaches TDC and continues during the first part of the power stroke. Therefore, the combustion process in these engines takes place over a longer interval. Because of this longer duration, the combustion process in the ideal diesel engine cycle is approximated to a constant-pressure heat addition process. In fact, this is the only process where the Otto and the Diesel cycles differ. The process diagrams on P—v and T—s coordinates for the ideal Diesel cycle are shown in Fig. 10.12. Like Otto cycle, the Diesel cycle is also executed in a piston-cylinder device, which forms a closed system. Therefore, the qi„ and lbw under the cold-air-standard conditions, can be expressed as

Gas Power Cycles 279 T

P

(b) T-s diagram

(a) P-v diagram Figure 10.12

P—v and T—s diagrams for Diesel cycle. gin = 923 = w23 (AU)23 = P2(113 -

v2) + (u3 — u2)

= h3 — h2 = CAT3 — T2)

(10.12)

and gout = g41 = —/V41 — 01041 = U4 — Ul = Cl(T4 — T1)

(10.13)

The thermal efficiency becomes rith,Diesa = wnet = 1 9h,

gout— 1

—

Y(T; — T2 )

9m

= 1

TRT4/70-1] T2[(T3/T2) — 1]

10.7.1 The Cutoff Ratio rc The cutoff ratio is defined as the ratio of the cylinder volumes after and before the combustion process, that is rc =

v V 3= 3 172 V2

(10.14)

Utilizing rc and ideal gas isentropic relations, we can express qtb.13iesel = 1

1 rr

[rcr — 1 y(rc — 1)

(10.15)

where r the compression ratio = vi/v2. From Eq. (10.15) it is evident that, under the cold-air-standard assumptions, qth.Diesei differs

280

Fundamentals of Engineering Thermodynamics

from gd,o, by the bracketed quantity, which is always > 1. Therefore, we can write (when both cycles operate at the same compression ratio) qth,Otto

(10.16)

qth.Diesel

As the cutoff ratio r, increases, qdtrojesei increases too, as shown in Fig. 10.13. For the limiting case of rc = 1, it can be shown that the quantity in brackets in Eq. (10.15) becomes unity and the gth for Otto and Diesel cycles becomes identical. 0.8

rc = 1(Otto)

0.6

2 3 4

0.4

Typical compression ratios for diesel engines

0.2

0.0

0 4 8 12 16 20 24 Compression ratio, r

Figure 10.13 Thermal efficiency versus cutoff ratio of the ideal Diesel cycle. It is essential to note that diesel engines operate at much higher compression ratios and, therefore, are more efficient than the SI engines. Further, the diesel engines also burn the fuel more completely since they usually operate at lower rpm than the SI engines. The 77th,Diesei ranges from 35 to 40 per cent. EXAMPLE 10.4 An ideal diesel cycle using air as the working fluid has a compression ratio of 16 and a cutoff ratio of 2. The intake conditions are 100 kPa, 20°C, and 2000 cm3. Using the cold-air-standard assumptions, determine (a) the T and P at the end of each process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure. Solution The P—V diagram of the cycle is shown below. All the four processes are closed-system processes since the cycle is executed in a pistoncylinder device with a fixed amount of air. (a) Under the cold-air-standard assumptions, air is assumed to be an ideal gas. Thus, R = 0.287 kJ/(kg K),

Gas Power Cycles 281

P

Qin

C, = 1.0045 kJ/(kg K) and C, = 0.7175 kJ/(kg K) V2 —

V1 2000 — 125 cm3 — 16 r

V3 = TcV2 = 2 x 125 = 250 cm3 V4 = Vi = 2000 cm3 Process 1-2 is isentropic compression. Therefore, T2 = 7;

P2 =

VI)7 I V,

= 293 .15(16)" = 888.66 K

P ( -11J- = moo" =

4850.3 kPa

2

Process 2-3 is constant pressure heat addition. Therefore, P3 = P2 = 4850.3 kPa V3 = 888.66 x 2 = 1777.32 K 7; = T2 (— v Process 3-4 is isentropic expansion. Therefore, V )7-1 250 74 = T3 (-1 = 1777.32 ( V4 2000

r

.1

773.62 K

14

P4 = P3 (21) V = 4850.3 ( 250 2000) V44)

I

1:1.4

= 263.9 kPa

282

Fundamentals of Engineering Thermodynamics

(b) The net work is w = m(qin — gout) where m is the mass of the air, given by m

=

_ (100 x 103 )(2000 x 10-6) — 2.38 x 10-3 kg RT 287 x 293.15

The heat added is h2) = m"T

Qin = Q23 = M(h3

T2,

= (2.38 x 10-3X1.0045X177732 — 888.66) =2.124 kJ The heat rejected is Qout = m(u4 — u1) = mCv(T4 — T1) = (2.38 x 10-3X0.7175X773.62 — 293.15) = 0.82 kJ Therefore,

W„n = Qi„ — a

t = 2.124 — 0.82 = 1.304 kJ

(c) The thermal efficiency of the cycle is = Wne, = 1.304 0.614 — 61.4 per cent 2.124 — The thermal efficiency of the Diesel cycle is also given by Eq. (10.15) as I r/th.Diesel — 1

rcr —1

rr —1 7(rc

=1—

—1)

1 21.4 — 1 — 0.614 = 61.4 per cent 16" 1.4(2 — 1)

(d) The mean effective pressure is given by MEP —

Wnet "max — "min

Wnet

V1 — V2

1.304 (2000 — 125)10-6 695.5 kPa

10.7.2 The Dual Cycle The dual cycle is a combination of Otto and Diesel cycles. It models the combustion process of both petrol and diesel engines as a combination of two heat transfer processes, one occurring at constant volume and the other at constant pressure. The P—v diagram of an ideal dual cycle is given in Fig. 10.14.

Gas Power Cycles 283 P

•v Figure 10.14 P—v diagram of an ideal dual cycle. 10.8 STIRLING AND ERICSSON CYCLES These are cycles involving an isothermal heat addition process at TH and an isothermal heat rejection process at TL. The differences between these and the Carrot cycle are that the two isentropic processes in the Carrot cycle are replaced by two constant-volume regeneration processes in the Stirling cycle, and by two constant-pressure regeneration processes in the Ericsson cycle, as shown in Fig. 10.15. Both Stirling and Ericsson cycles are totally reversible like the Carrot cycle, and, thus, have the same thermal efficiency as that of the Carrot cycle when operating between the same temperature limits. Therefore, T

T

A

A

T

gin

TH

TH 8

6° 8

II Cl)

II en

Ti.

71

3 Clout

1 S

P

A

v Figure 10.15 Process diagrams of Carnot, Stirling, and Ericsson cycles.

V

284 Fundamentals of Engineering Thermodynamics , hth,Stirting — hth,Ericsson = htli,Ca

t=

—

T TH

(10.17)

Both Stirling and Ericsson cycles involve heat transfer through a differential temperature difference in all their components including the regenerator, which makes realization of these cycles difficult in practice. Both the cycles, thus, are not practical, and are of only theoretical interest

10.9 THE BRAYTON CYCLE The Brayton cycle is the ideal cycle for gas-turbine engines, named eller George Brayton who proposed it around 1870. Gas turbines, usually operate on an open cycle, as shown in Fig. 10.16 and explained below. Fuel

Combustion chamber

Compressor

Fresh air

Turbine

Exhaust gas

Figure 10.16 Schematic of an open-cycle gas-turbine engine. • The T and P of the fresh ambient air drawn in is raised in the compressor. • The high pressure air is mixed with fuel and burnt at constant pressure in the combustion chamber. • The high temperature gases expand to the ambient pressure through the turbine, thus producing power. • The exhaust gases are thrown out through the turbine. The open gas-turbine cycle described above can be modified to a closed cycle, as shown in Fig. 10.17, using the air-standard assumptions. Here the compression and expansion processes remain the same, but the combustion process is replaced by a constant-pressure heat addition process from an external source, and the exhaust process is replaced by a constant pressure heat rejection process to the ambient air. In this ideal cycle, the working fluid undergoes the following proceses: 1-2 isentropic compression, 2-3 constant pressure heat addition, 3-4 isentropic expansion, 4-1 constant pressure heat rejection. This closed-loop ideal cycle is the Brayton cycle. All the four processes in the Brayton cycle are steady and reversible processes. The P—v and T—s diagrams of the Brayton cycle are given in Fig. 10.18. The conservation energy equation for a steady-flow process, in the absence of ke and pe, can be expressed as

Gas Power Cycles 285

Heat exchanger

Turbine

Compressor Heat exchanger

gout Figure 10.17 A closed-cycle gas-turbine engine. P

T

A

gout ► S

Figure 10.18 Process diagrams of the Brayton cycle. (10.18)

q — w = hextt — hiniet

Assuming constant specific heats at room temperature, the heat transfer to and from the working fluid becomes gin = q23 = h3 h2 =

(10.19)

T3 T2)

(1020)

90ta = —941 = h4 — h1 = CAT4 — T1) The qth of the ideal Brayton cycle becomes

nuo

— wne` — qin

goat — 1 gin

CP (T4 — T1) — 1 Cp (T3 — T2 )

T1RT4 / 71 — 1] T2 [(T3 T2)

286 Fundamentals of Engineering Thermodynamics Also, for the processes, P2 = P3, P4 = P1 and = (p r-lYr = T2

2

TI

( PI

( 8 )(7-0/7 =

( P4 )

J

. 3 T4 7

Therefore, we have hth Bray = .

1

1 r (7-017

(10.21)

P

where rp = P2/P1 is the pressure ratio and y is the specific heats ratio. From Eq. (10.21), it is evident that under the cold-air-standard assumptions, gdokayton depends on the pressure ratio of the gas turbine and the y of the working fluid. Further, ridonwton increases with both rp and y, which is true for actual gas turbines too. The variation of thermal efficiency of an ideal Brayton cycle with the pressure ratio, for y = 1.4, is shown in Fig. 10.19. The maximum temperature that the turbine blades can withstand

0.6 C

.:) s. e 0.4 co

Typical pressure ratios for gas turbine engines

0.2 I

I I 1 10 20 Pressure ratio, rp

i

Figure 10.19 Thermal efficiency of an ideal Brayton cycle vs. pressure ratio. limits the T„. of the cycle as well as the maximum value of rp that can be used in the cycle. For a fixed turbine inlet temperature T3, the net work output per cycle increases with rp, reaches a maximum, and then starts decreasing, as shown in Fig. 10.20. The ambient air drawn through the compressor not only supplies the required oxidant for combustion of the fuel, but also serves as a coolant to keep the temperature of various components within safe limits. The cooling function is accomplished by drawing more air than what is required for complete combustion of the fuel. An air-fuel ratio of 50 or above is common in gas turbines. Thus, treating the combustion gas as air is justified and will not cause any serious errors. The thermal efficiency of a gas-turbine engine depends on the allowable maximum gas temperature at the turbine inlet. With the modern advancements, today's gas turbines can withstand as high as 1425 GC at the turbine inlet, and gas-turbine power plants have efficiencies of the order of 30 per cent.

Gas Power Cycles 287 T

Trim 1000 K

Tmin

300 K Figure 10.20 Work output variation with pressure ratio of an ideal Brayton cycle. The major application areas of gas-turbine engines are aircraft propulsion and electric power generation. The ;_;as-turbine cycle can also be executed as a closed cycle for use in nuclear power plants. Here the working fluid is not limited to air, and any gas with more desirable characteristics, such as helium, can also be used as the working fluid. In gas-turbine power plants, the ratio of the compressor work to the turbine work, termed the back work ratio, is very high. More than one-half of the turbine work output is used to drive the compressor. A high back work ratio, therefore, requires a larger turbine to cater to the additional power requirements of the compressor. Therefore, for the same power rating, the turbines used in gas-turbine plants are larger in size than those used in steam power plants. EXAMPLE 105 An ideal air-standard Brayton cycle operates with air. At the compressor inlet, the air is at 300 K and at the turbine inlet the gas is at 1100 K. The pressure ratio of the cycle is 6. Accounting for the variation of specific heats with temperature, determine (a) the gas temperature at the compressor and turbine exits, (b) the back work ratio, and (c) the thermal efficiency of the cycle. Solution The T—s diagram of the cycle is shown below. Under air-standard assumptions, the working fluid is assumed to be air, which behaves as an ideal gas, and all the four processes of the cycle are internally reversible. Also, the combustion and exhaust processes are replaced with heat addition and heat rejection processes, respectively. T

Wturb

S

288 Fundamentals of Engineering Thermodynamics Process 1-2 is isentropic compression of an ideal gas. Therefore, for Ti = 300 K, from the ideal. gas properties table for air (Table 8 of the Appendix), we have hi = 300.19 kJ/kg, P,1 = 1.386 Thus, P,1 = 6 x L386 = &316

P2 =

Again from Table 8, for Pr2 = 8.316, we have T2 =

498.38 K (which is the temperature at the compressor exit)

and h2 = 501.35 kJ/kg Process 3-4 is isentropic expansion of an ideal gas. Therefore, for T3 = 1100 K, from Table 8, we get h3 = 1161.07 kJ/kg, P3 = 167.1 Thus, 1

P4 =

r3

= — x 167.1 = 27.85

3

6

This corresponds to T4=

693.7 K (which is the temperature at the turbine exit)

and h4 = 706.51 kJ/kg (b) The back work ratio is rbw

Wcomp,m •

wturbmut Now, w mi„ = h2 — hi = 501.35 — 300.19 = 201.16 kJ/kg and wturb,out = h3 h4 =

1161.07 — 706.51 = 454.56 kJ/kg

Thus, rbw —

201.16 — 0.442 = 44.2 per cent 454.56

That is, 44.2 per cent of the turbine output is required to drive the compressor. (c) The thermal efficiency is

,th _

_ gin

— win

h3 — h2

_

454.56 — 201.16 — 0.384 = 1161.07 — 501.35

38.4 per cent

Gas Power Cycles 289 The thermal efficiency is also given by 1

1

htiorartm — 1 r(r-lYr — 1 643.4/1.4 — 0.4006 = 40.06 per cent Note: The efficiency obtained with rp is slightly higher than the gfi, obtained after properly accounting for the variation of specific heats with temperature.

10.9.1 Deviation of Actual Gas-Turbine Cycles from Idealized Cycles The actual gas-turbine cycle differs from the ideal Brayton cycle because of the following reasons: • Some pressure drop in an actual gas-turbine cycle during heat addition and heat rejection processes is unavoidable. • Owing to irreversibilities, such as friction and non-quasi-equilibrium operation conditions of actual devices, the actual work input to the compressor will be more, and the actual work output from the turbine will be less. The deviation of the behaviour of the actual compressor and turbine from the idealized ones can be accurately accounted for, by utilizing the adiabatic efficiencies defined as qc

ws , =

7h.

— h2s — h2„

(1022)

h3 — 114, ws

h3 has

(10.23)

where the subscripts 2a and 4a refer to the actual exit states and 2s and 4s to the isentropic exit states of the compressor and turbine, respectively. The differences between the actual gasturbine cycle and the ideal Brayton cycle are shown in Fig. 10.21. Pressure drop during heat addition 3 2a 2s Pressure drop during heat rejection

Figure 10.21 Comparison between the actual and ideal gas-turbine cycles.

290

Fundamentals of Engineering Thermodynamics

EXAMPLE 10.6 An air-standard gas turbine cycle operates under the same conditions as those cited in Example 10.5, except that the compressor and turbine are only 80 and 82 per cent efficient, respectively. Determine (a) the back work ratio, (b) the temperature at the turbine exit, and (c) the thermal efficiency of the cycle. Solution The T—s diagram for the process described is shown below. T

s 201.16 (a) The actual work done by the compressor, w ws _- 0.80 — —251.45 kJ/kg a = qc — where ws is the isentropic work input to the compressor. Similarly, for the turbine wa = 11rws = 0.82 x 454.56 = 372.74 kJ/kg Thus, lbw =

Wcomp,to •

wantwat

=

251.45 = 0.675 372.74

Here the compressor consumes 67.5 per cent of the work produced by the turbine. This increase from the isentropic back work ratio of 0.442 is due to the irreversibilities that occur within the compressor and the turbine. .(b) The steady-flow first-law relation for the turbine is q34, — 11/34, = h3 — h4a

But q34, = 0, therefore, we have h4, = h3 — wii„b, = 1161.07 — 372.74 = 788.33 kJ/kg From the ideal gas properties table (Table 8 of the Appendix), for 14, = 788.33 kJ/kg, we have 7.4a = 769.66 K

Gas Power Cycles 291 (c) Wcemp,ia = h2a — hi

or h2a = hi +w iii = 300.19 + 251.45 = 551.64 kJ/kg

Therefore, gi, = h3 — hu = 1161.07 — 551.64 = 609.43 kJ/kg and w„, = wout — wi, = 372.74 — 251.45 = 121.29 kJ/kg Hence, wnet gin

qdk —

121.29 — 0.199 — 19.9 per cent 609.43

10.10 THE BRAYTON CYCLE WITH REGENERATION The regenerator used in the Brayton cycle is a counterflow heat exchanger which heats the high-pressure air leaving the compressor, by transferring heat to it from the hot exhaust gases leaving the turbine. The regenerator is also known as a recuperator. A gas-turbine engine with a regenerator and its process diagram are shown in Fig. 10.22. The temperature of the exhaust gases leaving the turbine and entering the regenerator is the highest temperature T4 occurring within the regenerator. Normally, the temperature 7'5 of the air leaving the regenerator will be Regenerator

Combustion chamber

Turbine

Compressor

T

4

Regeneration

qsaved = %men

aS

Figure 10.22 A gas turbine with a regenerator.

292 Fundamentals of Engineering Thermodynamics less than T4. In the limiting (ideal) case, T5 may be equal to T4. The actual and the maximum heat transfer from the exhaust gases to the air can be, respectively, expressed as q1.

= h5 — h2

(10.24)

and chg,„. = 175, — h2 = h4 — h2

(10.25)

The effectiveness e, which is the extent to which a regenerator approaches an ideal regenerator, is defined as e=

9,s,act _ h5 — h2 q,gnx — h4 — h2

(10.26)

With cold-air-standard assumptions, e becomes TZ e — Ts — T4

(10.27)

T2

Even though a regenerator with a higher value of effectiveness will preheat the air to a higher temperature prior to combustion and thus save a greater amount of fuel, it pays a higher penalty by causing a large pressure drop, since achieving a higher value of effectiveness requires the use of a larger regenerator. Most regenerators in use have e below 0.85. The thermal efficiency of an ideal Brayton cycle with regeneration, under cold-air-standard assumptions, is rithos = 1—() rp )(7-01r

(10.28)

The variation of thermal efficiency with rp and temperature ratio is shown in Fig. 10.23. It is seen from this figure that regeneration is most effective at lower pressure ratios and low minimum-tomaximum temperature ratios.

0.8

0.8

With regeneration

VOW:at tinn regenera.._ ----

8 0.

IIIT3 = 0.2

T1/T3 = 0.25 TIT3 = 0.33

0.2

10 20 Pressure ratio, rp

Figure 10.23 Comparison of riduiroton with and without regeneration.

Gas Power Cycles

293

EXAMPLE 10.7 Determine the thermal efficiency of the gas-turbine cycle described in Example 10.6 if a regenerator having 70 per cent effectiveness is installed. Solution The T—s diagram of the regenerative Brayton system is shown below. T

1100 K

300 K 1

The effectiveness can be expressed as

'&"`

E = q

grg.max

h5

h2a

h4a — ha

Therefore, 0.7 —

h5 — 551.64 788.33 — 551.64

or h5 = 717.32 kJ/kg Thus, my, = h3 — h5 = 1161.07 — 717.32 = 443.75 kJ/kg That is, 165.68 (= 609.43 — 443.75) kJ/kg of heat input is saved with the use of the regenerator. If the regenerator is assumed to be frictionless, the work output will not be affected because of the introduction of the regenerator. Thus, _ wnet _ 121.29 — 0.273 = 27.3 per cent q1 443.75 Note that, by recouping some of the excess energy of the exhaust gases, the thermal efficiency of the cycle has been increased to 27.3 per cent from 19.9 per cent.

294 Fundamentals of Engineering Thermodynamics

10.11 THE BRAYTON CYCLE WITH INTERCOOLING, REHEATING, AND REGENERATION The work input required to compress a gas between two specified pressures can be decreased by carrying out the compression process in stages and cooling the gas in between, that is, using multistage compression with intercooling. As the number of stages is increased, the compression becomes isothermal at the inlet temperature of the compressor. Similarly, the work output of a turbine operating between two pressure levels can be increased by expanding the gas in stages and reheating it in between, that is, utilizing multistage expansion with reheating. As the number of stages is increased, the expansion becomes isothermal. In a gas turbine with intercooling and reheating, the regeneration becomes more attractive since the working fluid leaves the compressor at a lower temperature and the turbine at a higher temperature. An ideal two-stage gas-turbine cycle, with intercooling, reheating, and regeneration and its process diagram are shown in Fig. 16.24. For the optimum operation, we have Regenerator 10

-'VVN.,-

Combustion chamber Repeater

/

_

Compressor

d 7

6

I

I

II

II

--a. Wnet

---------Turbine

Intercooler T

s Figure 10.24 An ideal gas-turbine with intercooling, reheating, and regeneration, and its process diagram.

Gas Power Cycles = P4 and P3

P s 6= P

295 (10.29)

"7

Even though the back work ratio of a gas-turbine cycle improves because of intercooling and reheating, it does not ensure increase of rhh. In fact, intercooling and reheating will always decrease the rith unless accompanied by regeneration. An ideal ,.:is-turbine cycle with intercooling, reheating, and regeneration will approach the Ericsson cycle, if the numbe• of compression and expansion stages is increased, as illustrated in Fig. 10.25. T

THav

TLay ► s Figure. 10.25 A gas-turbine cycle with intercooling, reheating, and regeneration approaching Ericsson cycle.

EXAMPLE 10.8 An ideal gas-turbine cycle has two stages of compression and two stages of expansion. The overall pressure ratio is 6. Air enters each stage of the compressor at 300 K and each stage of the turbine at 1100 K. Determine the back work ratio and the thermal efficiency of the cycle: (a) without regeneration, and (b) with an ideal regenerator with 90 per cent effectiveness. Assume that both the compressor and turbine are isentropic devices. Solution (a) The T—s diagram of the cycle is given below. The cycle is an ideal cycle and, therefore, all the processes are internally reversible. and no pressure drop occurs during intercooling and reheating. That is, P, = P3 and P, = P8. T

S

296 Fundamentals of Engineering Thermodynamics Also, it can be shown that, for two-stage compression and expansion, the work input is minimized and the work output is maximized when both the stages of the compressor and turbine have the same pressure ratio. Thus, PI

= 124- = 1/13 = 2.45 P3

and P P = 4$3 = 2.45 =

P7

P9

(a) There is no regeneration. Therefore, from Table 8 of the Appendix (ideal gas properties for air), we have at Ti = 300 K, hl = 300.19 kJ/kg,

Pa = 1.386

Therefore, Pr2 = — P2

P

i = 2.45 x L386 = 3.40

Now, for Pa = 3.40, from Table 8, we have T2 = 387.34 K, h2 = 388.19 kJ/kg Also, for T6 = 1100 K, from Table 8, we have h6 = 1161.07 kJ/kg, Pr6 = 167.1 Therefore,

1

= P7 Pf6 P6

2.45

x 167.1 = 68.2

Now, for P17 = 68.2, from Table 8, we have

T7 = 877.35 K, h7 = 907.61 kJ/kg Therefore, Wcomp,in =

2(Wcenv1,0 = 2(h2 hi)

= 2 (388.19 — 300.19) = 176 kJ/kg and Wturb,out = 2(Wanbl,out) = 2(h6

h7)

= 2(1161.07 — 907.61) = 506.92 kJ/kg Thus, wnet = wtbmg — w

m = 506.92 — 176 = 330.92 kJ/kg

and gin

= qp,;„„,), + q,cbm = (h6 — h4 ) + (h8 —

= (1161.07 — 388.19) + (1161.07 — 907.61) = 1026.34 kJ/kg

Gas Power Cycles

297

Therefore, rbw —

Wcomp,in _ 176 — 0.347 = 34.7 per cent 506.92 Wturb,out

„ qth

Wnet _ 330.92 — 0.322 = 32.2 per cent 1026.34 qth

Note: Comparing the present results with those of Example 10.5, we see that multistage compression with intercooling and multistage expansion with reheating improve the back work ratio. Here the rbW has dropped from 44.2 to 34.7 per cent. But the thermal efficiency has gone down considerably from 38.4 to 32.2 per cent. Therefore, unless accompanied by regeneration, intercooling and reheating should not be introduced in a gas-turbine power plant. (b) Introduction of an ideal regenerator results in considerable reduction of heat input. E = grg,aet

h5 h4

grg,max

h9 — h4

Or

0.9 —

h, —388 19 * 907.61— 388.19

h9 =h7)

Or

hs = 855.67 kJ/kg Therefore, qin = gprimary greheat = (h6 h5)

(h8

h7)

= (1161.07 — 855.67) + (1161.07 — 907.61) = 558.86 kJ/kg The compressor and turbine work are unaffected since both these devices are assumed to be isentropic, that is, 100 per cent efficient and the back work ratio is unchanged. Thus, 77th

Wnet _

qi„

330.92 — 0.592 = 59.2 per cent 558.86

The thermal efficiency of the cycle has gone up from 32.2 to 59.2 per cent. Note that as the number of compression and expansion stages is increased, the cycle will approach the Ericsson cycle, and the thermal efficiency will approach Rth,Eric n = hth.Camot

=1

TL = 1 Ty

300 _0.727 1100

10.12 IDEAL JET-PROPULSION CYCLES A jet-propulsion cycle is an open cycle on which aircraft gas turbines operate. It differs from the ideal Brayton cycle in the fact that the gases are not expanded to the ambient pressure in the turbine. Instead, the gases are expanded to a pressure such that the power produced by the

it

i

298

Fundamentals of Engineering Thermodynamics

turbine is just sufficient to run the compressor and the auxiliary equipment. That is, the net work output of a jet-propulsion cycle is zero. The thrust required to propel the aircraft is provided by a nozzle which accelerates the gases that exit the turbine at a relatively high pressure. The aircraft gas turbines operate at pressure ratios from 10 to 25. The propulsive thrust for aircraft is accomplished either by slightly accelerating a large mass of air in propeller engines or by largely accelerating a small mass of air in turbojet engines. Both these principles are combined in the turboprop engine. An ideal turbojet engine and its process diagram are shown in Fig. 10.26. The compressed air from the compressor is mixed with fuel and burned at constant pressure in the combustion chamber. The combustion gases at high P and T are expanded partially in the turbine, producing

1 Diffier

1 Compressor 04

Turbine

Burner Section

• • id. Nozzle •

T

Fig. 10.26 An ideal turbojet engine and its process diagram. enough power to drive the compressor and other auxiliary equipment. Finally, the gases expand in a nozzle to the ambient pressure and leave the nozzle at a high velocity. The unbalanced force owing to the difference between the momentum of the low-velocity air entering the compressor and the high-velocity exhaust gases leaving the nozzle is called the thrust. The net thrust developed by the engine is given by F =(thV), —(inV)i = rh(Ve — 11) (lcN)

(10.30)

where the subscripts i and e refer to the engine inlet and nozzle exit states, respectively. The m

Gas PoverICydes 299 is taken as the mass flow rate of air through the engine, since the air-fuel ratio in jet engines is usually very high and thus the fuel mass can be neglected.

10.12.1 Propulsive Power kfr, Propulsive power is the power that is developed from the thrust of the engine and is given by Wp = TV„c = th(Ve — Vi)V„ (kW)

(1031)

where Vac is the velocity of the aircraft. The net work developed by a turbojet engine is zero. Therefore, we cannot define the efficiency of this engine in the same way as that for a stationary gas-turbine engine. Here the efficiency is defined as the ratio of output to input. The desired output here is WI, and the required input Qm is the thermal energy of the fuel released during the combustion process. Thus, the propulsive efficiency rip is given by gp _ propulsive power _ gip energy input

(10.32)

EXAMPLE 10.9 A turbojet aircraft flying with a velocity of 260 m/s is at an altitude where the atmospheric pressure and temperature are 55 kPa and —18°C, respectively. The compressor has a pressure ratio of 8, and the temperature of the gases at the turbine inlet is 1000°C. Air enters at the rate of 40 kg/s. Using the cold-air-standard assumptions, determine (a) the temperature and pressure of the gases at the turbine exit, (b) the velocity of the gases at the nozzle exit, and (c) the propulsive efficiency of the cycle. Solution The T—s diagram of the cycle is shown below.

1273.15 K

255.15X

Under the cold-air-assumptions, all the processes are assumed to be internally reversible, and the working fluid air is assumed to be an ideal gas with constant specific heats evaluated at room temperature, that is, C, = 1.0045 kJ/(kg K) and y= 1.4. The combustion process is replaced by a heat addition process from an external source.

300 Fundamentals of Engineering Thermodynamics To begin with, let us fmd T and P at all the states. Let us assume the aircraft to be stationary and the air to be moving over the aircraft with velocity of 260 m/s. Also, assume the velocity of the air leaving the diffuser to be negligibly small. Process 1-2 is isentropic compression of an ideal gas in a diffuser. Therefore, by energy equation, we have 1712 — WI2 = h2

hl

v2 v2 2

2 where the subscripts 1 and 2 refer to inlet and exit states of the diffuser. Also, q 12 = 0, w12 = 0, and V2 23 0. Thus, we have

V2

0 = Cp(7; — Ti ) —

2

or T = 2

+

V?•

2Cp

= 255.15 +

2602— 288.8 K 2 x 1004.5

and

r -

T2 P2 = ( )

288.8

= 55

0.4/1.4)

(255.15 )

= 84.85 kPa

Process 2-3 is isentropic compression of an ideal gas in a compressor, therefore, P3 = rp X P2 = 8 x 84.85 = 678.8 kPa

Also, P4 = P3, thus, we get

(r-Wr T3 = T2 I

= 288.88"1.4 = 523.15 K 2

Process 4-5 is isentropic expansion of an ideal gas in a turbine. (a) Neglecting the changes in kinetic energy across the compressor and turbine and assuming the turbine work to be equal to the compressor work, we can find T and P at the turbine exit as follows: Wccnnp,in = Wturb,out

Or h3 — h2 = h4 — h5 Or Cp( T3 — T2) =

T4 —

T5)

Or T5 = T4 — T3 + T2 = 1273.15 — 523.15 + 288.8

1038.8 K

171(7-1)

P5 — P4

(R

4

1038.8 )1.4/0.4

= 678.8 (

1273.15

333.06 kPa

Gas Power Cycles 301 (b) Process 5-6 is isentropic expansion of an ideal gas in a nozzle. Thus, )(7-0/r T6 =

T5 HP5

= 10

55 38.8 — 333.06

)0.4/1.4

Note that the nozzle exit pressure P6 is assumed to be the ambient atmospheric pressure. This implies correct expansion of flow at the nozzle exit, which is always true for subsonic flows. When the flow at the nozzle exit is transonic (i.e. the flow velocity is equal to the speed of sound) or supersonic (i.e. the flow velck tv is greater than the speed of sound), the exit pressure can be less than or equal to or greater than the ambient pressure depending on whether the flow through the nozzle is overexpanded or correctly expanded or underexpanded, respectively. Now, by energy equation, we have 956 - w56 = h6 "5 +

- V2 5 2

V2

6

But 1756 = 0, W56 = 0, and Vs = 0. 11111 we get 0 = Cp( T6 - T5)

V22

or V6

= -12Ce(T5 - T6) = J2 x 1004.5(1038.8 - 620.63) 916.57 m/s

The speed of sound is a = ITTT = . 4 x 287 x 255.15 = 320.19 m/s Thus, the nozzle exit velocity is more than a and hence, the flow is supersonic at the nozzle exit. (c) The propulsive power developed by the engine is th(Ve - Vi )Vac = 40(916.57 - 260)260 = 68,28,328 J/s = 6,828.328 kJ/s Now,

= ti(h4 - h3) =

mCp(T4

- T3)

= (40 x 1004.5)(1273.15 - 523.15) = 30,135,000 J/s = 30,135 kJ/s Therefore, The propulsive efficiency is

71P

ik _ 6828 • 328 - 0.2266 = 22.66 per cent 30135.00

an

302 Fundamentals of Engineering Thermodynamics

10.12.2 Modifications to Turbojet Engines Modifications to turbojet engines aim at combining the desirable characteristics of propeller engines and jet engines. Two such modifications are the propjet engine and the turbofan engine. The turbofan engines are the most widely used ones in aircraft propulsion. In turbofan engines, a large fan driven by the turbine forces a large amount of air through a duct surrounding the engine, as shown in Fig. 10.27. The fan exhaust leaves the duct at a higher velocity, enhancing the total thrust of the engine considerably. The ratio of the mass flow rate of the air bypassing the combustion chamber to that of the air flowing through it is called the bypass ratio. Increasing the bypass ratio of a turbofan engine increases its thrust. With this aim, the fan in the propjet engine is without the cowl. Thus, the turbofan and propjet engines differ primarily in their bypass ratio. The bypass ratio for turbofans is 5 or 6 whereas that for propjets is as high as 100. Low-pressure Low-pressure turbine compressor Fan

Burner

Duct

L.------=1'. Fan exhaust E) --1. Turbine exhaust r•-•••••":„.:::-..-11P

High-pressure compressor

High-pressure turbine

Figure 10.27 Schematic diagram of a turbofan engine.

Afterburner it is an additional section between the turbine and the nozzle. Gas turbine engines with this modification are used in military aircraft. In the afterburner section, the oxygen-rich combustion gases leaving the turbine are burnt by injecting additional fuel. As a result of this added energy, the exhaust gases leave at a higher velocity, increasing the thrust.

10.12.3 Ramjet Engine The ramjet engine is a properly shaped duct, as shown in Fig. 10.28. The pressure rise in the Fuel nozzles

Jet nozzle

Air inlet

Flame holders

Figure 10.28 A ramjet engine.

,

1a

Power Cycles

303

engine is achieved by the ram effect of the incoming high-speed air being rammed against the barrier. Even though it has the advantage of being without any moving or rotating parts, a ramjet engine needs to be accelerated to a sufficiently high speed by an external source before it can be started. Ramjet is employed for propulsion at very high speeds. It operates on the same principle as the turbojet, except that, instead of compressor, the flow duct shape itself is used to convert the kinetic energy of the flow into pressure. The incoming flow velocity Vi is reduced in the inlet portion of the duct. Fuel is sprayed to the low-speed air and the fuel-air mixture is burned at constant pressure. The combustion product at high temperature and pressure is expanded through the nozzle to a high velocity Ve at the nozzle exit. The net thrust generated by the engine can be expressed as the change in momentum flux as Thrust =

( Ve —

Vi )(N)

where m is the mass flow rate through the duct.

10.12.4 Scramjet Engine The scramjet engine is essentially a ramjet engine in which the combustion takes place at supersonic speeds.

10.12.5 Rocket The rocket is a device where a solid or liquid fuel and an oxidizer react in the combustion chamber. The combustion gases at high pressure are then expanded through a nozzle. The gases leave the rocket at very high velocities, producing the thrust to propel the rocket. The schematic of a typical liquid-fuel rocket and its T—s diagram are shown in Fig. 10.29.

Ve Exhaust

NOZZL,'

P=

T

constant

3 S

Figure 10.29 A liquid-fuel rocket and its process diagram.

1

1 1

304 Fundamentals of Engineering Thermodynamics The combustion product at high pressure and temperature are expanded through a nozzle to very high velocities. The momentum thrust generated by a rocket engine can be expressed as Thrust = mVe (N) The engine shown in Fig. 10.29 is a liquid rocket engine, which uses fuel and oxidizer in liquid state. Rockets can also be propelled with solid fuel and oxidizer. In solid fuel rockets, both fuel and oxidizer are usually mixed in required proportion and stored in the chamber where combustion is to take place.

10.12.6 Specific Impulse Specific impulse, defined as the fuel weight flow rate specific thrust, is a popularly used parameter to compare propulsion systems. It is defined as uninstalled thrust fuel weight flow rate

&rif f

(second)

where the uninstalled thrust F is the total thrust generated by the engine, assuming ideal external flow (for airbreathing engines, it is equal and opposite to the difference in momentum fluxes between the inlet and exit flows), go is the standard sea level value of the gravitational acceleration. Aside from the conversion of fuel mass flow rate to sea level fuel weight flow rate, specific impulse is merely the inverse of specific fuel consumption. However, both are used in practice to bring out different qualities of airbreathing engines. Specific fuel consumption is preferred when the issue is fuel economy, and specific impulse is preferred when the issue is adequate thrust.

10.13 SECOND-LAW ANALYSIS OF GAS POWER CYCLES The second-law analysis of gas power cycles reveals where the largest irreversibilities occur in the internally reversible cycles, such as Otto, Diesel, and Brayton cycles. The irreversibility for a process of a closed system which may be exchanging heat QR with a reservoir at TR can be expressed as QR II _ Tx I =ToSgen = 74 (Se — Si )sys + — lui k R [

(10.33)

where To is the ambient temperature. A sim-lar relation for steady-flow systems can be expressed, as = To4,„ = To (E rites, —

E

1-1 (kW)

TR

(10.34)

or i = Toss,. = To (se — si + H 411 (kJ/kg) TR

(10.35)

where the subscripts i and e refer to the inlet and exit states, respectively. We know that entropy is a property, and its value depends on the state only. For a cycle, reversible or actual, the initial and final states are identical, therefore si = se. Thus, the

Gas Power Cycles

305

irreversibility of a cycle depends on the magnitude of the heat transfer with the high and low temperature reservoirs and on their temperatures. That is, i = To

E

(kJ/kg)

(10.36)

For a cycle which involves heat transfer with only one source at TH and one sink at T1, we have i = To 9 T1

— kl ) (kJ/kg) TH

(10.37)

and the availability of a fluid stream, ix at any state can The availability of a closed system, be determined using Eqs. (6.14) and (6.26). Thus, we have = (u — uo) — To(s — so) + Po(v — vo) (kJ/kg)

(10.38)

and v2 = (h — ho) — To(s — so) + — + gz (kJ/kg) 2 where the subscript 0 refers to the surroundings.

(10.39)

EXAMPLE 10.10 Calculate the irreversibilities associated with all the four processes of the Otto cycle discussed in Example 10.3. Assume that heat is transferred to the working fluid from a heat source at 1750 K and heat is rejected to the surrounding atmosphere at 22°C. Solution From Example 10.3, we have To = 295.15 K

r =8 P, = 1814.31 kPa

T1 = 295.15 K

P3 = 4.627 MPa

T2 = 662.74 K T3 = 1690.18 K

q23 = 900 kJ/kg

Net = 467.14 kJ/kg

q41 = —432.86 kJ/kg

Processes 1-2 and 3-4 are isentropic, and therefore do not involve any irreversibilities, that is, il2 = 0 and i34 = 0. Processes 2-3 and 4-1 are constant volume heat addition and heat rejection, respectively. The entropy change associated with process 2-3 is s,—s, _ 0

0

R ln P2

= 3.5908 — 2.5142 — 0.287 In

4.627 L814

= 0.8079 kJ/(kg K) Also, 98,23 =

I II

iI

(123 = - 900 kJ/kg and TR = 1750 K

Ill

306

Fundamentals of Engineering Thermodynamics

Thus,

i23 = To E(s3 — s2

qR 23 ] 6 + --= TR

= 295.15 [0.80799°1 — 1750

= 86.660 kJ/kg For process 4-1, we have si — s4 = s2 — s3 = — 0.8079 kJ/(kg K) Also, qR,41 = — q41

= 432.86 kJ/kg and

TR =

295.15 K

Thus, i41 = 295.15 [— 0.8079 + 432.86] 295.15

= 194.41 kJ/kg Therefore, the irreversibility of the cycle is given by icyck = il2 + 123 + 134 + 141 = 0 -I-

86.660 + 0 + 194.41

= 281.07 kJ/kg EXAMPLE 10.11 A turbojet engine operates between the pressure limits of 35 kPa and 350 kPa. The inlet air temperature is —40°C and the upper temperature limit is 1370 K. Calculate the specific momentum thrust of the engine assuming isentropic compression and expansion and an inlet velocity of 100 m/s. Also, determine the heat input and the power delivered per unit mass. Take the gas to be equivalent to air and the velocity at the nozzle inlet to be negligible. Solution The schematic of the engine and the T— s diagram are as shown in the figure below. For the compression process, we have fr-1)/Y

T2 =T (_172_ li

0.2136

50 = 233 (1-) = 450 K 35 Since the compressor and turbine work are equal, we have h2 — hi

=

h3 — h4

Gas Power Cydes 307

44.43"

Inlet air

Exhaust 5

1 Nozzle Turbine

Compressor

T

i.e. T2 — = T3 — T4

or T4 = 7'3 — T2

+ = 1370 — 450 + 233

= 1153 K For the isentropic expansion, r,N0.286

-

Ts = T3 (L.5 P3 35 =1370 (350

0.286

= 707 K

For the nozzle flow, V2 V2 h4 + = h5 + 2 2 Given that,

V4 =

0. Thus, V5 = (114 — h5 ) =

Cp (7'4 — Ts)

= J2 x 1005 x (1153 — 7091 = 944.7 m/s

308

Fundamentals of Engineering Thermodynamics

The momentum thrust is given by 7h=th(VI -1/5 )± The specific thrust becomes (100 — 944.7) = —844.7 N

Th

The negative sign indicates that the thrust is in the direction opposite to the inlet flow. The heat addition per unit mass is qh =(h3 — h2)=Cp(T3

T2)

=1.005(1153 — 450) 706.52 kW The power delivered by the engine is Power = (Th)li = 844.7 x 100 = 84470 W = 84470/746 = 113.2 hp

SUMMARY In this chapter, we have discussed the cycles of actual devices along with their idealized models. Air-standard assumptions reduce the complexities associated with the analysis of actual gas power cycles to a manageable level. In addition to the air-standard assumptions, if the air is assumed to have constant specific heats whose values are determined at 25°C, the air-standard assumptions are called the cold-air-standard assumptions. The resulting cycle with these assumptions is called an air-standard cycle. The reciprocating engine is basically a piston-cylinder device. The piston reciprocates in the cylinder between two fixed positions called the top dead centre (TDC) and the bottom dead centre (BDC). At TDC and BDC the piston forms the smallest and the largest volumes in the cylinder, respectively. The distance between the TDC and BDC is called the stroke of the engine, which is the largest distance that the piston can travel in one-direction. The diameter of the piston is called the bore. The compression ratio r of the reciprocating engine is the ratio of the maximum volume formed in the cylinder to the minimum volume. =

V _ VBDc Vniin

VTDc

The mean effective pressure (MEP) is a fictitious pressure and is defined as that pressure which when acting on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle. The MEP is given by MEP —

V,

Wnet — Vnin

Wnet

vmax —

Vmin

(kPa)

Gas Power Cycles

309

A spark-ignition (SI) engine is that internal combustion engine in which the combustion of the air-fuel mixture is initiated by a spark. A compression-ignition (CI) engine is that internal combustion engine in which the combustion of the air-fuel mixture is by self-ignition as a result of compression of the mixture above its self-ignition temperature. The Otto cycle is the ideal cycle for SI engines. An Otto cycle resulting from the airstandard assumptions is called the ideal Otto cycle. The thermal efficiency of the Otto cycle can be expressed as th,Otto = 1

Y -1

where r is the compression ratio and y the specific heats ratio. The rhh.0,,,, increases with both r and y. The Thh.oito typically ranges from 50 to 60 per cent. Autoignition is premature ignition of the fuel due to high compression. The noise produced by autoignition is called engine knock. For SI engines, the compression ratio lies typically between 7 and 10. The Diesel cycle is the ideal cycle for CI reciprocating engines. The diesel engines can have higher compression ratios, typically between 12 and 24. There is no possibility of autoignition in CI engines, since only air is compressed in the engine. The thermal efficiency of diesel engines can be expressed as 1 [ rr 1 qth,Diesel = 1 —

rr -I

)1(r,

1)

where r is the compression ratio and re the cutoff ratio. The cutoff ratio is defined as the ratio of the cylinder volumes after and before the combustion process. From Eqs. (10.11) and (10.1 I. it is evident that qth,Otto > qth.Diesel

The .,tith, Die„i ranges from 35 to 40 per cent. The Dual cycle is a combination of Otto and Diesel cycles. Both the Stirling and Ericsson cycles are cycles involving an isothermal heat addition process at TH and an isothermal heat rejection process at TL. Both these cycles are totally reversible like the Carrot cycle, and thus have the same thermal efficiency as that of the Carrot cycle when operating between the same temperature limits. Thus, hth,Stirting = hth,Ericsson = hth.Carnot =

I

TL TH

The Brayton cycle is the ideal cycle for gas-turbine engines. The efficiency of the Brayton cycle can be expressed as gth,Brayton = 1

r (r

1 - pir

where rp = P,/Pi is the pressure ratio and y the specific heats ratio.

310

Fundamentals of Engineering Thermodynamics

The regenerator used in the Brayton cycle is a counterflow heat exchanger which heats the high-pressure air leaving the compressor, by transferring heat to it from the hot exhaust gases leaving the turbine. The thermal efficiency of an ideal Brayton cycle with regeneration, under cold-air-standard assumptions is rith,,g = 1— (a)frP)(r -ivr T3 A jet-propulsion cycle is an open cycle on which aircraft gas-turbine engines operate. It differs from the ideal Brayton cycle in the fact that the combustion gases are not expanded to the ambient pressure in the turbine. Instead, the gases are expanded to a pressure such that the power produced by the turbine is just sufficient to run the compressor and the auxiliary equipment. The propulsive thrust for aircraft is accomplished either by slightly accelerating a large mass of air in propeller engines or by largely accelerating a small mass of air in turbojet engines. Both these principles are combined in the turboprop engine. Propjet and turbofan engines are modified turbojet engines. These modifications aim at combining the desirable characteristics of propeller engines and jet engines. An afterburner is an additional section between the turbine and the nozzle. In the afterburner section, the oxygen-rich combustion gases leaving the turbine are burnt by injecting additional fuel in order to gain some additional thrust. The ramjet engine is a properly shaped duct. The pressure rise in the engine is achieved by the ram effect of the incoming high-speed air being rammed against the barrier. The scramjet engine is essentially a ramjet engine in which the combustion takes place at supersonic speeds. The rocket is a device where a solid or liquid fuel and an oxidizer react in the combustion chamber. The high pressure combustion gases exiting at high speeds through a nozzle produce the thrust to propel the rocket.

PROBLEMS 10.1 An air-standard cycle executed in a closed system consists of the following processes: 1-2 isentropic compression from 101 kPa and 300 K to 606 kPa. 2-3 constant volume heat addition to 1600 K. 3-4 isentropic expansion to 101 kPa. 4-1 constant pressure heat rejection to 300 K. Show the cycle on T—s and P—v diagrams. Calculate the net work output per unit mass and the thermal efficiency of the cycle, assuming air to be an ideal gas with variable specific heats. [Ans. 452.38 kJ/kg, 48.1 per cent] 10.2 An air-standard Carrot cycle is executed in a closed system between the temperature limits 330 K and 1600 K. The minimum pressure of the cycle is 101 kPa and the heat added is 300 kJ/kg. Determine the thermal efficiency, the compression ratio, the pressure ratio, the maximum pressure, and the mean effective pressure of the cycle. Treat air to be an ideal gas with variable specific heats. [Ans. 79.4 per cent, 84.32, 408.85, 41293.85 kPa, 256.99 kPa]

I

Gas Power Cycles 311 10.3 The overall volume expansion ratio of a Carrot cycle is 25. If the temperature limits of the cycle are 700°C and 20°C, determine the volume ratios of the isothermal and isentropic processes. Also, determine the thermal efficiency of the cycle. Take y= 1.4. [Ans. 1.245, 20.08, 69.9 per cent] 10.4 Consider a Carrot cycle executed in a closed system with 1 kg of air. The initial temperature and pressure are 300°C and 1.8 MPa, respectively. From the initial state, the air is expanded isothermally to three times its initial volume and then further expanded adiabatically to six times its initial volume. Then, it is compressed in two stages by isothermal compression, followed by adiabatic compression, to the initial state. Determine (a) the pressure, volume, and temperature at the end of each process, (b) the thermal efficiency of the cycle, and (c) the work done by the cycle. [Ans. (a) P2 = 0.6 MPa, V2 = 0.2742 m3/kg, T3 = 161.22°C, P3 = 227.36 kPa, V3 = 0.5484 m3/kg, P4 = 682.08 kPa, V4 = 0.1828 m3/kg, (b) 24.2 per cent, (c) 43.73 Id] 10.5 An air-standard constant pressure cycle has the overall volume ratio of 8:1. The volume ratio of the adiabatic compression is 6:1. The initial state at the beginning of the compression is at 25°C, 100 kPa, and 0.85 m3. (a) Show the cycle on a P—v diagram. Calculate (b) the net work output per unit mass, and (c) the thermal efficiency. [Hint. The constant pressure cycle consists of the following processes: 1-2 adiabatic compression with PIP = constant; 2-3 constant pressure heat addition; 3-4 adiabatic expansion with PVr = constant; and 4-1 constant pressure heat rejection]. [Ans. (b) 103.86 kJ/kg, (c) 51.17 per cent] 10.6 A gas turbine operates on a simple air-standard constant pressure cycle. The pressure compression ratio is 8. The actual thermal efficiency of the turbine is 80 per cent of the ideal thermal efficiency. The fuel used has a calorific value of 40 MJ/kg. Assuming y= 1.4, determine (a) the actual thermal efficiency of the cycle, and (b) the specific fuel consumption of the gas turbine in kg/(kWh). [Ans. (a) 35.84 per cent (b) 0.251 kg/(kWh)] 10.7 An air-standard constant volume cycle is executed in a closed system. The cylinder has a bore of 80 mm and a stroke of 85 mm. The clearance volume is 0.06 litre. (a) Show the cycle on a P—v diagram, and (b) calculate the thermal efficiency of the cycle. Assume y = 1.4. [Hint. The constant volume cycle consists of adiabatic compression, constant volume heat addition, adiabatic expansion, and constant volume heat rejection.] [Ans. 56.7 per cent] 10.8 The pressures at the beginning and end of the adiabatic compression of a constant volume cycle are 101 kPa and 1.8 MPa, respectively. Determine the overall volume ratio and the thermal efficiency of the cycle. Take y= 1.4. [Ans. 7.83, 56.09 per cent] 10.9 An air-standard cycle executed in a closed system consists of the following processes: 1-2 constant volume heat addition of 800 kJ/kg from 100 kPa and 300 K 2-3 constant pressure heat addition to 1800 K. 3-4 isentropic expansion to 100 kPa. 4-1 constant pressure heat rejection to the initial state.

T1

312 Fundamentals of Engineering Thermodynamics (a) Show the cycle on P—v and T—s diagrams. Calculate (b) the total heat input per unit mass, and (c) the thermal efficiency of the cycle. Account for the variation of specific heats with temperature. [Ans. (b) 1548.96 kJ/kg, (c) 29.1 per cent] 10.10 In an air-standard cycle, heat is supplied at constant volume to raise the temperature of the air from T1 to T2. The air then expands isentropically until its temperature falls to T3. It is then returned to its original state through a reversible isothermal compression process. Draw the cycle on P—v and T—s diagrams and show that the efficiency can be expressed as lith

-1

TI

in T

T2 - T

T

Calculate the efficiency of the cycle if the pressure rises from 1 MPa to 3.5 MPa during the heat addition process. [Ans. 49.9 per cent] 10.11 An air-standard cycle is executed in a closed system and consists of the following four processes: 1-2 isentropic compression from 101 kPa and 15°C to 505 kPa. 2-3 heat addition of 2600 id/kg at constant volume. 3-4 isentropic expansion to initial volume. 4-1 heat rejection at constant volume. (a) Show the cycle on P—v and T—s diagrams. Calculate (b) the cycle efficiency, (c) the mean effective pressure, and (d) the peak pressure. Take y= 1.4. [Ans. (b) 36.8 per cent, (c) 1.71 MPa, (d) 4.511 MPa] 10.12 Consider a Carrot cycle in which 1 kg of air is compressed isothermally from 100 kPa to 400 kPa, at a constant temperature of 10°C. The maximum temperature reached in the cycle is 400°C. Determine the cycle efficiency, the peak pressure, and the mean effective pressure of the cycle. [Ans. 57.94 per cent, 8288 kPa, 372.96 kPa] 10.13 An air-standard cycle is executed in a closed system and consists of the following four processes: 1-2 isentropic compression from 102 kPa and 15°C to 612 kPa. 2-3 constant pressure heat addition of 485 kJ/kg. 3-4 isentropic expansion to 102 kPa. 4-1 constant pressure heat rejection to the initial state. (a) Show the cycle on a T—s diagram. Calculate (b) the net work output per unit mass, and (c) the thermal efficiency. [Ans. (b) 194.32 U/kg, (c) 40.07 per cent] 10.14 Consider a petrol engine working on an ideal air-standard Otto cycle. The cylinder bore is 60 mm, the stroke length is 85 mm, and the clearance volume is 25 cm3. Determine the cycle efficiency. [Ans. 61.12 per cent]

Gas Power Cycles 313

10.15 In an air-standard Otto cycle, the maximum and minimum temperatures are 1200°C and 20°C, respectively. The heat supplied is 700 kJ/kg. Determine the compression ratio, the cycle efficiency, and the ratio of the maximum to minimum pressures in the cycle. [Ans. 3.75, 41.06 per cent, 18.84] 10.16 An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, the air is at 101 kPa and 300 K. The maximum pressure in the cycle is 8 MPa. Determine the cycle efficiency, heat transferred to the air during the constant-volume heat addition process and the mean effective pressure. [Ans. 54.08 per cent, 1965.2 kJ/kg, 1455.37 kPa] 10.17 An ideal Otto cycle has a compression ratio of 9. At the beginning of the compression process the air is at 100 kPa and 300 K, and the cylinder volume is 8000 cm3, and 7.5 kJ of heat is added during the heat-addition process. Accounting for variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure which occur during the cycle, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure of the cycle. [Ans. (a) 1624.2 K, 4872.6 kPa, (b) 435.85 kJ/kg, (c) 54.05 per cent, (d) 569.5 kPa] 10.18 A four-cylinder petrol engine working on an air-standard Otto cycle has a swept volume of 2000 cm3, and the clearance volume in each cylinder is 60 cm3. Determine the cycle efficiency. If the air at the beginning of the compression stroke is at 100 kPa and 300 K, and the maximum cycle temperature is 1650 K, determine the mean effective pressure of the cycle. [Ans. 59.07 per cent, 505.6 kPa] 10.19 An air-standard Diesel cycle has a compression ratio of 16. At the beginning of the compression process, air is at 90 kPa and 40°C. The maximum temperature of the cycle is 1400°C. Determine (a) the cutoff ratio, (b) the thermal efficiency, and (c) the mean effective pressure of the cycle. [Ans. (a) 1.76, (b) 62.71 per cent, (c) 487.07 kPa] 10.20 An air-standard Diesel cycle has a compression ratio of 15. Air is at 100 kPa and 27°C at the beginning of the compression process. The volume ratio of the adiabatic expansion is 7.5:1. The pressure at the end of adiabatic expansion is 250 kPa. Determine (a) the maximum temperature attained during the cycle, and (b) the thermal efficiency of the cycle. [Ans. (a) 1680 K, (b) 59.46 per cent] 10.21 An ideal diesel engine has a compression ratio of 15 and uses air as the working fluid. The constant pressure energy addition ceases at 10 per cent of the stroke. The state of the air at the beginning of the compression process is 95 kPa and 20°C. The engine uses 0.05 m3 of air per second. Determine (a) the thermal efficiency of the cycle, and (b) the indicated power of the cycle. [Ans. (a) 58.44 per cent, (b) 40.21 kW] 10.22 The compression ratio of an ideal dual cycle is 15. Air is at 101 kPa and 22°C at the beginning of the compression process and at 2000 K at the end of the heat addition process. Heat transfer to air takes place partly at constant volume and partly at constant pressure, and it amounts to 1000 kJ/kg. Assuming constant specific heats for air, determine (a) the fraction of heat transferred at constant volume, and (b) the thermal efficiency of the cycle. [Ans. (a) 332.9 Id/kg, (b) 79.84 per cent]

314 Fundamentals of Engineering Thermodynamics 10.23 The temperature and pressure at the beginning of compression in an air-standard Diesel cycle are 300 K and 101 kPa, respectively. The compression ratio is 19 and the amount of heat addition is 2000 kJ/kg. Determine (a) the maximum pressure and temperature of the cycle, (b) the thermal efficiency, and (c) the mean effective pressure. [Ans. (a) 6231.27 kPa, 974.14 K, (b) 59.62 per cent, (c) 1476.54 kPa] 10.24 An ideal dual cycle has a compression ratio of 15. Air is at 97 kPa, and 0.084 m3, and 28°C at the beginning of compression. The maximum pressure and temperature of the cycle are 6.2 MPa and 1320°C, respectively. Determine (a) the net work done, (b) the thermal efficiency, and (c) the mean effective pressure of the cycle. [Ans. (a) 36.58. kJ, (b) 65.34 per cent, (c) 466.58 kPa] 10.25 An ideal Ericsson cycle, using air and including regeneration, has air at 100 kPa, 20°C, and 0.08 m3 at the beginning of isothermal compression. The volume becomes 1/5th of the initial value at the end of compression. After isothermal compression, the volume of the air is doubled during the constant pressure process. Determine (a) the maximum temperature, (b) the net work done, (c) the ideal thermal efficiency, and (d) the thermal efficiency if the process of regeneration was not included. [Ans. (a) 586.3 K, (b) 12.86 kJ, (c) 50 per cent, (d) 23.84 per cent] 10.26 An ideal Stirling cycle with regeneration, using air as the working fluid, has air at 110 kPa, 0.05 m3, and 30°C at the beginning of compression. The minimum volume of the cycle is 0.005 m3. The maximum temperature of the cycle is 700°C. Determine (a) the net work done, (b) the ideal thermal efficiency, and (c) the thermal efficiency if the process of regeneration was not included. [Ans. (a) 27.98 kJ, (b) 68.85 per cent, (c) 39.4 per cent] 10.27 An ideal Ericsson cycle using helium as the working fluid operates between the temperature limits of 290 K and 1600 K and pressure limits of 101 kPa and 800 kPa. Assuming a mass flow rate of 5 kg/s, determine (a) the thermal efficiency of the cycle, (b) the heat transfer rate in the regenerator, and (c) the power delivered. [Ans. (a) 81.88 per cent, (b) 33999 kW, (c) 28154.45 kW] 10.28 A gas-turbine engine working on an air-standard Brayton cycle operates between the temperature limits of 300 K and 1200 K and pressure limits of 101 kPa and 505 kPa. Determine (a) the thermal efficiency of the cycle, (b) the compressor work, (c) the turbine work, and (d) the air-flow rate required for 2.0 kW of net power output. [Ans. (a) 36.86 per cent, (b) -175.94 kJ/kg, (c) 444.33 kJ/kg, (d) 13.41 kg/h] 10.29 For the gas turbine in Problem 10.28, if the adiabatic efficiencies of the turbine and compressor are 85 per cent and 80 per cent, respectively, determine (a) the thermal efficiency of the cycle, (b) the compressor and turbine work, (c) the turbine temperature at the exit and the rate of air flow required for 2.0 kW of net power output. [Ans. (a) 26.4 per cent, (b) -219.93 kJ/kg and 377.68 kJ/kg, (c) 824 K and 45.64 kg/h] 10.30 Air enters the compressor of a gas-turbine engine at 30°C and 95 kPa, where it is compressed to 670 kPa and 300°C. Heat transferred to the air is 1000 kJ/kg before it enters the turbine. For a turbine efficiency of 82 per cent, determine (a) the fraction of the turbine work output used to drive the compressor, and (b) the thermal efficiency of the engine. [Ans. (a) 49.08 per cent, (b) 34.16 per cent]

Gas Power Cycles 315 10.31 An ideal Brayton cycle with regeneration has a pressure ratio of 8. The air enters the compressor at 20°C and the turbine at 1000°C. If the effectiveness of the regenerator is 100 per cent, determine the net work output and the thermal efficiency of the cycle (a) assuming constant specific heats for air at room temperature, and (b) taking the variation of specific heats with temperature into account. [Ans. (a) 335.11 kJ/kg, 58.45 per cent, (b) 354.5 kJ/kg, 59.74 per cent] 10.32 An air-standard Brayton cycle operates with a pressure ratio of 6. Air enters the compressor at 95 kPa and 290 K and leaves the combustion chamber at 1100 K. Determine the compressor work input, the turbine work output, and the thermal efficiency of the cycle. Take into account the variation of specific heats with temperature. [Ans. 194.52 kJ/kg, 454.57 kJ/kg, 38.45 per cent] 10.33 An air-standard gas-turbine engine working on a Brayton cycle operates under the conditions described in Problem 10.32, except that the compressor and the turbine are only 80 and 85 per cent efficient, respectively. Determine the compressor work input, the turbine work output, and the thermal efficiency of the cycle, taking into account the variation of specific heats with temperature. [Ans. 243.15 kJ/kg, 385.96 kJ/kg, 22.75 per cent] 10.34 For the air-standard Brayton cycle described in Problem 10.32, determine the thermal efficiency when an ideal regenerator is introduced into the cycle. [Ans. 51.20 per cent] 10.35 For the gas-turbine engine described in Problem 10.33, determine the effect on the thermal efficiency when a regenerator of 75 per cent effectiveness is introduced into the cycle. [Ans. 31.97 per cent] 10.36 Consider and ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each stage is 2.45. Air enters each stage of the compressor at 300 K and each stage of the turbine at 1100 K. Determine the back work ratio and the thermal efficiency of the cycle, (a) when no regenerator is used, and (b) when a regenerator of 100 per cent effectiveness is used. Take into account the variation of specific heats with temperature. [Ans. (a) 0.3557, 31.9 per cent, (b) 0.3557, 64.43 per cent] 10.37 If the efficiencies of the two compressor stages are 75 per cent each and those of the two turbine stages are 80 per cent each for the gas-turbine cycle in Problem 10.36, repeat the problem. [Ans. (a) 0.5928, 17.5 per cent, (b) 0.5928, 40.72 per cent] 10.38 A turbojet aircraft is flying with a velocity of 250 m/s at an altitude of 6000 m, where the ambient conditions are 47 kPa and -24°C. The pressure ratio across the compressor is 12, and the temperature at the turbine inlet is 1200 K. Assuming ideal operation for all components and constant specific heats for air at room temperature, determine (a) the pressure at the turbine exit, (b) the velocity of the exhaust gases, and (c) the propulsive efficiency. [Ans. (a) 323.61 kPa, (b) 880.3 m/s, (c) 24.9 per cent]

316

Fundamentals of Engineering Thermodynamics

10.39 Air at 15°C enters a turbojet engine at a rate of 22 kg/s and a velocity of 310 m/s. Air is heated in the combustion chamber at a rate of 25,000 Icl/s and it leaves the engine at 400°C. Determine the thrust produced by the engine considering the entire engine as the control volume. [Ans. 20966.22 N] 10.40 Determine the total irreversibilities associated with the Otto cycle described in Problem 10.16. Assume a source temperature of 3000 K and a sink temperature of 310 K. Also, determine the availability at the end of the power stroke. [Ans. 705.82 kJ/kg, 507.54 kJ/kg] 10.41 Determine the irreversibilities associated with each of the processes of the Otto cycle described in Problem 10.17, assuming a source temperature of 1600 K and a sink temperature of 300 K. Also, determine the availability at the end of the isentropic expansion process. [Ans. i12 = 0, i23 = 64.425 kJ/kg, i34 = 0, i41 = 100.95 kJ/kg, 100.945 kJ/kg] 10.42 Determine the total irreversibility associated with the Diesel cycle described in Problem 10.19. Assume a source temperature of 1800 K and a sink temperature of 40°C. Also, determine the availability at the end of the isentropic compression process. [Ans. 144.6 kJ/kg, 372.18 kJ/kg] 10.43 Determine the total irreversibility associated with the Brayton cycle described in Problem 10.28. Assume a source temperature of 1500 K and a sink temperature of 300 K. Also, determine the availability of the working fluid at the turbine exit. Assume Pexi, = Po = 101 kPa. [Ans. 314.1 kJ/kg, 172.81 kJ/kg] 10.44 An air-standard Brayton cycle receives air at 101 kPa and 20°C. The upper pressure and temperature limits of the cycle are 410 kPa and 820°C, respectively. The compressor and turbine efficiencies are 85 and 90 per cent, respectively. Calculate the thermal efficiency of the cycle, assuming constant specific heats. [Ans. 24.58 per cent] 10.45 A regenerator of efficiency 80 per cent is installed on the gas-turbine cycle of Problem 10.44. Calcualte the increase in thermal efficiency of the cycle achieved with the regenerator, assuming constant specific heats. [Ans. 18.6 per cent]

CHAPTER

11 Refrigeration Cycles 11.1 INTRODUCTION Refrigeration, which is the transfer of heat from a lower temperature region to a higher temperature one, is one of the major application areas of thermodynamics. Devices that produce refrigeration are called refrigerators or heat pumps, and the cycle on which they operate is called the refrigeration cycle. The vapour-compression refrigeration cycle is the most frequently used refrigeration cycle, in which the working fluid called the refrigerant is vaporized and condensed alternately and compressed in the vapour phase. The gas refrigeration cycle, in which the refrigerant remains in the gaseous phase throughout, is another popular refrigeration cycle. The other refrigeration cycles are cascade refrigeration, where more than one refrigeration cycle is used; absorption refrigeration, where the refrigerant is dissolved in a liquid before it is compressed, and thermoelectric refrigeration, where refrigeration is produced by the passage of the electric current through two dissimilar metals.

11.2 REFRIGERATORS AND HEAT PUMPS We know that heat flows from high-temperature regions to low-temperature ones, spontaneously. However, the reverse process cannot occur on its own, and requires refrigerators for its occurrence. Refrigerators are cyclic devices. The heat pump is another device which transfers heat from a low-temperature medium to a high-temperature one. The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it. Discharging heat to a higher temperature medium is merely a necessary part of the operation and not the purpose. However, the objective of a heat pump is to maintain a heated space at a high temperature. That is, the refrigerators and heat pumps are essentially the same devices; they differ only in their objectives. The coefficient of performance (COP) of refrigerators and heat pumps is respectively defined as = output _ cooling effect _ QL COPR input work input Wnet,in 317

318 Fundamentals of Engineering Thermodynamics COPHp =

output heating effect work input input

QH Wnetin

(112)

where QL is the heat absorbed from the refrigerated space and QH is the heat delivered to the heated space. Both COPR and COPHp can be greater than 1. From Eqs. (11.1) and (11.2), we can also infer that, for given values of QL and QH, COPHp = COPR + 1

(113)

This relation implies that COPie > 1, since COPR is positive. That is, a heat pump will function as a resistance heater, in the worst condition, supplying as much energy to the area to be kept warm, as it consumes. The cooling capacity of a refrigeration system that can freeze 1 ton of liquid water at 0°C into ice at 0°C in 24 hours is said to be 1 ton, and this is equivalent to 211 kJ/min.

11.3 THE REVERSED CARNOT CYCLE We know that all four processes of the Carrot cycle can be reversed, having heat and work interactions. The reversed cycle which operates in the counterclockwise direction is called the reversed Carnot cycle. A refrigerator or heat pump which operates on the reversed Carrot cycle is called a Carnot refrigerator or a Carnot heat pump. Let us examine a reversed Carrot cycle, executed within the saturation dome of a refrigeration cycle, as shown in Fig. 11.1. This is the most efficient refrigeration cycle operating between two specific temperature levels. In the actual refrigeration cycles, the two isothermal

T

Turbine

Compressor

Figure 11.1 Schematic of a reversed Carrot cycle.

Refrigeration Cycles 319

heat transfer processes 1-2 and 3-4 of the Camot refrigeration cycle can be achieved, since maintaining a constant pressure automatically fixes the temperature of a two-phase mixture at the saturation value. However, the process 2-3 involves the compression of a liquid-vapour mixture which requires a compressor that will handle two phases, and 4-1 involves the expansion of the high moisture-content refrigerant. However, the reversed Carnot cycle can serve as a standard for comparison of actual refrigeration cycles. The coefficients of performance of Carnot refrigerators and heat pumps were determined in Section 4.15 to be 1 C704..cmmm (11.4) (41 /7L) — 1 C°P1 IP,Camot

(11.5)

1 — (TL/TH )

Note that an increase in TL or a decrease in TH results in an increase in COP of both refrigerators and heat pumps.

11.4 THE IDEAL VAPOUR-COMPRESSION REFRIGERATION CYCLE In the ideal vapour-compression refrigeration cycle, the refrigerant is completely vaporized before it is compressed and the turbine is replaced with a throttling device. Because of these changes, many of the impracticalities associated with the reversed Carnot cycle are eliminated. This is the most widely used cycle in refrigerators, air-conditioning systems, and heat pumps. The cycle is composed of four processes, namely isentropic compression (1-2) in a compressor, constant pressure heat rejection (2-3) in a condenser, throttling (3-4) in an expansion device, and constant pressure heat absorption (4-1) in an evaporator, as shown in Fig. 11.2. The freezer Warm environment T Saturated liquid

Condenser

Valve Q 4—

Compressor

In

win

Evaporator C)1_

Cold refrigerated space

Figure 11.2

Saturated vapour

The ideal vapour-compression refrigeration cycle and its process diagram.

320 Fundamentals of Engineering Thermodynamics compartment, where heat is absorbed by the refrigerant, serves as the evaporator, and the coils behind the refrigerator where heat is dissipated to the air serve as the condenser, in a household refrigerator. In Fig. 11.2, the area under the process curve (4-1) and that under (2-3) represent the heat absorbed and heat rejected, respectively. Another diagram commonly used in analysis of vapour-compression refrigeration cycles is the P—h diagram, as shown in Fig. 11.3. It is P

Figure 11.3 The P—h diagram of an ideal vapour-compression refrigeration cycle. important to note that unlike other ideal cycles, the ideal vapour-compression refrigeration cycle is not an internally-reversible cycle, since it involves a throttling process, which is irreversible. This process is maintained in this model to make it a more realistic model for the actual vapourcompression refrigeration cycle. If the throttling device is replaced by an isentropic turbine, the refrigerant would enter the evaporator at state 4' instead of state 4 (Fig. 11.2). This would result in an increase in refrigeration capacity (area under 4-4') thereby decreasing the net work input. However, this change will prove to be more expensive and complex rather than an advantage. All the four devices in this cycle are steady-flow devices and Ake and Ape of the refrigerant can be neglected in comparison with the work and heat transfer terms. Thus, by energy equation we have q — w = he — hi (11.6) The COPs of the refrigerators and heat pumps operating on the vapour compression cycle can be expressed as ht COPR = qL— — 114 (11.7) Wnet,in h2 - hl and qH h2 — h3 COP" = (11.8) %col h2 where h1 = hg at P1 and h3 = hi. at P3 for the ideal case. EXAMPLE 11.1 A refrigerator working on an ideal vapour-compression refrigeration cycle uses refrigerant-12 as the working fluid. The minimum and the maximum pressures of the cycle are 0.15 MPa and

Refrigeration Cycles 321

0.9 MPa, respectively. If the mass flow rate of the refrigerant is 0.045 kg/s, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the COP of the refrigerator. Solution The refrigeration cycle is shown in the figure below. In an ideal vapour-compression refrigeration cycle, the compression process is isentropic, and the refrigerant enters the compressor as a saturated vapour at the evaporator pressure and leaves the condenser as saturated liquid at the condenser pressure. T

The enthalpies of the refrigerant at all four states obtained from the refrigerant-12 table (from Thermodynamics: An Engineering Approach, by Cengel, Y.A., and M.A. Boles.) are: At Pi = 0.15 MPa, h1 = [email protected] MPa = 178.64 kJ/kg, s1 = [email protected] MPa = 0.7089 kJ/(kg K) At P2= 0.9 MPa and s2 = s1) h2 = 210.43 kJ/kg At P3 = 0.9 MPa, h3 = [email protected] mpa = 71.93 kJ/kg h4 = h3 (throttling), that is, h4 = 71.93 kJ/kg (a) The rate of heat removal and the power input to the compressor, respectively, are given by QL = th

— h4) and

= m (h2 — h1)

Thus,

QL = 0.045(178.64 — 71.93) = 4.80 kW and = 0.045 (210.43 — 178.64) = 1.43 kW

322

Fundamentals of Engineering Thermodynamics

(b) The rate of heat rejection from the refrigerant to the environment is

OH = m (h2 Also,

- h3) = 0.045(210.43 — 71.93) = 6.23 kW

OH is given by QH = QL + gi = 4.80 + 1.43 = 6.23 kW

(c) COPR

_ 01, _ 4.80 Win 1.43

3.36

11.5 THE ACTUAL VAPOUR-COMPRESSION REFRIGERATION CYCLE The irreversibilities, owing to fluid friction and heat transfer to and from the surroundings and so on, that occur during the processes make the actual vapour-compression refrigeration cycle differ from the ideal one in several ways. The actual cycle and its process diagram are shown in Fig. 11.4. T

4

z Expansion valve —6

Evaporator

Figure 11.4 The actual vapour-compression refrigeration cycle and its process diagram. The refrigerant in the ideal cycle, is in saturated vapour state while leaving the evaporator as well as while entering the compressor. In practice, it is not possible to control the state of the refrigerant this precisely. Instead, the refrigerant is slightly superheated at the compressor inlet. This ensures that the refrigerant is completely in the vapour state while entering the compressor. Further, the connecting line between the evaporator and the compressor causes pressure drop because of friction. Also, the heat transfer from the surroundings to the refrigerant can be considerable. All these cause the irreversibilities.

Refrigeration Cycles

323

The compression process In an ideal cycle the compression process is internally reversible and adiabatic, that is, isentropic, but in an actual cycle it is not. But the nonisentropic compression process 1-2' may he desirable, since the specific volume of the refrigerant and thus the work input required are smaller in this case. Therefore, the refrigerant should be cooled during the compression process whenever possible. The refrigerant is assumed to leave the condenser as saturated liquid at the compressor exit pressure, in the ideal case. This is not possible in practice because of the pressure drop in the connecting lines. Further, it is easy to precisely execute the condensation process so that the refrigerant is a saturated liquid at the end. Also, it is undesirable to route the refrigerant to the throttling valve before it is completely condensed. Therefore, the refrigerant is subcooled before it enters the throttle valve. EXAMPLE 11.2 In a refrigerator, refrigerant-12 enters the compressor as superheated vapour at 0.18 MPa and —10°C at a rate of 0.045 kg/s and leaves at 0.9 MPa and 60°C. The refrigerant is cooled in the condenser to 30°C and 0.8 MPa, and is throttled to 0.20 MPa. Neglecting any heat transfer and pressure drops in the connecting lines between the components, determine (a) the rate of heat removal from the refrigerated space and power input to the compressor, (b) the adiabatic efficiency of the compressor, and (c) the coefficient of performance of the refrigerator. Solution The

T—s diagram of the cycle described is shown in the figure below.

Wn

S State 1:

P1

= 0.18 MPa, T1 = —10°C

From the superheated refrigerant-12 table, h1 = 181.03 kJ/kg State

2a: P2, = 0.9 MPa,

T2, = 60°C

Therefore, h2a = 219.37 kJ/kg State

3: P3 = 0.8 MPa, T3 = 30°C

324 Fundamentals of Engineering Thermodynamics Therefore, h3 = hpo3c 0c = 64.59 kJ/kg

and h4 -= h3 (throttling) = 64.59 kJ/kg

(a)QL = /is (hi — h4 ) = 0.045(181 . 03 — 64 . 59) = 5.24 kW *L. = m (hza — hi) = 0.045(219.37 — 181.03) = 1.73 kW (b) The adiabatic efficiency of the compressor is given by h c

h2 hza

-h1

where h2 is the enthalpy at State 2. At State 2, we have P2 = 0.9 MPa, s2 = si = 0.7181 kJ/(kg K) and therefore, h2 = 213.4 kJ/kg

Thus, 213.40 — 181.03 — 0.844 = 84.4 per cent 219.37 — 181.03 (c)

.

a = 5.24 W. 1.73

. COr-R = ,

3.03

11.6 THE REFRIGERANT SELECTION The important parameters that need to be considered in the selection of a refrigerant are the temperature of the refrigerated space and the environment. For a reasonable heat transfer rate, a temperature difference of 10°C or more should be maintained between the refrigerant and the environment. For example, if a refrigerated space is to be maintained at —10°C, the temperature of the refrigerant should be at about —20°C while it absorbs heat in the evaporator. The lowest pressure in a refrigeration cycle occurs in the evaporator, and this pressure should be above the atmospheric pressure to prevent any air leakage into the refrigeration system. Therefore, in this particular case, the refrigerant should have a saturation pressure of 1 atm or higher at —20°C. Refrigerant-12, called Freon-12, and ammonia are the two substances that have these qualities, and hence are the most widely used refrigerants.

11.7 HEAT PUMP SYSTEMS Heat pumps are used for cooling a large area, like a room, in summer, or for warming a large area in winter. That is, they are used in places which have a large cooling load or a relatively small heating load. However, in places where the heating load is significant and the cooling load is small, the heat pumps are not competitive. Heat pumps and air-conditioners have the same mechanical components. Therefore, it is economical to use the same system as a heat pump in winter and an air-conditioner in summer. This is accomplished by adding a reversing valve to the cycle, as shown in Fig. 11.5. The COP of heat pumps usually ranges between 1.5 and 4, depending on the system used and the temperature of the source.

Refrigeration Cycles 325

Heat pump operation: Heating mode Outdoor Reversing coil valve

Indoor coil

High pressure vapour Low pressure liquid-vapour -.—.— Low pressure vapour -0--ci— High pressure liquid

Expansion valve Cooling mode

Outdoor coil

Indoor coil

Figure 11.5 A heat pump cycle with a reversing valve.

11.8 INNOVATIVE VAPOUR-COMPRESSION REFRIGERATION SYSTEMS Even though the ordinary vapour-compression refrigeration systems are simple, reliable, less expensive, and practically maintenance-free, they are inadequate for large industrial applications. For such applications, the cycle need to be modified. Some such modifications are described in subsections below.

11.8.1 Cascade Refrigeration Cycle A cascade refrigeration cycle is that in which two or more refrigeration cycles operate in series to perform the refrigeration process in stages. Figure 11.6 shows a two-stage cascade refrigeration cycle. The heat exchanger that connects the cycles A and B serves as the evaporator for the upper cycle A and as the condenser for the lower cycle B. The heat transfer from the fluid in cycle A should be equal to that in cycle B, assuming that the heat exchanger is well insulated and Ake and Ape are negligible. Thus, the ratio of mass flow rates through the cycles should be fil A(hs - ha) = hi a(h2 - h3) Or

ihA _ h2 - h3 the

hs - ha

(11.9)

326 Fundamentals of Engineering Thermodynamics

T

Decrease in compressor work

Increase in refrigeration capacity Figure 11.6 A two-stage cascade refrigeration cycle and its T—s diagram. Therefore,

d.— . — COPR.c. mA Wnei.ili

msh4

— 10+

h2 — )

(11.10)

From the T—s diagram it is seen that the compressor work decreases and the amount of heat absorbed from the refrigerated space increases as a result of cascading. That is, cascading improves the COP of a refrigeration system. EXAMPLE 113 Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.15 MPa and 0.9 MPa. Each stage operates on an ideal vapour-compression refrigeration cycle with refrigerant-12 as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.4 MPa. If the mass flow rate of the refrigerant through the upper cycle is 0.045 kg/s, determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator.

Refrigeration Cycles 327 Solution The T—s diagram for the two stages A and B of the cycle is shown below. The enthalpies of the refrigerant at various states shown in the diagram are found to be as follows:

T.

h1 = 178.64 kJ/kg h2 = 195.59 kJ/kg h3 = 43.64. kJ/kg h4 = 43.64 kJ/kg h5 = 190.97 kJ/kg h6 = 205.31 kJ/kg h7 = 71.93 kJ/kg h8 = 71.93 kJ/kg

►s (a) The mass flow rate of the refrigerant through the lower cycle can be determined from energy balance on the heat exchanger. Thus, we have 0 0 JL Ethehe Ethihi or

)e=

IhA(h5 - hit) =

— h3)

or 0.045(190.97 — 71.93) = mB(195.59 — 43.64) or

mB = 0.0353 kg/s (b) The rate of heat removal by the cascade cycle is the rate of heat absorption in the evaporator of the lower cycle. The power input to the cascade cycle is the stun of the power inputs to both the compressors. Thus, we have Qi = thB(hi - h4) = 0.0353(178.64 — 43.64) = 4.77 kW and *in = Wcomp I.in + *comp Il in = thA (h6

h3) Phri(hz — hi)

= 0.045(205.31 — 190.97) + 0.0353(195.59 — 178.64) 1.24 kW

(c)

4.77 COP = .6, — = 3.85 Wain 1.24

Note: This problem was worked out in Example 11.1 for a single stage refrigeration system. In the present case, the COP has gone up from 3.36 to 3.85 as a result of cascading.

328 Fundamentals of Engineering Thermodynamics

11.8.2 Multistage Compression Refrigeration Systems A multistage compression refrigeration system is an innovative vapour-compression refrigeration system with a mixing chamber called the flash chamber connecting the stages, instead of a heat exchanger as in cascade systems, as shown in Fig. 11.7. In this system, the working fluid used in different stages is the same.

en environment vironment

T Condenser Expansion X valve Compressor (HP)

9

3 7

Expansion valve

Flash Compressor chamber (LP) Evaporator

1A

A

s

Cold refrigerated space

Figure 11.7 A two-stage compression refrigeration system with a flash chamber. EXAMPLE HA Consider a two-stage compression refrigeration system operating between the pressure limits of 0.15 and 0.9 MPa, with refrigerant-12 as the working fluid. The refrigerant leaves the condenser as saturated liquid and is throttled to a flash chamber operating at 0.4 MPa. Part of the refrigerant evaporates during the flash process, and the evaporated vapour is mixed with the refrigerant leaving the low-pressure compressor. The mixture is then compressed to the condenser pressure by the high-pressure compressor. The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. Assuming that the refrigerant leaves the evaporator as saturated vapour and both the compressors are isentropic, determine (a) the fraction of the refrigerant which evaporates as it is throttled to the flash chamber, (b) the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser, and (c) the coefficient of performance of this two-stage compression refrigeration system.

i

329

Refrigeration Cycles

Solution The T—s diagram of the cycle described is shown below. The enthalpies of the refrigerant at various states are found to be as follows: TA hi = 178.64 kJ/kg h2 = 195.59 kJ/kg h3 = 190.97 kJ/kg h4 = its = 71.93 kJ/kg h6 = 71.93 kJ/kg h2 = 43.64 kJ/kg h8 = 43.64 kJ/kg h9 = 195.12 kJ/kg

► s

(a) The fraction of the refrigerant which evaporates as it is throttled to the flash chamber is simply the quality at state 6. Therefore, we have x6

= h6 — /11 _ 71.93 —43.64 147.33 his

0.192

(b) The amount of heat removed from the refrigerated space is = (1 — x6Xhi — h8) = (1 — 0.192X178.64 — 43.64) 109.08 kJ /kg The compressor work input per unit mass of refrigerant flowing through the condenser is win = "Icon, Lin + "Icon, min = (1 — x6Xh2 — h1) (1Xh4 — h9) The enthalpy at state 9 can be obtained from energy balance on the mixing chamber. Thus, we have Q-W= E lit ehe —Eml p j But Q = 0 and W = 0, therefore, (1)h9 = x6h3 + (1 — x6)h2 = 0.192 x 190.97 + (1 — 0.192)195.59 = 194.70 kJ/kg Also, s9 = 0.7058 kJ/(kg K) Therefore, at P4 = 0.9 MPa and 54 = s9 = 0.7058 kJ/(kg K), we have h4 = 209.44 kJ/kg

330 Fundamentals of Engineering Thermodynamics Thus,

win = (1 — 0.192X195.59 — 178.64) + (209.44 — 194.70) 28.44 kJ/kg

109.08 3.84 (c) COP — 9t. = 28.44 — wit, Note that the COP of the multistage refrigeration system has increased considerably compared to that of a single-stage system (Example 11.1), but it is almost the same as that of the cascade system (Example 11.3).

11.8.3 Multipurpose Refrigeration System with a Single Compressor This system is capable of producing refrigeration at more than one temperature.

11.9 LIQUEFACTION OF GASES This is an extremely important area of refrigeration. Many processes at cryogenic temperatures (less than about —100°C) depend on liquefied gases. For example, superconductivity, separation of oxygen and nitrogen from air, preparation of liquid propellants and so on depend on processes at cryogenic temperatures. We know that liquefaction of gases is possible at temperatures below their critical-point value. The critical temperatures of helium, hydrogen, and nitrogen, which are commonly used liquefied gases, are —268°C, —240°C, and —147°C, respectively. Also, we know that with ordinary refrigeration techniques, these magnitudes are unattainable. Therefore, it is mandatory to have special techniques to accomplish the liquefaction of gases. Let us see the Linde-Hampson cycle, which is one such special case, shown in Fig. 11.8. The special feature here is that the O

Heat exchanger

T

Figure 11.8 Linde-Hampson system for gas liquefaction.

Refrigeration Cycles 331 high-temperature gas is cooled in an aftercooler by a cooling medium or by a separate external refrigeration system to state 4. The gas is further cooled to state 5 in a refrigerative counterflow heat exchanger by the uncondensed portion of the gas from the previous cycle and then throttled to state 6, which is a saturated liquid-vapour mixture state. The liquid is collected, and the vapour is routed through the refrigerator to cool the high-pressure gas approaching the throttle valve. This cycle may also be used for solidification of gases.

11.10 GAS REFRIGERATION CYCLES A gas refrigeration cycle is nothing but the reversed Brayton cycle. It is schematically shown in Fig. 11.9. All the processes in the ideal gas refrigeration cycle are internally reversible. In an

environment

TA

Heat

Turbine Compressor

Wnet, in —t

Heat exchanger Cold refrigerated spate

Figure 11.9 A simple gas refrigeration cycle. actual cycle, the compression and expansion processes will deviate from the isentropic ones, and T3 will be more than To unless the heat exchanger is infinitely large. The COP for this cycle may be expressed as COPR —

9L

qt. _ Wnet,in

Wcomp,in

Wturb,out

where qL = hi — ha, whirb,out = h3 — h4, and wcomon = h2 — ht. EXAMPLE 115 An ideal gas refrigeration cycle using air as the working medium maintains a refrigerated space

332 Fundamentals of Engineering Thermodynamics at —3°C while rejecting heat to the surrounding medium at 30°C. The pressure ratio of the compressor is 3. Determine (a) the maximum and minimum temperatures in the cycle, and (b) the coefficient of performance of the cycle. Assume air to be an ideal gas. Solution The T—s diagram of the cycle is described below.

(a) For the isentropic process 1-2, we have ( )70-1) T2

Therefore, T = 2

)(7-1)/T = 270.15 (3)0.4/1.4 = 369.8 K

That is, the maximum temperature in the cycle is 96.65 °C Similarly, the minimum temperature T4 is given by N(r• - Yr )0.4/1.4 = 303.15 (2 1 = 221.48 K = —51.67 °C T4 = T3 ( j 3 3 f

(b)

COPR —

qL %run .

wcomp,in

Wturkots1

Now, 91. = hi — ha = CAT! — Ta) = 1.0045(270.15 — 221.48) = 48.9 kJ/kg wamv,i„ = h2 — hi = 1.0045(369.8 — 270.15) = 100.1 kJ/kg and = h3 — h4 = 1.0045(303.15 — 221.48) = 82.04 kJ/kg Thus, COPR =

48.9 100.1 — 82.04

2.7

Refrigeration Cycles 333

11.11 ABSORPTION REFRIGERATION SYSTEMS An absorption refrigeration system is a system involving absorption of a refrigerant by a transport medium. It is competitive in places where there is a source of inexpensive heat energy at 100 to 200°C, such as solar energy and geothermal energy. The most commonly used absorption refrigeration system is the ammonia-water system, where NH3 serves as the refrigerant and H2O as the transport medium. The other popular systems are water-lithium bromide and water-lithium chloride systems, where water serves as the refrigerant. An ammonia-water absorption refrigeration system is shown in Fig. 11.10. This system looks similar to the vapour-compression system, except that the compressor has been replaced with an air absorption mechanism made up of an absorber, a pump, a generator, a refrigerator, a valve, and a rectifier. Ammonia vapour from the evaporator enters the absorber and reacts with water exothermally to form NH3 and H20. Thus, heat is released during this process. The liquid NH3 and H2O mixture is pumped to the generator. Heat is transferred to the solution from a source to vaporize some of the solution. The vapour which is rich in NH3 passes through a rectifier, which separates the water and returns it to the generator. The high-pressure pure NH3 vapour passes through the rest of the cycle. The hot ammonia-water solution, which is weak in NH3, then passes through a regenerator, where it transfers some heat to the rich solution leaving the pump, and is throttled to the absorber pressure. Even though these systems are bulky,

Regenerator

Figure 11.10 An ammonia-water absorption refrigeration system.

334

Fundamentals of Engineering Thermodynamics

complex, and expensive, they have the major advantage of compressing a liquid instead of a vapour, as in vapour-compression systems. For absorption systems, the COP is defined as COPR — Qgen + Winunp,in

(11.12)

Qgen

When the system is assumed to be reversible, the COP becomes COPR.,„, =

--L— 11en

=

7711kCarnot COPR,Carnot = 1

To Ts

T

TL _

TL.

(11.13)

where TL, To, and T, are the temperatures of the refrigerated space, environment and heat source, respectively. The COP of an actual absorption system is usually less than 1. EXAMPLE 11.6 An absorption refrigeration system receives heat from a source at 130°C and maintains the refrigeration space at —18°C. If the environment temperature is 30°C, determine the maximum possible COP for this absorption refrigeration system. Solution The maximum possible COP for the absorption refrigeration system is that COP which assumes the system to be reversible. That is, COPR.ret, = —- — 77th.camot COPR,Camot = s=

1

To )( TL. T To — TL . s

Given To = 30°C = 303.15 IC, Ts = 130°C = 403.15 K and TL = —18°C = 255.15 IC, we get 03.15 - 255.1 5 ) COPRoev = (1 _ _... 403.15)) ( 303.15 — 255.15)

1.32

11.12 THERMOELECTRIC POWER GENERATION When two wires of dissimilar metals are joined at both ends to form a closed circuit, as shown in Fig. 11.11, and one of the junctions is heated, a current flows continuously in the circuit. This

Figure 11.11 Seebeck effect.

Refrigeration Cycles

335

is called the Seebeck effect. The circuit is called thermoelectric circuit and a device that operates on this circuit is called a thermoelectric device. The Seebeck effect finds applications in two areas, namely temperature measurement and power generation. When the circuit is broken, as shown in Fig. 11.12, the current ceases to flow and an electromotive force or voltage Metal A / • Metal B V Figure 11.12 Voltage generated in a thermoelectric circuit. is generated. This voltage is a function of the temperature difference and the materials of the two wires used. Therefore, temperatures can be measured by simply measuring voltages. The two wires used to measure temperature in this way form a thermocouple, which is the most widely used device for temperature measurement. A thermoelectric generator is shown in Fig. 11.13. The net electric work produced is We = QH - QL. It resembles the heat engine cycle, with electrons serving as the working fluid. Therefore, in the absence of any irreversibilities, the thermoelectric generator should have the Carnot efficiency. These generators are still in the research stage.

Hot junction

/ Cold junction Low-temperature sink

Figure 11.13 A thermoelectric generator.

336

Fundamentals of Engineering Thermodynamics

11.12.1 Thermocouple Thermocouple is a device which operates on the Seebeck principle, which states that, a flow of heat in a metal always accompanies a flow of an electromotive force (emf). In other words, the Seebeck principle states that, "heat flow in a metal is always accompanied by an emf flow". This is also referred to as Seebeck effect. This forms the basis for the working of a thermocouple. In a vast number of metals and metal alloys such as copper, platinum, chromal, tungsten, and iron the flow of heat and emf are in the same direction. But in another group of metals and metal alloys, such as constantan, alumel, rhenium, and rhodium, the direction of heat flow is opposite to that of emf. These two groups, namely that in which the heat and emf flow in the same direction and the other in which the heat and emf flow in opposite directions are popularly known as dissimilar metals. EXAMPLE 11.7 A thermoelectric generator receives heat from a source at 130°C and rejects the waste heat to the environment at 30°C. What is the maximum possible efficiency for this thermoelectric generator? Solution The maximum possible efficiency is the Carnot efficiency. Thus, ?hoax = rith,Carnot = I

TL TH

303.15 403.15

— 0.248

= 24.8 per cent

SUMMARY A Carrot refrigerator or a Carnot heat pump is a refrigerator or a heat pump which operates on the reversed Carrot cycle. The ideal vapour-compression refrigeration cycle is that in which the refrigerant is completely vaporized before it is compressed and the turbine is replaced with a throttling device. The COPS of the refrigerators and heat pumps operating on the vapour compression cycle can be expressed as — 114 COPR wnet,in

h2 —

hl

For a reversible heat transfer rate, a temperature difference of 10°C or more should be maintained between the refrigerant and the environment. Refrigerant-12 (called Freon-12) and ammonia are the two most widely used refrigerants. A cascade refrigeration cycle is that in which two or more refrigeration cycles operate in series to perform the refrigeration process in stages. A multistage compression refrigeration system is an innovative vapour-compression refrigeration system with a mixing chamber called the flash chamber connecting the stages, instead of a heat exchanger as in cascade systems. Liquefaction of gases is an extremely important area of refrigeration. Many processes at cryogenic temperatures (less than about —100°C) depend on liquefied gases. Liquefaction of gases is possible at temperatures below their critical point values.

Refrigeration Cycles 337 A gas refrigeration cycle is the reversed Brayton cycle. The COP for this cycle may be expressed as 9t

COPR —

Wnet,in

9t. Wcomp,in

Wturb,our

An absorption refrigeration system is a system involving absorption of a refrigerant by a transport medium. For absorption systems, the COP is defined as COPR — Qgen Wpump,in Qgen

PROBLEMS 11.1 A refrigerator has working temperatures of —25°C and 35°C in the evaporator and condenser coil, respectively. Determine the maximum coefficient of performance possible for the refrigerator. If the coefficient of performance of the actual refrigerator is 80 per cent of the maximum, calculate the power input required for the refrigerating effect of 4 kW. [Ans. 4.14, 1.2 kW] 11.2 Ammonia enters the condenser of a steady-flow Carrot refrigerator as saturated vapour at 1200 kPa and leaves with a quality of 0.05. The heat absorption from the refrigerated space takes place at 190 kPa. Show the cycle on a T—s diagram relative to saturation lines, and determine (a) the coefficient of performance, (b) the quality at the beginning of the heat absorption process, and (c) the net work input. [Ans. (a) 4.97, (b) 0.198, (c) 181.4 kJ/kg] 11.3 Solve Problem 11.2 for refrigerant R134a irstead of ammonia. [Ans. (a) 4.53, (b) 0.381, (c) 26.65 kJ/kg] 11.4 A refrigerator working on an ideal vapour-compression refrigeration cycle uses refrigerant-12 as the working fluid. The cycle operates between 0.1 and 0.7 MPa. The mass flow rate of the refrigerant is 0.045 kg/s. Determine (a) the rate of heat removal from the refrigerated space, (b) the power input to the compressor, (c) the rate of heat rejection to the environment, and (d) the coefficient of performance. [Ans. (a) 5.034 kW, (b) 1.54 kW, (c) 6.57 kW, (d) 3.27] 11.5 A refrigerator working on an ideal vapour-compression refrigeration cycle uses Freon-12 as the working fluid. The condensing temperature is 36°C and the evaporation temperature is —10°C. Determine (a) the coefficient of performance of the cycle, (b) the power required to produce 1 ton of refrigeration, and (c) the required mass flow rate of the refrigerant for each ton of refrigeration. [Ans. (a) 4.63, (b) 0.760 kW/ton, (c) 1.873 kg/min] 11.6 A reversed Carnot cycle is designed for heating and cooling. The power supplied is 30 kW. The coefficient of performance of cooling is 3.9. Determine (a) the ratio TH/TL of the two heat reservoirs involved, (b) the refrigeration effect in tons of refrigeration, and (c) the coefficient of performance for heating. [Ans. (a) 1.256, (b) 33.27, (c) 4.9]

338 Fundamentals of Engineering Thermodynamics 11.7 A refrigerator working on an ideal vapour-compression refrigeration cycle using Freon-12 as the working fluid, operates with an evaporator temperature of —20°C and a condenser temperature of 40°C. The fluid enters the compressor as saturated vapour and enters the expansion valve as saturated liquid. The desired refrigeration effect is 20 tons. Assuming the adiabatic compressor efficiency of 100 per cent, determine (a) the power input to the compressor, (b) the coefficient of performance, and (c) the mass flow rate of Freon-12. [Ans. (a) 36.26 kJ/kg, (b) 2.872, (c) 40.52 kg/min] 11.8 An ice-producing machine operating on an ideal vapour-compression cycle uses refrigerant-12 as the working fluid. The refrigerant enters the compressor as saturated vapour at 140 kPa and leaves the condenser as saturated liquid at 600 kPa. Liquid water enters the ice machine at 15°C and leaves as ice at —5°C. If the ice production rate is 100 kg/h, determine the power input to the ice-machine. (Hint. 384 kJ of heat has to be removed from each kilogram of water at 15°C to turn it into ice at —5°C.) [Ans. 2.242 kW] 11.9 Refrigerant-12 enters the compressor of a refrigerator at 140 kPa and —10°C at a rate of 0.15 m3/min, and leaves at 1 MPa. The adiabatic efficiency of the compressor is 80 per cent. The refrigerant enters the throttling valve at 0.9 MPa and 30°C and leaves the evaporator as saturated vapour at —20.16°C. Determine (a) the power input to the compressor, (b) the rate of heat removal from the refrigerated space, and (c) the pressure drop and the rate of heat gain in the line between the evaporator and the compressor. [Ans. (a) 0.938 kW, (b) 2.31 kW, (c) 10 kPa, 0.128 kW] 11.10 A heat pump operating on the ideal vapour-compression cycle with refrigerant-134a is used to heat a house and maintain it at 22°C using underground water at 15°C as the heat source. The heat loss from the house is 20 kW. The condenser and evaporator pressures are 1 MPa and 300 kPa, respectively. Determine (a) the power input required to the heat pump and (b) the electric power saved by using a heat pump instead of a resistance heater. [Ans. (a) 2.99 kW, (b) 17.01 kW] 11.11 A heat pump operating on an ideal vapour-compression cycle with refrigerant-12 as the working fluid is used for water heating. The water flows at the rate of 0.25 kg/s and its temperature has to be increased from 10°C to 50°C. The condenser and evaporator pressures are 1 and 0.3 MPa, respectively. Determine the power input to the heat pump. [Ans. 6:76 kW] 11.12 A heat pump working on an ideal vapour-compression cycle using Freon-12 as the working fluid is used to heat a room. The condenser and evaporator pressures are 1.6 and 0.32 MPa, respectively. The heat transfer required for the condenser unit is 100 MJ/h. The Freon-12 is saturated at the beginning of compression and is at 55°C at the end of compression. Determine (a) the mass flow rate of the refrigerant, (b) the quality of Freon-12 at the entry to the evaporator, and (c) the power input to the compressor. [Ans. (a) 843.31 kg/h, (b) 0.405, (c) 6.74 kW] 11.13 A heat pump working on an ideal vapour-compression cycle uses ammonia as the working fluid. The pressure limits of the cycle are 0.5347 MPa and 1.8 MPa. The mass flow rate of ammonia is 0.5 kg/s and the ammonia is 97 per cent dry at entry to the

Refrigeration Cycles

339

compressor. At the end of isentropic compression, the temperature is 86°C. The temperature at the end of condensation is 34°C. The specific heat capacity of ammonia is 5 kJ/(kg K). Determine (a) the heat transfer available at the condenser per hour, and (b) the power required to run the pump if the overall efficiency of the compressor and driving motor is 75 per cent. [Ans. (a) 2119.10 M.1/11, (b) 109.58 kW] 11.14 A two-stage cascade refrigeration system operating between the pressure limits of 0.6 MPa and 0.1 MPa uses refrigerant-12 as the working fluid. Each stage operates on the ideal vapour-compression refrigeration cycle. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.28 MPa. The mass flow rate of the refrigerant through the upper cycle is 0.1 kg/s. Determine (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space and power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator. [Ans. (a) 0.0818 kg/s, (b) 11.52 kW, 2.758 kW, (c) 4.18] 11.15 An ideal-gas refrigeration cycle uses air as the working fluid. A refrigerator working on this cycle has to maintain a refrigerated space at —20°C while rejecting heat to the surroundings at 300 K. If the pressure ratio of the compressor is 2.7, determine (a) the maximum and minimum temperatures of the cycle, (b) the coefficient of performance, and (c) the rate of refrigeration for a mass flow rate of 0.1 kg/s. Assume air to be an ideal gas with variable specific heats. [Ans. (a) 336.35 K, 225.63 K, (b) 3.075, (c) 2.758 kJ/s] 11.16 A gas refrigeration system having a pressure ratio of 3 uses air as the working fluid. At the entry to the compressor, the air is at —10°C. The high pressure air is cooled to 25°C by rejecting heat to the surroundings. It is further cooled to —18°C by regenerative cooling before it enters the turbine. The adiabatic efficiencies of the compressor and turbine are 80 per cent and 85 per cent, respectively. Assuming constant specific heats for air at room temperature, determine (a) the lowest temperature that can be obtained in the cycle, (b) the coefficient of performance of the cycle, and (c) the mass flow rate of air for a refrigeration rate of 10 kW. [Ans. (a) —76.43°C, (b) 0.5368, (c) 0.487 kg/s] 11.17 A geothermal well at 130°C supplies heat at a rate of 100,500 kJ/h to an absorption refrigeration system. The environment is at 30°C and the refrigerated space is maintained at —22°C. Determine the maximum possible heat removal from the refrigerated space. [Ans. 120,600 kJ/h] 11.18 A thermoelectric refrigerator removes heat at a rate of 100 W from a refrigerated space at —10°C and rejects it to the surroundings at 30°C. Determine the maximum possible coefficient of performance of this refrigerator and the minimum power input required for its operation. [Ans. 6.58, 15.2 W] 11.19 An ideal air refrigeration system operates with pressures 101 kPa and 404 kPa at the compressor inlet and outlet, respectively. The temperature at compressor inlet is —6°C and the temperature at turbine inlet is 27°C. Determine (a) the coefficient of performance of the cycle, (b) the power required to produce 1 ton of refrigeration, and (c) the aircirculation rate for each ton of refrigeration, if the pressure at the turbine exit is 101 kPa. [Ans. (a) 2.053, (b) 1.7 kW/ton, (c) 3.22 kg/min]

CH AP TER

12 Psychrometrics 12.1 INTRODUCTION In our discussion on gas mixtures, we did not consider the possibility of gases condensing during a process, because the mixtures of gases were assumed to be above their critical temperatures. However, as regards gas-vapour mixtures, the vapour may condense out of the mixture during a process, forming a two-phase mixture. Therefore, a gas-vapour mixture has to be treated differently from an ordinary gas mixture. Even though several gas-vapour mixtures are encountered in engineering, in our study here we will consider only the air-water-vapour mixture, which is the most commonly used gas-vapour mixture in practice. The study of the air-watervapour mixture is called psychrometrics.

12.2 PROPERTIES OF ATMOSPHERIC AIR Air is a mixture of nitrogen, oxygen, and small amounts of other gases. In the absence of water vapour, this mixture is called dry air. The air in the atmosphere normally contains some water vapour, also called moisture, and is referred to as atmospheric air. Even though the amount of water vapour in atmospheric air is small, it is convenient to treat the air as a mixture of water vapour and dry air. In air-conditioning applications, the air temperature ranges from —10°C to 50°C. In this temperature range, the dry air can be treated as an ideal gas with Cp and C, as constants. The change in enthalpy for dry air can be determined from (taking 0°C as the reference temperature) hdrmi = CPT = 1.0045 T (kJ/kg)

(12.1)

and hdryair = CpAT = 1.0045 it T (kJ/kg)

(12.2)

where T is the air temperature in °C. In air-conditioning applications, the change in enthalpy O h is of importance, which is independent of the reference point chosen. Now, it is natural to question whether the vapour can also be treated as an ideal gas? The answer to this question is 340

Psychrometrics 341 the following: At 50°C, the saturation pressure of water is 12.3 kPa. At pressures below this value, water vapour can be treated as an ideal gas with an error only of the order of 0.2 per cent, even when it is a saturated vapour (see compressibility chart, Fig. 7.12). Therefore, water vapour in air also obeys the ideal-gis relation Pv = RT. Thus, the atmospheric air can be treated as an ideal gas mixture whose pressure P is the sum of the partial pressures of dry air Pa and that of the water vapour P,, that is P = Pa + Pv(kPa) (12.3) The partial pressure of water vapour is often called the vapour pressure. Since water vapour is an ideal gas, the enthalpy of water vapour in air can be taken to be equal to the enthalpy of the saturated vapour at the same temperature, that is, we can write h,(T, low P) = hg(T)

(12.4)

The enthalpy of water vapour at 0°C is 2501.3 kJ/kg. The average Cp of water vapour in the temperature range from —10°C to 50°C is 1.82 kJ/(kg °C). Therefore, the enthalpy of water vapour at any temperature T is given by hg (T) -= 2501.3 + 1.82T (kJ/kg)

(12.5)

where T is in °C.

12.3 HUMIDITY OF AIR The mass of water vapour mv present in a unit mass of dry air ma is called the absolute or specific humidity and is denoted by u. as: 0) =

Mm,,

ma

(12.6)

This is also referred to as the humidity ratio. It may also be expressed as in,,, ma

P,VIR,T _ P,IR, _ 0.622 Pa V/Ra T Pa /Ra Pa

(12.7)

or co

— 0.6224 P—

(12.8)

where P is the total pressure [see Eq. (12.3)] and 0.622 is obtained with Ra = 0.287 kJ/(kg K) and = 0.4615 kJ/(kg K). The specific humidity of dry air is zero. When water vapour is added to dry air its specific humidity will increase. With the addition .of water vapour, the humidity can be increased up to the limit where the air can hold no more moisture. At this point, the air is saturated with moisture, and is called the saturated air. Any moisture introduced into saturated air will condense. The amount of water vapour in saturated air at a given T and P can be determined from Eq. (12.8) by replacing P, by Ps, the saturation pressure of water at that T.

342 Fundamentals of Engineering Thermodynamics

12.3.1 Relative Humidity It is the ratio of the amount of moisture (m,) that the air holds to the maximum amount of moisture (mg) that the air can hold at the same T, that is, _ m _ 1v VIR T _Pv mg PgVIR,T Pg

(12.9)

From Eqs. (12.8) and (12.9), we get —

0.6220Pg coP and to — (Q622 + co)Pg P — OPg

(1110)

where 0 varies from 0 for dry air to 1 for saturated air. For atmospheric air, the total enthalpy is the sum of the enthalpies of dry air and the water vapour, that is, H = Ha + Hy= maha mvhv or H h = — = ha + n h, = ha + coh, ma ma Or

h = ha + cohg (kJ/kg dry air)

(... h, = h5)

(12.11)

Note that the enthalpy of atmospheric air is expressed per unit mass of dry air instead of per unit mass of the air-water-vapour mixture. This is because in most applications, the amount of dry air in the air-water-vapour mixture remains constant, and only the vapour changes. Generally, the temperature of the atmospheric air is referred to as the dry-bulb temperature to differentiate it from other forms of temperatures which we will discuss shortly. EXAMPLE 12.1 An air-water-vapour mixture at 25°C and 101 kPa has a relative humidity of 60 per cent. Compute (a) the partial pressure of the dry air, (b) the specific humidity of air, and (c) the specific enthalpy per unit mass of dry air. Solution (a) The partial pressure of the dry air can be determined using Eq. (12.3), Pa = P — Pv where Pv = 0Pg = OPsat@25°C = 0.60 x 3.169 = 1.901 kPa Thus, P.= 101 — 1.901 = 99.1 kPa (b) The specific humidity of the air is given by co

— 0.622 P, _ 0.622 x 1.901 P — P„ 101 — 1.901 0.01193 (kg H20)/(kg dry air)

Psychrometrics 343 (c) The specific enthalpy per unit mass of the dry air is given by h=

+ whg = CpT + whg

= (1.0045 x 25) + (0.01193 x 2547.2) 55.5 kJ/(kg dry air)

12.4 TEMPERATURES OF ATMOSPHERIC AIR 12.4.1 Dew-Point Temperature The dew-point temperature Tdp of an air-vapour mixture is the temperature at which condensation begins if the air is cooled at constant pressure. It is essentially the saturation temperature of water corresponding to the vapour pressure, that is, TTdp = Tsat at Pv

(12.12)

The dew-point temperature is illustrated on the T—s diagram of water in Fig. 12.1.

Figure 12.1

T—s diagram of water.

As the air cools at constant pressure (P, remains constant), the vapour in the air (state 1) undergoes a constant-pressure cooling process until it strikes the saturated vapour line (state 2). The corresponding T at this point is Tdp. EXAMPLE 122 The temperature of the air in a room is 25°C and the relative humidity is 60 per cent. Determine the dew-point temperature of the air. Solution The dew-point temperature of the air is determined from Tdp =

Tsat ® pv

where P„ = ø"g@25°C = 0.60 x 3.169 = 1.901 kPa

344 Fundamentals of Engineering Thermodynamics Therefore, T, = Tsat @ 1.901 kPa = 16.61 °C

12.4.2 Adiabatic Saturation Temperature The adiabatic saturation process is another way of determining the absolute or relative humidity. The process is illustrated in Fig. 12.2. A steady stream of unsaturated air at w and Ti passes Saturated air

Unsaturated air I

1

w1 ,*

T2 , 0/2 , 02 = 00% C)

It Liquid water at T2 T

Adiabatic saturation temperature

Dew-point temperature

Figure 12.2 The adiabatic saturation process. through the channel. The moisture content of air will increase during the process, and its temperature will decrease, since part of the latent heat of vaporization of the water that evaporates will come from the air. If the channel is long enough, the air stream will exit as saturated air with 0 = 100 per cent, and at T2 which is called the adiabatic saturation temperature. For the steady-flow process considered, we have

By conservation of mass thai =

ma2 = tha (the mass flow rate of dry air remains invariant)

and wi + fitI = tti w2 where thf is the rate of evaporation of water, or

?how l

thf

= th.c02

Psychrometrics 345 Thus, nif= iha (c02 - oh)

By energy conservation Since 141 = 0 and Q = 0, we get

F.mihi =F. moho

That is,

thalhl

ihfhf2 = tha2h2

Pack + thei(c02 — coi)hf2 = thah2

That is,

hi + (w2 — coi)h/2 = h2 or (Cpri

ng i) +

O)i

(02 COI)hp = (CpT2

CO2hg2)

or

cal —

Cp(T2 —

+ ahhfg2 (12.13)

hgl — h f2

where 0.622Pg2 =

(12.14)

P2 — Pg2

With 02 = 100 per cent, oh and co2 of air can be determined from Eqs. (12.13) and (12.14) respectively, by measuring the P and T of the air at the inlet and exit of an adiabatic saturater.

12.4.3 Wet-Bulb Temperature For determining the absolute or relative humidity of air by the adiabatic saturation process discussed above, we need a long channel or a spray mechanism to achieve saturation conditions at the exit. A simple and practical approach to measure the saturation temperature is by the use of a thermometer whose bulb is wrapped with a cotton wick saturated with water and air is blown over the wick, as shown in Fig. 12.3. The temperature measured in this way is called the wet-bulb temperature. Ordinary thermometer Wet-bulb temperature

Air-flow

Wick Water

Figure 12.3 Wet-bulb temperature measurement.

346 Fundamentals of Engineering Thermodynamics

12.4.4 Dry-Bulb Temperature The dry-bulb temperature is the temperature of the air as measured by an ordinary thermometer placed in the air. This term is used to distinguish it from the wet-bulb temperature which is obtained from a thermometer whose bulb is wrapped with a cotton wick saturated with water.

12.4.5 Psychrometer It is an instrument which measures both the dry-bulb and the wet-bulb temperatures of air. The schematic of a psychrometer is shown in Fig. 12.4. When unsaturated air passes over the wick, some of the water in the wick evaporates, resulting in a temperature difference between the air and the water. After a while, the heat loss from the water by evaporation equals the heat gain from the air, and the water temperature stabilizes. The thermometer reading at this point is the wet-bulb temperature. The dry-bulb thermometer measures the dry-bulb temperature simultaneously.

I I

Dry-bulb thermometer Wet-bulb thermometer

Wet-bulb thermometer wick Figure 12.4 Schematic of a psychrometer. EXAMPLE 123 The dry- and wet-bulb temperatures of atmospheric air at 1 atm pressure measured with a sling psychrometer are determined to be 25°C and 10°C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air. Solution (a) The specific humidity w is determined from

;

Psychrometrics 347

—

where

T2

Cp(T2 — Ti ) + co2hfg2 he — hf2

is the wet-bulb temperature, and (02 is determined from 0.622Pg (1)2

2

= P 2

g2

P

0.622 x 1.2276 — 101.325 — 1.2276 = 0.00763 (kg H20)/(kg dry air)

Thus, 1.0045 (10 — 25) + (0.00763 x 2477.7) 2547.2 — 42.01 0.00153 (kg H20)/(kg dry air) (b) The relative humidity O is determined from 01 =

0.00153 x 101.325 (0.622 + 0.00153)3.169

toiP2 (0.622 + toi )Pe

= 0.0785 = 7.85 per cent (c) The enthalpy of the air per unit mass of dry air is given by hi =

+ (Me Cpri + Col he

= (1.0045 x 25) + (0.00153 x 2547.2) 29.01 kJ/(kg dry air)

12.5 PSYCHROMETRIC CHARTS A psychrometric chart is a graphical representation of the various properties of moist air. A thermodynamic state of moist air is uniquely fixed if the mixture pressure and two independent properties, such as the dry-bulb temperature and the specific humidity, are known. This implies that a psychrometric chart may be constructed for a given mixture pressure, using Tdb and a) as coordinates, as shown in Fig. 12.5(a) which illustrates the basic features of the chart. At a given mixture pressure, the vapour pressure P, is a function of the specific humidity only. Therefore, there is only one value of P,, for each value of to. For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures are identical, as illustrated in Fig. 12.5(b). Therefore, the dew-point temperature of atmospheric air at any point on the chart can be determined by drawing a vertical line (line of P, = constant or 0) = constant) from the point to the saturated curve. The psychrometric chart serves as a tool for visualizing the air-conditioning processes. For example, a heating or cooling process will appear as a horizontal line on this chart if no humidification or dehumidification is involved (i.e. w = constant). Any deviation from a horizontal line indicates that moisture is added or removed from the air during the process.

lilt

348

Fundamentals of Engineering Thermodynamics

Saturation line 0 = 100%

Saturation line

.-

.a

=

-9

a"

0

Specific humidity, w (b) (a) Figure 12.5 Schematic of a psychrometric chart. EXAMPLE 12.4 Consider atmospheric air at 1 atm, 32°C, and 50 per cent relative humidity. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy, (c) the wet-bulb temperature, and (d) the specific volume of the air (in m3/(kg dry air)). Solution At a given pressure, two independent properties such as the dry-bulb temperature and the relative humidity completely specify the state of atmospheric air. Other properties are determined by directly reading their values at the specified state. (a) The specific humidity is determined by drawing a horizontal line from the specific state to the right until it intersects the w axis, as shown schematically in the figure below.

T= 32°C At the intersection point, we read from the Psychromatic chart W = 0.015 (kg H20)/(kg dry air)

Psychrometrics

349

(b) The enthalpy of air per unit mass of dry air is determined by drawing a line parallel to the h = constant line from the specific state until it intersects the specific scale. At the intersection point, we get from the chart h= 70.05 kJ/(kg dry air) (c) The wet-bulb temperature is determined by drawing a line parallel to the T„.1, = constant line from the specified state until it intersect the saturation line. At the intersection point, we get 23.7 °C (d) The specific volume per unit mass of dry air is determined by noting the distance between the specified state and the v = constant lines on both sides of the point. The specific volume is determined by interpolation (visual) to be V = 0.885 m3/(kg dry air)

12.6 AIR-CONDITIONING PROCESS An air-conditioning process is meant for maintaining a space at the desired temperature and humidity. For example, simple heating, simple cooling, humidifying and dehumidifying are some of the air-conditioning processes. Sometimes two or more of these processes are required to bring the air to the desired temperature and humidity level. Most air-conditioning processes can be modelled as steady-flow processes. Therefore, for such processes applying the principles of conservation of mass and conservation of energy, we have (12.15) Dry air mass: Ethaa = Irk.° Wet mass: Energy:

Ethwo

(12.16)

= Eineho — Ethihi

(12.17)

Q —W

12.6.1 Simple Heating or Cooling During simple heating, the specific humidity remains constant, but the relative humidity decreases. On a psychrometric chart, a simple heating process will proceed in the direction of the increasing dry-bulb temperature following a specific humidity co, which appears as a vertical line (Fig. 12.5). Heating can be accomplished by passing the air over a heated coil. A cooling process at constant co is similar to the heating process, except that the dry-bulb temperature decreases and the relative humidity increases during such a process. Cooling can be accomplished by passing the air over some coils through which a refrigerant or cold water flows. For heating and cooling processes, the energy conservation equation reduces to = m 0(h, — hi ) or q = h2 — h,

(12.18)

where hi and h, are the enthalpies per unit mass of dry air at the inlet and the exit of the heating or cooling section, respectively.

I '

I 1.

• •I

350

Fundamentals of Engineering Thermodynamics

12.6.2 Heating with Humidification This is accomplished by an arrangement shown in Fig. 12.6. The exit state 3 depends on the

Heating coils

J Humidifier

.1 Air rat = Humidifying section

Heating section

Figure 12.6 Heating with humidification. amount of heating and humidification. This process is used to eliminate the problems associated with the low relative humidity resulting from simple heating. EXAMPLE 12.5 An air-conditioning system takes in atmospheric air at 12°C and 25 per cent relative humidity at a steady rate of 50 m3/min and conditions it to 22°C and 50 per cent relative humidity. The atmospheric air is first heated to 20°C in the heating section and then humidified by the injection of hot steam in the humidifying section. Assuming that the entire process takes place at a pressure of 100 kPa, determine (a) the rate of heat supply in the heating section and (b) the required mass flow rate of the steam in the humidifying section. Solution The system and the psychrometric chart of the process are shown schematically in the following figure. The mass flow rate of the dry air remains constant during the entire process. The amount of moisture in the air remains constant as it flows through the heating section (0)1 = (02), but increases in the humidifying section (0)3 > 0)2). (a) By conservation of mass and conservation of energy equations for the heating section, we have Dry air mass: Water mass:

E !hall = ni es,e

thwi =

?hall =

tha2 = ma

---> ?howl = tha2(02

or foi = (02

0

Energy:

E ntehe —

=

a(h2 — hi)

It is essential to realize that, even though the psychrometric chart is extremely valuable in determining the properties of moist air, its use is limited to a specified pressure only, which is 1 atm (101.325 kPa) for the chart given in the Appendix. At pressures other than 1 atm, either other charts for that pressure or the relations developed earlier should be used.

Psychrometrics 351

12 Heating coils

20

22°C Humidifier

= 12°C 01 =25% V1 = 50

1,

AIR T2 = 20°C

1

T3 = 22°C 03 = 50°/0 3

Now, = fhPgi = APsit@irc = 0.25 x 1.4186 = 0.3547 kPa Pal = P1 — Pvi = 100 — 0.3547 = 99.645 kPa =

RaTI _ 0.287 x 285.15 _ — 0.821 m3/kg dry air 99.645

• —

=

50 — 60.90 kg/min

nia v1 0.821

col =

0.62241 = 0.622 x 0.3547 = 0.00221 (kg H 20)/(kg dry air) — /3,4 100 —0.3547

h1 = CPT, + w1hgi = (1.0045 x 12) + (0.00221 x 2523.44) = 17.63 (kJ/kg dry air) h2 = CpT2 + ahhg2= (1.0045 x 20) + (0.00221 x 2538.10) = 25.70 kJ/(kg dry air) With col = co2, the rate of heat transfer to the air in the heating section becomes

Q = nia(h2 — h1) = 60.90(25.70 — 17.63) 491.46 kJ/min

352

Fundamentals of Engineering Thermodynamics

(b) The mass conservation for water in the humidifying section can be expressed as ma2c02 + mw = ma3W3 Or

iii. = tha(w3 - w2) where

W3 —

0.622 03 Pg3 P3 — 03Pg3

_

0.622 x 0.50 x 2.671 100 — (0.50 X 2.671)

= 0.00842 (kg H20)/(kg dry air) Thus,

/it.= 60.90(0.00842 — 0.00221) =

0.378 kg/min

12.6.3 Cooling with Dehumidification This process is meant for removing moisture from air. It requires cooling the air below its dewpoint temperature. The process is illustrated in Fig. 12.7.

14°C

30°C Cooling coils

T3 = 14°C

Ti = 30°C 01 =80% Air-.---r— V1 = 10 m3/min

WWW000000

-4—,— •:••4.--• r.s 0.3 = 100% L .—:. .:4; : . 3 \ / U / 14°C

1

Condensate removal

Figure 12.7 Cooling with dehumidification.

Psych roni el rit

353

Evaporative cooling When water evaporates, the latent heat of vaporization is absorbed from the water container body and the surrounding air. Evaporative cooling ks based on this principle. Both the water and. the air are cooled during the evaporative cooling process. Observe a porous pot filled with water and left in an open, shaded area. A small amount of water leaks out through the porous holes, as shown in Fig. 12.8, and the pot sweats. In a dry environment, this water evaporates and cools the remaining water in the pot. An evaporative cooler \\. lush works on this principle is shown in Fig. 12.9.

Figure 12.8 Illustration of evaporative cooling.

Liquid water

Cool, moist air

-41-- Hot, dry air Figure 12.9 Evaporative cooler.

Hot, dry air at state 1 entering the cooler exits at state 2 with increased humidity and decreased temperature. In the limiting case, the air will exit at saturated state 2', with the lowest temperature which can be achieved by this process. This is essentially an adiabatic saturation process since the heat transfer between the air stream and the surroundings is usually negligible. The evaporative cooling process, therefore, follows a line of constant on the psychrometric chart. But the constant TWb lines and constant h lines almost coincide on the chart, thus, we have Twb 1,g constant

(12.19)

354 Fundamentals of Engineering Thermodynamics and h = constant

(12.20)

during an evaporative cooling process. This approximation is commonly used in air-conditioning calculations. EXAMPLE 12.6 Air at 1 atm, 35°C, and 70 per cent relative humidity enters an air-conditioner at a rate of 9.5 m3/min, and leaves as saturated air at 15°C. Part of the moisture in the air which condenses during the process is removed at 15°C. Determine the rate of heat and moisture removal from the air. Solution The system and psychrometric chart of the process are schematically shown in the figure below.

15

35°C Cooling coils

T3 = 15°C

UW110000000

Ti = 35°C 9$1 = 70%

AIR.*"1"-- Vi = 9.5 m3/min

15°C

Condensate removal

The mass flow rate of the dry air remains constant during the entire process, but the amount of moisture in the air decreases because of dehumidification (012 > oh). For the combined cooling and dehumidification section, by the mass and energy conservation equations, we have Dry air mass:

?hal = 'ha = ma

Psychrometrics

Water mass: or

355

met al = tho,602 thw =

Energy:

c°2)

Q = inn (h, — hi ) + »O,,,

The inlet and the exit states of the air are completely specified and the total pressure is 1 atm. Therefore, we can use the psychrometric chart of the Appendix to determine the properties of the air at both states. Thus, h i = 100 kJ/kg dry air wt = 0.0255 kg H20/kg dry air v1 = 0.907 m3/kg dry air and h 2 = 42 kJ/kg dry air cot = 0.0110 kg H20/kg dry ail h„. = hf ,, i5oc = 62.99 kJ/kg Then V1 v1

9.5 0.907

— 10. -4 kg/min

and rh,„ = 10.474(0.0255 — 0.0110) = 0.152 kg/min] and 0= 10.474(42 — 100) + 0.152 x 62.99 — 597.92 kJ/min Therefore, the air-conditioning unit removes moisture and heat from the air at the rate of 0.152 kg/min and 597.92 kJ/min, respectively. EXAMPLE 12.7 Air at 1 atm, 35°C, and 30 per cent relative humidity enters an evaporative cooler and exits at 90 per cent relative humidity. Determine (a) the exit temperature of the air, and (b) the lowest temperature to which the air can be cooled by this evaporative cooler. Solution The evaporative cooler and the psychrometric chart of the process are schematically shown in the following figure.

356

Fundamentals of Engineering Thermodynamics

1=

AIR .442' 2

35°C

*44— = 30% P = 1 atm 1

Tinin T2

35°C

(a) If the temperature of the liquid water supplied is assumed to be the same as the exit temperature of the air stream, the evaporative cooling process follows a line of constant wetbulb temperature on the psychrometric chart. That is, Twb = constant. The wet-bulb temperature at 35°C and 30 per cent relative humidity (from the psychrometric chart) is 21.5°C. The intersection point of the Tub = 21.5°C and the 0 = 90 per cent lines is the exit state of the air. The temperature at this point is the exit tcavemiture of the air, and from the psychrometric chart it is determined to be T2 = 22.5°C (b) In the limiting case, the air will leave the evaporative cooler saturated (0 = 100 per cent), and the exit state of the air in this case will be the state where To, = 21.5°C intersects the saturation line. For saturated air, the dry and wet-bulb temperatures are identical. Therefore, the lowest temperature to which the air can be cooled is the wet-bulb temperature, which is 21.5°C

= =

12.6.4 Adiabatic Mixing of Air Streams Mixing of two air streams is required in many air-conditioning applications. The mixing can be accomplished by simply merging the two air streams, as shown in Fig. 12.10. The process can he assumed to be adiabatic, since the heat transfer with the surroundings is usually small. For this process, the conservation of mass and the conservation of energy equations become Mass of dry air:

thol

nio2

= nia3

Mass of water vapour: cipii.1 cogha2 = Energy: niaih

nia2/12 =

(12.21) (1222) (1223)

Elimination of /ha from the relations above results in mal _ (02 — toi = h2 — h3 mat

W3 — 0)1

h3 — h1

(1224)

Psychrometrics 357

h1

Mixing , section

0)3 h3

Figure 12.10 Mixing of two air streams. The states that satisfy the first part of the Eq. (12.24) are shown by the line AB and those that satisfy the second part are shown by the line CD. Therefore, the point of intersection of AB and CD is the only state which satisfies both the conditions in Eq. (12.24), and is located on the straight line connecting the states 1 and 2. Thus, we can summarize that: • When two air streams at different states (1 and 2) are mixed adiabatically, the state of the mixture will lie on the straight line connecting the states of the air streams on the psychrometric chart, and the ratio of the distances 2-3 and 3-1 will be equal to the ratio of the mass flow rates Thai and ina2 of the two streams. EXAMPLE 12.8 Saturated air at 20°C at a rate of 70 m3/min is mixed adiabatically with the outside air at 35°C and 50 per cent relative humidity at a rate 30 m3/min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. Solution The system and the psychrometric chart of the process are shown schematically in the figure below.

Saturated air T1 = 20° C V1= 70 m3/min 0 > Mixing , \> section / P=1 s atm T2 = 35°C sds/ 02 = 50% 1/2= 30 m3/min

V3 0

0/3

03 T3 20°C

35°C

358 Fundamentals of Engineering Thermodynamics The properties of each inlet stream (from the psychrometric chart) are: h i = 57.9 kJ/kg dry air = 0.015 kg H20/kg dry air

v1 = 0.85 m3/kg dry air h2 = 81 kJ/kg dry air 0)2 =

0.0175 kg H20/kg dry air

v2 = 0.898 m3/kg dry air The mass flow rates of dry air in each stream are ?hal

70 = = — — 82.35 kg/min v1 0.85 30 = — 33 . 41 kg/min v2 0.898

mat =

By mass conservation, we have tha3 = thal tha2 = 82.35 + 33.41 = 115.76 kg/min

The specific humidity and the enthalpy of the mixture can be determined from the equation mai _

0)2 — 0)3

rha2

613 —

_ h2 — h3 h3

or 82.35 _ 0.0175 — 0)3 _ 81 — h3 33.41 0)3 — 0.015 h3 —57.9 Solving, we get (03 = 0.0157 (kg H20)/(kg dry air) and h3 = 64.57 (kJ/kg dry air) The h3 and 0)3 fix the state of the mixture. The other properties of the mixture, from the psychrometric chart, are

T= 26 °C 03 = 80 per cent v3 = 0.865 m3/(kg dry air) The volume flow rate of the mixture is given by = a3 V3 =

115.76 x 0.865 = 100.13 m3/min

Psychrometrics 359 It is interesting to note that the volume flow rate of the mixture is approximately equal to the sum of the volume flow rates of the two incoming streams. This is typical in air-conditioning applications.

12.6.5 Wet Cooling Towers These are devices that recirculate cooling water to serve as a transport medium for heat between the source and the sink. It is essentially a semi-enclosed evaporative cooler. An induced-draft counterflow wet cooling tower is shown schematically in Fig. 12.11. As the water droplets fall Air exit t++ tft

Spray

nozzles

Warm water

Air inlet

Air inlet

Cooled water -4

Figure 12.11 An induced-draft cotmterflow wet cooling tower. through the tower, a small amount of water evaporates and cools the remaining water. The temperature and the moisture content of the air increase during this process. The water so cooled collects at the bottom of the tower and is pumped back to the condenser to pick up additional waste heat. The cooling tower sketched in Fig. 12.11 is classified as a forced-draft cooling tower, since the air circulation for the cooling process is provided by a fan.

12.6.6 Natural-Draft Cooling Towers Air in these towers is induced naturally, without any external agency. The natural-draft cooling towers are hyperbolic in profile, as shown schematically in Fig. 12.12. They are usually very tall Air exit +SW+

Warm water Cooled water

Figure 12.12 A natural-draft cooling tower.

360 Fundamentals of Engineering Thermodynamics (about 100 m). The profile given is only from the strength point of view. The air in the tower has a higher water-vapour content, and thus being lighter than the outside air, rises and the heavier outside air fills the vacant space, creating an airflow from the bottom to the top of the tower. The air flow rate is controlled by the conditions of the atmospheric air.

12.6.7 Spray Ponds In spray ponds, the warm water is sprayed into the air. It then gets cooled by the air as it falls into the pond (Fig. 12.13). Even though spray ponds are simpler and less expensive than cooling Air in

Air out

Warm water Cooled water Figure 12.13 A spray pond. towers, they require 25 to 50 times larger area than the cooling towers. Also, the loss of water owing to air drift is very high and there is no protection against dust.

Cooling ponds They are basically large water reservoirs open to the atmosphere to which the waste heat is dumped. But, because of the poor natural heat transfer from the water surface to the atmosphere, the area required is about 20 times that of a spray pond. EXAMPLE 12.9 Water at 40°C enters a cooling tower at a rate of 200 kg/s. The water is cooled to 25°C in the cooling tower by the air which enters the tower at 1 atm, 20°C, 60 per cent relative humidity and leaves saturated at 30°C. Neglecting the power input to the fan, determine (a) the volume flow rate of the air entering the cooling tower, and (b) the required mass flow rate of the make-up water. Solution The cooling tower is schematically shown in the following figure. The mass flow rate of the dry air through the tower remains constant Oita1 = ma2 = ma), but the mass flow rate of the liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain a steady operation.

Psychrometrics 361 30°C 0 0 = 100%

tft fft Warm water

40°C 200 kg/s Air

System boundary

1 atm 20°C 01 = 60% V1 = ?

200 kg/s Cooled water

t 25°C Make-up water

(a) By conservation of mass and conservation of energy equations, we have Dry air mass: tho = nia2 = ?ha Water mass: ii13 + ?kiwi = m4 thaco2 th3 — m4 =

— col) =

make-up

0 0

Energy:l2

= E *eh, — E phihi 0 = Pha(h2 — hi) + (rims — Mrneke_up)h4 — 1h3h3

This gives (h2 —

m3(h3 - h4) — (02 — (00/14

From the psychrometric chart, we have h i = 42 kJ/(kg dry air) col = 0.0088 (kg 1-120)/0cg dry air) v1 = 0.84 m3/(kg dry air) h2 = 99.9 kJ/(kg dry air) (02=

0.0275 (kg H20)/0(8 dry air)

From Table 3 of the Appendix, we have h3 = hf@ grc = 167.57 kJ/(kg H2O) h4 = hf@ 25.c = 104.89 kJ/(kg H2O)

362 Fundamentals of Engineering Thermodynamics Thus, ma

200(167.57 —104.89) (99.9 — 42) — (0.0275 — 0.0088) 104.89 224.1 kg/s

The volume flow rate of the air entering the cooling tower is given by =inav1 = 224.1 x 0.84 = 188.24 m3/s (b) The required mass flow rate of the make-up water is determined from = Ii a(w2 — col ) = 224.1(0.0275 — 0.0088) 4.19 kg/s That is, only 2.1 per cent of the cooling water is carried away by the hot air going out of the cooling tower.

SUMMARY Psychmmetrics is the study of the air-water-vapour mixture. Air is a mixture of nitrogen, oxygen, and small amounts of other gases. In the absence of water vapour, this mixture is called dry air. The air in the atmosphere normally contains some water vapour, also called moisture and is referred to as atmospheric air. The atmospheric air can be treated as an ideal gas mixture whose pressure P is the sum of the partial pressures of the dry air Pa and that of the water vapour P,„ that is = Pa + Pv (kPa) The partial pressure of water vapour is often called the vapour pressure. Since water vapour is an ideal gas, the enthalpy of water vapour in air can be taken to be equal to the enthalpy of the saturated vapour at the same temperature, that is, we can write 14,(T, low P) = hg(T) The absolute or specific humidity w is the mass of water vapour present in a unit mass of dry air. That is,

This is also referred to as the humidity ratio. It may also be expressed as w—

P IR P,,VIR„T mb, — — v — 0.622 --IL ma PaVIRa T PalRa Pa

Psych torn etrics

363

or =

0.622F., P— P,.

where P is the total pressure. Relative humidity is the ratio of the amount of moisture (ms,) that the air holds to the maximum amount of moisture (mg) tha the air can hold at the same T. 111,11 is, m,, P,. — m— g g P VIR,,T Pg From Eqs. (12.8) and (12.9), we get —

coP (0.622 + co)Pg

and to —

0.6220P P — 0Pg

where 0 varies from 0 for dry air to 1 for saturated air. The dry-bulb temperature usually refers to the temperature of the atmospheric air. The dew-point temperature is the temperature of an air-vapour mixture at which condensation begins if the air is cooled at constant pressure. The adiabatic saturation temperature is the temperature of the saturated air with 100 per cent relative humidity. The wet-bulb temperature is the temperature measured by a thermometer whose bulb is wrapped with a cotton wick saturated with water and air is blown over the wick. A psychrometer is an instrument which measures both the dry- and wet-bulb temperatures of air. A psychrometric chart is a graphical representation of various properties of moist air.

PROBLEMS 12.1 The pressure and (emperature of the air in a room are 100 kPa and 30°C. If the relative humidity is 30 per cent, find the partial pressure of the water vapour and the dew point, the specific volume of the air and water vapour, and the specific humidity. [Ans. 1.274 kPa, 10.3°C, 0.881 m3/(kg dry air), 109.81 m3/(kg H20), 0.00802] 12.2 100 m3 of air-water mixture in a tank is at 0.1 MPa, 35°C and 70 per cent humidity. Determine the humidity ratio, the dew-point, the mass of air, and the mass of water vapour. Also, calculate the amount of water vapour that will condense if the mixture is cooled to 5°C in a constant pressure process. [Ans. 0.0255, 28.67°C, 108.67 kg, 2.771 kg, 2.771 kg] 12.3 A tank contains 10 kg of dry air and 0.1 kg of water vapour at 30°C and total pressure of 100 kPa. Determine (a) the specific humidity, (b) the relative humidity, and (c) the volume of the tank. [Ans. (a) 0.01 (kg H20)/(kg dry air), (b) 37.26 per cent, (c) 8.84 m3] 12.4 The pressure and temperature of the air in a room are 101.325 kPa and 25°C, respectively. The relative humidity is 40 per cent. Determine (a) the saturation pressure of the water vapour at the dry-bulb temperature and (b) the dew-point temperature. [Ans. (a) 3.169 kPa, (b) 10.22°C]

364 Fundamentals of Engineering Thermodynamics 12.5 An air-water-vapour mixture at 30°C and 100 kPa has a relative humidity of 60 per cent. Determine (a) the humidity ratio, (b) the dew-point, (c) the enthalpy, in kJ/kg dry air (where h = 0 at 0°C), and (d) the specific volume, in m3/kg dry air. [Ans. (a) 0.01626 (kg H20)/(kg dry air), (b) 21.37°C, (c) 71.7 kJ/(kg dry air), (d) 0.870 m3/(kg dry air)] 12.6 Atmospheric air inside a house is at 20°C and 60 per cent relative humidity. Will any moisture condense on the inner surfaces of the glass windows when the temperature at the windows drops to 15°C? [Ans. No] 12.7 The dry- and wet-bulb temperatures of the air in a room are 25°C and 16°C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the dew-point temperature. Assume the air pressure to be 100 kPa. [Ans. (a) 0.00788 (kg H20)/(kg dry air), (b) 39.5 per cent, (c) 10.03°C] 12.8 A man wearing glasses walks through an environment at 10°C and enters a house at 20°C and 40 per cent relative humidity. Determine whether the glasses will become fogged. [Ans. No] 12.9 For atmospheric air at 1 atm, 30°C, and 50 per cent relative humidity, using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy (in kJ/ (kg dry air), (c) the wet-bulb temperature, (d) the dew-point temperature, and (e) the specific volume of the air (in m3/kg dry air). [Ans. (a) 0.0134 (kg H20)/(kg dry air), (b) 65 kJ/(kg dry air), (c) 22°C, (d) 18.38°C, (e) 0.877 m3/(kg dry air)] 12.10 For air at 1 atm pressure the dry- and wet-bulb temperatures are 25°C and 15°C, respectively. Using the psychrometric chart, determine (a) the specific humidity, (b) the enthalpy, (c) the relative humidity, (d) the dew-point temperature, and (e) the specific volume of the air (in m3/kg dry air). [Ans. (a) 0.0065 (kg H20)/(kg dry air), (b) 42 kJ/(kg dry air), (c) 33 per cent (d) 7.53°C, (e) 0.865 m3/(kg dry air)] 12.11 Air at 100 kPa, 20°C, and 40 per cent relative humidity enters a heating section at a rate of 10 m3/min, and leaves at 30°C. Determine (a) the rate of heat transfer in the heating section, and (b) the relative humidity of the air at the exit. [Ans. (a) 119.57 kJ/min, (b) 22.03 per cent] 12.12 Air is heated in a heating section of diameter 25 cm, containing a 5 kW electric resistance heater. At the entrance, the air is at 1 atm, 15°C, and 30 per cent relative humidity and travels at 10 m/s. Determine (a) the exit temperature, (b) the exit relative humidity of the air, and (c) the exit velocity. [Ans. (a) 23°C, (b) 16 per cent, (c) 10.27 m/s] 12.13 Air at 1 atm, 10°C, and 50 per cent relative humidity is first heated to 30°C in a heating section and then humidified by introducing water vapour. The air leaves the humidifying section at 25°C and 70 per cent relative humidity. Determine (a) the amount of steam added to the air in kg H20/kg dry air, and (b) the amount of heat transfer to the air in the heating section, in kJ/kg dry air. [Ans. (a) 0.0102 (kg H20)/(kg dry air), (b) 45.5 kJ/(kg dry air)]

Psychrometrics 365 12.14 Air at 1 atm, 45°C, and 30 per cent relative humidity has to be cooled to 30°C by an adiabatic humidification process. Determine (a) the amount of water to be added to the air, and (b) the final relative humidity. [Ans. (a) 0.0061 (kg H20)/(kg dry air), (b) 89.73 per cent] 12.15 An air-conditioning system consists of a heating section and a humidifier which supplies wet steam (saturated water vapour) at 100°C. The air at 15°C and 60 per cent relative humidity enters the heating section at a rate of 55 m3/min, and leaves the humidifying section at 20°C and 50 per cent relative humidity. If the total pressure of the airconditioning system is 1 atm, determine (a) the temperature and relative humidity of the air when it leaves the heating section, (b) the rate of heat transfer in the heating section, and (c) the rate at which water is added to the air in the humidifying section. [Ans. (a) 21.5°C, 35 per cent (b) 304.38 kJ/min, (c) 0.0734 kg/min] 12.16 Air at 1 atm, 35°C. and 70 per cent relative humidity enters a window air-conditioner, at a rate of 10 m3/min and leaves as saturated air at 10°C. Part of the moisture in the air which condenses during the process is also removed at 10°C. Determine the rate of heat and moisture removal from the air. [Ans. —766.60 kJ/min, 0.195 kg/min] 12.17 An air-conditioning system takes in air at 1 atm, 35°C and 70 per cent relative humidity and delivers it at 20°C and 55 per cent relative humidity. The air flows first over the cooling coils, where it is cooled and dehumidified, and then over the resistance heating wires where it is heated to the desired temperature. Assuming that the condensate is removed from the cooling section at 10°C, determine (a) the temperature of the air before it enters the heating section, (b) the amount of heat removed in the cooling section, and (c) the amount of heat transferred in the heating section. [Ans. (a) 14°C, (b) — 63.30 kJ/(kg dry air), (c) 9.5 kJ/(kg dry air)] 12.18 Air at a dry-bulb temperature of 40°C and a relative humidity of 50 per cent has to be conditioned to a final dry-bulb temperature of 20°C and a final relative humidity of 40 per cent by a dehumidification process followed by a reheat process. Assuming a constant pressure of 1 atm for the process, determine (a) the amount of water removed from the air, and (b) the temperature of the air leaving the dehumidifier. [Ans. (a) 0.0177 kg H,0/(kg dry air), (b) 6.7°C] 12.19 Air passes through a 25 cm diameter cooling section at 100 m/min. At the entrance of the passage, the air is at 1 atm, 40°C, and 50 per cent relative humidity. In the cooling section, the air is cooled by passing it over a cooling coil through which cold water is flowing. The temperature of the water is increased by 10°C. The air leaves the cooling section saturated at 25°C. Determine (a) the rate of heat transfer, (b) the mass flow rate of water, and (c) the air velocity at the exit of the cooling section. [Ans. (a) —124.46 kJ/min, (b) 2.98 kg/min, (c) 94.58 rn/min] 12.20 Air at 40°C, 1 atm and 10 per cent relative humidity passes through an evaporative cooler. Water is added at 20°C. Determine (a) the amount of water added per kg of dry air if the final temperature of the air is 20°C, (b) the final relative humidity, and (c) the minimum temperature that could be achieved by this process. [Ans. (a) 0.00825 (kg H20)/(kg dry air), (b) 87.7 per cent, (c) 18°C]

366 Fundamentals of Engineering Thermodynamics 12.21 Air at 1 atm, 40°C, and 15 per cent relative humidity enters an evaporative cooler at a rate of 12 dimin, and leaves with a relative humidity of 80 per cent. Determine (a) the exit temperature of the air, and (b) the required rate of water supply to the evaporative cooler. [Ans. (a) 22.5°C, (b) 0.098 kg/min] 12.22 What is the lowest temperature which the air can attain in an evaporative cooler if it enters at 1 atm, 30°C and 30 per cent relative humidity? [Ans. 18°C] 12.23 Two air streams are mixed steadily and adiabatically. The first stream at 10°C and 80 per cent relative humidity enters at a rate of 150 m3/min, while the second stream at 32°C and 60 per cent relative humidity enters at a rate of 100 m3/min. Assuming that the mixing process occurs at a pressure of 1 atm, determine the specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture. [Ans. 0.0107 (kg H20)/(kg dry air), 80.5 per cent, 16°C, 248.75 m3/min] 12.24 Two kilograms of air at 200 kPa, 5°C, and 80 per cent relative humidity is mixed steadily and adiabatically with 4 kg of air at 200 kPa, 40°C, and 20 per cent relative humidity. Determine the specific humidity, the temperature, and the relative humidity of the mixed stream. [Ans. 0.003812 (kg H20)/(kg dry air), 28.35°C, 31.31 per cent] 12.25 The cooling water from the condenser of a power plant enters at wet cooling tower at 40°C at a rate of 52 kg/s. The water is cooled to 25°C in the cooling tower by the air which enters the tower at 1 atm, 20°C, and 50 per cent relative humidity and leaves saturated at 30°C. Neglecting the power input to the fan, determine (a) the volume flow rate of the air flowing into the cooling tower, and (b) the required mass flow rate of the make-up water. [Ans. (a) 47.32 m3/s, (b) 1.15 kg/s] 12.26 A small cooling tower is designed to cool 6 kg of water per second. The inlet temperature of water is 40°C. The motor-driven fan induces 8 m3/s of air through the tower and the power required for the fan is 5 kW. The air enters the tower at 20°C and 60 per cent relative humidity. The air leaves the tower saturated at 25°C. Assuming the pressure during the cooling process to be 1 atm, determine (a) the required mass flow rate of the make-up water, and (b) the temperature of the water leaving the tower. [Ans. (a) 0.1107 kg/s, (b) 27.5°C] 12.27 Air-water vapour mixture at 1055 kPa, 30°C, and 80 per cent relative humidity entering an air-conditioning unit exits at 100 kPa, 15°C and 95 per cent relative humidity. At the exit, water vapour condensed to liquid water at 15°C is drained out. Calculate the heat transfer per kilogram of dry air, assuming that the change in kinetic energy is negligible and treating the water vapour as an ideal gas. [Ans. —41.78 kJ/(kg dry air)]

CHAPTER

13 General Thermodynamic Property Relations

13.1 INTRODUCTION We know that some properties such as T, P, V, and mass can be measured directly. However, some other properties, such as p and v cannot be measured directly, and are determined using simple relations. But properties such as U, H, and S are not so easy to determine because they can neither be measured directly nor be related to easily measurable properties through some simple relations. Therefore, it is essential to develop some fundamental relations between the commonly encountered thermodynamic properties, and express the properties that cannot be measured directly in terms of easily measurable properties. In previous chapters, we made use of the property tables to get the values of many thermodynamic properties. We will now see that the property relations developed in this chapter form the basis for the preparation of property tables.

13.2 MATHEMATICAL PRELIMINARIES ON PARTIAL DERIVATIVES AND ASSOCIATED RELATIONS Many of the relations developed in this chapter are based on the state postulate which states that the state of a simple, compressible substance is completely specified by any two of its independent, intensive properties. All other properties at that state can be expressed in terms of these two properties. This statement can be expressed mathematically, as z = z(x,y) where x and y are the two independent properties that specify the state and z represents any other property. 367

368 Fundamentals of Engineering Thermodynamics The total differential of a dependent variable respect to the independent variables x and y is dz

z in terms of its partial derivatives with

=( az) dr + (az) dy Cax-j r —33 )x

(13.1)

or (13.2) dz = Mdx + Ndy where M = (dzIdx)y and N= (dzlay),.. Also, (dM/dy)x = d2z/(dxdy) and (aV/dx)y = crgdxaY). The order of differentiation is immaterial for properties since they are continuous point functions and have exact differentials. Therefore,

aM) =( al41) (

aY

L

(133)

x

All z(x, y) that obey Eq. (13.3) are point functions and are called path functions if they do not satisfy this equation. In thermodynamics, this relation forms the basis for the development of the Maxwell relations. EXAMPLE 13.1 Consider an ideal gas at 700 kPa and 400 K. During a process, the state of the gas changes to 695 kPa and 397 K. Estimate the change in specific volume of the gas, using Eq. (13.1). Solution Equation (13.1) can be used with reasonable accuracy if the changes are small, even though the equation is valid only for differential changes in variables. The changes in P and T can be expressed, respectively, as dP = AP= 700 — 695 = 5 kPa and dr= AT= 400 — 397 = 3 K By the ideal-gas state equation, we have RT

P As R is a constant, we have, v = v(P, 7). Applying Eq. (13.1) and using the average values for P and T, we get

dv =

ap

dP T

ar

dT P

-RT dP RdT +

= 0.287 F -398.5 x 5 L 697.52 5.9 x 10-5 m3/kg

3

697.5 j

General Thermodynamic Property Relations

369

13.2.1 Reciprocity and Cyclic Relations The function z = z(x, y) can also be expressed as x = x(y, z) if y and z are taken to be independent variables. Then, we have dx =

dy + (%) dz (3Y

(13.4)

z

Combining Eqs. (13.1) and (13.4), we get

(%)y laxly dz

d z = Rt)y (t)z Or

(133) [ ( My

(fY1

(° YU dY = - (My (My] dz

The variables y and z can be varied independently, since they are independent, that is, y can be held constant (dy = 0) and z can be varied over a range of values (dz 0). For Eq. (13.5) to be valid at all times, the terms in the brackets must be equal to zero, regardless of the values of y and z. Equating the brackets to zero, we obtain (13.6) ( aZ y

(a4

az J X

(aZiaX)y

ax Ca

= —1

(13.7)

Equation (13.6) is called the reciprocity relation and Eq. (13.7) is called the cyclic relation. EXAMPLE 13.2 Verify the cyclic relation using the ideal-gas state equation. Solution The ideal-gas state equation Pv = RT consists of variables P, v and T. At any time, two of them can be treated as independent, with the remaining one being the dependent variable. The cyclic relation is

aX

z

il y

l aZ

1

Replacing x, y, and z in the above equation by P, v, and T, respectively, we get

( ap) (ay) (ar aV T ar p ap)

370 Fundamentals of Engineering Thermodynamics But RT P = P(v, n = v Thus, ( all _ —RT .dv )7. v2 Similarly, ( MI =

R P

and

( aT = v— 491')v

R

Substituting the partial derivatives in the cyclic relation, we get —RT

R v RT

H v ) (;) (i) = — — P v =-1 Thus, the cyclic relation is satisfied.

13.3 THE MAXWELL RELATIONS The Maxwell relations are equations that relate the partial derivatives of P, v, T, and s of a simple compressible substance to each other. They are obtained from the four Gibbs equations by exploiting the exactness of the differentials of thermodynamic properties.

13.3.1 The Gibbs and Helmholtz Relations Two of the Gibbs relations were derived in Chapter 5 and expressed as du = Tds — Pdv

(13.8)

dh = Tds + vdP

(13.9)

The other two Gibbs relations are obtained with the Helmholtz function a and the Gibbs function g, which are two new combination properties and defined as a = u — Ts

(13.10)

g = h — Ts

(13.11)

The Helmholtz function is also called the Helmholtz free energy. For an infinitesimal reversible process, we can express Eq. (13.10) as dA = dU — TdS — SdT

General

Thermodynamic Property

Relations 371

From Eq. (13.8), we have

TdS = dU + PdV Thus,

dA =—SdT—PdV From this relation it is evident that 1. For a reversible isothermal process:

dA = — PdV or

Af — Ai = — Pdil where the subscripts i and f denote the initial and final states, respectively. Hence, the change of the Helmholtz function during a reversible isothermal process equals the work done on the system. 2. For a reversible isothermal and isochoric(constant volume) process:

dA = 0 and

A = constant These properties are often of interest in chemistry and are useful in considering chemical reactions that take place isothermally and isochorically. The Gibbs function is also called the Gibbs free energy. For an infinitesimal reversible process, we can write the Gibbs free energy [Eq. (13.11)] as

dG = dH — TdS — SdT But dH can be expressed using Eq. (13.9) as

dH = TdS + VdP Therefore, we have

dG = — SdT + VdP For a reversible isothermal and isobaric process, we have

dG = 0 and G = constant This is an important result in combustion in respect of processes involving a change of phase. Sublimation, fusion, and vaporization take place isothermally and isobarically and can be conceived as occurring reversibly. Hence, during such processes, the Gibbs function of the system remains constant. If the molar Gibbs functions of a saturated solid, saturated liquid, and saturated vapour are denoted by g', g", and g', respectively, then the equation of the fusion curve is given by g , = g„ and the equation of the vaporization curve by g t1 =

gIll

i l I'

372

Fundamentals of Engineering Thermodynamics

and the equation of the sublimation curve by g f = el

All the g's can be regarded as functions of P and T only, and hence the above two equations serve to determine the P and T of the triple point uniquely. The Gibbs function is of utmost importance in chemistry, since chemical reactions can be conceived of as taking place at constant P and T. It is also of some use in engineering. Differentiation of Eqs. (13.10) and (13.11) results in da = du — Tds — sdT dg = dh — Tds — sdT Simplification of these two relations, using Eqs. (13.8) and (13.9), yields the other two Gibbs relations for simple compressible systems as da = — sdT — Pdv

(13.12)

dg = — sdT + vdP

(13.13)

The above four relations in Eqs. (13.10) to (13.13) are of the form dz = Mdx + Ndy with

am) = ( aN) (4 x

ax)y

The quantities u, h, a, and g are properties and, therefore, they should have exact differentials. Applying Eq. (13.3) to each of them results in

( ar = ( av ), ) — as )),

(13.14)

(ar ) = () op s as )p

(13.15)

(t), =(;'), (ap), = — (ar)p

(13.16)

(13.17)

Equations (13.14) to (13.17) are called the Maxwell relations. They are extremely useful relations in thermodynamics since they provide a means of determining the changes in entropy, which cannot be measured directly, by simply measuring the change in properties P, v, and T. It is important to note that the above Maxwell relations are valid only for simple compressible substances. However, other similar relations can be written, following the same method, for non-simple systems such as those involving electrical, magnetic, and other effects.

373

General Thermodynamic Property Relations

13.4 THE CLAPEYRON EQUATION It is a thermodynamic relation that enables the determination of change in enthalpy associated with a phase change process, such as the enthalpy of vaporization hfg, from a knowledge of P, v, and T data alone. Consider the Maxwell relation, Eq. (13.16),

aP) (as) = lav )T During a phase-change process, the pressure is the saturation pressure which depends on T only and is independent of v. Thus, Psai =firma). Therefore, (dPlaT),, can be expressed as a total derivative (dPIdT).t. The derivative (dPIdT).t is the slope of the saturation curve in the P—T diagram at a specified saturation state, as shown in Fig. 13.1, and is independent of v. Thus, it can be treated as a constant during the integration of Eq. (13.16) between two saturation states

P Liquid (dP/dT)sat = constant

Solid

Vapour

T

T

P—T diagram of a saturation state.

Figure 13.1

at the same temperature. For example, for an isothermal liquid-vapour phase change process, integration of Eq. (13.16) yields

— sf = dP

(13.18)

(F [ ./a (Vg Vf)

Or

dP) _ sfg ( dT Lat vfg

(13.19)

The pressure also remains constant during this process. Therefore, from Eq. (13.9), we

dh= Tds,

or

Jj

dh

= Jj

Tds,

or

get

hfg = Tsfg

Substituting this into Eq. (13.19), we get

dP) _ hfg (17. Tvfg

(13.20)

374

Fundamentals of Engineering Thermodynamics

This is called the Clapeyron equation. It is valid for any phase-change process that occurs at constant T and P. It can be expressed in a general form as (dP) __ h12 dT sat Tv12

(13.21)

where the subscripts 1 and 2 refer to the two phases. EXAMPLE 133 Using the Clapeyron equation, estimate the enthalpy of vaporization of water at 300°C, and compare it with the tabulated value. Solution From Eq. (13.20), we have dP hfg = Tvfg (— dTlat From Table 3 of the Appendix, we get vfg = (vg — vi) 300.c = 0.02167 — 0.001404 = 0.02027 m3/kg Also, _ Psat ® 305°C — Pmt @ 295*C 1 dP) = (A P) k.dr Jsat.300.,c AT Jsat.3pp0c 305 — 295 =

9.202 — 7.993 — 0.1209 MPa/K 10

Since AMC) = ATTIC), substituting, we get hfg = 573.15 x 0.02027 x 120.9 = 1404.59 kJ/(kg K) The volume of ht at 300°C for water from Table 3 is 1404.9 kJ/(kg K). The difference between the two values is due to the approximation used in determining the slope of the saturation curve at 300°C. For liquid-vapour and solid-vapour phase-change processes, the Clapeyron equation can be simplified by utilizing some approximations. At low pressures vg >> vf, and thus vfg . vg. By treating the vapour as an ideal gas, we can write vg = RT/P. Using these in Eq. (13.20), we obtain Phfg (d dP TIat — RT 2 Or

) (P

= 1115 ( dr) ,, R T2 )sat

4

General Thermodynamic Property Relations 375 For small temperature intervals, hfg can be treated as a constant at some average value. Then, integrating the above equation between the two saturation states, we get hfg ( 1

P2 lat

1

R

T2 1t

(1322)

This is called the Clapeyron-Clausius equation. For solid-vapour regions, ft has to be replaced with the enthalpy of sublimation kg in Eq. (13.22). EXAMPLE 13A Estimate the saturation pressure of refrigerant-12 at 20°C, using the data available in the refrigerant tables. Solution By Eq. (13.22), we have In ( P2 )sat

h fg 1 _ R 7i T2 )sat

Let T1 = 16°C and T2 = 20°C. For refrigerant-12, R = 0.06876 kJ/(kg K) At Ti = 16°C, hik = 143.14 kJ/kg, and P1 = Psat © 160c = 0.50591 MPa. Substituting these values in the above equation, we get In

( P2 143.14 ( 1 1 — 0.09824 0.50591) 0.06876 289.15 293.15

Or P2 = 0.55813 MPa This predicted value of P2 = 0.55813 MPa differs from the tabulated value of 0.56729 by 1.64 per cent, which is acceptable for most applications.

13.5 GENERAL RELATIONS FOR du, dh, ds, Cv AND Cp 13.5.1 Internal Energy Changes du Let us assume u to be a function of T and v. That is, u = u(T, v). In differential form, we can write du =l -u—) d T +

az v

av T

dv

or

au ay

du = C,„ dT + (—) dv T

(13.23)

376 Fundamentals of Engineering Thermodynamics Now let us express s, as s = s(T, v). Therefore,

ds=(A) dT+1) dv .9i. ), av h.

(13.24)

Using Eq. (13.24), the relation du = Tds - Pdv can be written as

du = T (I) dT + [T (t) - Pl d v T " Equating the coefficients of dT and dv in Eqs. (13.23) and (13.25), we obtain OT), = T (Ls

av r = T and ()

(1325)

avI - P VIT

Using the Maxwell relation of Eq. (13.16) and the second relation above, we get

(ail =7. 4212- - P av )) 7. (a7')),, With this, Eq. (13.23) yields

du = CvdT +[T

( aP) - ddv ar

(1326)

The change in u of a simple compressible system associated with a change of state from (T1, v1) to (7'2, v2) is given by

u2 _ u, = r C, a + r .1

[T(— aP ) ar ,,

pi dv

(1327)

13.5.2 Enthalpy Changes dh Let h = h(T, P). Therefore, dh = ( ah

(3T )Pp

dT + ( ah ) dP OP T

or

dh = CpdT + (

ah

ap

) dP r

(1328)

Assuming s = s(T, P), we have

ds=(23 ) dr +( as ) dP aT p aP r

(13.29)

Using this into the relation dh = Tds + vdP results in

dh=T( as ) dT+[T( as ) +v]c/P ap r ar p

(13.30)

I

General Thermodynamic Property Relations 377 From Eqs. (13.28) and (13.30), we get ( as ) _

Cp

a7' )1,

T

and

( ah) = T ( as ) + v ap )r aP

Using the Maxwell relation of Eq. (13.17) and the second relation above, we get aV ) 11P 'T (ar JP

dh=CpdT +[v -

(13.31)

The dh of a simple compressible system associated with a change of state from (Ti , P1 ) to (7'2, P2) is given by h2

-

r CpdT +

v_T(&)]dp ar

r[

(13.32)

p

Note that if either h2 - h, or u2 - u i is known, the other can be determined from

h, -h, =u2 - u1 + (P2v2 - Pivi )

(13.33)

13.5.3 Entropy Changes ds We have seen that ds=rs ) dT +( s ) dv and a = ay T T ( as r, Also, (-Pll =(i1 al by Eq. (13.16). Therefore, av

ds =

(aP

T

dT + --) dv

ar

(13.34)

We have also seen that CP ds=( as ) dr +( as ) dP and ( as = — T ap r ar ar P By Maxwell's relation of Eq. (13.17), we have

as )

ap

av ) )P

Therefore, ds =

dr -( T

ar)

dP

(1335)

p

13.5.4 Specific Heats Cv and Cp At low pressures, gases behave as ideal gases and their C,, and Cp depend on T only. These are

II

378

Fundamentals of Engineering

Thermodynamics

called zero-pressure or ideal-gas, specific heats (Co and Cpo). But for calculating Cy and Cp at high pressures, we need general relations for them. Such relations are developed by applying the exactness test on Eqs. (13.34) and (13.35), which gives

racv ) = T av )7.

f a2 P) OT 2 )v

(13.36)

and (aCp) =—T o2v)

aP Jr

(1337)

ar2

Another useful relation is that relating Cp and Cv, obtained by equating the two ds relations (13.34) and (13.35) and solving for dT, that is dT =

T(OPlar)„ Cp

dv+ T(avlar)p Cp —C,

dP

Assuming T = 7(v, P) and differentiating, we get +

dT =() 49T )

av

dP

Equating the coefficient of either dP or dv, in the above relations, we obtain

( ay) cp — c.= T (5 I rap

. )paT)1,

(1338)

The alternate form of this relation is obtained as follows: By cyclic relation,

ap aT av (Ii),(0;)p (IV2)T This gives

(My=

-(fa(Z17.

Using this in Eq. (13.38), we get 2

Cp - Cv = - T

(f;,) ( aP) , ,

(1339)

This relation can also be expressed in terms of two other thermodynamic properties called the isothermal compressibility a and volume expansivity /3, defined as a=

1 (av —

v ap),

(13.40)

(13.41)

General Thermodynamic Property Relations 379 Note that Eqs. (13.40) and (13.41) remind us of the coefficient of linear expansion S. which we studied in strength of materials, defined as

l ar, 8 1 L OT )p This coefficient indicates how the length of a solid body is influenced by a change in temperature while the pressure remains constant. Similarly, isothermal compressibility a is an indication of the change in volume that results from a change in pressure while the temperature remains constant. The volume expansivity also called the coefficient of volume expansion, is an indication of the change in volume that results from a change in temperature while the pressure remains constant. Substituting a and fi into Eq. (13.39), we obtain

Cp — C„ —

vTfi2 a

(13.42)

This is called the Mayer relation, named after J.R. Mayer. The following conclusions can be drawn from Eq. (13.42): • The a is a positive quantity for all substances in all phases. The fi could be negative for some substances, such as liquid water below 4°C, but its square is always positive or zero. The absolute temperature 7' is also positive. Therefore, we can conclude that Cp • The difference between approaches zero.

C„

(13.43)

c,, and C,, approaches zero as the absolute temperature

• For incompressible substances, Cp and C, are identical, since v = constant. Therefore, for liquids and solids, which are almost incompressible, the difference between Cp and is very small and usually disregarded. The reciprocal of the isothermal compressibility is called the isothermal bulk modulus Br.

BT =

—V

ap) av Jr

The adiabatic compressibility cc is an indication of the change in volume that results from a change in pressure while the entropy remains constant, and is defined as

_ as

1 ( av ) ap),

The adiabatic bulk modulus B, is the reciprocal of the adiabatic compressibility.

op =- -v (— Both the isothermal compressibility a and volume expansivity fi are thermodynamic properties of

380

Fundamentals of Engineering Thermodynamics

a substance, and for a simple compressible substance they are functions of two independent properties. The values of these properties are found in the standard handbooks on physical properties. EXAMPLE 13.5 Show that the internal energy of an ideal gas is a function of temperature only, that is, u = u(T). Solution The differential change in the internal energy of a simple compressible system is given by Eq. (13.26): du = Cv dT +[T ( aP) - Pidv

ar

v

For an ideal gas, Pt, = RT. Therefore, Pdv + vdP = RdT

or

ap) = R ar )„, v Thus, R du= Cv dT +[T(—) - Pidv v ,, = CvdT + (P - P)dv

That is, du = CAT

Now, by Eq. (13.36), we have 1 ac) L ay

)r

a 2 P) =TI

are v

For an ideal gas, P = RT/v. Therefore, (aP) = A ( a2P) _ ( a(R/v)) _ and U .9r ), v ar )v art v Thus,

av )r That is, cv does not change with specific volume and hence is not a function of specific volume. Therefore, the relation du = CvdT

implies that the internal energy of an ideal gas is a function of temperature only.

1

Comsat Thermodynamic Property Relations 381 EXAMPLE 13.6 For an ideal gas, show that

CI, — C, = R

Solution By Eq. (13.39), we have Op

av

Cp — Cv = —T (F, /P ( av)T For an ideal gas,

P = RT/v. Therefore, ( OP) —RT —P av )7v2 v

and from v =

RT/P, we have

ar p= 11P (N Substituting, we get

(12 (__)_R 0

Cp

Cv =

T

.... p

P

Thus,

Cp — C,, = R EXAMPLE 13.7 For an ideal gas, show that (a) the coefficient of volume expansion is a function of temperature only and (b) the isothermal compressibility is a function of pressure only. Solution (a) The coefficient of volume expansion or volume expansivity fi is given by Eq. (13.41) as

P= r) v -ar p For an ideal gas, v = RT/P, therefore,

OT = p R (N P Thus,

1R vP

1 T

That is, the coefficient of volume expansion of an ideal gas is a function of temperature only. (b) The thermal compressibility is expressed by Eq. (13.40) as

382

Fundamentals of Engineering Thermodynamics

1

P This shows that the isothermal compressibility of an ideal gas is a function of a pressure only.

13.6 THE JOULE-THOMPSON COEFFICIENT The Joule-Thompson coefficient is a measure of the change in T with P during a constant enthalpy process. When a fluid passes through a restriction such as a porous plug or an ordinary valve, its P decreases. The enthalpy of the fluid remains approximately constant during such a throttling process. The T of a fluid may increase, decrease, or remain unchanged during a throttling process, as shown in Fig. 13.2. The temperature behaviour of a fluid during a throttling (h = constant) process is described by the Joule-Thompson coefficient, which is

• = 20°C • = 800 kPa

T2 {.= 20°C P2 = 200 kPa

Figure 13.2 Flow through a throttle valve. defined as

_(ar) 11- ap)„

(13.44)

It is evident from Eq. (13.44) that for < 0 T increases p = 0 T remains unchanged > 0 T decreases The Joule-Thompson coefficient represents the slope of h = constant lines on a T—P diagram. Such diagrams can be easily constructed from the T and P measurements alone during a throttling process. The development of a h = constant line on a P—T diagram is shown in Fig. 13.3(a). Constant-enthalpy lines of a substance on a P—T diagram are shown in Fig. 13.3(b). As seen from Fig. 13.3(b), some h = constant lines pass through a point of zero slope or p = 0. The locus of these points is called the inversion line. The T at a point when h = constant line intersects the inversion line is called the inversion temperature. The T at the intersection of P = 0 line and the upper part of inversion line is called the maximum inversion temperature. From Fig. 13.3(b) it is seen that a throttling process proceeds along a h = constant line in the direction of decreasing P. Therefore, depending on whether the process is taking place to

General Thermodynamic Property Relations

T

383

Maximum inversion temperature

A

P2, T2

131, Tt

(varied)

(Fixed)

Exit states

Inlet state h = constant line

(a)

(b)

Figure 13.3 P-T diagram with h = constant lines. the left or right of the inversion Ilre. the T will decrease or increase during the process. Also, a cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion temperature. For example, hydrogen must be cooled below its maximum inversion temperature of -68°C if any further cooling is to be achieved by throttling.

A general relation for Joule-Thompson coefficient By Eq. (13.31), we have

dh=CpdT +[v - T(P) ]dP T p Therefore, by taking the partial derivative of each term with respect to P and holding h = constant and rearranging, we obtain =(-or - ) 6, h

l] =-1 - Ti — Cp

iv

a

(13.45)

This is the general !elation for it. Thus, p can be determined from a knowledge of Cp and the P-v-T behaviour of the substance. EXAMPLE 13.8 Show that for an ideal gas, the Joule-Thompson coefficient is zero. Solution The general relation for the hlt) le Thompson coefficient given by Eq. (13.45) is

384 Fundamentals of Engineering Thermodynamics

ji —

ar Jp j

c, L

For an ideal gas, v = RT/P. Therefore, P (ar aL )p = R

Thus, Cp

[V - T P =- 1(1, - = ] Cp

Note that for an ideal gas, the enthalpy remains unchanged during throttling. Further, enthalpy is a function of temperature only. That is, h = h(7) for ideal gases, which requires that the temperature should remain constant unless the enthalpy changes. Therefore, a throttling process cannot cool an ideal gas.

13.7 PROPERTY RELATIONS FOR REAL GASES Let us extend the analysis to evaluate Ah, Au, and As of real (non-ideal) gases, using the general relations developed in the preceding sections.

13.7.1 Enthalpy Changes of Real Gases The Ah of a real gas depends, generally, on P as well as T, and can be evaluated from Eq. (13.32), h2

=

Cp dr +

r[v_r(av) ] dP ar p

where the subscripts 1 and 2 refer to initial and final states, respectively. The fact that change in a property between two specified states is the same no matter which process path is followed can be exploited to simplify the integration of Eq. (13.32). Examine the process diagram shown in Fig. 13.4. The AA during this process can be obtained by integrating Eq. (13.32) along Ti = constant and T2 = constant lines, and Po = constant line instead of the actual process path shown in the figure. The pressure Po can be chosen to be very low or zero, so that the gases can be treated as ideal during the isobaric process. Denoting the ideal-gas state with the superscript *, we can express the Ah of a real gas during process 1-2 as h2 — hi = (h2 — h2*) + (h2* — hi*) + (hi* — hi)

(13.46)

From Eq. (13.32), we have h2 —14 = 0 +

r [ v_ T

(61 CST ot=

=r [v_Trvi L

dP T2

dP

=

T T=2 jP1-7.

(13.47)

General Thermodynamic Property Relations 385 T Actual process path

P

2 P0 = 0

T2 Tt Alternative process path

Figure 13.4 Process paths to evaluate

4 — hit =

&i of real gases.

IT? Cp dT +0 = S?T Cp0(T) dT

(13.48)

hi —hi = 0 + 'PI; [v—T ( av ) ] dP * aT p T.4 =—

0

[v _ T ( ay )]

ar

dP

(13.49)

P r=7;

The difference between h and ht, which represents the variation of the enthalpy of a gas with P at a fixed T, is called the enthalpy departure. The calculation of enthalpy departure requires a knowledge of P—v--T behaviour of the gas. In the absence of such information, Pv = ZRT can be used, where Z is the compressibility factor. Substituting v = ZRT/P in Eq. (13.49) and simplifying, we get (h* — h)T = — RT2 r 0

(az)dP aT

P

This equation can be generalized using the reduced coordinates: T = TcyiTR and P = PaiPR. After some mathematical arrangement, the enthalpy departure in a nondimensional form becomes

_ (IT* —TOT = R.Teri

rPR (az ) don PR) Jo arR

(13.50)

•R

where Zh is called the enthalpy departure factor. The integral in Eq. (13.50) can be evaluated graphically or numerically using data from compressibility charts for different values of PR and TR. The graphical representation of Zh as a function of PR and TR is called the generalized enthalpy departure chart. The chart is given in the Appendix. It is used to determine the enthalpy deviation of a gas at a given P and T from the enthalpy of an ideal gas at the same T.

386 Fundamentals of Engineering Thermodynamics By replacing h* by hikA, Eq. (13.46) can be written as h2

= RN Tai(Zh, — Z42 ) + (h2 — 171)ideal

(13.51)

or h2 — h1 = RT„i (Zh, —

)+ (h2 — ha )idea

(13.52)

13.7.2 Internal Energy Changes of Real Gases The change in internal energy of a real gas is determined by relating it to the Ah through the definition h = u + Pi7 = u + ZRUT, to result in (13.53)

—ua=( — 171) — Ru(z2T2 — zi7i)

13.7.3 Entropy Changes of Real Gases The general relation for ds, given by Eq. (13.34), is s2 si =

r

_.tdr_r ar

T

dP p

The path chosen for integrating the enthalpy change equation will not be suitable here since it involves the value of entropy at P = 0, which is infinity. To overcome this problem, let us choose a different path between the two states, as shown in Fig. 13.5. T

ActUal process path

2*

Alternative process path

Figure 13.5 Process path for evaluating ds for real gases. The ds can be expressed as s2

— 4) +(sb — s;) +(sl — 4) +

—

(1334)

States 1 and 1* and 2 and 2* are identical. States 1* and 2* are imaginary, and the gas is assurped to behave as an ideal gas from 1* to 2*. Therefore, the ds during process 1*-2* can be determined from the ds relations for ideal gases.

General Thermodynamic Property Relations 387 Consider a gas at P and T. To determine the ds of this gas and that of an ideal gas at the same P and T, consider an isothermal process from P, T to zero pressure and back to the imaginary ideal-gas state P*, 7*, as shown in Fig. 13.5. The ds during this T = constant process can be expressed as

(SP — spa )r —(SP — so )r +

( ay )

aT

dP—

P

_ 6arv* p dP

= RT/P. This equation yields

where v = ZRT/P and v* =

(SP

sp)r =— Jo

SP'T

fP [(1 — Z)R P

Jo

RT ( az) P

dP

But T = Tc.,;TR and P = P„;PR, therefore, z — (it s

312"P —

[Z — 1 + TRi —a-2± ) id011 PR) R

ar

Jo

(13.55)

The difference (' — s )7:p is called the entropy departure and Z, is called the entropy departure factor. Here again, the generalized entropy departure chart showing Z, as a function of PR and TR can be made. The chart is given in the Appendix. Replacing e by swab we can write Eq. (13.54), for the ds of a gas during a process 1-2, as =

Ru ( Zs, —Zs,)

(1336)

or

s2

= R(Zsi — Z52 ) + (52 — sdideal

(1337)

13.8 FUGACITY The fugacity f is a thermodynamic property which is useful in studying mixtures. It is essentially a pseudo-pressure. When fugacity is substituted instead of pressure, we can, in effect, use the same equations for real gases that we normally use for ideal gases. The concept of fugacity can be described as follows: Consider the Gibbs relation given by Eq. (13.13),

dg = —sdT + vdP At constant temperature,

dgr = vdPr

(13.58)

For an ideal gas, Eq. (13.58) may be written as

dgr = RT

P

= RT d(ln P)r

(13.59)

For a real gas, with the equation of state Pv = ZRT, Eq. (13.58) becomes

dgr = ZRT-- --r- = ZRT d(ln P)r

P

(13.60)

388 Fundamentals of Engineering Thermodynamics This equation enables the evaluation of the chemical potential of a pure gas which approximates ideal-gas behaviour, relative to some arbitrary reference state. As P increases at a given T, the equation Pv = RT may be in considerable error. Thus Eq. (13.60) also may not be acceptable for predicting real-gas behaviour if reasonable accuracy is desired. One way to overcome this difficulty is to use a more accurate equation of state relating v to P and T for the real-gas state. Such an equation could be substituted directly into the vdP term in Eq. (13.58), as it was same for the ideal-gas state. That is, it is essential to use different equations of state for various substances to achieve the desired accuracy. Thus the equation for the chemical potential would have different forms, depending on the equation of state used. It would be more convenient if the mathematical form of Eq. (13.60) is retained for real-gas behaviour also. To accomplish this, the pressure in Eq. (13.60) is replaced by a new thermodynamic property called the fugacity, f Fugacity f is defined as dgT =

RT d(111f )7.

(13.61)

From Eqs. (13.60) and (13.61), we get ain ln Ph.

=z

(13.62)

But Z --) 1 as P --) 0. The only state where all real gases behave like ideal gas is at zero pressure. As we approach this state, f and P should become identical. This condition can be expressed as limp-.o

P

-1 —

Note that the fugacity is expressed in pressure units. In practice, fugacity has essentially the same value as the pressure at system pressures well above zero. As long as the gas behaves like ideal gas, the fugacity equals the pressure. But in reality, many gases are not ideal gases even at atmospheric pressure and a given temperature. Therefore, even at pressures close to atmospheric, we may have to use the fugacity concept for accurate evaluation of chemical potential. Now, consider the change in the Gibbs function of a real gas during an isothermal process at temperature T that involves a change in pressure from a very low pressure P* (where ideal gas behaviour can be assumed) to a high pressure P. Let the Gibbs function corresponding to P* be g*. The value of the Gibbs function at this temperature T and P can be found in terms of the fugacity at P and T and g*, by integrating Eq. (13.61) from P* to P Thus, P.

figi* dgT =

RT (d in f)T

flap

Or

g = g* + RT In (fIP*)

(13.63)

EXAMPLE 13.9 Determine the change in enthalpy and the change in entropy of nitrogen per unit mole as it undergoes a change of state from 101.325 kPa and 300 K to 20.0 MPa and 220 K (a) by assuming ideal-gas behaviour, and (b) by accounting for the deviation from the ideal-gas behaviour.

General Thermodynamic Property Relations 389

Solution The critical temperature and pressure of nitrogen are Tai = 126.2 K and P„; = 3.39 MPa (Table 1 of the Appendix), respectively. Here, nitrogen remains above its critical temperature, therefore, it is in the gas phase, but its pressure is quite high. It will hence deviate from idealgas behaviour and should be treated as a real gas. (a) If nitrogen is assumed to behave as an ideal gas, its enthalpy will depend on T only, and the enthalpy values at the initial and final temperatures can be determined from the ideal-gas table of N2 (Table 9 of the Appendix). —

172, ideal

ideal

= 6391 — 8723 = —2332 kJ/lanol The entropy depends upon both T and P even for ideal gases. Under the ideal-gas assumption, the change in entropy of nitrogen is determined from —

g2

-Ruin 4 1

= 182.639 — 191.682 — 8.314 In 20000 101.325 — 52.984 kJ/(kmol K) (b) The deviation from ideal-gas behaviour can be accounted for by determining the enthalpy and entropy departures from the generalized charts at each state. State 1:

—

—

— 300 — 2 38 Tcy; — 126.2 *

220 T., = = 126.2 — 1.74 — Tai

State 2: PRI —

PI 0.101325 P2 20 c — — 0.03 and PR2 = = — — J.7 Pcri 3.39 Psi 3.39

From the generalized enthalpy departure chart, we get Zhi = 0 and

42 1.45

From the generalized entropy departure chart, we get Zs' = 0 and

Zs2 = 0.75

The enthalpy and entropy changes of nitrogen during this process can be determined by substituting the above values into Eqs. (13.51) and (13.56). Thus, we have

42

— = Rurcri(zhi — Zh2)

(h2 — 171)idm

= (8.314X126.2X0 — 1.45) + (— 2332) —3853.38 kJ/kmol

390

Fundamentals of Engineering Thermodynamics

and — S1 = /2,(Zsi — Z$2) + (12 — = (8314X0 — 0.75) + (— 52.984) —59.22 kJ/(kmol K) Note that the ideal-gas assumption underestimates the enthalpy change by 39.5 per cent and the entropy change by 10.53 per cent.

SUMMARY General thermodynamic property relations express the properties that cannot be measured directly in terms of easily measurable properties. The Maxwell relations are equations that relate the partial derivatives of P, v, T and s of a simple compressible substance to each other. The Gibbs relations are du = Tds — Pdv dh = Tds + vdP a = u — Ts g = h — Ts where a is the Helmholtz function and g the Gibbs function. The Maxwell relations are

(

=

av 2, las )v (ar = (a_z) ap as T

= i— aaT), P

ap5 )T =

(-

ar)p

4 °2-

The Clapeyron equation is a thermodynamic relation that enables the determination of change in enthalpy associated with a phase-change process. It is expressed as d11 ) dT

hfg Tvfg

It is valid for any phase-change process that occurs at constant T and P. The Clapeyron-Clausius equation is the simplified form of the Clapeyron equation valid for liquid-vapour and solid-vapour phase-change processes. It is expressed as p2 ) I )sat

hfg R

T

T2 )sat

General Thermodynamic Property Relations 391 For solid-vapour regions, hfg has to be replaced with the enthalpy of sublimation kg, in the above equation. The general relations of du, dh, ds, Cv, and Cp can be expressed as du= CydT +[T( aP ) - P]dv ar

,

-f dh=CpdT +[v - T (-°V a

)]dP p

ds=S-dT +r) dv T aT „ ds=adT-(-±) 6 dP T ar p

( av ) cp-c,=T(H- )1,

p

) a7' ), av )2 pp cp -c,=-T(or (av )T The isothermal compressibility a and volume expansivity 13, respectively, are defined as a=-

fi =

1 (av —

v ap)r ( av

v

OT )p

The Mayer relation between Cp and C„ can be expressed as Cp

C,

• The difference between Cp and C, approaches zero as the absolute temperature approaches zero. • For incompressible substances Cp and C, are identical, since v = constant. Therefore, for liquids and solids, which are almost incompressible, the difference between Cp and C, is very small and usually disregarded. The Joule-Thompson coefficient is a measure of the change in T with P during a constant enthalpy process. It is defined as =

ar) aP

392 Fundamentals of Engineering Thermodynamics It is evident from the preceding equation that for < 0 T increases p = 0 T remains unchanged > 0 T decreases The general relation for # is

fiTs7;1=-- +[1, -TN ar

P

]

The Ai: of a real gas depends, generally, on P as well as T, and can be expressed as h2 where

Zh

= RTai(Zhi — Zh2) +

— hOideal

is called the enthalpy departure factor, given by

Zh

-

— TOT RT „ „;

az PR - ,,,2 jo ( aT —) cikui pit R

The change in internal energy of a real gas is given by —ul = cri2 — T/ o — R(Z2T2 —47;) The change in entropy of a real gas during a process 1-2 can be expressed as s2 — sl = R(41 — 42 ) + (s2 — si )idem

The Fugacity f is a thermodynamic property which is useful in studying mixtures. It is essentially a pseudo-pressure. When fugacity is substituted instead of pressure, we can, in effect, use the same equations for real gases that we normally use for ideal gases.

PROBLEMS 13.1 Consider air at 400 K and 0.88 m3/kg. Using Eq. (13.1), determine the change in pressure corresponding to an increase of 5 per cent in temperature at constant specific volume. [Ans. 6.52 kPa] 13.2 Find an expression for the change in entropy with volume for an isothermal process involving an ideal gas, using the Maxwell's relations. [Ans. (oldav)r = 13.3 Compare the shape of the solid-liquid equilibrium curve on a P—T diagram for a substance that contracts on freezing with the curve for a substance that expands on freezing. [Ans. For substances that contract on freezing, (v1 — v3) = v51 > 0 and for substances that expand on freezing, v., < 0] 13.4 Derive an expression for the slope of the v = constant lines on a T—P diagram for a gas that obeys the van der Waals equation of state. [Ans. (aT/dP), = (v — b)111

General Thermodynamic Property Relations

393

13.5 Verify the validity of the Maxwell relation given by Eq. (13.17) for steam at 350°C and 500 kPa. 13.6 Determine the relation for (a/dv)T for an ideal gas, using the Maxwell relations and the ideal-gas state equation. [Ans. (61./av)T = R/v] 13.7 Estimate the enthalpy of vaporization of water at 50°C, using the Clapeyron equation, and compare it with the value in the steam table. [Ans. 2396.44 kJ/kg] 13.8 Estimate the enthalpy of vaporization of steam at 500 kPa, using the Clapeyron equation and compare it with the tabulated value. [Ans. 2104.26 kJ/kg] 13.9 Calculate the values of hfg and sfg of steam at 100°C from the Clapeyron equation, and compare them with the values in the steam table. [Ans. 2262.72 kJ/kg, 6.0638 kJ/(kg K)] 13.10 Calculate the big of refrigerant-12 at —10°C on the basis of (a) the Clapeyron equation, and (b) the Clapeyron-Clausius equation. Compare the results with the tabulated value. [ Ans. (a) 159.59 kJ/k2, (b) 169.96 kJ/kg] 13.11 Show that the enthalpy of an ideal gas is a function of temperature only. Also, show that for an incompressible substance the enthalpy depends on temperature and pressure. 13.12 Express the specific heats difference Cp — C„ for (a) an ideal gas, (b) an incompressible substance, and (c) a van der Waals gas in terms of P, v and T. [Ans. (a) C,, —

C. = R, (b)

Cp -

=

0,

-2

(c) Cp — C„ = T

[ RT

2a(b —v) Rv3

T v—b

(v — b)2

13.13 Estimate the volume expansivity 13 and the isothermal compressibility a of steam at 200 kPa and 200°C. [Ans. 0.00221 ICI , 0.00674 kPa'] 13.14 The state equation of a gas is P(v — a) = RT, where a is a positive constant. Will it be possible to cool this gas by throttling? [Ans. No] 13.15 Estimate the Joule-Thompson coefficient of refrigerant-12 at 1 MPa and 140°C. [Ans. 9.375°C/MPa] 13.16 Steam is throttled from 3 MPa and 600°C to 2.5 MPa. Determine the temperature of the steam at the end of the throttling process. [Ans. 598.2°C] 13.17 Determine the enthalpy of nitrogen at 200 K and 10 MPa using the generalized enthalpy chart. Compare this value with the value from the ideal-gas nitrogen table. [Ans. 162.53 kJ/kg]

II

394 Fundamentals of Engineering Thermodynamics 13.18 Estimate the errors involved in the enthalpy and internal energy of oxygen in assuming it as an ideal gas at 200 K and 12 MPa. [Ans. 66.1 per cent, 55.09 per cent] 13.19 Show that C,H --T(ay, )s (8;1 13.20 Oxygen enters an adiabatic nozzle at 10 MPa and 400 K with a low velocity and leaves at 4 MPa and 300 K. Determine the exit velocity of oxygen, using the generalized enthalpy departure chart. [Ans. 426.52 m/s] 13.21 Nitrogen is compressed adiabatically by a steady-flow compressor from 2 MPa and 10°C to 10 MPa and 100°C at a rate of 2 kg/s. Using the generalized chart, determine the required power input to the compressor. [Ans. —176.08 kW]

CHAPTER

14 Reactive Systems

14.1 INTRODUCTION So far, we have considered only nonreacting thermodynamic systems. That is, the chemical composition of all the systems considered remained unchanged during a process. For nonreacting systems, only the sensible energy (energy associated with changes in T and P) and the latent internal energy (energy associated with phase changes) need to be considered. But while dealing with reacting systems, in addition to sensible energy and latent internal energy, we need to consider the energy associated with the destruction and the formation of chemical bonds between the atoms, called ilr chemical internal energy. In this chapter, we will considci only one type of chemical reaction, known as combustion, because of its importance in engineering applications. However, the principles developed are applicable to any chemical reaction.

14.2 FUELS AND COMBUSTION Fuels are those materials which can be burnt to release energy. Most of the commonly used fuels consist primarily of hydrogen and carbon, hence called hydrocarbon fuels, denoted by C,1-1„,. These fuels exist in all phases. for example, coal, gasoline, and natural gas are all used as fuels. Combustion is a chemical reaction during which a fuel is oxidized and a large quantity of energy is released. Air is the commonly used oxidizer in combustion processes. Pure oxygen is used only in specialized applications. A mole of dry air is composed of 20.9% 02, 78.1% N2, 0.9% Ar, and small amounts of CO2, He, Ne, and H2. Usually, in combustion analysis the argon in air is treated as nitrogen, and the gases that exist in small amounts are disregarded. Thus, the composition of dry air can be approximated to 21% of 02 and 79% of N2 by mole numbers. Therefore, each mole of 02 entering a combustion chamber will 1,‘: accompanied by 0.79/0.21 = 3.76 moles of N2. That is, 395

396 Fundamentals of Engineering Thermodynamics 1101101 02 + 3.76 kmol N2 = 4.76 lanol air

(14.1)

At ordinary combustion temperatures, N2 behaves as an inert gas and does not react with other chemical elements. Reactants are the components that exist before the reaction of a combustion process. Products are the components that exist after the reaction. For example, in the reaction C + 02 -4 CO2, C and 02 are reactants and CO2 is the product. It is important to note that merely bringing a fuel into contact with oxygen is not sufficient to start a combustion process. The fuel must be brought above its ignition temperature to start the combustion process. Further, the proportions of the fuel and air must be in the proper range for combustion to begin. For example, natural gas will not burn in air in concentrations less than 5 per cent or greater than 15 per cent.

Mass conservation principle The total mass of each element is conserved during a chemical reaction. Also, the total mass of reactants is equal to the total mass of products. However, the total number of moles is not conserved during a chemical reaction.

Air-fuel ratio AF It is the ratio of the mass of air to the mass of fuel for a combustion process. That is, AF —

(142)

where m = N x M (number of moles x molar mass).

14.3 THEORETICAL AND ACTUAL COMBUSTION PROCESSES A combustion process is complete if all the carbon in the fuel burns to CO2, all the hydrogen bums to H2O, and all the sulphur (if any) burns to SO2. If the.combustion products contain any unburnt fuel or components such as C, H2, CO, or OH, then the combustion process is incomplete. The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. The ratio of actual amount of air used to the stoichiometric amount of air is called the equivalence ratio. The Orsat gas analyser is a commonly used device to analyse the composition of combustion gases. A sample of the combustion gases is collected in this device and cooled to room temperature and pressure. The volume is measured at this state. The sample is then brought into contact with a chemical that absorbs the CO2. The remaining gases are restored to room P and T and the new volume they occupy is measured. The ratio of the reduction in volume to the original volume is the volume fraction of CO2, which is equal to the mole fraction if ideal-gas behaviour is assumed.

Reactive Systems

397

EXAMPLE 14.1 Calculate the air-fuel ratio required for the burning of propane (C3H8) with 150 per cent theoretical air. Solution The chemical equation for this combustion process can be written as C3H8 +

502 ---+ 3CO2 + 4H20

With 150 per cent theoretical air, the combustion equation becomes C3118 + 1.5 [502 + (5X3.76)N2 --+ 3CO2 + 4H20 + 2.502 + (1.5X5X3.76)N2 Note that the air consists of oxygen and nitrogen. The air-fuel ratio (AF) is determined by taking the ratio of the mass of the air and the mass of the fuel. AF

= mak _

(NM)air (NM)c + (NM)

H2

1.5 [5 + 5 x 3.76 kmol] (29 kg/lanol) (3 kmolX12 kg/kmol) + (4 kmolX2 kg/ktnol)

23.53 (kg air)/(kg fuel) That is, 23.53 kg of air is used to burn each kilogram of fuel during this combustion process. EXAMPLE 14.2 Octane (C8H18) is burnt with 50 per cent excess air during a combustion process. The product group contains CO2, H2O, 02 and N2. Assuming complete combustion and a total pressure of 100 kPa, determine (a) the air-fuel ratio, and (b) the dew-point temperature of the products. Solution It is mentioned that the octane is burnt completely. Therefore, the products will contain only CO2, H2O, the unused 02, and N2. The combustion equation for this can be written as C81118 + 1.5ad,(02 + 3.76N2 ) --+ 8CO2 + 9H20 + 0.5a,802 + (1.5 x 3.76)atbN2 where ad, is the theoretical or stoichiometric coefficient of air. The use of 1.5ad, instead of ad, automatically accounts for the 50 per cent excess air. The stoichiometric amount of oxygen (ad,02) will be used to oxidize the fuel and the remaining excess amount (0.5ad,02) will appear as the unused oxygen in the products. The coefficient ad, is determined from the 02 balance: 02: 1.5ad, = 8 + 4.5 + 0.5ad,

398

Fundamentals of Engineering Thermodynamics

or ad, = 12.5 This gives the combustion equation as C81118 + 18.75(02 + 3.76N2 ) ----* 8CO2 + 9H20 + 6.2502 + 70.5N2 (a) The air-fuel ratio is given by AF

(18.75 x 4.76 lcmolX29 kg/kmol) _ mak _ (8 lcmol)(12 kg/kmol) + (9 lcmolX2 kg/kmol) No

=

22.7 (kg air)/(kg fuel)

That is, 22.7 kg of air is supplied for each kilogram of fuel during the combustion process of the present case. (b) The dew-point temperature of the products is the temperature at which the water vapour in the products begins to condense as the products are cooled. Also, we know that the dew-point temperature of a gas-vapour mixture is the saturation temperature of the water vapour corresponding to its partial pressure. Assuming ideal-gas behaviour for the combustion gases, we have P = (-2-Vv (PP rod) = lam' )0 001cPa) = 9.61cPa v N d ( 937 5 kmol The corresponding temperature from Table 4 is Tdp = 44.93°C EXAMPLE 143 A gaseous fuel has the following volumetric analysis: 60 per cent methane (CH4), 30 per cent ethane (C2H6), and 10 per cent hydrogen. The fuel is burnt with the stoichiometric amount of air which enters the combustion chamber at 25°C, 1 atm, and 75 per cent relative humidity. Assuming complete combustion and total pressure of 1 atm, determine the dew-point temperature of the products. Solution The combustion process is assumed to be complete, therefore, all the carbon in the fuel will bum to CO2, and all the hydrogen to H2O. Also, the fuel is burnt with the stoichiometric amount of air, therefore, there will be no free 02 in the products. The moisture in the air does not react with anything, it simply shows up as additional H2O in the products. For simplicity, let us balance the combustion equation by using dry air and then add the moisture to both sides of the equation. The combustion equation for 1 kmol of fuel can be written as (0.6C114 + 0.3C2H6 + 0.1H2) + ath(02 + 3.76N2) -- xCO2 + yH20 + z1•12

Reactive Systems 399 The unknown coefficients in the equation can be determined as follows: C

0.6 + 0.6 =xx=1.2

H:

2.4 +1.8 + 0.2 =4.4 =2yy=2.2

02:

ad, = x +

N2:

3.76aa, = z

ath = 2.3

2

z = 8.648

The partial pressure of the moisture in the air is Pv,air = 4frair Psat @ 25°C =

0.75 x 3.169 = 2.377 kPa

Assuming ideal-gas behaviour, the number of moles of the moisture in the air

Ny, air

is

(P

( ;air Moat • total

2.377 ) (Ndry,atr 101.325

N v,air • )

Ndry,i,. = 4 .76ath = 4.76 x 2.3 = 10.948 lanol Therefore, Nv,sur

(

2.377 )(10.948 + Nv,air 101.325

This gives, NvAir = 0.263 kmol

The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.263 kmol of H2O on both sides of the equation. (0.6CH4 + 0.3C2H6 + 0.1H2) + 2.3(02 + 3.76N2) + 0.263H20 + 2.2H20 + 8.648N2 + 0.2631120 Assuming ideal-gas behaviour, the partial pressure of the water vapour in the combustion gases is "v,prod r

N„,,pmd " r "Prod

rprod

2.463 kmol )

(12.311 kmol

(101.325 kPa) = 20.27 kPa

Thus, Tdp= Tsat @ 20.27 kPa = 60.33 °C EXAMPLE 14.4 An unknown hydrocarbon is burnt with dry air. The volumetric analysis of the products on a dry basis is 12.5 per cent CO2, 0.5 per cent CO, 3.0 per cent 02, and 84.0 per cent N2. Determine (a) the air-fuel ratio, (b) the percentage of the theoretical air used, and (c) the fraction of the H2O which condenses as the products are cooled to 20°C at 100 kPa. Solution From the data, only the relative composition of the products is known. If the -combustion gases

400

Fundamentals of Engineering Thermodynamics

are assumed to be ideal gases, the volume fractions given above become the mole fractions. By considering 100 kmol of dry products (for convenience) the combustion equation can be written as cHy + a(02 + 3.76N2) --)12.5CO2 + 0.5C0 + 302 + 84N2 + bH20 The unknown qualities x, y, a, and b can be determined from mass balances: N2:

3.76a = 84 a = 22.34

C

x = 12.5 + 0.5

R

y = 2b

02:

x = 13

a = 12.5 + 0.25 + 3 + 2

b = 13.18

Thus, y = 26.36 The combustion equation becomes C131126.36 +

22.34(02 + 3.76N2) --) 12.5CO2 + 0.5C0 + 302 + 84N2 + 13.18H20

(a) The air-fuel ratio is given by AF

= mwr _ (22.34 x 4.76)29 _ 16.91 kg air/kg fuel mfiwi (13 x 12) + (13.18 x 2)

(b) To find the percentage of the theoretical air used, we need to know the theoretical amount of air. This is determined from the theoretical combustion equation of the fuel: C13H26.36

ad,(02 + 3.76N2) 02:

ad, = 13 +

13CO2 + 13.18H20 + 3.76ad,N2 13.18 2

Thus, the percentage of the theoretical air used =

ad, = 19.59 trct

Nair act_

mairth

Nair,th

22.34 x 4.76 - 1.14 19.59 x 4.76 114 per cent That is, 14 per cent excess air was used during this combustion process. (c) For each kmol of the fuel burnt, 113.18 kmol of products are formed. Let us assume that the dew-point temperature of the products is above 20°C and some of the water vapour will condense as the products are cooled to 20°C. If NW kmol of H2O condenses, there will be (13.18 - NN) kmol of water vapour left in the products. The mole number of the products in the gas phase will decrease to (113.18 - NW). By treating the product gases (including the water

Reactive Systems 401 vapour) as ideal gases, Na, is determined by equating the mole fraction of the water vapour to its pressure fraction. That is, N, _ P,„ NProd4as PProd Or

13.18 — N,„ _ 2.339 113.18 — Na, 100

Pv = gat @ 20•C = 2.339 kPa)

Therefore, N,,, = 10.78 kmol

14.4 ENTHALPY OF FORMATION AND ENTHALPY OF REACTION The molecules of a system possess various forms of energy, such as sensible energy, latent energy, chemical energy, and nuclear energy. A process which involves chemical reactions will involve changes in chemical energy, which must be accounted for in an energy balance. In the absence of nuclear reactions and Ake and Ape, the energy change of a system during a chemical reaction will only be due to change of state and change in chemical composition. That is, Esys = A Estate + AEc con,.

(143)

In our analysis, since we are concerned with the changes in the energy of a system during a process, and not with the absolute values of energy at a particular state, any state can be chosen as the reference state and assigned a value of zero to U or H of a substance at that state. To have the advantage of a common reference state for all substances, it is a common practice to choose 25°C and 1 atm as the reference state. This is known as the standard reference state, and the properties at this state are denoted by a superscript "0".

14.4.1 Enthalpy of Reaction hR It is the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. The Ah will be different for different reactions, and it would be desirable to have a property to represent AEche, comp. during a process. The hR is used for such representation. For combustion processes, the hR is usually referred to as the enthalpy of combustion h„ which represents the amount of heat released during a steady-flow combustion process when 1 kg (or 1 kmol) of fuel is burnt completely at a specified temperature and pressure. It is expressed as

k = hp -

(14.4)

where hp is the enthalpy of the products and hr is the enthalpy of the reactants. From Eq. (14.4) it is evident that, he is a useful property for analysing the combustion processes. However, it is impractical to list the values of he for all possible fuels and fuel mixtures, since there are

402 Fundamentals of Engineering Thermodynamics numerous fuels. Also, for incomplete combustion, h, is not of much use. Therefore, a more practical approach would be to have a fundamental property that would represent the chemical energy of an element or a compound at some reference state. The enthalpy of formation hi- is such a fundamental property, which is the enthalpy of a substance at a specified state owing to its chemical composition. All stable elements, such as 02, N2, H2, and C have a value zero as their enthalpy of = 0 for all stable elements. formation at the standard reference state (25°C and 1 atm). That is, The stable form of an element is simply the chemically stable form of it at 25°C and 1 atm. For example, the stable form of hydrogen at the standard reference state is diatomic hydrogen H2, not the monatomic hydrogen H. When an element exists in more than one stable form at the standard reference state, one of the forms should be specified as the stable form. Let us consider the reaction of burning carbon with oxygen to yield carbon dioxide as the product, C(s) + 02(g) CO2(g) If the reaction is carried at the standard reference state, then the heat transferred from the system is the standard enthalpy of reaction given by q = h° = 11° —

_' 0:

- °-/7,32

and ii,3 are zero. Therefore, the quantity of heat transferred may be considered as the But heat involved in the forming of 1 mol of CO2. From the enthalpy of formation table (Table 17 of the Appendix), kb, = —393,520 kJ/kmol. The negative sign is due to the fact that the enthalpy of 1 kmol of CO2 at 25°C and 1 atm is 393,520 Id less than the enthalpy of 1 kmol of C and 1 kmol of 02 at the same state. In other words, 393,520 Id of chemical energy is released when C and 02 combine to form 1 kmol of CO2. Thus, a negative hf for a compound indicates that heat is released during the formation of the compound from its stable elements. Heating value, defined as "the amount of energy released when a fuel is burnt completely in a steady-flow process and the products are returned to the state of the reactants," is another term commonly used in the study of combustion of fuels. The heating value (HV) of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel, that is, HV = Ihcl [(ki/(kg fuel)] The HV depends on the phase of the H2O in the products. It is called the higher heating value HHV when the H2O in the products is in the liquid form, and it is called the lower heating value LHV when the H20 in the products is in the vapOur form. These two heating values are related by HHV =

( I4fg )H 2o

(kJ/(kg

fuel)]

where N is the number of moles of H2O in the products and hfg is the enthalpy of vaporization of water at the specified temperature. EXAMPLE 14.5 Determine the lower heating value (LHV) of methane gas at 25°C and 1 atm.

Reactive Systems 403 Solution The stoichiometric equation for this reaction is CH4 + 202 ----> CO2 + 2H20 Thus the amount of H2O formed is 2 x 18 - -2.25 kg/(kg fuel) 6

MH 0 = 2

From the enthalpy of combustion table (Table 18 of the Appendix), we have, for methane Ak°

= —890,360 kJ/kmol —

— 890,360 kJ/(kg fuel) 16

= —55647.5 kJ/(kg fuel) This value of heat of combustion corresponds to HHV since the H2O in the product is in liquid form. Thus, HHV = IN = 55647.5 kJ/(kg fuel) At 25°C, the enthalpy of vaporization of water ilk = 2442.3 kJ/kg (from Table 3). Also, HHV = LHV + (Mi.& ) kJ/(kg fuel) 11,0 Therefore, LHV = HHV — mH20/7fg = 55,647.5 — (2.25 x 2442.3) kJ/(kg fuel) 50,152.3 kJ/(kg fuel)

14.5 FIRST-LAW ANALYSIS OF REACTING SYSTEMS It will be more convenient, for the analysis of reacting systems, to rewrite the first-law relation so that the changes in chemical energies are expressed explicitly.

14.5.1 Steady-Flow Systems When the AKE and APE are negligible, the conservation of energy relation for a chemically reacting steady-flow system can be expressed as Q — W = Hp — H,. (kJ) where H p =EN p (17, +I — 17) (kJ) P

(14.5)

404 Fundamentals of Engineering Thermodynamics and

H, =Eisc(riy+T: —h°), (kJ) Therefore,

Q — w =EN(h, +17 - TO) — EN,.(ray +Ts

- 17)

(14.6)

where II/ is the chemical enthalpy of a substance at the standard reference state and the term h — TO represents the sensible and latent enthalpy relative to the standard reference state. The Np and N, represent the number of moles of the product and the reactants, respectively. In terms of the enthalpy of combustion re, Eq. (14.6) can be expressed as

2 _ w = ko

EN, _ 71 °1 (1. 1)

Np

ir

IP

(14.7)

EXAMPLE 14.6 Determine the enthalpy of combustion of gaseous propane (C3118) at 25°C and 1 atm. Assume that water in the product is in the liquid form. Solution The stoichiometric equation for this reaction is C3H8 + ad,(02 + 3.76N2)

3CO2 + 4H20(1) + 3.76ad,N2

Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also,

N2 and 02 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C3H8 becomes

k = hp — hr

Or

= (Nhpco,

k =I NPI;;P —

(Av).20 —

Using the enthalpy of formation Table 17, we get Tic = (3 lano1X-393,520 kJ/kmol) + (4 kmo1X-285,830 kJ/kmol) — (1 kmo1X-103,850 kJ/Icmol) —2,220,030 kJ/kmol C31-18 Note: The listed value in the enthalpy of combustion table is —2,220,000 kJ/kmol C3H8. Since the water in the products is assumed to be in liquid phase, this he value corresponds to the HHV of cgi..

14.5.2 Closed Systems For a chemically reacting closed system, the conservation of energy relation can be expressed as Q — W = Up — Ur (kJ)

(14.8)

Reactive Systems

405

where the subscripts p and r denote products and reactants, respectively. To avoid the usage of another property, namely the internal energy of formation t7, , let us use to express Eq. (14.8). By definition, = 17 — PT,

or

tif) + Ft. — 17° =

+—°— P

Thus, we can express Eq. (14.8) as Q

_w=E

E Air (171 + —

P

—

(kJ) (14.9)

For solids and liquids, the Pv terms are negligible and for gases which behave as an ideal gas the Pri terms can be replaced by Ru T. EXAMPLE 14.7 Methane (CH4) gas is burnt with the stoichiometric amount of oxygen gas. The water in the products is in the gas phase. Determine the heat released or absorbed if the reaction occurs at 25°C and 1 atm Solution The combustion equation is CH4 + 202

CO2 + 2H20

Both the reactants and products are at the standard reference state of 25°C and 1 atm. Also, 02 is a stable element and thus its enthalpy of formation is zero. The heat transfer is a steady-flow combustion process and is, therefore, determined from Eq. (14.6) with W = 0. Thus, we have Q

= I NpuTy + h - 17% -

Aviv +

Or

Q

=E N

p

l

h

E

Nrhl

= (1 kmol CO2X-393,520 kJ/(kmol CO2) + (2 kmol H2O)

(-241,820 kJ/krnol H2O) — (1 kmol CH4)( —74,850 kJ/(kmol CH4) — (2 kmol 02X0 kJ/kmol 02) —802,310 k.1/(kJmol CH4 ) Note that the Q is independent of the amount of oxidant supplied to the reaction. Even if excess amount of air is supplied, the 02 and N2 enter and leave at the standard reference temperature. Hence, the value of /IT — h798.15 is zero for these elements regardless of the quantity of each. EXAMPLE 14.8 One kmol of methane (CH4) gas is burnt completely with 10 per cent excess air in a constant-

406

Fundamentals of Engineering Thermodynamics

volume container. Before combustion, both methane and air are at 25°C and 1 atm. The combustion products are at 150°C at the end of the combustion. Determine the heat transfer during the process. Solution Since the combustion is complete, the combustion equation with 110 per cent theoretical air is CH4 + 1.1(202 + 2 x 3.76N2) --o CO2 + 2H20 + 0.202 + 1.1 x 2 x 3.76N2 Note that in the event of complete combustion, all the carbon in the methane burns to CO2 and all the hydrogen to H2O. The heat transfer during this constant-volume combustion process can be determined from Eq. (14.9):

Q—w=E

+ — — PV)p — E

+Ts —

— Pv"),

If we assume both the reactants and products as ideal gases, all the internal energies and enthalpies depend on temperature only, and the Pi; terms in the above equation can be replaced by R.T. Thus, we get

Q = E N(h,

/74 2 3 K

172 9 8 K

RuT) p

E Nr(171 —

Runr

since W = 0 and the reactants are at the standard reference temperature of 25°C. From ideal gas tables, we have Substance

k°f (kJ/kmol)

CH4

—74,850 0 0 —393,520 —241,820

02 N2

CO2 H20(g)

17423

K (kJ/kmol)

h298 K (kJ/kmol)

—

12,405.2 12,312.9 14,332.6 14,146.5

8682 8669 9364 9904

Substituting, we get Q = (1 kmol CO2)[(-393,520 + 14,332.6 — 9364 — 8.314 x 423) kJ/(lcrnol CO2)] + (2 kmol H20)[(-241,820 + 14,146.5 — 9904 — 8.314 x 423) kJ/(lanol H2O)] + (0.2 kmol 02)[(0 + 12,405.2 — 8682 — 8.314 x 423) kJ/(lcmol 02)1 + (8.272 kmol N2)[(0 + 12,312.9 — 8669 — 8.314 x 423) kJ/(kmol N2)] —(1 kmol CH4)[(-74,850 — 8.314 x 298) Ic.1/(lanol CH4)] —(2.2 kmol 02)[(0 — 8.314 x 298) kJ/(kmol 02)1 —(8.272 kmol N2)[(0 — 8.314 x 298) kJ/(kmol N2)] —769,888.42 1c1/(kmol CH4)

and

Reactive Systems 407 The negative sign indicates that heat is transferred from the combustion chamber to the surroundings. The heat transfer per unit mass of methane is Q = —769,888.42/16 = —48,118.026 kJ/(kg CI-1 4 )

14.6 ADIABATIC FLAME TEMPERATURE The chemical energy released- during a combustion process is lost as heat to the surroundings or is used internally to raise the temperature of the combustion products, when work interactions and AKE and APE are absent. In the limiting case of no heat loss to the surroundings (Q = 0), the temperature of the products will reach a maximum, which is called the adiabatic flame temperature of the reaction. In other words, the adiabatic flame temperature is the maximum temperature that the combustion products can attain when there is no heat loss from the system to the surroundings, during the combustion process. For a steady-flow combustion process, the adiabatic flame temperature is determined from Eq. (14.5) with Q = 0 and W = 0, as Hp = H,. (14.10) Or

E /kip (Ts,

= E N, (1:1- + h —h° )

(14.11)

For the case of no excess air, the adiabatic flame temperature is called the theoretical adiabatic flame temperature for the fuel. It is the highest temperature that can be obtained from the fuel used in the combustion reaction (unless we preheat some of the reactants to temperatures above the standard reference state), since no heat transfer is allowed to occur to the surroundings, and no excess air is present to absorb any of the enthalpy of combustion. EXAMPLE 14.9 Liquid butane (C41110) at 25°C and 1 atm is burnt with air at the same state in a steady-flow combustion chamber. Neglecting the changes in kinetic and potential energies, determine the adiabatic flame temperature (a) for complete combustion with 100 per cent theoretical air, (b) for complete combustion with 400 per cent theoretical air, and (c) for incomplete combustion with 80 per cent theoretical air. Solution Under adiabatic conditions, there is no heat transfer from the combustion chamber, and the combustion products will exit at the highest possible temperature, known as the adiabatic flame temperature. (a) The combustion equation for the process with 100 per cent theoretical air is C41110(/) + 63(02 + 3.76142) —> 4CO2 + 5E120 + 24.44N2 The adiabatic flame temperature relation given by Eq. (14.11) is

408

Fundamentals of Engineering Thermodynamics

E Np (17; + — h°)= E N,(11, + ti

(Niq)c4.,.

-711 r

Since all the reactants are at the standard reference state and ky = 0 for 02 and N2, the h values of the different components at 298 K are Substance

rif (kJ/kmol)

Qin) (I)

— 147,210 0

02 N2

11298 (kjilami)

0

H2O (g) CO2

and

— 241,820 — 393,520

8682 8669 9904 9364

Note that the tabulated value of 1:1,298 for n-butane in the table is for the gas phase. But we need h, for the liquid phase. This can be obtained by adding hik to hj-) for the gas phase. That is,

Ti(/) = 1,7 (g) + hfg

Substituting, we get (4 kmol CO2)[(-393,520 + lic02 — 9364) kJ/(kmol CO2)] + (5 kmol H20)[(-241,820 + / 420 — 9904) kJ/(kmol H2O)] + (24.44 kmol N2)[(0 + T14 2 - 8669) k.Klanol N2)] = (1 kmol C4HioX-147,210 kJ/(lanol C4H1 o) This gives 41k-02 + 514420 + 24.44/N2 = 2,934,816.36 Now, we have one equation with three unknowns. But note that we have only one unknown— the temperature of the products Tp, since h = h(7) for ideal gases. Therefore, we will have to solve this equation by trial and error. • A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which gives 2,934,816.36/33.44 = 87,763.65 kElanol. • This enthalpy value will correspond to about 2625 K for N2, 2100 K for H2O and 1780 K for CO2. • Noting that the major portion of the moles belongs to N2, we see that Tp will be close to about 2600 K. Therefore, let Tp = 2600 K. This yields 4(137,449) + 5(114,273) + 24.44(86,650) = 3,238,887 This is higher than the actual value of 2,934,816.36.

Reactive Systems 409 Now, let Tp = 2400 K. This gives 4(125,152) + 5(103,508) + 24.44(79,320) = 2,956,728.8 This is slightly more than the actual value. Thus, the adiabatic flame temperature is slightly less than 2400 K. Hence, let Tp = 2350 K. This gives 4(122,091) + 5(100,846) + 24.44(77,496) = 2,886,596.24 This is slightly less than the actual value. Therefore, let Tp = 2380 K. This gives 4(123,927.6) + 5(102,443.2) + 24.44(78,590.4) = 2,928,675.776 This is also slightly less than the actual value. So, let Tp = 2390 K. This gives 4(124,539.8) + 5(102,975.6) + 24.44(78,955.2) = 2,942,702.288 This is slightly more than the actual value. Thus, the adiabatic flame temperature lies between 2390 K and 2380 K That is, Tp = 2385 K (approximately) (b) The balanced equation for the complete combustion process with 400 per cent theoretical air is C4H10(/) + 26(02 + 3.76N2) —) 4CO2 + 5H20 + 19.502 + 97.76N2 By following the above procedure, the adiabatic flame temperature in this case is found to be Tp = 1215 K (approximately). Note that the temperature of the products decreases considerably as a result of using excess air in the combustion process. (c) The equation for the incomplete combustion process with 80 per cent theoretical air is C41-110(/) + 5.2(02 + 3.76N2) —+ 1.4CO2 + 2.6C0 + 5H,0 + 19.55N2 By following the above procedure, the adiabatic flame temperature in this case is found to be Tp = 2150 K Note that the adiabatic flame temperature decreases as a result of incomplete combustion, or complete combustion using excess air. Also, the maximum adiabatic flame temperature is achieved only when complete combustion occurs with the theoretical amount of air.

14.7 ENTROPY CHANGE By the second law of thermodynamics, we have Sgen =ASsr +AS=Fr

(kJ/K)

(14.12)

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Fundamentals of Engineering Thermodynamics

whether the system is of chemically reacting type or not. For reacting systems, the ASsys can be taken to represent the entropy change associated with the reaction within the reaction chamber boundaries, which is equal to (Sp — Sr). That is% Ssys = Sp — =

E Npip — E Nrir OcJIK)

(14.13)

and (kJ/K) (14.14) To where Q,is the heat transferred from the system to the surroundings and To is the temperature of the surroundings, which is assumed to remain constant. S,,, —

14.7.1 Third-Law of Thermodynamics The Sp and Si in Eq. (14.13) involve the entropy of the components, and not the entropy changes which is the case for nonreacting systems. Thus, it is necessary to find a common base for the S of all substances, as we did with enthalpy. The search for such a common base led to the establishment of the third law of thermodynamics, which states that: The change in entropy of a pure crystalline substance at absolute zero temperature is zero. The third law provides an absolute base for the entropy values of all substances. The entropy values relative to this base are called the absolute entropies. The i° values listed in Tables 9-16 of the Appendix for various gases such as N2, 02, CO2, CO, H2, H20, 0, and OH are the ideal-gas absolute entropy values at the specified temperature and at a pressure of 1 atm. Equation (14.13) is a general relation for the entropy change of a reacting system. It requires the determination of entropy of each individual component of the reactants and products, which is a laborious task. The entropy(S) calculations can be simplified significantly if the gaseous components of the reactants and products are approximated as ideal gases. However, the calculations of S are always difficult compared to those of H or 1E, since S is a function of both T and P even for ideal gases. For evaluating S of a component of an ideal-gas mixture, we should use the temperature and the partial pressure of the component. The absolute S at P other than P0 = 1 atm for any T can be obtained from the ideal-gas entropy change relation written for an imaginary isothermal process between states (T, P0) and (T, P), as shown in Fig. 14.1. T

P0 = 1 atm

Or. P) T

3°( T Po) (Tabulated)

s As = —R In — Po

Figure 14.1

T—s diagram.

Reactive Systems 411 (T, P) = (T, Po ) — R„ In — [kJ/(kmol

(14.15)

Po For a component i of an ideal gas mixture, Eq. (14.15) becomes

S; (T, Pi ) = gi° (T, Po ) —

ln YiP — m [kJ/(kmol Po

(14.16)

where Pi is the partial pressure, Y, the mole fraction of the component, and P„, is the total pressure of the mixture.

14.8 SECOND-LAW ANALYSIS OF REACTING SYSTEMS The irreversibility I associated with a chemical reaction can be determined from

1= ToSgeo (kJ)

(14.17)

where To is the temperature of the surroundings and Sgen is the total entropy change or the entropy generation. In the analysis of reacting systems, we are interested in the change in the availability, known as the reversible work Wre„ rather than the values of availability at various states. In the absence of AKE and APE, Wre„ for a steady-flow combustion process, which involves heat transfer only with the surroundings at To, can be obtained by replacing the enthalpy terms in Eq. (6.21) by 17, + h + Ts° , which gives W„„ = E

E

— Ts°

+

T° — T0.1)

Np (T1, +

For a combustion chamber involving heat transfer QR with a reservoir at W

= E Air (rs, +

— Ts° — ro s——)

— E Np (4 + _ — _ ° — 4-7)

P

(14.18)

TR, Eq. (14.18) becomes

To ) — QR (1 — TO

(14.19)

Gibbs function For the • situation when both the reactants and the products are at the temperature of the surroundings To, we get from Eq. (13.11) — To.I =

— T07)4 = So

where go is called the Gibbs function of a unit mole of a substance at To. The Wre„ for this case becomes wrev

=E

E

Nigo.R

NP8o, p

(14.20)

Or

wrev = E+gTo where

—

_

E Np (4) + kr. — k° )

(14.21)

is the Gibbs function of formation, which is zero for stable elements at the standard

412 Fundamentals of Engineering Thermodynamics reference state. The term (4 — go) represents the sensible Gibbs function of a substance at To relative to the standard reference state. For the special case with T, = Tp = To = 25°C and the partial pressure P1 = 1 atm, for each component of the reactants and the products, Eq. (14.21) simplifies to

Wrev = The

4

Nrnr

(14.22)

I Np.O.p

values of several substances are given in Table 17 in the Appendix.

EXAMPLE 14.10 One kmol of carbon monoxide at 25°C and 1 atm is burnt steadily with 1 kmol of oxygen at the same state. The carbon dioxide formed is then brought to 25°C and 1 atm, the conditions of surroundings. Assuming the combustion to be complete, determine the reversible work for this process. Solution The combustion equation is CO + 2 02

-+ CO2

The CO, 02, and CO2 are at 25°C and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products (Eq. 14.22). That is, we have Key

=

Acg, —

Npkl,p

1 m 0,0

NCO4C0 + — 2 74gf,o2

Nc024:c02

= Nc04.co — Na2.2 4,c02 = —137,150 — (-394,360) 257,210 kJ/kmol since the 4 of the stable elements at 25°C and 1 atm is zero. Therefore, 257,210 Id of work is done as 1 kmol of CO is burnt with 1/2 kmol of 02 at 25°C and 1 atm in an environment at the same state. The reversible work in this case represents the availability of the reactants, since the product (the CO2) is at the state of the surroundings. The reversible work can also be determined, without involving the Gibbs function. By using Eq. (14.19), we have Wre, =

E Air (Ti, T — he — Toy) — E Np (17, E N„ — T0.7), — E N T0.7) p

— h° — Toy)

p

= Nco (T11 — To7 )co+ NO2 (11-1- ToT)o,

N co, 0-, Toi )co,

Reactive Systems 413 Or Wre, = (1 kmol CO)[-110,530 kJ/kmol - 288.15(197.65) kJ/kmol] 1 + (- kmol 02)[0 - 288.15(205.04) kJ/lanol] 2 - (1 kmol CO2)[-393,520 kJ/kmol - 288.15(213.80) kJ/kmol] 258,102.5 kJ/kmol EXAMPLE 14.11 Butane (C4Hio) gas enters a steady-flow adiabatic combustion chamber at 25°C and 1 atm. It is burnt with 30 per cent excess air which also enters at 25°C and 1 atm. Assuming complete combustion, determine (a) the temperature of the products, (b) the entropy generation, and (c) the reversible work and irreversibility. Assume that the products leave the combustion chamber at 1 atm pressure and To = 298 K. Solution The combustion process equation with 30 per cent excess air is C41-110(g) + 8.45(02 + 3.76N2)

4CO2 + 5H20 + 1.9502 + 31.772N2

The adiabatic flame temperature relation, under steady-flow conditions, reduces to

E

= (N Tin C4Hio

Np

since all the reactants are at the standard reference state and Ili = 0 for 02 and N2. The 17, and h values of various components at 298 K (under ideal-gas assumption for air and for the products) are: Substance C4H10 (g)

02 N2 H2O (g) CO2

hj

(kJ/lanol)

11298K OCRICMOD

- 126,150 0 0

- 241,820 - 393,520

Therefore, (4 kmol CO2)R-393,520 + hco, - 9364) kJ/(kmol CO2)] + (5 kmol H20)[(-241,820 + /7H20 - 9904) kJ/(kmol H20)] + (31.772 kmol N2)[(0 + rtN2 - 8669) kJ/(krnol N2)] + (1.95 kmol 02 )[(0 + 1702 - 8682) kJ/(kmol OD] = (1 kmol C41-110)(-126,150 kJ/kmol C41-10

8682 8669 9904 9364

414

Fundamentals of Engineering Thermodynamics

This gives 4/7co, + 5& 0 + 1.95h0 + 31.772IN, = 3,036,367.368 By trial and error, the temperature of the products is found to be Tp = 2030 K (b) The entropy generation during the process is given by S8 = ASsys + But ASs,„, = 0, since the process is adiabatic. Thus, we get •

= ASsys = Sp — S,. =

E

Npip -

E

Nrir

The C4Hio is at 25°C and 1 atm, and thus its absolute entropy is 1c41-1 10

= 310.12 kgkinol K), from Table 17 of the Appendix

The entropy of any component i is given by •

(YiPm)]

= Nisi(TM) = Nire(T,Po)

where Pi = YiPlutui is the partial pressure, Y, is the mole fraction of the component i, and Ru is universal gas constant. Component

Ni

Yi

.11 (T,

C4H10

1 8.45 31.772

1 021 0.79

310.12 205.04 191.61

02 N2

1 atm)

— Ru ln(Yign ) 12.98 1.96

Niii 310.12 184227 6150.11

Sr = 8302.50 CO2 H2O 02 N2

4 5 1.95 31.772

0.0936 0.1170 0.0456 0.7437

310.105 26533 269.22 252.50

19.694 17.838 25.672 2.462

131920 1415.84 575.04 8100.65 Sp = 11,410.73

Thus, Sgen = Sp —Sr = 11 410•73 — 8302.50 = 3108.23 kJ/(kmol K)C4 Hio (c) The irreversibility associated with this process is given by 1 = ToSsen = 298 x 3108.23 926,252.54 kJ/kmol C41-11,3 That is, 926,252.54 Id of work potential is wasted during the combustion process for each mole of butane burnt.

Reactive Systems 415 This process involves no actual work. Therefore, the reversible work and the irreversibilities are identical. Thus, we have

w=

926,252.54 kJ/(kmol C41110 )

EXAMPLE 14.12 In a steam power plant 325,000 kg of water per hour enters the boiler at 12.5 MPa and 200°C. Steam leaves the boiler at 9 MPa and 500°C. The power input to the turbine is 81 MW. Coal with a higher heating value of 33,250 kJ/kg is used at the rate of 26,700 kg/h. Determine the efficiency of the steam generator and the overall thermal efficiency of the plant. Solution In a power plant the efficiency of the boiler and the overall efficiency of the plant are based on the HHV of the fuel. The efficiency of the boiler is defined as %team generator —

heat transferred to H20/kg of fuel HHV of fuel

Therefore, _ 325,000(3386.1— 852.45) 100 liste*m Maid" 26,700 x 33,250 x 92.75 per cent The thermal efficiency is defined as rith

HHV

Thus, nth —

81,000 x 3600 x 100 26,700 x 33,250 32.85 per cent

EXAMPLE 14.13 A gas-turbine uses a liquid hydrocarbon fuel that has an approximate composition of octane C81-118. The fuel at 50°C is mixed with air at 400 K and 100 m/s. If the actual air-fuel ratio is 48 (kg air)/(kg fuel) and the combustion product temperature and velocity are 1100 K and 150 m/s, determine the combustion efficiency of this process, treating the air and combustion products as ideal gases. Solution For the ideal chemical reaction the heat transfer is zero. Therefore, by first-law, we have HR + KER = Hp + KEp

416 Fundamentals of Engineering Thermodynamics Now,

HR

KER =

(T11 + + jn-111) 2

= [1:1 + mEp(50 — 25)] + No, colis 0) — m V2 + 3.76 NO2 (Ah +

22 ) mV

142

= [-249,950 +1.7113 x 114 x (50 — 25)] + NO2 ( 3034 +

+ 3.76 N0 (2971 + 2

32 x1002 ) 2x1000

28 x1002 ) 2x1000

= — 245,073 +14,891NO2 and Hp + KEp =8 Li; +A/7+ V2 Ico,

+ 9 (iif° +Ali+ ni12 ) 2 H2O

+ (NO2 —12.5) (Ai + ??12 2 = 8 (—393,520 + 34,886 +

•-•2

+ 3.76 No, A/7 + 1V1 2 m4.2

44x1502 x1502 x1000 + 9 I —241,820 + 26,715 + 18 x 1000 2 2

+ (NO2 —12.5)(23,188 +

32 x1502 28 x1502 ) + 3.76NO2 (21,786 + 2 x 1000 2 x 1000

= —2,865,112 — 1,934,122.5 — 294,350 + 23,548 N0, + 83,100/402 Therefore, or or

= —509,384 +106,648NO2 —245,073 + 14,891NO2 = —5,093,584 + 106,648NO2 91,757NO2 = 4,848,511 NO2 = 52.84 (kmol 02)/(1cmol fuel)

Thus, (lanol air)/(lanol fuel) = 4.76 x 52.84 = 251.52 and AFickai

28.97 x 251.52 = 63.92 114

Reactive Systems 417 The combustion efficiency is given by 'bomb

= AF.tua Akin] = 48 — x 100= 75 per cent 63.92

SUMMARY While dealing with reacting systems, in addition to sensible and latent internal energy, the chemical internal energy should also be taken into account. Fuels are those materials which can be burnt to release energy. Combustion is a chemical reaction during which a fuel is oxidized and a large quantity of energy is released. Reactants are the components that exist before the reaction of a combustion process. Products are the components that exist after the reaction. The air-fuel ratio AF is the ratio of the mass of air to the mass of fuel for a combustion process. That is, —

main a

The standard reference state is the state with 25°C and 1 atm. The enthalpy of reaction hR is the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. For combustion processes, the hR is usually referred to as the enthalpy of combustion he, which represents the amount of heat released during a steady-flow combustion process when 1 kg (or 1 kmol) of fuel is burnt completely at a specified temperature and pressure. It is expressed as he = hp hr The enthalpy of formation hf is the enthalpy of a substance at a specified state owing to its chemical composition. The enthalpy of formation of all stable elements is zero at the standard reference state (25°C and 1 atm). The stable form of an element is simply the chemically stable form of it at 25°C and 1 atm Heating value is defined as "the amount of energy released when a fuel is burnt completely in a steady-flow process and the products are returned to the state of the reactants." The heating value (HV) of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. The HV depends on the phase of the H2O in the products. It is called the higher heating value HHV when the H2O in the products is in the liquid form, and is termed the lower heating value LHV when the H2O in the products is in the vapour form. These two heating values are related by HHV = LHV + (Niif)20 [(kJ/(kg fuel)] where N is the number of moles of H2O in the products and Ts is the enthalpy of vaporization of water at the specified temperature. The adiabatic flame temperature of a reaction is the maximum temperature that the products can attain during a combustion process when there is no heat loss to the surroundings

418

Fundamentals of Engineering Thermodynamics

(Q = 0) in the absence of work interaction and AKE and APE. For the case of no excess air, the adiabatic flame temperature is called the theoretical adiabatic flame temperature for the fuel. It is the highest temperature that can be obtained from the fuel used in the combustion reaction. For reacting systems, the ASsy, can be taken to represent the entropy associated with the reaction within the reaction chamber boundaries, which is equal to (Sp — S1). That is, Ssys = Sp — S, =

Npip —

(kJ/K)

and

QSUIT

(U/K)

where Q'sun is the heat transferred from the system to the surroundings and To is the temperature of the surroundings, which is assumed to remain constant. The Sp and S,. in the above equation involve the entropy of the components, and not the entropy changes which is the case for nonreacting systems. Thus, it is necessary to fmd a common base for the S of all substances, as we did with enthalpy. The search for such a common base led to the establishment of the third law of thermodynamics, which states that: The change in entropy of a pure crystalline substance at absolute zero temperature is zero. The third law provides an absolute base for the entropy values of all substances. The entropy values relative to this base are called the absolute values. The irreversibility I associated with a chemical reaction can be determined from I = ToSgen (U) where To is the temperature of the surroundings and Sr„ is the total entropy change or the entropry generation.

PROBLEMS 14.1 A fuel approximately equivalent to hexane (C61114) is burnt with the stoichiometric amount of air during a combustion process. Assuming complete combustion, determine the airfuel ratio. [Ans. 15.25 (kg air)/(kg fuel)] 14.2 Methane (CH4) is burnt with 50 per cent excess air in a combustion process. Assuming complete combustion, determine the air-fuel and fuel-air ratios. [Ans. 25.58 (kg air)/(kg fuel), 0.0386 (kg fuel)/(kg air)] 143 Propane (C3H8) is burnt with the stoichiometric amount of air during a combustion process. Assuming complete combustion, determine the air-fuel ratio on a mass basis as well as on a mole basis. [Ans. 15.69 (kg air)/(kg fuel), 23.8 (kmol air)/(kmol fuel)] 14.4 One kmol of propane (C3H8) is burnt with an unknown amount of air during a combustion process. An analysis of the combustion products shows that the combustion is complete, and there is three kmol of free 02 in the products. Determine (a) the air-fuel ratio, and (b) the percentage of the theoretical air used during this process. [Ans. (a) 25.1 (kg air)/(kg fuel), (b) 160 per cent]

Reactive Systems

419

14.5 Methane (CH4) is burnt with 150 per cent theoretical air during a combustion process. Assuming complete combustion and a total pressure of 150 kPa, determine (a) the air-fuel ratio, and (b) the dew-point temperature of the products. [Ans. 25.58 (kg air)/(kg fuel), 59.61°C] 14.6 The volumetric analysis of a natural gas is 79 per cent methane (CH4) and 21 per cent ethane (C2H6). If this fuel is burnt with air, what is the theoretical air-fuel ratio on a mass basis and also on a mole basis? Assume complete combustion. [Ans. 16.87 (kg air)/(kg fuel), 11.02 (kmol air)/(kmol fuel)] 14.7 Octane (C81-118) is burnt steadily with air in a jet engine. If the air-fuel ratio is 20 kg air/kg fuel, determine the percentage of the theoretical air used during the process. [Ans. 132.13 per cent] 14.8 Propane (C3H8) is burnt with 130 per cent theoretical air at a pressure of 95 kPa. If the air entering is dry, determine (a) the mole analysis of the product gas, assuming complete combustion, (b) the dew-point temperature of the products, and (c) the percentage of the H2O formed, that is, condensed if the product gases are cooled to 20°C. [Ans. (a) 9.11 per cent CO2, 12.14 per cent H2O, 4.55 per cent 02, 74.20 per cent N2, (b) 48.90°C, (c) 81.75 per cent] 14.9 A fuel has 83.7 per cent of C and 16.3 per cent of H. If the dry products analysis by vohnne is CO2 11.8 per cent, 02 3.7 per cent, and N2 84.5 per cent, determine the air-fuel ratio. [Ans. 18.33 (kg air)/(kg fuel)] 14.10 Determine the enthalpy of combustion of methane (CH4) at 25°C and 1 atm. Assume that the water in the products is in the liquid form. [Ans. - 890,330 LT/(kmol CH4)] 14.11 Determine the enthalpy of combustion of benzene (C6H6) at 25°C and 1 atm, using the enthalpy of formation table. Assume that the water in the products is in gaseous form. [Ans. -3,169,510 kJ/(kmol C6H6)] 14.12 Propane (C3H8) is burnt completely with the stoichiometric amount of air during a steadyflow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer for this process. How much will be the change in the heat transfer if the combustion is achieved with 20 per cent excess air? [Ans. -2,220,030 kJ/(1cmol C3H8) 0] 14.13 Liquid butane (C4H 10) enters a combustion chamber at 25°C and 0.5 kg/min, where it is burnt with 200 per cent theoretical air which enters the combustion chamber at 17°C. If the combustion is complete and the exit temperature of the combustion gases is 1000 K, determine (a) the mass flow rate of the air, and (b) the rate of heat transfer from the combustion chamber. [Ans. (a) 15.47 (kg air)/min, (b) -10025.381]/min] 14.14 Liquid ethyl alcohol (C2H5OH) at 25°C is burnt completely during a steady-flow combustion process with 125 per cent theoretical air which enters the combustion chamber at 25°C. If the products leave the chamber at 1200 K, determine (a) the air-fuel ratio, and (b) the heat transfer for this process. [Ans. (a) 11.25 (kg air)/(kg fuel), (b) - 623,744.95 (1a/lanol C2H5OH)]

420 Fundamentals of Engineering Thermodynamics 14.15 Diesel (C12H26) at 25°C is burnt in a steady-flow combustion chamber with 10 per cent excess air. The combustion products leave the chamber at 460 K. Assuming complete combustion, determine the required mass flow rate of diesel to supply heat at a rate of 2200 Id/s. Also, assume the air to be at 25°C. [Ans. 53 g/s] 14.16 A constant-volume rigid tank contains 250 g of ethane gas (C2H6) at 25°C and 1000 g of 02 at 25°C and 150 kPa. The contents of the tank are ignited, and the ethane gas burns completely. If the final temperature reached is 1000 K, determine the final pressure obtained in the tank. [Ans. 55632 kPa] 14.17 Liquid octane (C8H18) at 25°C is burnt with 200 per cent theoretical air (which is also at 25°C) during an adiabatic steady-flow combustion process. Assuming complete combustion, determine the adiabatic flame temperature of the process. [Ans. 1506.6 K] 14.18 Methane gas (CH4) at 25°C is burnt during a steady-flow combustion process with theoretical air at 25°C. Assuming complete combustion, determine the exit temperature of the product gases for this process. [Ans. 2331 K] 14.19 Hydrogen (H2) at 25°C is burnt steadily with 20 per cent excess air which is at 25°C, 1 atm and 50 per cent relative humidity. Assuming that the combustion is complete and adiabatic, compute the exit temperature of the product gases. [Ans. 1450.8 K] 14.20 One kmol of methane (CH4) at 25°C and 1 atm is burnt steadily with 3 kmol of 02 at the same state. The combustion products formed during the process are then brought to 25°C and 1 atm, i.e. to the condition of the surroundings. Assuming complete combustion, determine the reversible work and the irreversibility for this process. [Ans. 817,930 kJ, 817,930 kJ] 14.21 Liquid Octane (C8H18) enters a combustion chamber at 25°C and 1 atm at a rate of 0.5 kg/ min where it is mixed and burnt with 130 per cent excess air which enters the combustion chamber at 12°C. If the combustion poroducts leave at 1000 K and 1 atm, determine (a) the mass flow rate of the air, (b) the rate of heat transfer from the combustion chamber, and (c) the rate of entropy generation during this process. Take T0 to be 25°C. [Ans. (a) 17.4 (kg air)/min, (b) — 8036.41 kJ/min, (c) 55.29

K)]

14.22 Octane (C8H18) is burned with 200% theoretical air. Determine the mole fractions of the combustion products. Also, find the dew-point temperature if the pressure is 0.1 MPa. [Ans. CO2 = 6.48%, H2O = 7.29%, 02 = 10.12%, N2 = 76.11%, 39.7°C]

CHAPTER

15 Chemical and Phase Equilibrium 15.1 INTRODUCTION A system is said to be in equilibrium if no changes occur within the system when it is isolated from its surroundings. The conditions for mechanical and thermal equilibrium are straightforward; they can be ensured if no changes occur in pressure and temperature, respectively. However, the conditions for chemical and phase equilibrium can be rather involved. The equilibrium criterion for reacting systems is based on the second law. For adiabatic systems, chemical equilibrium is established when the entropy of the reacting system reaches a maximum.

15.2 CHEMICAL EQUILIBRIUM CRITERION By the increase-of-entropy principle, we have 62 dSsys — 7.

(15.1)

For adiabatic systems, this principle reduces to dSsy, ?. 0. That is, a chemical reaction in an adiabatic chamber proceeds in the direction of increasing entropy and when the entropy reaches a maximum, the reaction stops. However, Eq. (15.1) becomes impractical when a reacting system involves heat transfer, since it requires a knowledge of the heat transfer between the system and its surroundings. Therefore, developing a relation for the equilibrium criterion in terms of the properties of the reacting system becomes essential for practical applications. Let us examine a reacting (or non-reacting), simple compressible system of fixed mass with only quasi-equilibrium work modes at a specified T and P, as shown in Fig. 15.1. Using the first and second laws, we can express

(52 T

5Q —PdV=dU, dS ?_ — 421

422 Fundamentals of Engineering Thermodynamics

SWb

SO Figure 15.1 Reaction chamber. or dU + PdV — TdS 5 0 At constant T and P, by Gibbs function G = H —

(152)

is, we have

(dG)T. p = dH — TdS — SdT = (dU + PdV + VdP)— TdS — SdT = dU + PdV — TdS

(15.3)

since dP = 0 and (IT = 0. From Eqs. (15.2) and (15.3), we get (dG)r. p 5 0

(15.4)

Equation (15.4) implies that a chemical reaction at a specified T and P will proceed in the direction of decreasing Gibbs function. The reaction will stop and chemical equilibrium will be established when the Gibbs function attains a minimum value, as shown in Fig. 15.2. Therefore, the criterion for chemical equilibrium can be expressed as

(dG)"

=0

(153)

G

dG < 0 dG > 0 ,. .A: A,1k

dG = ..,..0 ..-/

Violation of the second law

100 % Equilibrium 100% reactants composition products Figure 15.2 Criterion for chemical equilibrium at a given T and P.

Chemical and Phase Equilibrium 423 That is, a chemical reaction at a specified T and P cannot proceed in the direction of increasing Gibbs function, since this will be a violation of the second law of thermodynamics.

15.2.1 Relation for Chemical Equilibrium To obtain a relation for chemical equilibrium in terms of the properties of the individual components, consider a mixture of four chemical components A, B, C, and D which exist in equilibrium at a specified T and P. Now consider a reaction which occurs to an infmitesimal extent during which differential amounts of reactants A and B are converted to products C and D at T and P remaining constant. That is, dNAA + dNBB dNcC + dNDD where NA, NB, Nc, and ND are the number of moles of the components A, B, C, and D, respectively. By Eq. (15.5), we have for chemical equilibrium (d07: P =

E

(dGi)T. P=

E

dN, +

dNc + k D dND = 0

(i

P=0

(15.6)

Or

g A dNA +

(15.7)

where the g - s are called the chemical potentials or the molar Gibbs functions at the specified T and P and dNs are the differential changes in the number of moles of the components. The stoichiometric or theoretical reaction for the system considered is vAA + vBB

vcC + vDD

(15.8)

where the vs are the stoichiometric coefficients. The stoichiometric reaction plays an important role in the determination of the equilibrium composition of reacting mixtures because the changes in the number of moles of the components are proportional to the stoichiometric coefficients. That is, dNA = - EVA, dNB =

evB, dNc = evc,

AID = evp

(15.9)

where c is the proportionality constant and represents the extent of a reaction, the minus sign indicates that the number of moles of reactants A and B decreases as the reaction progresses. Combining Eqs. (15.7) and (15.9), we get VCEC- + VDG - VAgA VB:gB =

0

(15.10)

This is the criterion for chemical equilibrium and is valid for any chemical reaction regardless of the phases involved. Equation (15.10) can be modified to handle reactions consisting of any number of reactants and products.

15.3 THE EQUILIBRIUM CONSTANT FOR IDEAL-GAS MIXTURES Like entropy, the Gibbs function of an ideal gas depends on both T and P. The Gibbs function values are usually listed as a function of T at a fixed reference pressure P0 (usually taken to be 1 atm). At a fixed T, the variation of the Gibbs function of an ideal gas with P is determined from the relation

424 Fundamentals of Engineering Thermodynamics

(Ak)T = Ah — Ti.? = —Ti A =

In

P2 ri

Note that Ah = 0, for constant T, for an ideal gas. Then, the Gibbs function of the component i of an ideal-gas mixture at its partial pressure P1 and mixture temperature T can be expressed as

(T, Pi) =

(1) + RUT ln

(15.11)

where gi (7) is the Gibbs function of the component i at 1 atm pressure and temperature T, and P, is in atmosphere. With Eq. (15.11), Eq. (15.10) becomes vakca + RUT In Pc] + viAg + RUT In vA[gA + RuT In PA] — vB[gB + RUT In

— =0

Let us express this as

G*(7) = —RuT(Vc In Pc + vp In PD — VA In PA — VB In PB) or (Pc)°c (PD)" (PA)v" (Ps)"

AG*(7) = —

(15.12)

where

AG*(T)= vckc. (7) +

(T) — vAkAa (T) — vBka. (T)

is called the standard-state Gibbs function change.

15.3.1 Equilibrium Constant

Kp

The equilibrium constant for the chemical equilibrium of ideal-gas mixtures is defined as

Kp —

(Pc)°` (Ps)" (PA)° 4 (PB)"a

(15.13)

Using Eq. (15.13) in Eq. (15.12), we get

Kp = e-AG*(T AR°T)

(15.14)

The partial pressure of component i can be written as

=Y——P Altotal where P is the total pressure and Mot.' is the number of moles present in the reaction chamber, including any inert gas. Using the above relation in Eq. (15.13), we obtain

c t'v - vA - vB ' P jv (Nat vc (Nc) — K,, (N A)°"(NB)°e Alma

(15.15)

Chemical and Phase Equilibrium

425

EXAMPLE 15.1 Determine the equilibrium constant Kp at 1600 K for the gas-phase reaction CO2 CO + + 02. Compare .the result obtained with the value of Kp given in the equilibrium constant Table 19. Solution The natural logarithm of the equilibrium constant Kp at 1600 K for the given reaction is listed in Table 19 as hi Kp = —10.830. The Kp can also be determined from the Gibbs function data and Eq. (15.14). That is, Kp = e AG'ITY(Ron

In the above equation, AG* is the standard-state Gibbs function change. It can be expressed as AG*(T) = E

+ jt — — 7754))p — E

+ — — Tic),

= 1[— 110,530 + 51,053 — 8669 — 1600(250.592)] 1 + — [0 + 52,961 — 8682 —1600(260333)] 2 —1 [— 393,520 + 76,944 — 9364 — 1600(295.901)] = 144,161.5 kJ/kmol Thus, 144,161.5

ln Kp —

8.314 x 1600

—10.837

Or Kp

= 1.966 x 10-5

This value of Kp is in agreement with the value of Kp given in Table 19, namely = e-10.830 = 1.98 x 10'5 Kp

EXAMPLE 15.2 Determine the equilibrium constant Kp at 298 K for the dissociation process H2 —4_1"-± 2H. Compare the result with the tabulated value of Kp. Solution By Eq. (15.14), we have Kp = e-AGN1Ymn

where AGV) = viirti(1) —

42(T)

= 2(203,290) — 1(0) Note that k*(T) when T = 298 become gl and the values of gf for different gases are given in Table 17.

426 Fundamentals of Engineering Thermodynamics Therefore, AG*(7) = 406,580 kJ/kmol Substituting, we get ln Kp — —

406,580 164.1 8.314 x 298 --

Or

icp = 5.4 x 10-22 The tabulated value of In icp = —164.005 (Table 19) agrees closely with the present result. Note that the icp value for this reaction is negligibly small, indicating that this reaction will not occur at this temperature. EXAMPLE 15.3 Determine the temperature at which 20 per cent of diatomic oxygen (02) dissociates into monatomic oxygen (0) at a pressure of 5 atm. Solution For making the calculations simple, we consider 1 kmol of 02. The stoichiometric and actual reactions in this case are as follows: Stoichiometric: 02 =_ 20 (thus u02 = 1 and uo = 2) Actual: 02 -- 0.802 + 0.40 "--v—' Reactant leftover

Product

This reaction involves one reactant and one product. The equilibrium composition consists of 0.8 kmol of 02 (the leftover reactant) and 0.4 kmol of 0 (the product formed). Therefore, Noe = 0.8 and No = 0.4. Assuming ideal-gas behaviour for both 02 and 0, the equilibrium constant icp can be determined as follows from Eq. (15.15). Kp—

0 43

(N

(No2 )4'2

14) - 432

( P

N 'call )

. (0.4)2 ( 5 (0.8)1 0.8 + 0.4

)2-1 = 0.833

From Table 19, the temperature corresponding to Kp = 0.833 is T= 3763.9 K That is, at 3763.9 K, 20 per cent of 02 will dissociate into 0 when the pressure is maintained at 5 atm

Chemical and Phase Equilibrium 427

EXAMPLE 15A A mixture of 1 kmol of CO and 2 kmol of 02 is heated to 2000 K at a pressure of 4 atm. Determine the equilibrium composition, assuming that the mixture consists of CO2, CO and 02. Solution The stoichiometric and actual reactions for the given mixture are: Stoichiometric: CO + -102 F e--s CO2 (Thus ve0 = 1, t = 1 and vo2 = -I ) 1- z0; Actual: CO + 202 ---) xCO2 + r.:1 -__ . Reactants leftover

Product

Now by C and 0 balances, we have C balance: 1 = x + y 0 balance: 5 = 2x + y + 2z In terms of x, we can express y=1—x and 5 1 y 5 x z = -i- — x — -i= -i- — x — -i-(1— x) = 2 — i x Total number of moles, Mood =x+y+z=x+1 — x + 2 — -x- = 3 — 2 2 Pressure: P = 4 atm For the stoichiometric reaction taken in the reverse direction, from Table 19, we have In Kp = + 6.635 at 2000 K. This gives Kp= 761.279. Assuming ideal-gas behaviour for all components, the Kp becomes

Kp —

( slco,)vc°2

( P r°2 - vc° (Noo)va) (No, )1b2 ( Mout )

Substituting, we get 4

(x)1

761.279 — (y) (z)v2 i

x 2

[

I1

2

4 11/ 2

(1 _

X)

(2

—7

2

[

x 2

428 Fundamentals of Engineering Thermodynamics Let 761.279 = c. Then, simplifying we get x3(1 — 2c2) + 6x2(2c2 — 1) + x(-18c2) + 8c2 = 0 Or X3 — 6x2

18C2X 1 — 2C2

8c2 1 — 2C2

—0

or

x3 -6x2 +9x-4= 0 Solving for x, we obtain x=4

Then Y = —3 and z=0

Therefore, the equilibrium composition of the mixture at 2000 K and 4 atm is CO + 202 —) 4CO2 — 3C0 That is, 4C0 + 202 --) 4CO2 EXAMPLE 15.5 Water vapour (1120) and carbon monoxide (CO) react according to the reaction H2O + CO H2 + CO2 (This equation is known as the water-gas reaction equation) at 2000 K. Determine the equilibrium constant Kp for this reaction. Solution The given equation may be viewed as the overall reaction of the following two reactions: H2O

H2 + 1 02

(a)

2 CO2 CO + 1 02 T_"-

(b)

The equilibrium constant Kp. and Kpb for the reactions (i) and (ii), respectively, from Table 19 are: In Kpa = — 8.145 In Kpb = 6.635 From the definition of the equilibrium constant Kp, we have, in terms of partial pressures, Kp —

PH PCO PCO PH20 D D1/2

Kp — PH20

Chemical and Phase Equilibrium

—

K

429

Pc02

Pco PO:2 From which we see that Kp = Kp. x Kpb. Thus, we have ln Kp = ln Kpb + ln Kpb Therefore, In Kp= — 8.145 + 6.635 = —1.51 or Kp =

0.2209

Note that for any reaction at a given temperature and pressure the change in the Gibbs function of the reaction is independent of the path of the process. Thus, AG* = AG: + AG; For which we also get —AG'

—

AG. a AG.

RuT RuT RuT

But for any reaction involving ideal gases, we have ln Kp = —

AG' Ru r

Thus, In

= In Kpa + ln Kpb

Further, the change in enthalpy and change in entropy of any reaction are also independent of the path of the process. Consequently, we have

= Ali; + AH; and

AS' =

+ AS;

This implies that the thermochemical data of a complex reaction may be obtained from the thermochemical data of simple reactions.

15.4 SIMULTANEOUS REACTIONS It was shown in Section 15.2 that (dG)r p = 0 for chemical equilibrium. This is irrespective of the number of reactions that may be occurring. When more than one reaction is involved, this condition is satisfied only when (dG)r p = 0 for each reaction. Determination of the equilibrium composition of a reacting mixture requires that there be as many equations as unknowns. The mass balance of each element involved provides one equation and the rest of equations must come from the Kp relations written for each reaction.

430 Fundamentals of Engineering Thermodynamics EXAMPLE 154 A mixture of 1 kmol of carbon monoxide (CO) and 1 kmol of water (H2O) is heated to 2500 K at a pressure of 1 atm. Determine the equilibrium composition of the mixture, assuming that only CO, CO2, H2, H2O and 02 are present. Assume all these gases to be ideal gases. Solution The chemical reaction during this process can be expressed as CO + H20 --) xCO + yCO2 + z02 + /H2 + mH20 Mass balances for carbon, oxygen, and hydrogen yield C balance: 1 = x + y 0 balance: 2 = x + 2y + 2z + m H balance: 2 = 2/ + 2m

(iii)

The mass balances provide us with only three equations but with five unknowns, therefore, we need to have two more equations (to be obtained from icp relations) to determine the equilibrium composition of the mixture. It seems that part of CO2 in the products is dissociated into CO and 02 during this process according to the stoichiomeiric reactions CO + i O2 4.=---t CO2

(reaction 1)

CO + H2O ± CO2 + H2

(reaction 2)

The equilibrium constants for these two reactions at 2500 K are determined from Table 19 as In Km = 3.331

= 27.97

In Kp2 = -1.803 --Kp2 = 0.165 The Kp relations for these two simultaneous reactions are K

-

t)CO2

VICO DC°2 2

°CO

tb2

(N co)uc° (No2 )432 iN total Kp2 -

(NCO' )1P°2 (NH2 )1412 (NCO )VC° (N1120 )VH243

P I 2 + 412 -

- °HP

N O3i J

where Ntow = Nco + Nco2 + No2 + NH2 + NH20 =x+y+z+1+m. Substituting, we get 27.97 - (01 %1/2 (X)IttZ)

1 )1 -1 - 2 1 X +y+z-1-11-m)

(iv)

and )1 4-1-1-1 1 0.165 - (0 (x)I(m)I x+y+z+1+m)

(v)

431

Chemical and Phase Equilibrium

Solving for x, y, z, 1, and m, we get x= 0.71, y= 0.29, z = 0.00043, 1= 0.29, and m = 0.71 Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is 0.71C0 + 0.29CO2 + 0.0004302 + 0.29H2 + 0.71H20 Note that: 1. Nearly 30 per cent of both CO and H2O have reacted at the state of equilibrium. 2. Solving a system of simultaneous nonlinear equations manually is extremely difficult and time consuming. Therefore, it is often necessary to solve these kinds of problems, by using an iterative algorithm in a computer.

15.5

Kp VARIATION WITH TEMPERATURE

By Eq. (15.14), we have In K p

— AG*(T) RuT

But AG*(T) = AH*(T) — TAS*(T), therefore, we can write d(ln K,,) _ AH*(T) d[AH*(T)] d[AS*(T)] dT RuTdr R„ dT Ru T2 At constant pressure, the Tds relation Tds = dh — vdP becomes Tds = dh. Also, Td(AS*) = d(AH*). Thus the above relation becomes d(ln

H(T) — hR(T) dT Ru T2 Ru T2

(15.16)

where h R(T) is the enthalpy of reaction at temperature T. Note that for enthalpy at a constant pressure of 1 atm we do not use the superscript * since the enthalpy depends only on T for ideal gases. For small temperature intervals, h R can be treated as a constant and thus Eq. (15.16) yields, on integration In

K

R (1 _ K 1,1 RI, Ti T2

(15.17)

From Eq. (15.17) it is evident that hR can be calculated if Kp is known. Further, exothermic reaction (hR < 0) processes will be less complete at higher temperatures since Kp decreases with T for such reactions. EXAMPLE 15.7 Estimate the enthalpy of reaction hR for the combustion process of carbon monoxide CO + z 02 CO2 at 3000 K, using (a) the enthalpy data and (b) the Kp data.

432

Fundamentals of Engineering Thermodynamics

Solution (a) The hR of the combustion process of CO at 300 K is the amount of energy released when 1 lanol of CO is burnt in a steady-flow combustion chamber at a temperature of 3000 K. It can be determined from the equation: FIR = Q =

E

N p(17; .

17) p

E Nr(Til + -h°),

= NCO= (h, + 1 3000 K ri298 K)CO2 NCO(hJ? 173000 K — NO2 (ii; + T83000 K

17298 K )C0

T1298 K )02

Or

hR = (1 kmol CO,J(-393,520 + 162,226 — 9364) —(1 kmol COX-110,530 + 102,210 — 8669) —(1/2 laical 02)(0 + 106,780 — 8682) — 272,718 kJ/lanol (b) The hR value at 3000 K can also be estimated by using the Kp values at 2800 K and 3200 K from Table 19 as = 6.646

at T1 = 2800 K

Kp2 = 1.536

at T2 = 3200 K

Substituting these into Eq. (15.17), we get K In =-= Kpi

- 2-1 Ru

T2

or In

hR ( 1 1.536 1 6.646 8.314 28® 3200)

or hR uu —272,801 kJ/kmol Note that in spite of the temperature difference of 400 K between T1 and T2 the two results are almost identical.

15.6 PHASE EQUILIBRIUM A phase is defined as a homogeneous system having uniform intensive properties throughout. A multiphase system is known as a heterogeneous system. Two phases of a substance are said to be in phase equilibrium when there is no

Chemical and Phase Equilibrium 433 transformation from one phase to other. In other words, if the driving force between the two phases of a substance which forces the mass to transform from one phase to another is absent then the phases exist in equilibrium.

15.6.1 A Single-Component System Consider a mixture of a saturated liquid and saturated vapour of a pure substance in equilibrium at a specified T and P, as shown in Fig. 15.3. The total Gibbs function for this mixture is G = mf gf + mg gg, where gf and gg are the Gibbs functions of the liquid and vapour phases per unit mass, respectively. Now, assume that a differential amount of liquid dmf evaporates at constant T and P owing to a disturbance. The change in the Gibbs function owing to this disturbance is given by (dG)T. p = gfdmf + gg dmg

mg

P Vapour Liquid

Figure 153 A liquid-vapour mixture system. We know that gf and gg remain constant at constant T and P, therefore, at equilibrium, (dG)T. p = 0. From mass conservation, we have dmg = —dmf. Therefore, (dG)T. p = (gf — gg)dmf For equilibrium (dG)T. p = 0, which yields (15.18) gf = gg This implies that the two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function. At the triple point, the specific Gibbs function of each one of the three phases is equal. Further, it can be inferred from Eq. (15.18) that, just as the temperature difference is the driving force for heat transfer, the Gibbs function difference is the driving force for phase change. EXAMPLE 15.8 Show that a mixture of saturated liquid water and saturated water vapour at 100°C satisfies the condition for phase equilibrium. Solution The condition for phase equilibrium, given by Eq. (15.18), is = gg

434 Fundamentals of Engineering Thermodynamics Also, gf = hf — Tsf and gg = hg — Tsg Using the enthalpy and entropy data from Table 3, we have gf = 419.04 — (373.15 x 1.3069) = — 68.63 kJ/kg and gg = 2676.1 — (373.15 x 7.3549) = —68.38 kJ/kg The values gf and gg here are nearly equal, showing that the mixture is in phase equilibrium.

15.6.2 The Gibbs Phase Rule At the phase boundaries of a multipurpose system, there will be sudden changes in properties. For example, the density of liquid water is quite different from the density of ice in a two-phase mixture of liquid water and ice. Certain important relationships for a multiphase system of several components or species are given in the phase rule derived by J. Willard Gibbs in 1875. The Gibbs phase rule states that the number of independent variables associated with a multicomponent, multiphase system is given by IV = C — PH + 2

(15.19)

where IV is the number of independent variables, C the number of components, and PH the number of phases present in equilibrium.

15.6.3 Multicomponent Systems For phase equilibrium of a multicomponent system, the specific Gibbs function of each component must be the same in all phases. That is, gp = gg,i =

85,1

for component 1

= ggi =

85,2

for component 2

= ggN = &iv for component N

EXAMPLE 15.9 Consider a mixture of 80 per cent liquid ammonia and 20 per cent water by mole numbers, at 20°C. Determine the composition of the vapour phase of this mixture. Solution The saturation pressures of H2O and NH3 at 20°C are PH20 = 2.339 kPa

(see Table 3)

Chemical and Phase Equilibrium PNH3

= 857.5 kPa

435

(from Engineering Thermodynamics by Francis F. Huang)

The total pressure of the mixture is Ptigai = PH20 + PNH3 = 2.339 + 857.5 = 859.84 kPa The mole fractions in the vapour phase are Y0-120 =

PH o — 2 339

Plow

859.84

PNH,

857.5 859.84

"- 0.00272

and Yg.NH, = Ptotal

0.99728

SUMMARY A system is said to be in equilibrium if no changes occur within the system when it is isolated from its surroundings. The criterion for chemical equilibrium can be expressed as (dG)T.p = 0 That is, chemical reaction at a specified T and P cannot proceed in the direction of increasing Gibbs function, since this will be a violation of the second law of thermodynamics. The criterion for chemical equilibrium valid for any chemical reaction regardless of the phases involved can be expressed as 1-)CiC + )Dkl)

v AgA — v BSB =

The equilibrium constant for the chemical equilibrium of ideal-gas mixtures is defined as K

p

(Pc)

°c

(PD)"

(PA)" ( Pa)"

A phase is defined as a homogeneous system having uniform intensive properties throughout. Two phases of a substance are said to be in phase equilibrium when there is no transformation from one phase to another. Two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function. The Gibbs phase rule states that the number of independent variables associated with a multicomponent, multiphase system is given by IV = C — PH + 2 where IV is the number of independent variables, C the number of components, and PH the number of phases present in equilibrium.

436 Fundamentals of Engineering Thermodynamics

PROBLEMS 15.1 Determine the equilibrium constant Kp for the reaction H20(g) ---._± H2 + i 02 at (a) 298 K, and (b) 2000 K, using the Gibbs function data. Compare your results with the Kp values listed in Table 19. [Ans. (a) 8.55 x lei, (b) 3.45 x 101 15.2 Determine the equilibrium constant Kp for the reaction C2H4 + 302 -A— _± 2CO2 + 2H20 at 25°C. [Ans. 2.17 x 10230] 15.3 Using the Gibbs function data, determine the equilibrium constant Kp for the reaction N2 CO + 1 02 at 1800 K. [Ans. —279,018 la/lanol] 15.17 Estimate the Kp value of the combustion process H2 + 202 the enthalpy of reaction

rift data and the

—4 H2O at 2200 K, using

Kp value at 2000 K.

[Ans. 870.53] 15.18 Estimate the enthalpy of reaction h R for the dissociation process 02 ‹..= 20 at 3100 K, using (a) the enthalpy data and (b) the K,. data.. [Ans. (a) 513,614 kJ/kmol, (b) 512,808 kJ/kmol] 15.19 Consider a mixture of water and ammonia solution in liquid phase. How many independent properties are needed to fix the state of the system? [Ans. 3] 15.20 Determine the Gibbs function of formation of carbon dioxide, considering the reaction C + 02 —> CO2 and assuming that C and 02 are initially at 25°C and 0.1 MPa. [Ans. —394,390 kJ/lanol]

CHAPTER

16 Thermodynamics of Compressible Flow

16.1 INTRODUCTION Entropy and temperature are the two fundamental concepts of thermodynamics. The energy changes associated with compressible flow, unlike low-speed or incompressible flow, are substantial enough to strongly interact with other properties of the flow. Hence, the energy concepts play an important role in the study of compressible flow. In other words, the study of thermodynamics which deals with energy (and entropy) is an essential component in the study of compressible flow. The following divisions are the broad classifications of the fluid flow studies, based on thermodynamic concepts: • Fluid mechanics of perfect fluids, i.e. fluids without viscosity and heat (transfer) conductivity, is an extension of equilibrium thermodynamics to moving fluids. The kinetic energy of the fluid has to be considered in addition to the internal energy which the fluid possesses when at rest. • Fluid mechanics of real fluids goes beyond the scope of classical thermodynamics. The transport processes of momentum and heat are of primary interest here. But, even though thermodynamics is not fully and directly applicable to all phases of real fluid flow, it is often helpful in relating the initial and final conditions. • In modern high-speed problems, the kinetic energy content of the fluid can be very large compared to its heat content. Therefore, the variations in temperature associated with these high speed flows can be substantial. Hence, the emphasis on thermodynamics concepts.

16.2 STAGNATION PROPERTIES For high-speed flows, the potential energy of the fluid is usually negligible, but the kinetic 438

Thermodynamics of Compressible Flow

439

energy is significant. For such flows, it is convenient to combine the enthalpy and the kinetic energy of the fluid into a single term called the total or stagnation enthalpy ho, defined as V2 ho = h + — (kJ/kg) 2

(16.1)

where h is referred to as the static enthalpy. Consider steady adiabatic flow through a duct, shown in Fig. 16.1. Assuming Ape to be negligible, we can express the energy conservation relation as q — w = Ah + Ake + Ape . r.................... ,

Control volume

h2 —i—► v2 hoe = hol

Figure 16.1 Steady, adiabatic flow through a duct. But for the flow process under consideration, q = 0, w = 0, and Ape = 0. Thus, we have V2 V2 0=(h2 — 10+ (--1- — 1 2 2 Or

(162)

hoi = hoz

where the stagnation enthalpy represents the enthalpy of a fluid stream when it is brought to rest adiabatically.

16.2.1 Stagnation State It is a state of zero flow velocity. That is, it is the state achieved by decelerating a flow to zero velocity. The properties of a fluid at the stagnation state are called stagnation properties, and are usually denoted by the subscript "0". When the stagnation state is attained by decelerating the flow isentropically, it is called the isentropic stagnation state. The isentropic and actual stagnation processes are illustrated on the h—s diagram in Fig. 16.2. It is interesting to note that /to is the same for both the processes. But the Po, act < Po, and this is due to the fluid friction effect. For ideal gases, we know that h = CDT, therefore, Eq. (16.2) becomes 2

CpTo = CDT + — 2 Or

To = T +

V2 2Cp

(163)

440 Fundamentals of Engineering Thermodynamics h Isentropic stagnation state

po O. ad

ho

V

Actual stagnation state

2

2

h Actual state

Figure 16.2 Isentropic and actual stagnation processes. where To, the temperature attained by decelerating a flow to rest adiabatically is called the stagnation or total temperature. The term V2/2Cp which represents the temperature rise during such a process is called the dynamic temperature, which is just the difference between the total temperature To and the static temperature T. For low-speed flows, it can be easily shown that To and T are practically identical. However, for high speed flows, To may be significantly higher than T.

16.2.2 Stagnation Pressure P0 This is the pressure that a fluid attains when brought to rest isentropically. The corresponding density is called the stagnation density po For ideal gases with constant specific heats, the stagnation and static properties are related through Eqs. (5.42)—(5.44), as given below:

(16.4)

EXAMPLE 16.1 Air flowing isentropically through a nozzle is at 10°C and 0.5 atm and has a velocity of 425 m/s. Determine the stagnation enthalpy, temperature, pressure and density of the air. Solution By energy equation, we have L

ho = n +

2

2

Thermodynamics of Compressible Flow 441 Treating air to be an ideal gas, we have h = CpT, therefore,

h()

v2 + 4252 = 1004.5 x 283.15 2 2

r = o

374.74 kJ/kg

Also, To = T + V2 2Cp = 283.15 +

4252 2 x 1004.5

373.06 K

By isentropic relation, we have Po ( To r r P T Therefore, T )70-I)

po = p(9)

=0.5

(373.06)1.441A-1) 283.15)

1.313 atm

Also, Po = (Tor -1) p T) Thus, 1/(7 -1)

4) Po= 4-

,

but p = — by state equation RT

Therefore, _ P (T0 1(7-1) P° RT T ) Or

Po -

0.5 x 101325 (373.06 287 x 283.15 283.15

)10.4-1) since P = (0.5 x 101325) Pa and R = 287 m2/(s1 K)

1.242 kg/m3

16.3 SPEED OF SOUND AND MACH NUMBER An understanding of the mechanism by which disturbance waves are propagated through a fluid medium and the relation for the velocity of propagation of the waves will be extremely useful in deriving significant conclusions concerning the fundamental difference between incompressible and compressible flows.

442 Fundamentals of Engineering Thermodynamics Any disturbance created in a fluid medium propagates through sound waves, which are infinitely small pressure disturbances. The speed with which sound propagates in a medium is called speed of sound and is denoted by a. Thus, the velocity of sound (or the sonic velocity) becomes an important parameter in the study of compressible flows. To obtain a relation for the speed of sound in a medium, consider a pipe which is filled with a fluid at rest, as shown in Fig. 16.3. If an infinitesimal disturbance is created by the piston,

77777Y--.'

Pipe with crosssection area A

Pressure wave

Piston /

V

/ /-4"

AP

-„..,.°

-f-

A

Pi ir

Figure 163 Propagation of pressure disturbance. as shown, the wave propagates through the gas at the velocity of sound relative to the gas into which the disturbance is moving. Let the stationary gas in the pipe at pressure Pi and density A be set in motion by the piston. The infinitesimal pressure wave created by the piston movement travels with a speed a, leaving the medium behind it at a pressure P1 and density pi to move with velocity V. As a result of the compression created by the piston, the pressure and density next to the piston are infinitesimally greater than the pressure and density of the gas at rest ahead of the wave. Therefore, and AP = — Pi AP = Ps — Pi are small. Choose a control volume of length b as shown in Fig. 16.3. Compression of volume Ab causes the density to rise from pi to pi in time t = b/a. The mass flow rate into the volume Ab

is nt = AAV

(16.5)

For the conservation of mass, m must also be equal to the mass flow rate through the control volume, Ab(p5 — p;)/t. That is,

Ab(A— pi) _

AAV

or

a(Pi — Pi) = Ps V

(16.6)

Thermodynamics of Compressible Flow

443

since b/t = a. When the piston moves, the compression wave thus caused travels and accelerates the gas from zero velocity to V. The acceleration is given by V

=

„a

t

b

The mass in the control volume Ab is m= AbP where p = (pi + pi)/2. The force acting on the control volume is F = A(P1 — Pi). Therefore, by Newton's law A(P1 — P1) = Atli) V (

(16.7) .42 b.)

Since the disturbance is very weak, pi on the right-hand side of Eq. (16.6) may be replaced by P to give a(PI — = P V Using this relation, Eq. (16.7) can be written as a2

-

AP Pi — P; AP - f;

In the limiting case as AP and Ap approach zero, the above relation leads to a2

— dP

dp

(16.8)

This is Laplace's equation and is valid for any fluid. The sound wave is a weak compression wave, across which only infinitesimal changes in fluid properties occur. Further, the wave itself is extremely thin, and changes in properties occur very rapidly. The rapidity of the process rules out the possibility of any heat transfer between the system of fluid particles and its surroundings. For very strong pressure waves, the travelling speed of disturbance may be greater than that of sound. The pressure can then be expressed as P = P(P)

(16.9)

For isentropic process of a gas,

P

— = constant

Pr where the isentropic index y is the ratio of specific heats and is a constant for a perfect gas. Using Eq. (16.8) and the above isentropic relation, we get a2 =

YP

(16.10) P For a perfect gas, by state equation, we have P = pRT. Therefore, Eq. (16.10) can be expressed as a = ryif

(16.11)

444

Fundamentals of Engineering Thermodynamics

The assumption of perfect gas is valid as long as the speed of the gas stream is not too high. However, at hypersonic speeds the assumption of perfect gas is not valid and we must consider Eq. (16.10) to calculate the speed of sound. Note that in Eq. (16.11), the gas constant R has a fixed value for a specified gas and the specific heats ratio y of an ideal gas is, at most, a function of temperature, therefore, it is evident that the velocity of sound in a specified ideal gas is a function of temperature alone.

16.3.1 Mach Number The Mach number M is defined as the ratio of the local flow speed to the local speed of sound. M—

V local flow speed local speed of sound a

(16.12)

It is thus a dimensionless quantity. In general, both V and a are functions of position and time, so that the Mach number is not just the flow speed made nondimensional by dividing by a constant. Thus, we cannot write M o. V. It is, however, almost always true that M increases monotonically with V. A flow for which the Mach number is greater than unity is termed supersonic flow, i.e. for which V > a. This means that the upstream flow remains unaffected by changes in conditions at a given point in a supersonic flow field. Fluid flow regions are often described in terms of the flow Mach number. The flow with M < 1 is called subsonic, flow with M = 1 is called sonic, flow with M > 1 but less than 5 is called supersonic, flow with M > 5 is called hypersonic, and flow with M = 1 is called transonic. EXAMPLE 16.2 Air is accelerated isentropically from 100 m/s to 400 m/s in a nozzle. If the temperature at the. initial state is 400 K and the Mach number at the final state is 1.5, determine (a) the initial Mach number, and (b) the final temperature. Solution At the given conditions, air can be treated as an ideal gas with R = 287 J/(kg K) and specific heats ratio y= 1.4. (a) The velocity of sound at the initial state is = orifi = 4 x 287 x 400 = 400.9 m/s Thus, the initial Mach number is = = 100 al 400.9

0.249

The stagnation temperature T01 is given by Toi =T +

vI 2

2C p

= 400 +

1002 — 405 K 2 x 1004.5

Thermodynamics of Compressible Flow

445

(b) The final temperature T2 is given by T2 = To2

V22 = _ v22 , Toi 2Cp 2Cp

since T01 = T02 for isentropic flow.

Also, V, = M2 a2 =M20FRT2 =1.541.4 x 287T2 = 30.06747 2 Therefore,

T

30.0672 7j 2Cp

=

or (,

12 I +

30.0672 2 x 1004.5

— 405

Therefore, =

279.3 K

EXAMPLE 163 Steam at 2.0 MPa and 400°C flows through a constant area straight pipe with a velocity of 525 m/s. Determine (a) the velocity of sound and (b) the flow Mach number for the steam. Compare these results with those obtained by treating the superheated steam as an ideal gas with y = 1.3. Solution (a) The velocity of sound in steam at the given state can be determined exactly from Eq. (16.8), but we need to have a relation for pressure P in terms of density p. In the absence of such a relation, a reasonably accurate value for the velocity of sound can be obtained by replacing the differentials in that equation by differences about the specified pressure. That is,

a2

=

lap) op

[A A(1/v) is

The entropy of the steam at the given state of 2 MPa and 400°C is s = 7.1217 kJ/(kg K) (see Table 5). The specific volumes of the steam at this entropy and the listed pressures just below and just above the specified pressure (1.8 and 2.5 MPa) are 0.16418 and 0.12682 m3/kg, respectively (from steam table by interpolation). Substituting, we get 2500 — 1800 .— J(1/0.12682) — (1/0.16418)

x

1000 = 624.6 m/s

11 I

446 Fundamentals of Engineering Thermodynamics (b) The Mach number is M = V = 525 — a 624.6 0.840 If ideal-gas behaviour is assumed for steam, the velocity of sound can be determined from a=

yRT

= J1.3 x 0.4615 x 673.15 x 1000 = 635.5 m/s and M=

V 525 = a 635.5

0.826

From the results it is seen that the ideal-gas assumption is a reasonable one for steam at the specified state.

16.4 ONE-DIMENSIONAL ISENTROPIC FLOW Flow through several devices such as nozzles, diffusers, and turbine blade passages can be approximated to one-dimensional isentropic flow without introducing any significant error.

16.4.1 Streamtube Area-Velocity Relation Consider a quasi-one-dimensional flow, allowing the streamtube area A to vary with x, as shown in Fig. 16.4.

Figure 16.4 Quasi-one-dimensional flow. Let us assume that all flow properties are uniform across any given cross-section of the streamtube, and hence are functions of x only for steady flows. Such a flow, where A = A(x), P = P(x), p = p(x), and V = V(x) for steady flow, is defined as quasi-one-dimensional flow. For any streamtube of area A, the continuity equation is given by pA V = constant

(16.13)

Thermodynamics of Compressible Flow 447 Differentiating with respect to

V, we obtain d(pAV) _

di + A d(pV) _ 0 pv dV dV dV

The second term of the right-hand side in the above relation can be written as

A d(pV) d p dP) —Ap+V—— dV dP dV By Bernoulli equation, we have 2

+ jdP — = constant p

Differentiating and rearranging, we get

dP pV = -dV dP

Using this, and replacing ;F p; by a2, we obtain

" Lon' — A p(1— M2)

A

dV

Therefore,

pV— di + A p(1— M2) = 0 dV Or

dA dV

=_

0 _ m 2) V

(16.14)

Equation (16.14) is an important result. It is called the area-velocity relation. The following can be inferred from the area-velocity relation: • For incompressible flow limit, i.e. for M --> 0, Eq. (16.14) shows that AV = constant. This is the famous volume conservation or continuity equation for incompressible flow. • For 0 < M < 1, a decrease in area results in an increase in velocity and vice versa. Therefore, the velocity increases in a convergent duct and decreases in a divergent duct. This result for compressible flows is the same as that for incompressible flow. • For M > 1, an increase in area results in an increase in velocity and vice versa, i.e. the velocity increases in a divergent duct and decreases in a convergent duct. This is directly opposite to the behaviour of subsonic flow in divergent and convergent ducts. • For M = 1, by Eq. (16.14), dAl A = 0, which implies that the location where the Mach number is unity, the area of the passage is either minimum or maximum. We can easily show that the minimum in area is the only physically realistic solution. The above results are shown schematically in Fig. 16.5. From the above discussion, it is

448

Fundamentals of Engineering Thermodynamics

Velocity, V

Subsonic, M < 1

Supersonic, M> 1

'44444 444444444Q Flow

Increases

irrrrirrrrrtrrrrrr

Decreases

1114111j1111jjj114 ---► Flow ?'7,77777, 777,,y77.

10J 4111j1jjj Flow 77'77777, 77,7777 '4444 44444444444e ----► Flow irrirrrtrrrrrffrrr

Figure 16.5 Flow in convergent and divergent ducts. clear that for a gas to expand isentropically from subsonic to supersonic speeds, it must flow through a convergent-divergent duct, as shown in Fig. 16.6. The minimum area that divides the

M< 1

V increasing

Figure 16.6 Flow in a convergent-divergent duct. convergent and divergent sections of the duct is called the throat. From inference 4 above, we see that the flow at the throat must be sonic with M = 1. Conversely, for a gas to get compressed isentropically from supersonic to subsonic speeds, it must again flow through a convergent-divergent duct, with a throat where the flow speed reduces to sonic level.

16.5 PROPERTY RELATIONS FOR ISENTROPIC FLOW OF PERFECT GASES A gas is said to be thermally perfect when its internal energy and enthalpy are functions of temperature alone. For thermally perfect gases, Cp and C„ are functions of temperature only. The perfect gas is still more a specialization than a thermally perfect gas. For a perfect gas,

Thermodynamics of Compressible Flow 449 both Cp and C, are constants and independent of temperature. Such a gas with constant Cp and C„ is called a calorically perfect gas. Therefore, a perfect gas should be thermally as well as calorically perfect. The perfect gases are also called ideal gases. It is possible to express the ratios of stagnation to static values of temperature, pressure, and density of an ideal gas in terms of the Mach number M and specific heats ratio y The energy equation for an adiabatic process can be expressed as V2 ho = h + — 2 For perfect gases, h = CPT and ho = CpTo, where the subscript 0 refers to the stagnation state. Therefore, we have 7,; = T +

V2 2Cp

or

TO = 1 + V2

2CpT

T

For perfect gases, C,, = yRI(y — 1) and a2 = yRT. With these relations, the above energy equation can be written as

TO = 1 + y-1 2

V2 yRT

1+

y— 1 M2 2

(16.15)

since, M2 = V2/a2 = vi/yRr. Using Eq. (16.15) in Eq. (16.4), we get

y — Mel

=

2

Po

(16.16)

and Po = (1÷ 1 — 1 Apr i) 2 P

(16.17)

The numerical values of T/To, P/Po, and plpo versus the Mach number for different y are listed in Gas Tables (Rathakrishnan, E , 1993).

16.5.1 Critical Properties The properties of a fluid at a location where M = 1 (at throat) are called critical properties, denoted by the superscript asterisk (*). With M = 1, Eqs. (16.15)—(16.17) become 7'11 To

2 y+ 1

P. *

( 2

(16.18)

y47-1) 1)

(16.19)

450 Fundamentals of Engineering Thermodynamics P* _ ( Po

2 1"r-I) (16.20)

(Y ÷ 1)

The above ratios in Eqs. (16.18)—(16.20) are called critical ratios. EXAMPLE 16.4 Helium flows steadily through a varying cross-sectional area duct, shown in the figure below. Helium enters the duct at 10 atm and 160°C with a low velocity, and it expands in the duct to a pressure of 270 kPa. Treating the flow in the duct to be isentropic, calculate the critical pressure and temperature of helium. Po = 10 atm

1

He

To = 160°C P"

r Solution For helium, the specific heats ratio y= 1.667 at room temperature. Using this value of y, the ratios of critical to stagnation temperature and pressure, respectively, from Eqs. (16.18) and (16.19) become T* 7,;

2 2 — — 0.75 y+1 2.667

P* = 2 Po (Y+ 1

2 V.667/0.667

j r1(1-1) =

(2.667)

— 0 487

Given that P0 = 10 atm, and To = 160°C = 160 + 273.15 = 433.15 K, we have T* = 0.75 x 433.15 = 324.86 K and P* = 0.487 x 10 = 4.87 atm

16.6 ISENTROPIC FLOW THROUGH NOZZLES Nozzles, in general, are passages to accelerate the flow of fluids. The flow condition in the nozzle, for any given upstream pressure, is dictated by the pressure applied at the nozzle discharge region, called the back pressure. We will now discuss the effects of back pressure on the exit velocity, the mass flow rate, and the pressure distribution along the nozzle.

Thermodynamics of Compressible Flow

451

16.6.1 Convergent Nozzles Consider a fluid at stagnation conditions Po and To from a large reservoir expanded through a convergent nozzle, shown in Fig. 16.7. Since the reservoir is large, the flow velocity at the nozzle inlet can be taken as negligible. Now let us begin to reduce the back pressure Pb. When Pb = = Po, _there will be no flow and the pressure distribution will be uniform along the nozzle. When Pb is reduced to P2. the exit pressure Pe also comes down to P2, causing the pressure along the nozzle to decrease in the direction of flow, as shown in Fig. 16.7. Up to pressure level P3 at the nozzle exit, the flow in the nozzle continues to be subsonic. Reservoir Po To

------- s Pb ----4

(Back pressure)

Lowest exit pressure

Figure 16.7 Pressure distribution along a converging nozzle. When Pe is reduced to P3, which is the choking condition, the mass flow reaches a maximum value and the flow is said to be choked, and the flow inside the nozzle is said to be frozen. Further reduction in Pb below P3 does not influence the nozzle flow field. Under steady-flow conditions, the mass flow rate rh is constant and given by

th = pAV = (— )A(M oFICT) = PAM iF RT RT In terms of To and Po, this can be expressed, using Eqs. (16.15) and (16.16), as _

) ) AMpo‘. 17(ru o (1+ 7-1m2r+ 2

(16.21)

452 Fundamentals of Engineering Thermodynamics

For choked flows, M = 1 and A= A*, and if the fluid is air with y= 1.4, Eq. (16.21) simplifies to 0.6847P0A* thmax

Off;

(16.22)

Thus, for any given ideal gas, m through a nozzle is fixed by Po and To. The flow rate through a given nozzle can be controlled by changing Po and To, and thus a convergent nozzle can be used as a flowmeter. The effect of Po and To on m through a convergent nozzle is plotted in Fig. 16.8. Mi = 1

Mt < 1 014

•1

Increase in Po, decrease in To or both

Decrease in Po, increase in T0 or both

0

Pd Po

1.0

PIP°

Figure 16.8 Effect of Po and To on mass flow rate through a convergent nozzle.

16.6.2 Area-Mach Number Relation A relation showing the variation of flow area A through a nozzle relative to throat area A* can be obtained in terms of the Mach number M. Another parameter of interest in the analysis of one-dimensional isentropic flow of perfect gases is the ratio of the local velocity V to the velocity of sound at the throat a*, denoted as M*. It can be expressed as V M* = — = M a*

or ms2 —

7+ 1M 2 2 1+ M2 2

Thermodynamics of Compressible Flow 453 where M is the local Mach number. Consider the convergent-divergent nozzle shown in Fig. 16.9. Throat

M=1

A*

M* = 1 = a*

A

M Figure 16.9 Convergent-divergent nozzle. By continuity equation, we have

p*V*A* = pVA Therefore,

A = p*a* _ P*Poa* A* PV PoPV

(... vs = a*)

2

*l 2

Squaring this equation and substituting for (12—

Po

(A

)2

1 2 + ApLy +1 .

*)2

, (L °) , and (5—

V

P

y-1M 2 cr +00 --I) zll 2

)]

we obtain

(16.24)

Equation (16.24) is called the Area-Mach number relation. From this equation we get the striking result M = f(A/A*), i.e. the Mach number at any location in the duct is a function of the ratio of the local area of the duct to the sonic throat area. As seen from Eq. (16.24), the local area A must be larger than or at least equal to A*. The case in which A < A* is physically impossible in an isentropic flow. Further, from Eq. (16.24), for any given AIA* > 1, two values of M are obtained: a subsonic value and a supersonic value. The plot given in Fig. 16.10 is the solution of Eq. (16.24) showing the subsonic and supersonic branches. The values of A/A* as a function of M are tabulated in Gas Tables (Rathakrishnan E., 1993), for different values of y Once the variation of the Mach number through the nozzle is known, the variation of static temperature, pressure, and density follow from Eqs. (16.15)—(16.17), respectively. The P, T, and p decrease continuously throughout the nozzle. Also, Pe/Po, pal po, and Te/To depend only on the area ratio Ae/A*.

454

Fundamentals of Engineering Thermodynamics

4.0

2.0

1.0

2 0.6 0.4

0.2

0.1

4.0

1 0 2.0

6.0

8.0

10.0

Area ratio, NA*

Figure 16.10 Area-Mach number relation. EXAMPLE 16.5 Air at 500 kPa and 305 K enters a convergent nozzle with a velocity of 100 m/s. Determine the mass flow rate through the nozzle when the back pressure is (a) 300 kPa and (b) 100 kPa. The exit area of the nozzle is 100 crn2. Treat air to be an ideal gas. Solution Let the subscripts i and e represent the inlet and exit of the nozzle, respectively. For the convergent nozzle, stagnation pressure and temperature are given by To; = 7; +

1002

2'p

= 305 + 2 x 1004.5 — 310 K

and ( 7, Poi

=

Pi

(3101.410.4-1)

r4r -I)

01

= 500

305)

= 529.28 kPa

If the flow is assumed to be isentropic, then P0, and To; remain constant throughout the nozzle. That is, To = To; = 310 K and P0 = Poi = 529.28 kPa. The critical pressure ratio for air with y = 1.4 is determined to be 2 P*/P0 = (+1) 7

= 0.5283

Thermodynamics of Compressible Flow 455 (a) The back pressure ratio for this case is Pb = 300 — 0.567 Po 529.28 which is greater than the critical pressure ratio. Thus, the flow is not choked and the exit pressure (or the throat pressure) is equal to the back pressure in this case. From isentropic table, for Pb/P0 = 0.567, we get M, = 0.94 and T,/To = 0.8498. The mass flow rate through the nozzle can be calculated from Eq. (16.21) as

AA630.170

fit —

J(7+ IYI2/(7

(1 +

— 1 Af 2 2 (100 x le )(0.94) (529.28 x 103)J1.4/(287 x 310) (1 + 0.2 x 0.942)2.4/0.8 12.11 kg/s

The mass flow rate can also be determined as follows: 7', = 0.8498T0 = 0.8498 x 310 = 263.44 K

P, RT,

300 x 103 — 3.968 kg/m3 287 x 263.44

V, = M,a, = M,FIF7; = 0.94 41.4 x 287 x 263.44 = 305.83 m/s Thus,

in = p,A,V,= 3.968 x 100 x 10-4 x 305.83 = 12.13 kg/s (b) The pressure ratio for this case is Pb _ 100 Po 529.28

0.189

which is less than the critical pressure ratio, 0.5283. Therefore, the flow is choked at the nozzle exit. The mass flow rate for this case is given by Eq. (16.22) as Amax 0.6847P0A* 0.6847 x 529.28 x 103 x 100 x 10-4 4287 x 310

RTo

12.15 kg/s This is the maximum mass flow rate through the nozzle for the specified inlet conditions and the nozzle throat area. EXAMPLE 16.6 Oxygen at 330 K and 200 kPa enters a duct with varying flow area. If the Mach number at the

11 1

456 Fundamentals of Engineering Thermodynamics entrance is 0.25, determine the temperature, pressure, and the Mach number at a location where the flow area has reduced to 60 per cent of the entrance area. Assume the flow to be steady and isentropic and oxygen at the given conditions to be an ideal gas. Solution For isentropic flow through a duct, the area ratio A/A* (the local area over the area of the throat where M = 1) as a function of M is listed in isentropic table (Gas Tables, Rathakrishnan E., 1993). Let the subscripts 1 and 2 refer to the entrance state and the required state, respectively. At M1 = 0.25, from isentropic table, we have A A*

P

—• L = 2.403,

To

= 0.9877, --1 = 0.9575 Po

At the desired location, A2 = 0.6A1, and A2 = A2 Al = 0.6 x 2.403 = 1.4418 A* Ai A* A, For this value ofA=* , from isentropic table, we have T = 0.9606 To P I = 0.8687 Po 0.453

M2 =

Note that we have chosen the subsonic Mach number for the calculated A2/A* instead of the supersonic one. This is because the duct is converging in the flow direction and the initial flow is subsonic. Since the stagnation properties are constant for isentropic flow, we can write T 2 _T2 /To Ti

Tao

Or = (330) (0.9606) T2

0.9877

- 320.95 K

Also, P2 _ P2 14 Pi 11/P0 Or (200) P2

=

(0.8687) 0.9575

- 181.45 kPa

The above values of M2, P2, and T2 are the desired values of the Mach number, pressure, and temperature, respectively.

i I 1

Thermodynamics of Compressible Flow

457

16.6.3 Converging-Diverging Nozzles To accelerate the flow through a nozzle, a favourable pressure gradient must be exerted across the nozzle. Therefore, in order to establish a flow through any duct, the pressure at the exit must be lower than the inlet pressure, i.e. Pe/P0 < 1. Supersonic velocity can be reached only if Pe/P0 < 0.528. For such pressure ratios, the convergent portion of the nozzle accelerates the flow up to M = 1, and the divergent portion further accelerates the flow beyond M = 1. A variety of flow fields can be generated in a convergent-divergent nozzle by independently governing Pb. Consider the flow through a convergent-divergent nozzle as shown in Fig. 16.11.

PO

At

Flow

1.0

Pel Pee

P3

Pe 4

Q.° 0.528

Pe5

PB C

X

0.0 Figure 16.11 Flow in a convergent-divergent nozzle.

When P = Po, there will be no flow through the nozzle. Let the exit pressure be reduced to a value (41 ) slightly below Po. This small favourable pressure gradient will cause a flow through the nozzle at low subsonic speeds. The local Mach number will increase continuously through the convergent portion of the nozzle, reaching a maximum at the throat. In other words, the static pressure will decrease continuously in the convergent portion of the nozzle, reaching a minimum

458 Fundamentals of Engineering Thermodynamics at the throat, as shown by the curve 'a' in the figure. Assume that P is reduced further to 42. Then the pressure gradient will become stronger, flow acceleration will be faster, and variation of Mach number and static pressure through the duct will become larger, as shown by the curve `b'. Similarly, if P is reduced continuously, at some value of Pe, the flow will reach sonic velocity at the throat, as shown by the curve 'c'. For this case, A, = A*. Now, the sonic flow at the throat will expand further in the divergent portion of the nozzle as supersonic flow if Pe/P, < 1, and will decelerate as a subsonic flow as shown by the curve 'c', for 43/P, > 1. For the cases discussed above, the mass flow through the duct increases as Pe decreases. This mass flow can be calculated by evaluating Eq. (16.21) at the throat, A = p,A,V,. When Pe is equal to 43, where sonic flow is attained at the throat, m = p*A*a*. Also, the Mach number at the throat is unity. Hence, the flow properties at the throat, and indeed throughout the subsonic section of the duct, become "frozen" when P < Pe3, i.e. the subsonic flow in the convergent portion of the nozzle remains unaffected and the mass flow remains constant for Pe < I. This condition for sonic flow at the throat is called choked flow. For any further reduction in Pe below 43, after the flow becomes choked, the mass flow remains constant. From the above discussion, it is clear that in the convergent portion of the duct, the flow remains unchanged for back pressures below I. But in the divergent portion of the duct the flow expands as a supersonic flow for Pe < I. However, P should be adequately reduced to a specified low value, PC1 , for an isentropic expansion of flow in the divergent portion of the nozzle, resulting in a shock-free supersonic flow, the variation of pressure for such an isentropic expansion is shown by the curve 'd' in Fig. 16.11. For values of exit pressures between P„ and 43, a normal shock wave exists inside the divergent portion of the nozzle. The flow rear of the normal shock is subsonic, hence the static pressure increases to 44 at the exit. The normal shock moves downstream with reduction in below 44 and will stand precisely at the exit when P = 45, where Pe5 is the static pressure rear of the normal shock at the design Mach number of the nozzle. When P is further reduced such that Pet < Pb < 45, the flow inside the nozzle is fully supersonic and isentropic. Any further reduction in back pressure below 4, results in equilibration of the flow across compression and expansion waves outside the duct. For 4, < Pb, the flow is said to be overexpanded, since the pressure at the exit has expanded below the back pressure. When > Pb, the flow is said to be underexpanded, since the exit pressure is higher than the back pressure. EXAMPLE 16.7 Air enters a converging-diverging nozzle at 600 kPa and 330 K with negligible velocity. If the exit Mach number is 2.2 and the throat area is 0.003 m3, determine (a) the exit plane conditions, (b) the throat conditions, and (c) the mass flow rate through the nozzle. Assume the flow to be steady, isentropic, and one-dimensional. Solution The exit Mach number is supersonic, therefore, the flow must be sonic at the throat. At the inlet the velocity is negligible, hence the pressure and temperature at the inlet can be treated as stagnation values. That is, Pc, = 600 kPa and To = 330 K. With ideal-gas state equation, the stagnation density is obtained as Po 600 — 6.335 kg/m3 Po — RT0 0.287 x 330

Thermodynamics of Compressible Flow 459 (a) At the exit, Me = 2.2. From the isentropic table (Gas Tables, Rathalcrishnan E , 1993), for Me = 2.2, we have --t = 0 . 0935, = 0 . 5081, —/- = 0.1841 Po Po To Thus, Pe = 0.0935Po = 0.0935 x 600 = 56.1 kPa = 0.5081T0 = 0.5081 x 330 = 167.67 K •

0.1841po = 0.1841 x 6.335 = 1.166 kg/m3

V, = Mea, = M, ffIFT, =121114 x 287 x 167.67 = 571 m/s (b) At the throat, Mt = 1.0. Thus, from the insentropic table, we have

Po

= 0.5283, —L = 0.8333, — Pr = 0.6339 To Po

Therefore, Pr = P* = 0.5283Po = 0.5283 x 600 = 316.98 kPa T, = = 0.8333T0 = 0.8333 x 330 = 274.99 K Pt= P* = 0.6339p0 = 0.6339 x 6.335 = 4.016 kg/m3 = V* = a* = .171lT* = .11.4 x 287 x 274.99 = 332.4 m/s (c) The mass flow rate of the fluid is the same at all sections of the nozzle, for the given case. Thus, it may be calculated by using the properties at any cross-section of the nozzle. Therefore, rim = p*A*V* = 4.016 x 0.003 x 332.4 = 4 kg/s

The mass flow rate for air with y= 1.4 is also given by 0.6847 4 4* 4RT0 0.6847 x 600 x 103 x 0.003 1/287 x 330

4 kg/s

16.7 NORMAL SHOCKS Shock may be described as a compression front in a supersonic flow field and the flow process across which results an abrupt change in flow properties. The thickness of the shock is

460 Fundamentals of Engineering Thermodynamics comparable to the mean free path of the gas molecules in the fluid. When the shock is normal to the flow direction, it is called a normal shock and when it is inclined at an angle to the flow it is termed oblique shock. The flow process through the shock wave is highly irreversible and cannot be approximated as being isentropic. For a perfect gas, it can be shown that all the flow property ratios across a normal shock are unique functions of y and the upstream Mach number. We will now develop some relations which relate the flow properties on either side of the normal shock. Consider a stationary control volume which contains the shock, as shown in Fig. 16.12. Assume that the flow is steady and there are no heat and work interactions, and there is no potential energy change. V1

V2

Pi

P2 P2

AA1 > 1

Shock wave

M2< 1

-11\

Control volume

Figure 16.12 Flow through a normal shock. By conservation of mass, we have (1625)

PI VI = P2 V2 By conservation of momentum, we have A(Pi — P2) =

V1 )

(1626)

The conservation of energy equation gives +

V2 V2 = h2 + 2

(16.27)

hog = hoz

(1628)

s2 — si 0

(1629)

or The second law of thermodynamics gives An insight into the nature of changes in flow conditions which occur across a shock wave, where the area can be considered to be constant, can be obtained by examining the relationship graphically. If the upstream conditions are taken to be fixed, curves can be drawn showing all the corresponding possible conditions downstream of the shock wave. It is possible to draw one set of curves, known as Fanno lines, in which each curve represents conditions which, for a particular mass flow, satisfy the continuity and energy equations. These conditions are Mass flow rate per unit area, G = MIA = pV = constant

(16.30)

v2 Stagnation enthalpy, Ho = H + — = constant 2

(1631)

and

Thermodynamics of Compressible Flow

461

It is instructive to plot the Fanno lines as a graph of enthalpy H against entropy S. The entropy equation for a perfect gas is

S — SI = Cv ln

(LI} P

(16.32)

and H = CpT

Cp-

(16.33)

Rp

Combining Eqs. (16.30H16.33), we get

S = S, + C, In {H(Ho — HP-1Y2) + constant

(16.34)

where the constant is determined by the mass flow per unit area G and the upstream conditions. The Fanno line for a particular value of G is shown in Fig. 16.13. The maximum entropy occurs H

A Ha .._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._.

Subsonic

Norma shock ,

Rayleigh line

S

Figure 16.13 Fanno and Rayleigh lines. at the point P, this condition being found by differentiating Eq. (16.34) with respect to H and equating dS/dH = 0 as, for H = Hp the value at the point P.

dS = 1 — y — 1 1 dH 2 Ho — Hp Hp or

Ho = or

y+1 V2 Hp = Hp + 2 2

VI; =(y — 1)Hp =(y — 1)CpTp =(y —

RTp = y RTp

Therefore, Vp = oF/FT !, = velocity of sound Thus, the Fanno line shows that maximum entropy occurs at the point P which is a sonic point.

462 Fundamentals of Engineering Thermodynamics If H > Hp, the flow is subsonic and, if H < Hp the flow is supersonic. For a shock wave, the conditions before and after the shock must both lie on the Fanno line for the mass flow and the area of the section at which the shock occurs. To determine the position of these points, we now consider the requirement that both the continuity and momentum equations must be satisfied for a given mass flow. A curve known as the Rayleigh line, which is obtained by considering the conservation of mass and momentum equations into a single equation and plotting it on the H—S diagram, can be drawn as shown in Fig. 16.13. The momentum equation is pit 11 — P = — (V — Vi ) A But AM = pV = pi V1 = G, therefore, we can write the above equation as P + pV2 = p + pivi2 = constant i Or

G2 P + — = constant = B P Substituting for P in Eq. (16.32), we get S = Si + C, In ( (B — G2IP) ) + constant 137

(1635)

where the constant depends upon the upstream conditions. Now, H = CpT = C B— 22 ) Rp p

(1636)

From Eqs. (16.35) and (1636), Rayleigh line, for the given mass flow, can be plotted as shown in Fig. 16.13. On the Rayleigh line the point Q corresponds to S.,. We can write ds &I

dS dp . 2L, RpQ [G21{pQ (B — G2/PQ )}] — Y _ 0 dp dH Cp 2G2/pQ — B

where the subscript "Q" refers to the point Q. If the denominator is not zero, this gives G2

pQ(B — G2/pQ )

y=0

But G = pQVQ and B = PQ + G2/pQ, therefore, VQ =

= velocity of sound

Thus, sonic conditions occur at the point of maximum entropy. The upper limit of the curve corresponds to subsonic flow and the lower limit to supersonic flow.

Thermodynamics of Compressible Flow

463

Across a shock wave the continuity, momentum, and energy equations must be satisfied. Therefore, it is clear that the conditions before and after the shock wave must lie on both the appropriate Fanno and Rayleigh lines. The intersection points x and y on Fig. 16.13 represent the states at which all the three conservation equations are satisfied. One of these (state x) corresponds to the state upstream of the shock, and the other (state y) to the state downstream of the shock. Note that for a normal shock, the flow is supersonic before the shock and subsonic after the shock. Therefore, the flow must change from supersonic to subsonic if a normal shock is to occur. The larger the Mach number before the shock, the stronger the shock will be. In the limiting case of M = 1, the shock wave simply becomes the sonic wave. Further, it is seen from Fig. 16.13 that Sy > Sx. This is expected since the flow through the shock is adiabatic and irreversible. For perfect gases, Eq. (16.28), with he, = Cpro, becomes (16.37)

Tot = TO2

That is, the stagnation temperature of a perfect gas remains constant across the shock. Note that the stagnation pressure decreases across the shock, because of irreversibilities.

16.7.1 Properties Relations across the Shock Even though the flow process inside the shock is nonisentropic, the flow processes ahead of and after the shock can be treated as isentropic. Therefore, by Eq. (16.15), we have ° 11- = 1 + ---:1 f - M' and -L 7 Q- = 1 + r---- 1- M; T1 2 T2 2 Therefore, 7'2 —

4

1 + Y — 1 MI2

2

(*.' Toi = To2)

1+ Y- 1 m2 2 2

(16.38)

Equation (16.25) can be expressed as P1 = T1 V2

1; VI

P2

or )2 i

(

(

v2

P2 i

( T2 J

p

it T2

2

=

/112

)

(1639)

M1

since M = V/a = VO ,RW' . Combining Eqs. (16.38) and (16.39), we obtain the static pressure ratio across the shock as

P2

P1

l+

= Mill

Y— 1

2 Mi

M2 1 + 11M2

2

2

(16.40)

464 Fundamentals of Engineering Thermodynamics This is a combination of the conservation of mass and energy equations; thus it is also the equation of the Fanno line for a perfect gas. A similar relation for the Rayleigh line can be obtained by combining the mass and momentum equations. By Eq. (16.26), we have PI — P2 = -7! 41 (1/2 — VI) = P2V22 — PIVI2

= RP27. v22 - -L RTI I - V2 = 7P2m2 - ',Pim? 2 or P1(1 + yM?) = P2(1 + yM22) Therefore, P2 _ 1+ y Nil Pi 1+ y MI

(16.41)

Combination of Eqs. (16.40) and (16 41) yields

1 + y-1 412 2 I M1 — (16.42) rmi2 7 —1 2 This represents the intersections of Fanno and Rayleigh lines and relates the Mach number ahead of and after the shock. The entropy change across the shock can be expressed using Eq. (5.13), as s2 —s1 =Cp in—Rln P2 Pi 7i

(16.43)

A plot of dimensionless entropy change across the normal shock (s2 — s1 )/R versus M1 is given in Fig. 16.14. Since the flow across the shock is adiabatic and irreversible, the second law

(s2 — sI)/R

s2 —s1 >0 0 S2 — 81

M1

Subsonic flow after the shock

M1 = 1 Supersonic flow before the shock

Figure 16.14 Entropy change across the normal shock.

Thermodynamics of Compressible Flow 465 requires that the entropy should increase across the shock wave. Thus, a shock wave cannot exist for values of M1 less than unity where the entropy change would be negative. At this stage, it is natural to question as to how the flow across a shock which experiences a large increase in static temperature can be adiabatic. The answer to this question is that since the thickness of the shock is of the order of mean free path (for air at standard atmospheric conditions the shock thickness is approximately 10-5 cm) and the fluid particles travelling with supersonic speed pass through the shock within a very small time interval, the fluid particles do not have enough contact area or contact time to transfer heat to the surroundings. Thus, the total temperature of the flow remains invariant, rendering the flow adiabatic. EXAMPLE 16.8 If the air flow through the converging-diverging nozzle of Example 16.7 experiences a normal shock wave at the nozzle exit plane, as shown in the figure below, determine the following after the shock: (a) the static pressure, temperature, and density, and the stagnation pressure, (b) the entropy change, across the shock, (c) the exit velocity, and (d) the mass flow rate through the nozzle. Assume' the flow from the nozzle inlet to the shock location to be steady, onedimensional, and isentropic. Solution Shock wave

(a) Given P01 = 600 kPa, P1 = 56.1 kPa; T1 = 167.67 K, pi = 1.166 kg/m3 (from Example 16.7) and y= 1.4 for air. The shock relations relate the fluid properties downstream of the shock to the flow Mach number ahead of the shock. In this example, the Mach number in front of the shock is M1 = 2.2. For M1 = 2.2, from normal shock table (Gas Tables, Rathakrishnan E., 1993), we get M2 = 0.5471, — °2 = 0.6281, — P2 = 5.48, TZ = 1.8569, = 2.9512 P Tj Pol A Thus, the static pressure, temperature, density, and the stagnation pressure P02 after the shock are P2 = 5.48P1 = 5.48 x 56.1 = 307.43 kPa T2 = 1.8569T1 = 1.8569 x 167.67 = 311.35 K p2 = 2.9512p1 = 2.9512 x 1.166 = 3.44 kg/m3 P0, = 0.6281P01 = 0.6281 x 600 = 376.86 kPa

466 Fundamentals of Engineering Thermodynamics (b) The entropy change across the shock is given by

s2 — si = Cp in

T

F' — R In 27-i Pi

= 1.0045 In (1.8569) — 0.287 In (5.48)

= 0.1335 kJ/(kg K) That is, the entropy of the flow increases across the normal shock wave, thereby causing the flow to be highly irreversible. (c) The exit velocity V2 is given by V2 = M2a2

7RT =041 = .54 24 7141.4 2 x 287 x 311.35

= 193.5 m/s (d) The mass flow rate through the nozzle with sonic conditions at the throat is unaffected by the presence of shock waves downstream of the throat. Thus, the mass flow rate in this case is the same as that calculated in Example 16.7. Thus, we have m=

4 kg/s

16.8 FLOW THROUGH ACTUAL NOZZLES AND DIFFUSERS The assumption of isentropic process made so far in this chapter is only an approximation. However, the flow through actual nozzles and diffusers is not isentropic. Therefore, it is essential to check how realistic is this approximation. In our discussion on adiabatic efficiency of nozzles in Section 5.11, we have done this by defining the nozzle efficiency as TIN —

VV 2 V2s/2

—

actual ke at the nozzle exit ke at the nozzle exit for isentropic flow from the same inlet state to the same exit pressure

(16.44)

In terms of enthalpy we can express this as 71N

hoi — h2

(16.45)

hoi — h2s

where km is the stagnation enthalpy at the nozzle inlet, h, is the actual static enthalpy at the nozzle exit and h2s is the exit static enthalpy for isentropic processes illustrated on the h—s diagram in Fig. 16.15. The efficiencies of nozzles range from 90 to 99 per cent. The larger nozzles with straight-line axes have the higher efficiencies. Thus, the approximation of the actual process to isentropic one is justified for nozzles.

Thermodynamics of Compressible Flow 467 h

1301

01 t

t

2

V2s 2

V2 Isentropic

2

•s Figure 16.15 Isentropic and actual flow processes through a nozzle.

16.8.1 Velocity Coefficient Cv This

is another parameter used to express the performance of a nozzle. It is defined as Cv —

V2

V2s

actual velocity at the nozzle exit velocity at the exit for isentropic flow from the same inlet state to the same exit pressure

(16.46)

From Eqs. (16.44) and (16.46), it is seen that = 3;17 s1

(16.47)

16.8.2 Discharge Coefficient Co This is a quantity of interest in nozzle design and is associated with the mass flow rate. It is defined as CD —

Ois

actual mass flow rate through the nozzle mass flow rate under isentropic assumption from the same inlet state to the same exit pressures

(16A8)

EXAMPLE 16.9 Hot air at 700 K enters a convergent nozzle with negligible velocity. If the stagnation pressure is (a) 220 kPa and (b) 150 kPa, determine the velocity, temperature, and pressure at the nozzle exit for a back pressure of 100 kPa and a nozzle efficiency of 90 per cent. Assume the air to be an ideal gas with y = 1.4. Solution A schematic of the nozzle and the T—s diagram of the process are given as follows:

468 Fundamentals of Engineering Thermodynamics

P01 Tot

T

01

From the given data of the stagnation pressure and back pressure, we can identify whether the flow at the nozzle exit is choked or not. Also, we know that for choked flow the velocity at the nozzle exit is independent of the back pressure and equal to the velocity of sound. If the nozzle is not choked, then the exit pressure will be the same as the back pressure. Let us assume the flow to be adiabatic, so that the stagnation enthalpy of the fluid remains invariant throughout the nozzle. Thus, hoi = That is, T01 = T02 = 700 K, since the air is assumed to be an ideal gas. We know that if the critical pressure P* is more than the back pressure Pb, the nozzle is choked. The critical pressure P* of a converging nozzle is the actual pressure at the nozzle exit when the velocity at the nozzle exit has reached the sonic value. For M2 = 1, the actual T2 and P2 at the nozzle exit can be determined as follows: By Eq. (16.18), we have 7'2 2 T02 7+ 1 Or = T2

To2

(

I) _ 700 +

) _ 583.33K (0.42+ 1

For an ideal gas with constant specific heats, the nozzle efficiency can be expressed as

Thermodynamics of Compressible Flow

qN

_

hot

h2

h01

h2s

CP( Tot — T2) _

CP(701

T2s)

469

T2 Tot

ris

Thus, 0.9 — 700 — 583.33 700 — T2s Therefore, T2s = 570.37 K The actual temperature at the nozzle exit is T* = T2 = 583.33 K (for sonic conditions), and T2s = 570.37 K is the exit temperature that would be achieved at sonic conditions if the process were isentropic. The exit pressure for both the actual and the isentropic cases is the same, and given by .4/0.4( 7 , )710' -11) ( 570.3 2s = 0.4883 7007 Poi — (To. Thus, P* = P2 = 0.4833P01

)t

(a) For stagnation pressure Poi = 220 kPa, the critical pressure is = 0.4833 x 220 = 106.33 kPa which is larger than Pb, therefore, the nozzle is choked. The actual pressure and temperature at the nozzle exit are thus P* and T*. That is, P2 = P* =

106.33 kPa

T2 = T* = 583.33 K The velocity at the nozzle exit is the sonic velocity and is given by V2 = a2 = oF/FT 2 = 111.4 x 287 x 583.33 = 484.13 m/s (b) For P01 = 150 kPa, the critical pressure is P* = 0.4833 x 150 = 72.495 kPa which is less than Pb. Thus, the nozzle is not choked and the pressure at the nozzle exit is the same as the back pressure. P2 = Pb = 100 kPa

The actual temperature at the nozzle exit can be determined as follows: 7,

25

=

;1

p2s (Poi

- Dir

470

Fundamentals of Engineering Thermodynamics

or (100(1.4 —1)/1.4

T2s = 700 (

= 623 . 36 K

150)

and

—

;1 - T2 Tot - T2,

or 0.9 —

700 — 7; 700 — 623.36

or T2 = 631.02 K The actual velocity is determined from V2 = V2Cp(To1 — T2) = J(2 x 1004.5 (700 — 631.02) 372.26 m/s

16.8.3 Diffusers Diffusers are passages through which flow decelerates. In this sense, they perform the opposite task of a nozzle. A typical subsonic diffuser is shown schematically in Fig. 16.16. The

Figure 16.16 A subsonic diffuser. performance of a diffuser is usually expressed in terms of the diffuser efficiency ris, defined as n

A hs VI /2

"ID — 2 —

hoes —

ht

hot — h,

(16.49)

where V1/2 is the maximum ke available for converting to pressure rise and Ohs. The definition of 77 is better understood with the help of the h—s diagram given in Fig. 16.17. Diffuser efficiencies vary from 90 per cent to close to 100 per cent for subsonic diffusers, and decrease with increasing Mach number.

Thermodynamics of Compressible Flow 471

P

S

Figure 16.17 h—s diagram for the definition of 77D.

16.8.4 Pressure Recovery Factor Fp It is a measure of the diffuser's ability to increase the pressure of the fluid stream. It represents the actual stagnation pressure of a fluid stream at the diffuser exit relative to the maximum possible stagnation pressure. That is, FP —

actual stagnation pressure at the diffuser exit isentropic stagnation pressure

PO2 P01

(16.50)

For an isentropic diffuser, P02 = Poi, and thus FP = 1.

16.8.5 Pressure Rise Coefficient

CpR

It is yet another parameter frequently used to evaluate the performance of a diffuser. It is defined as CPR

actual pressure rise isentropic pressure rise

P2 — PI

P01 —

(16.51)

EXAMPLE 16.10 Air enters a subsonic diffuser with a velocity of 320 m/s, the stagnation pressure and temperature of the stream are 600 kPa and 700 K, respectively. The diffuser exit velocity is 90 in/s. For a diffuser efficiency of 85 per cent, determine (a) the pressure rise coefficient, and (b) the required exit-to-inlet area ratio. Assume the air to be an ideal gas with constant specific heats. Solution The diffuser and the T—s diagram of the process are shown as follows.

472 Fundamentals of Engineering Thermodynamics

T

01

02 V22

2Cp

1/12 2Cp

Assuming the flow process through the diffuser to be adiabatic, the stagnation enthalpy of the air remains constant as it flows steadily through the diffuser. Thus, T02 = TO1 = 700 K

(a) The static temperature at the diffuser inlet can be obtained from the energy equation as 2

3202 — 649 K ii = Tot — 2C = 700 2 x 1004.5 P The static pressure at the inlet P1 is given by T P =Poi (—T1—)

y/(7-1) =

(649)1.41(1.4-1)

(700)

.ot

=46°.4 kPa

For an ideal gas with constant specific heats, the diffuser efficiency can be expressed as „ 'I

D=

hots ht — Tou hOl — 7O1

Then, 0.85 — T02s — 649 700 — 649

Thermodynamics of Compressible Flow 473 or To2s = 692.35 K Also, P02 = Po2s, therefore, from the isentropic relation (between the states 1 and 2) we can write ( 692.35 )t.4/0.4- i) 7 17 - I ) =460.4 = 577.3 kPa T ) p02 = P02, = P. (1,2L 649 ) -1 The exit static temperature T2 is given by 2 902 v2 71 = TM — 2C = 7® 2 x 1004.5 - 696 K The exit static pressure P2 can be determined from ,, rr-I) P2 = P02 (

696

1.410.4 - I)

= 577.3 (— 700 JJJ

To2

= 565.8 kPa

Thus, the pressure rise coefficient is CPR -

P2 — Pi _ 565.8 - 460.4 _

Po, —

P,

600 - 460.4

0.755

(b) The exit-to-inlet area ratio .92/.91 can be determined by applying the mass conservation relation. Thus, we have m = PIA i VI = P2A2V2

Or

A2 = P 1V1 — (Pl i RTI)VI — FIT2VI P2TI V2 P 2V2 (P2II RT2 )V2 Al .

460.4 x 696 x 320 - 3.1 565.8 x 649 x 90

16.9 STEAM NOZZLES We know that water vapour at moderate or high pressures deviates considerably from the idealgas behaviour. Therefore, most of the relations developed based on ideal-gas assumption are not applicable to the flow of steam through nozzles. It becomes necessary to use steam tables or an h-s diagram for the properties of steam. Examine the h-s diagram given in Fig. 16.18. As steam expands in the nozzle, its P and T drop, and it is expected that steam would start condensing when it strikes the saturation line. But this does not happen always. Because of high velocities, the residence time of the steam in the nozzle is small, and therefore, sufficient time is not available for the necessary heat transfer and the formation of liquid droplets. Thus, the condensation is delayed for a little while. This phenomenon is known as supersaturation. The steam which exists in the wet region without containing any liquid is called the supersaturated steam. Supersaturation states are nonequilibrium or metastable states.

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Fundamentals of Engineering Thermodynamics h

"2 Saturation line Wilson line (x= 0.96) Figure 16.18 h-s diagram for the isentropic expansion of steam in a nozzle. During the expansion, the T reached is lower than that normally required for the condensation process to begin. Once T drops sufficiently below the saturation T corresponding to the local P, moisture droplets are formed, and condensation occurs rapidly. The locus of the points where condensation takes place irrespective of the initial T and P at the nozzle inlet is called the Nilson line. The critical pressure ratio P*/P0 for steam depends on the nozzle inlet state and whether the steam is superheated or saturated at the nozzle inlet. However, the ideal-gas Eq. (16.19) gives reasonably good results over a wide range of inlet states. Taking r= 1.3 for superheated steam, we get

1747-1)

P* = ( P0

+ 1)

= 0.546

When steam enters the nozzle as a saturated vapour, the value of which gives P*

r has to be taken as 1.14,

— = 0.576 Po

EXAMPLE 16.11 Steam at 1 MPa and 500°C enters a converging-diverging nozzle with negligible velocity. The mass flow rate of the steam is 3 kg/s and the pressure at the nozzle exit is 200 kPa. The flow is isentropic between the nozzle entrance and the throat, and the overall efficiency is 90 per cent. Determine (a) the throat area and the exit area, and (b) the Mach number at the throat and also at the nozzle exit. Solution The schematic of the nozzle and the h-s diagram of the process are shown in the figures below. (a) The ratio of the exit static pressure to the inlet stagnation pressure is p22 = = 1000 P01

- 0.2

Thermodynamics of Compressible Flow 475

V1 =

h

= pot

This ratio is much less than the critical pressure ratio of P*/Po = 0.546 since steam is superheated at the inlet. Therefore, the flow at the nozzle exit must be supersonic and the velocity at the nozzle throat is sonic. The pressure at the throat is given by P, = 0.546Po3 = 0.546 x 1 = 0.546 MPa At the inlet

P1 =Poi =1MPa = Tot = 500°C For this state, from steam tables, we have hi = hoi = 3478.5 kJ/kg s1 = s, = s2s = 7.7622 kJ/(kg K) At the throat, P, = 0.546 MPa and s, = 7.7622 kJ/(kg K) Thus, h, = 3278.4 kJ/kg and = 0.5713 m3/kg The velocity at the throat is given by V, = 42(ho1 — h,) = 42(3478 .5 — 3278 .4)1000 = 632.6 m/s

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Fundamentals of Engineering Thermodynamics

The throat area

4

My V,

3 x 0.5713 632.6

27.09 cm2

At state 2s P, = F12 = 200 1cPa

and s2, = s1 = 7.7622 kJ/(kg K) For these values of P, and s2„ the corresponding h2, from steam tables is h2s = 3000.36 kJ/(kg K) The enthalpy of the steam at the actual exit state is determined from „ kw — h2 TIN — L L

"IN — "b

Or 0.9 —

3478.5 — h2 3478.5 — 3000.36

or h2 = 3048.17 kJ/(kg K) For P2 = 200 kPa and h2 = 3048.17 kJ/(kg K), the corresponding v2 and s2 are v2 = 1.2887 m3/kg, s2 = 7.8495 kJ/(kg K) The velocity at the nozzle exit is V2 =42(h01 — h2) = J2(3478.5 — 3048.17)1000 = 927.72 m/s The exit area

1 3 x 1.2887 thy A2 = — 4.1673 x 10-3 m2 2— 927.72 V2

= 41.67 cm2 (b) The speed of sound and the Mach number at the throat of the nozzle can be determined as follows: The speed of sound can be expressed as a

= Ikap) op 5

_

AP 1/v), 11 64

The speed of sound at the throat can be determined by evaluating the specific volume at s, = 7.7622 kJ/(kg K) and at pressure 0.571 MPa and 0.521 MPa (P, ± 25 kPa): a, —

571 —521 x 1000 = 639.1 mis (1/0.5513) — (1/0.5912)

Thermodynamics of Compressible Flow 477 The Mach number at the throat is Y 632.6 M =— ' = a, 639.1

0.99

Thus the flow at the throat is sonic, as expected (the slight deviation of M, from 1.0 is due to the approximation involved in the expression for a,). The speed of sound and the Mach number at the nozzle exit can be obtained by evaluating the specific volume at s2 = 7.8495 kJ/(kg K) and at pressures of 225 kPa and 175 kPa (P2 ± 25 kPa). Therefore, a2

l 225 — 175 x 1000 = 539.95 m/s il (1/1.2006) — (1/1.5119)

and M2 =

V 2 a2

-

927.72 _ 1.72 539.95

That is, the flow at the nozzle exit is supersonic.

SUMMARY The energy changes associated with compressible flow, unlike low-speed or incompressible flow, are substantial enough to strongly interact with other properties of the flow. Therefore, the study of thermodynamics which deals with energy (and entropy) becomes an essential component of the study of compressible flow. In modem high-speed problems, the kinetic energy content of the fluid can be very large compared to its heat content leading to substantial variations in temperature associated with these high speed flows. Stagnation state is a state of zero flow velocity, achieved by decelerating the flow. The properties of a fluid at the stagnation state are called stagnation properties. For ideal gases with constant specific heats, the stagnation and static properties are related as given below: Po = Par Tr") P p) _w Speed of sound is the speed with which sound propagatei in a medium. For perfect gases, the speed of sound a can be expressed as a = 157 R7 Mach number M is defined as the ratio of the local flow speed to the local speed of sound. M—

local flow speed local speed of sound

V a

478 Fundamentals of Engineering Thermodynamics The area-velocity relation for one-dimensional isotropic flow is given by dA

A

dV

V

A thermally perfect gas is that for which the internal energy and enthalpy are functions of temperature alone. A calorically perfect gas is that for which the internal energy and enthalpy are constant and independent of temperature. A perfect gas is that which is thermally as well as calorically perfect. A perfect gas is also called an ideal gas. The ratios of stagnation to static values of T, P, and p of an ideal gas in terms of M and y can be expressed as Y

=

o =

1 m2 2

T

Y — 1 A2r r

2 Y — 1 M2 Po = (1 + 2 Critical properties are the properties of a fluid at a location where M = 1.

Nozzles are passages to accelerate the flow of fluids. Choked flow is the maximum mass flow possible through a passage for a given upstream stagnation state. The mass flow rate through a nozzle can be expressed in terms of To and P0 as

AMPoor,/(1 0 )

rii — 1+

1 —1M2

)(r -1)112(7 -1)1

2 ( The area-Mach number relation for a convergent—divergent nozzle can be expressed as

Overexpanded flow is that for which the nozzle exit pressure Pe < the back pressure Pb. Underexpanded flow is that for which Pe > Pb. Shock may be defined as a compression front in a supersonic flow field and the flow process across which results an abrupt change in fluid properties. When the shock is normal to the flow direction, it is called a normal shock; when it is inclined at an angle to the flow, it is termed oblique shock The temperatures and pressures on either side of the normal shock can be expressed as r2

1 4-

I

1+

y— 2 2 1 MI Y— 1 2

2

M2

Thermodynamics of Compressible Flow 479

P, 1 + TM? Pi 1+yMi The Mach number downstream of a normal shock, M2, is given by 1 + 7— 1 A/2 MZ

—

rive

2

I

r- 1 2

The entropy change across the shock can be written as s2 —

2 — R In -= = Cp In —

The velocity coefficient Cy is a parameter used to express the performance of a nozzle. It is defmed as —

V 2 — V2s

actual velocity at the nozzle exit velocity at the exit for isentropic flow from the same inlet state to the same exit pressure

The velocity coefficient is related to nozzle efficiency as Cv = Afr7; The nozzle discharge coefficient CD is defmed as

CD

ins

actual mass flow rate through the nozzle mass flow under isentropic assumption from the same inlet state to the same exit pressure

Diffusers are passages through which flow decelerates. The diffuser efficiency is defined as 77D

Ohs 112 /2

hoes — hoi —

The pressure recovery factor Fp is a measure of the diffuser's ability to increase the pressure of the fluid stream. That is, FP —

actual stagnation pressure at the diffuser exit isentropic stagnation pressure

P02 Poi

PROBLEMS 16.1

The static temperature and pressure of an air stream flowing at 300 m/s are 300 K and 95 kPa, respectively. Determine the stagnation temperature and stagnation pressure of the air. Assume the flow to be isentropic. [Ans. 344.8 K, 154.63 kPa]

480

Fundamentals of Engineering Thermodynamics

16.2 Determine the stagnation temperature and pressure for the following cases of flow through a duct: (a) steam at 3.0 MPa, 400°C and 420 m/s; (b) nitrogen at 200 kPa, 40°C, and 250 m/s, and (c) helium at 200 kPa, 40°C, and 250 m/s. [Ans. (a) 445.5°C, 4.0 MPa, (b) 70°C, 275.48 kPa, (c) 46°C, 209.72 kPa] 16.3 Air flows through a duct. The pressure and temperature at station 1 are P1 = 0.7 atm and = 30°C. At a second station, the pressure is 0.5 atm. Calculate the temperature and density at the second station. Assume the flow to be isentropic. [Ans. 275.36 K, 0.641 kg/m3] 16.4 Steam flows through a duct with a velocity of 220 m/s. If the stagnation temperature and pressure of steam are 400°C and 1 MPa, respectively, determine the static temperature and pressure of the flowing steam. Assume the flow to be isentropic. [Ans. 386.3°C, 0.95 MPa] 16.5 Atmospheric air at 101 kPa and 303 K enters a compressor and gets compressed to a stagnation pressure of 1 MPa. Assuming the compression process to be isentropic, determine the power input to the compressor for a mass flow rate of 0.035 kg/s. [Ans. - 9.85 kW] 16.6 Combustion products at stagnation pressure of 800 kPa and stagnation temperature of 590°C expand through a gas turbine to a stagnation pressure of 101 kPa. Assuming y= 1.33 and R = 287 kJ/(kg K) for the combustion products, and treating the expansion process to be isentropic, determine the power output of the gas turbine per unit mass flow. [Ans. 445.73 kJ/kg] 16.7 For an air stream at 630 m/s, if the stagnation pressure and temperature are 720 kPa and 430°C, respectively, determine the static temperature and pressure of the stream. [Ans. 505.6 K, 227 kPa] 16.8 Determine the velocity of sound in air at (a) 30°C, (b) 60°C and (c) 90°C. Also, calculate the Mach number of an aircraft flying with a velocity of 400 m/s, at an attitude where the outside air temperature is 30°C. [Ans. (a) 349 m/s, (b) 365.9 m/s, (c) 382 m/s, 1.15] 16.9 Helium gas at 300 K enters an adiabatic nozzle with a velocity of 40.m/s and leaves at 200 K. Determine the Mach number (a) at the inlet, and (b) at the exit of the nozzle. [Ans. (a) 0.04, (b) 1.225] 16.10 Air at 200 kPa and 20°C, enters a steady-flow heat exchanger with a velocity of 70 in/s. It receives 100 kJ/kg of heat as it flows through the heat exchanger. Air leaves the heat exchanger at 101 kPa with a velocity of 150 m/s. Determine the Mach number of the air at (a) the inlet, and (b) the exit of the heat exchanger. [Ans. (a) 0.204, (b) 0.382] 16.11 The static pressure and static temperature of steam flowing with a velocity of 260 m/s are 0.6 MPa and 300°C, respectively. Determine the Mach number of the steam at this state by (a) assuming ideal-gas behaviour with y = 1.3 and (b) using data from steam tables. [Ans. (a) 0.443, (b) 0.448]

Thermodynamics of Compressible Flow 481 16.12 An aircraft is flying at a constant Mach number 2.2 at an altitude where the atmospheric temperature is 220 K. Determine the stagnation temperature at the aircraft's nose. [Ans. 432.96 K] 16.13 Nitrogen at a pressure of 400 kPa enters a converging-diverging nozzle with negligible velocity. Determine the lowest pressure which can be obtained at the throat of the nozzle. [Ans. 211.31 kPa] 16.14 Air at 320 kPa, 300 K and 0.6 Mach flows through a duct. Determine the velocity and stagnation temperature, pressure, and density of the air. [Ans. 208.3 m/s, 321.6 K, 408.16 kPa, 4.423 kg/m3] 16.15 Air at 200 kPa and 300 K, enters a nozzle with a velocity of 100 in/s. Determine the pressure and temperature of the air at a location where its velocity equals the speed of sound. Also, calculate the ratio of the area at this location to the entrance area. Assume the flow to be isentrpic. [Ans. 254.14 K, 111.93 kPa, 0.477] 16.16 A converging-diverging nozzle is designed to operate with an exit Mach number of 1.75. The nozzle is supplied from an air reservoir at 6.8 MPa. Assuming one-dimensional flow, calculate (a) the maximum back pressure to choke the flow, and (b) the back pressure for the nozzle to be correctly expanded to the designed Mach number. [Ans. (a) 5.8 MPa, (b) 1.278 MPa] 16.17 Air at 100 kPa and 300 K expands in an ideal diffuser. The entrance speed is 180 m/s. Calculate the maximum pressure that can be achieved if the air is diffused to zero speed. [Ans. 120.1 kPa] 16.18 Air flows across a normal shock. The flow Mach number, pressure, and temperature ahead of the shock are given as 2.0, 0.5 atm, and 300 K, respectively. Determine M2, T2, and V2 downstream of the shock wave. [Ans. 0.5774, 2.25 atm, 506.1 K, 260.37 m/s] 16.19 Air at 600 kPa and 400 K enters a converging-diverging nozzle with low velocity. If the nozzle exit area is 2 times the throat area, what should be the back pressure to produce a normal shock at the exit plane of the nozzle? [Ans. 307.43 kPa] 16.20 Calculate the entropy change experienced by the air passing through the normal shock in Problem 16.18. [Ans. 0.0939 kJ/(kg K)] 16.21 Air at 100 kPa and 300 K enters the diffuser of an engine at 220 m/s and at a rate of 13 kg/s. If the efficiency of the diffuser is 90 per cent, determine the diffuser exit temperature, pressure, and the area for an exit velocity of 50 m/s. [Ans. 322.9 K, 126.05 kPa, 0.191 m2] 16.22 Determine the pressure recovery factor and the pressure rise coefficient for the diffuser of Problem 16.21. [Ans. 0.974, 0.839]

482

Fundamentals of Engineering Thermodynamics

16.23 Steam at 2.0 MPa, 500°C enters a convergent nozzle with negligible velocity. At the nozzle exit, the pressure is 1 MPa and the area is 18 cm2. Determine the exit velocity, mass flow rate, and the exit Mach number if the nozzle (a) is isentropic, and (b) has an efficiency of 95 per cent. [Ans. (a) 671.3 m/s, 4.005 kg/s, 1.08, (b) 654.2 m/s, 3.87 kg/s, 1.04] 16.24 Steam at 2 MPa and 500°C enters a converging-diverging nozzle with negligible velocity. The exit pressure is 300 kPa and the mass flow rate is 2 kg/s. Assuming the nozzle flow to be isentropic, determine the exit velocity and the exit Mach number. [Ans. 1041.36 m/s, 1.91]

CHAPTER

17 Kinetic Theory of an Ideal Gas 17.1 INTRODUCTION We explained that the macroscopic approach to the study of thermodynamics which does not require any knowledge of the behaviour of individual particles of the substance is called classical thermodynamics. In other words, the point of view of classical thermodynamics is entirely macroscopic. In classical thermodynamics, the systems are described with the aid of their gross, or large-scale, properties. The first law of thermodynamics is a relation among the fundamental physical quantities of work, internal energy, and heat. When the first law is applied to a class of systems, a general relation is obtained which holds for all members of the class but which contains no qualities or properties of a particular system that would distinguish it from another. For example, the following relation for specific heat at constant volume, given by Eq. (2.22), is true for all hydrostatic systems whether liquid or gas.

r _ au

),

—v

It enables us to calculate the Cy of a hydrostatic system, provided that we know the internal energy as a function of T and v. The heat transferred during an isochoric process, T qv= L:., CvdT may be calculated once the Cy of the particular system under consideration is known as a function of T. But classical thermodynamics does not provide detailed information concerning u or Cv. Another example highlighting the limitation of classical thermodynamics is its failure to provide the equation of state of any desired system. To make use of any thermodynamic (dV d7)p, and (dT1dP), we must have an equation involving P, V, T and the derivatives (dPI equation of state. Experimental values are very often useful, but there are situations where it is not possible to conduct the necessary experiments. Let us assume that we perform an experiment 483

484

Fundamentals of Engineering Thermodynamics

on, say, helium, where the numerical constants in the equation of state of helium only are obtained, and we do not get any clue concerning the values of constants for any other gas. To obtain detailed information concerning the thermodynamics coordinates from the thermal properties of systems without resorting to experimental measurements, we need calculations based on the properties and behaviour of the molecules of the systems. The kinetic theory and statistical mechanics are two such microscopic theories. Both these theories deal with molecules, their internal and external motions, their collisions with one another and with any existing walls, and their forces of interaction. Basically, the kinetic theory deals with the details of motion of molecules and their impacts, and is also capable of dealing with the following nonequilibrium processes: 1. Effusion of molecules through a hole in a container. 2. Laminar flow of molecules through a pipe under the action of a pressure difference. 3. Molecular momentum exchange—a molecular process responsible for viscosity. 4. Molecular energy exchange—a molecular process responsible for heat conduction. 5. Molecular mass exchange—a process known as diffusion. 6. Chemical kinetics—chemical combination between two or more kinds of molecules, which takes place at a finite rate. 7. Brownian motion—a haphazard zigzag motion of the molecules in a fluid. Statistical mechanics deals only with the energy aspects of the molecules and avoids the mechanical details concerning molecular motions. Even though it relies heavily on the theory of probability, it is mathematically simpler than the kinetic theory. But it is more subtle conceptually and only the equilibrium states can be handled. However, statistical mechanics can handle equilibrium states in a uniform and straightforward manner so that once the energy levels of the molecules or of systems of molecules are understood, the equations of state, the energy, and thermodynamics functions may be obtained by carrying out a program of calculations. In this chapter, we will study only a small portion of the kinetic theory of an ideal gas.

17.2 EQUATION OF STATE OF AN IDEAL GAS The fundamental assumptions made in the development of kinetic theory of an ideal gas are: • Any small volume in a gaseous system consists of a large number of molecules N. For any one chemical species, all molecules are identical. If m is the mass of each molecule, then the total mass is mN. If M denotes the molecular mass (molecular weight) in grams, then the number of gram-moles n is given by n=

mN M

The number of molecules per mole of gas is called the Avogadro number NA, where N M NA = — = — = 6.0225 x 1023 molecules per gram-mole n m

1 I

Kinetic Theory of an Ideal Gas

485

Since a mole of an ideal gas at 273 K and at 1 atm pressure occupies a volume of 2.24 x 104 cm3, there are approximately 3 x 1019 molecules in a volume of 1 cm3, and even a volume as small as cubic micron contains as many as 3 x 1010 molecules. • The molecules of an ideal gas are assumed to resemble small hard spheres that are in perpetual random motion. Within the temperature and pressure range of an ideal gas, the average distance between neighbouring molecules is large compared with the size of the molecule. The diameter of the molecule is of the order of 2 or 3 x 10-10 m. Under standard conditions, the average distance between molecules is about 50 times their diameter. • The molecules of an ideal gas are assumed to exert no forces of attraction or repulsion on other molecules except when they collide with one another and with a wall. Between collisions, they, therefore, move with uniform rectilinear motion. • The wall surface with which a molecule collides is considered to be smooth, and the collision is assumed to be perfectly elastic. If V is the speed of a molecule approaching a wall, only the normal component of velocity Vn is changed upon collision with the wall, from V, to —V., or a total change of —2 V, in velocity is experienced by the molecule. • In the absence of any external force or field, the molecules are distributed uniformly throughout a container. The molecular density N/V is assumed constant, so that in any small element of volume dV there are dN molecules, where N dN =— d V V • There is no preferred direction for the velocity of any molecule, so that at any moment there are as many molecules moving in one direction as in another. • Not all molecules have the same speed. A few molecules at any moment move slowly and a few move very rapidly, so that speed may be considered to cover the range from zero to the speed of light. Since most molecular speeds are so far below the speed of light, no error is introduced in integrating the speed from 0 to co. If dN v represent the number of molecules with speeds between V and V + dV, it is assumed that dN1, remains constant at equilibrium even though the molecules are perpetually colliding and changing their speeds. With the above assumptions, let us derive a form of the equation of state, using a simplified picture of molecular motion. The purpose of revisiting the equation of state here is that the derivation provides some useful insights into the molecular properties of gases. Consider a gas contained within a cubical box shown in Fig. 17.1. Identify a specific gas particle at some instant of time and at some location, Pi . This particle has a translational velocity C, with x, y, and z components of velocity denoted by Cx, Co and Cl„ respectively. From the above assumption it is evident that a gas particle is treated as a structureless "billiard ball" translating in space and frequently colliding with neighbouring particles. In fact, it is such molecular collisions that give enough time to establish a state of equilibrium in the system. Let us assume that the gas in the box is in equilibrium. This implies that at any given point P1 , if a given particle causes a change in velocity, then there is another collision between other particles in the same neighbourhood which causes one of those other particles to have the

486

Fundamentals of Engineering Thermodynamics

Figure 17.1 Movement of a gas particle in a cubical box. velocity C1 at the point Pi . The net result resembles the situation as if the original particle simply continued with the velocity C. With this picture, we can visualize a particle traversing the box with a constant velocity in the x-direction, given by Cx. When the particle reaches the right face of the box in Fig. 17.1, it is assumed to specularly reflect from the surface at the point P2. Specular reflection is perfectly elastic with the molecular velocity component normal to the surface being reversed, while that parallel to the surface remains unchanged. The other kind of reflection in which the velocity of each molecule after reflection is independent of its incident velocity is termed diffuse reflection. That is, if C1 is the velocity just before impacting the surface at the point P2, and C2 is the velocity immediately after the impact, then IC1 1 = IC2I, = —Cxl, Cy2 = Cyl, and Ca = Czi. During the impact, the particle experiences a change in momentum in the x-direction given by 2mc, where m is the mass of the particle. During a unit time (say, one second), the particle makes a number of traverses back and forth across the box in the x-direction. Counting a complete traverse as going and coming to and from the right-hand face, the number of complete traverses per unit time is Ci/2L, where L is the length of the box along the x-axis. The time rate of change of momentum experienced by the particle when impacting the right-hand face is given by (2mc)(Cy2L)= mC1IL We know that the time rate of change of momentum is equal to force (by Newton's second law). Hence the force exerted by the particle on the right-hand face is also mc2/L. But, as we know, pressure is force per unit area, and the area of the face is L2, thus, the pressure exerted by the particle on the right-hand face is given by mCx2IL3 = mC 2IV where V is the volume of the system.

Kinetic Theory of an Ideal Gas 487 Let us now assume that there is a large number of particles in the box, each with a different mass m and different velocity Cr The pressure exerted on the right-hand face by the particles in the system is given by

1 = —Ern•C? V "'x

(17.1)

where the summation is taken over all the particles. By constructing an expression for the pressure exerted on the upper face (perpendicular to the y-axis), using identical arguments, we get

P = —1 E

?y

(172)

s

Similarly, the pressure exerted on the face perpendicular to the z-axis can be expressed as

1 P = — Em.C? "x

(17.3)

Adding Eqs. (17.1) to (17.3), we get

P= _1 mi (c?,+C.1 + 3V 1—‘ '" 'Y

z

)= 1_Emicl 3V

(17.4)

where C, is the velocity of the ith particle. The total kinetic energy of the system, Erns, is given by

Eirans =

1

miCf

(175)

From Eqs. (17.4) and (17.5), we obtain PV =

3

E' s

(17.6)

Equation (17.6) is the kinetic theory equivalent of the perfect gas equation of state. To relate this to temperature we need to seek the help of classical thermodynamics, since T is a variable which originated from classical thermodynamics. For example, assume that there is one mole of particles in the system. Then V in Eq. (17.6) becomes a mole volume ig , and Et. is the kinetic energy per mole. Therefore, we have

= — Emus 3

(17.7)

But we know that

Pt, = RuT

(17.8)

where Ru is the universal gas constant. Combining Eqs. (17.7) and (17.8), we get

3 Emus = — RuT

(17.9)

488 Fundamentals of Engineering Thermodynamics Dividing both sides of Eq. (17.9) by Avogadro's number NA (the number of atoms in a gramatom, or the .number of molecules in a gram-molecule or mole is called the Avogadro's number NA where NA = 6.0225 x 1023), we get _ 3 R„ T NA 2 NA or E =

(17.10)

R" = 1.3805 x 10-23 J/K is called the Boltzmann constant. where k = — NA This equation establishes the physical link between the thermodynamic variable T and the molecular picture, i.e. temperature is a direct index for the mean kinetic energy of a particle in the system. The higher the temperature, the higher is the mean molecular kinetic energy. From Eqs. (17.6) and (17.7) we see that they establish a relation between the product of pressure and volume, and the molecular kinetic energy of the system. Therefore, the pressure volume (PV) product can be interpreted as a measure of energy of the system. Now, let us divide both sides of Eq. (17.4) by the total mass of the system M, where M = E mr PV _ 1 M 3

Emiq

E m;

But MN = p. Let us define a mean square velocity

E 2 as

E C2 =

E

(17.12)

Equation (17.11) may now be expressed as P =

C2 1

P

3

(17.13)

This is another form of kinetic theory equivalent of the perfect gas equation of state. From Eqs. (7.7) and (17.13), we obtain E2 = 3RT i.e., the root mean square (rms) velocity is given by

C77-2

=

15FT

(17.14)

Let us look at Eq. (17.10) more closely. The translational kinetic energy for a particle, Etrans, is given by irniCi2, where mi is the mass of the particle. But, from Eq. (17.10) we have Et. as lkT 2 ' independent of the mass of the particle. Hence, for a gas mixture at temperature 7', the heavy particles will move more slowly, on the average, than the light particles.

Kinetic Theory of an Ideal Gas 489

17.3 COLLISION FREQUENCY AND MEAN FREE PATH Consider a hard sphere particle of molecular diameter d moving at the mean molecular velocity, E . Whenever this molecule comes into contact with a like molecule, the distance of separation of the centres of the two molecules is also d, as shown in Fig. 17.2. This separation can be viewed as the radius of influence, in that any colliding molecule whose centre comes within a distance d of the given molecule is going to cause a collision. Therefore, as the given molecule moves in space, its radius of influence will sweep out a cylindrical volume per unit time equal to ird2.E, as sketched in Fig. 17.3.

Figure 17.2 Illustration of radius of influence.

.....''''

........,..-„---

,,'•

.

I

%

i

Figure 17.3 Cylinder volume swept in one second by a particle of radius d moving with mean velocity E. .

490

Fundamentals of Engineering Thermodynamics

If the number density (the number of particles per unit volume) is n, then the given particle will experience nxd2 E collisions per second. This is defmed as the single particle collision frequency denoted by Z'. Hence, (17.15) Z' = nxd 2 E' The mean free path, 2„ is defmed as the mean distance travelled by a particle between the successive collisions. In unit time the particle travels a distance E. and experiences Z' collisions during this time, therefore, we can express the mean free path A as — Z' — nnd2

(17.16)

The above analysis is elementary in nature and the results obtained are very much simplified, and more accurate results for collision frequency and mean free path for a gas are slightly different from those given by Eqs. (17.15) and (17.16). In the above analysis, only one particle with radius of influence d sweeping out a volume in space was imagined and the other molecules were simply assumed to be present in the volume. But in reality, the other molecules are also moving, and for more accuracy we should, therefore, take into account the relative velocity between the molecules rather than the mean velocity of just one molecule. This requires a more sophisticated analysis as that given in Chapter 2 of Vincenti and Kruger, 1965. For our discussion here, we will observe the results of such analysis without going into the details. The single-particle collision frequency between a single molecule of a chemical species A and the molecules of another chemical species B is given by (17.17)

ZAB = nBird2 AB

where CAB is the mean relative velocity between molecules A and B, given by _ CnB

8kT trmAB

(17.18)

Hence, we can write = nBlaiA2B11 8kr,, irmAB

(17.19)

In Eqs. (17.18) and (17.19), mAB is the reduced mass, defmed as • mAma mAa — mA + ma

(1720)

where mA and mB are the masses of A and B particles, respectively. For a single species gas, the single particle collision frequency is given by Z=

-42

Ird

2c —=

42

ffd

2

8kT ffni

(17.21)

Note that the difference between the simple result given by Eq. (17.15) and the more accurate result given by Eq. (17.21) is the factor which accounts for the relative velocities between

Kinetic Theory of an Ideal Gas 491

particles. Also, note that Eq. (17.21) for a single species gas does not come out directly by simply inserting mA = mB in Eqs. (17.19) and (17.20). To apply Eq. (17.19) to a single species gas, it must be divided by an additional factor of 2 because of the collision counting procedure used to derive Eq. (17.19) (Vincenti and Kruger, 1965). For the mean free path of a single species gas, taking into account the relative velocities of the molecules, we can show that 1 A_T-T— 2 zn

(1722)

The quantity iir/2 in the above equations is called the collision cross-section, denoted by a For an accurate evaluation of collision frequency and mean free path, we need appropriate values of a These are obtained in different ways from experiments. For our purpose, we will assume that a is a known quantity that can be obtained from literature. The main objective here for displaying the results shown in Eqs. (17.21) and (17.22) is to indicate how collision frequency and mean free path vary with pressure and temperature of the gas. For instance, using the relation (state equation) n= P kT

we can express Eq. (17.21) as

P

(1723)

and from Eq. (17.22), we get T

A oc — P

(1724)

From Eqs. (17.23) and (17.24) it is seen that gases at high temperatures and low pressures are characterized by low collision frequency and large mean free paths.

17.4 VELOCITY AND SPEED DISTRIBUTION FUNCTIONS Consider a system of N particles distributed in some manner (not necessarily uniformly) throughout a physical system. The location of any one particle at any time t (instantaneous position) in physical space is given by the location vector r ((Fig. 17.4(a)). For each particle, there is a corresponding point in the x-y-z physical space. The system of N particles is then represented by a cloud of N points in the physical space. Also, at the same instant a given particle has a velocity C, as shown in the velocity space (Fig. 17.4(b)). For each particle in the system, there is a corresponding point in the velocity space. Therefore, the system of N particles is also represented by a cloud of N points in the velocity space. Now, let us examine the point in Fig. 17.4(a) denoted by r, and a unit volume in physical space centred around that point. Simultaneously, consider a point in Fig. 17.4(b) denoted by C, and a unit volume in velocity space centred around that point. Then by definition, the distribution function, f(r, C), is defined as the number of particles per unit volume of physical space at r with velocities per unit volume of velocity space at C. In other words, let dxdydz be

c-cy-c;

492 Fundamentals of Engineering Thermodynamics

Physical space

(a)

Cz

(b)

Figure 17.4 Volume elements in physical and velocity spaces. an element volume in physical space, and dCzdCydCz be an element volume in velocity space, then the function f(x, y, z, Cz, Cy, Cz)drdydzdCzdCydC, represents the number of particles located between x and x + dx, y and y + dy, and z and z + dz with velocities that range from Cx to Cr + dCz, Cy to Cy + dC ,, and CZ to CZ + dc,. Note that the , gaseous system is composed of particles in constant motion m space, and that they collide with neighbouring molecules, thus changing their velocities both in magnitude and direction. Therefore, in the most general case of a nonequilibrium gas, the particles will be distributed non-uniformly throughout space and time, that is, the number of points within the element dxdydz in Fig. 17.4(a) will be a function of r and t, and the number of points at any instant (of time) within the element dCzdCydCz in Fig. 17.4(b) may be changing with time. In classical kinetic theory, the concept of the distribution function is a fundamental tool. Integration of f (as in Eq. 17.25) over all space and all velocities gives

re cco cco cf(x,y, z, C,, Cy , C,) dx dy dz dCx dCy dC, = N

(17.25)

One of the intrinsic values of the distribution function f is, that the average value of any Cy, CZ), can physical quantity Q which is a function of space and/or velocity, Q = Q (x, y, z, be obtained from

= _N_

r. f dt dy dz dC,, dCy dC,

(17.26)

where 6 is the average value of the property Q. We will now examine the special case of a gas in translational equilibrium. In terms of kinetic energy, a gas in equilibrium has the particles distributed uniformly throughout space and the number of molecular collisions that tend to decrease the number of points in the volume dCzdCydCz in velocity space (Fig. 17.4(b)) is exactly balanced by other molecular collisions that increase the number of points in this elemental volume. For this case, f becomes essentially a velocity distribution function, f = f(Cz, Cs,, CZ).

Kinetic Theory of an Ideal Gas 493 For a gas in translational equilibrium, f takes on a specific form which can be rigorously derived by studying the detailed collision processes within the gas. For detailed information about such processes the reader is encouraged to consult Chapter 2 of Vincenti and Kruger (1965). The equilibrium velocity distribution function is expressed as

I (cx Cy, Cz)= "

3/2 r m exp m x2 + — 2kT (C (2irkT)

+ CO]

(1727)

This is called the Maxwellian distribution. Physically, this represents the number of particles per unit volume of velocity space located by the velocity vector C in Fig. 17.4(b). For an equilibrium system, f is a symmetric function. That is, f(Cx , Cy, Cz)= f(—Cx, Cy, CZ ) = f(Cz, —Cy, CZ), etc. Thus, for a system under equilibrium the velocity direction is meaningless. In other words, we need to consider only the magnitude of the particle velocities, i.e. the speed of the particles. Hence we can introduce a speed distribution function 2(C) for the equilibrium system.

17.4.1 Speed Distribution Function Examine the velocity space shown in Fig. 17.5. All particles on the surface of the sphere of radius C have the same speed.

Outer spherical surface with radius C+ dC

CZ

Inner spherical surface with radius C

Figure 17.5 Concentric spherical surfaces with radii C and C + dC. The volume of the space between the sphere of radius C and another sphere of radius C + dC, is 4,1C2dC, where dC is an incremental change in speed. The number of particles per unit volume of velocity space is given by Eq. (17.27). Therefore, for the number of particles between the two spheres with the speed between C and C + dC, we have 3/2 4/tN( m ) C2 exp rkT

mc2 T dC 2k

494 Fundamentals of Engineering Thermodynamics Hence, the speed distribution function z defmed as the number of particles with speed C per unit velocity change, is given by 3/2

mC2

c2 e 2kT

= 47W

(17.28)

bar

Equation (17.28) is shown graphically in Fig. 17.6. It is seen from the figure that, for a system in equilibrium at a given temperature, all the particles do not move at the same speed. In fact, some of the particles move more rapidly, and the others slowly. Equation (17.28) gives the distribution of these speeds over all the particles in the system.

r

x

mN

I

I

I

I

1.0

2.0

CI 4(f/m 2.1 Figure 17.6 The speed distribution function and the values of the most probable speed Cmp, mean speed E7 , and root mean square speed Ire'. Note that there are three speeds shown in Fig. 17.6. The three speeds are defmed as • Most probable speed, is the speed corresponding to the maximum value of z, and it can be obtained by differentiating Eq. (17.28), as (17.29) • Average speed is obtained from Eq. (17.26) by inserting Q = C, as C

8RT if

(1730)

• Root mean square speed is obtained by inserting Q = C 2 in Eq. (17.26), as = 3RT

(17.31)

Kinetic Theory plan Ideal Gas 495

SUMMARY Some preliminary concepts of kinetic theory of an ideal gas have been introduced in this chapter. The equation of state for a perfect gas has been derived with kinetic theory as

P= 2 P 3 The single-particle collision frequency between a particle of species A and those of species B is given by ZAB =

2

nEvag AB

8kT .

gymAB

The mean free path of a single species gas is expressed as

,-2rai2n 'The collision frequency Z and the mean free path A vary with P and T as

Z cc and

T

A cc — P For a system of gas in equilibrium: The Maxwellian distribution function for particle velocities is )3/ 2

f (Cx ,

C.: )= N

exp —

(2:kr

+ Cy +

The speed distribution function, defined as the number of particles with speed C per unit velocity change, is given by )312

X =4NN

c2e 2kT

2ifIcT

The speed corresponding to the maximum value of x is defined as the most probable

speed, and is given by =

42RT

The average speed is given by E,- =

18RT 1

The root mean square speed is given by

Ire7 = NIZT

CHAPTER

18 Elements of Statistical Thermodynamics 18.1 INTRODUCTION In our discussion on kinetic theory of an ideal gas in Chapter 17, we did not regard the molecules of an ideal gas as completely independent of one another, since they could not arrive at an equilibrium distribution of velocities. Therefore, we assumed that interaction between molecules did take place, but only during collisions with each other and with the walls. To describe this limited form of interaction, we refer to the molecules as "weakly interacting" or "quasi-independent". The treatment of strongly interacting particles is not included in our discussion here. Besides their quasi-independent nature, the molecules of an ideal gas have another characteristic, i.e. they are indistinguishable from one another, because they are not localized in space. It was emphasized in Chapter 17 that the molecules have neither a preferred location nor a preferred velocity. The particles occupying regular lattice sites in a crystal are indistinguishable, however, because they are constrained to oscillate about fixed positions, a particle is distinguished from its neighbours by its location. In our discussion in the present chapter as well, we will confine our attention to indistinguishable, quasi-independent particles of an ideal gas only. Let us assume that a monatomic ideal gas consists of N particles, where N is a large number, say, about 1020. Let the gas be contained in a cubical enclosure whose edge has a length L, and let the energy e of any particle, as a first step, be entirely the kinetic energy of translation. In the x-direction, we have 1 .2= (n2±)2 PT2 = _mx — 2 2m 2m

(18.1)

where Ps is the x-component of the momentum. If the particle is assumed to move freely back and forth between two planes of distance L apart, the quantum mechanics in its simplest form provides that, in a complete cycle, the constant momentum P, multiplied by the total path 2L is 496

Elements of Statistical Thermodynamics 497 an integer nx times Planck's constant hP (= 6.6256 x 10-34 J s). Therefore, Pr(2L) = nxhp

(18.2)

where 2L is the total distance travelled by a particle in a complete cycle, that is, from one wall to the other and back. Substituting this result into Eq. (18.1), we get Ex

= nz

x 8mL2

or nx = — Ii8tne7

hp The values of kinetic energy 4 are discrete, corresponding to integer values of n, but when n, changes by unity, the corresponding change in ex is very small, because n, itself is very large. For example, consider a cubical box, with edge 100 mm, containing gaseous helium at 300 K. From our discussion on kinetic theory of gases in Chapter 17, we know that the average energy associated with each translational degree of freedom is i kT where k is the Boltzmann constant (= 1.3805 x 10-23 J/K). Thus, Ex

1 1 = — kT = — x 1.3805 x 10-23 [J/K] x 300 [K] 2 2 = 2.07 x 10-21 J

and n, —

0.01[m] 6.6256 x 10-34 [J 5]

x ‘18 x 6.6 x 10-24[g] x 2.07 x 10-21 [J]

10-2 x 10.45 x 3.16 x 10-23 6.6256 x 10-34 z 109 Therefore, the change in 4 when nx changes by unity is so small that, for most practical systems, the energy may be assumed to vary continuously. Considering all the three components of momentum, we get the total kinetic energy of a particle as e—

p2 x

p2 x

2m

hP 2 ( 2 n., 8m1;

n

n z2 )

(183)

The specification of an integer for each nx, ny, and nz is a specification of a quantum state of a particle. All states characterized by the values of the n's such that n,r2 + + = constant will have the same energy. EXAMPLE 18.1 Examine the states corresponding to the values of nx, n,,, n, in Table 18.1. It is seen that there are twelve quantum states associated with the same energy level, and therefore we can say that this energy level has a degeneracy of 12. Also, it is seen that n2 + n2, + 11.? = 66.

498 Fundamentals of Engineering Thermodynamics Table 18.1 1

2

3

4

5

6

7

8

9

10

11

12

n,

1

1

1

1

4

4

4

5

5

7

7

8

ny

8

1

7

4

1

7

5

5

4

1

4

1

8

4

7

7

1

5

4

5

4

1

1

nz

In any practical system, nX + ny2 nz2 is an extremely large number so that the degeneracy of an actual energy level is extremely large. That is, for any given energy level e, there can be a number of different states that all have the same energy. The number of states is called the degeneracy or statistical weight of the given energy level e, and is denoted by g,. This concept is illustrated in Fig. 18.1.

k. 1 =3 =4 90 = 5 Figure 18.1 Illustration of statistical weight. Figure 18.1 shows energy levels in the vertical direction, with the corresponding states as individual horizontal lines arrayed to the right at the proper energy value. For example, the second array from the bottom is shown with 4 states, all with an energy value equal to el ; hence g1 = 4. The values of gi for a given molecule are obtained from quantum theory and/or. spectroscopic measurements. Consider a system consisting of N molecules. Let Nj be the number of molecules in a given energy level ej. The value Ni is defined as the population of the energy level. Thus, N = ZNi

(18.4)

where the summation is taken over all energy levels. The set of numbers Nj is also referred to as macrostate. Due to the molecular collisions, some molecules will change from one energy level to another. Thus, when we look at the system under consideration at some later instant of time, there may be different sets of Nj's, and hence a different population distribution, or macrostate. Now, let us denote the total energy of the system as E, where (185) The above definitions of N and E in Eqs. (18.4) and (18.5), respectively, are illustrated schematically in Fig. 18.2. As shown in Fig. 18.2, for a system of N molecules and energy E, there is a series of

Elements of Statistical Thermodynamics

Energy levels:

to

et

E2

ei

Statistical weights:

go

91

92

91

Populations at one instant

CNo = 2

Populations at the next instant

( N° = 3

499

—

N1 = 3

/V2 = 5 ... NI = 2 ... )One macrostate

N1 =1

N2 = 3

... N1= 6 ) Another macrostate

Figure 18.2 Illustration of macrostates. quantized energy levels, eo, e,, e2, ..., with corresponding statistical weight, go, g,, g2, . At some given instant, the molecules are distributed over the energy level in a distinct way, No, N,, ..., N.; ..., constituting a distinct macrostate. In the next instant, due to molecular collisions, the populations of some levels may change, creating a different set of Nis and hence a different macrostate. Over a time period, one particular macrostate, i.e. one specific set of Nis, will occur much more frequently than any other. This particular macrostate is called the most probable macrostate (or the most probable distribution). It is the macrostate which occurs when the system is in thermodynamics equilibrium. In fact, this is the definition of the thermodynamics equilibrium within the framework of statistical mechanics. The central problem of statistical thermodynamics is as follows:

gi,

How to find the most probable macrostate for a given system with a fixed number of identical particles, N = E and a fixed energy E = Zei Ni? In order to solve this problem, let us consider the schematic shown in Fig. 18.3, which illustrates a given macrostate. In this schematic, we choose No = 3, N, = 2, N2 = 4, etc. Each statistical weight for each energy level is displayed as a vertical array of boxes in Fig. 18.3. For example, under el, we have g, = 3, and hence three boxes, one for each different energy state with the energy el. In the energy level el, we have two molecules (N1 = 2). At some instant of time, these two molecules individually occupy the top and bottom boxes under g,, with the middle box left vacant (i.e. no molecules at that instant have the energy state represented by the middle box). The way that the molecules are distributed over the available boxes defines a microstate of the system, say microstate I as shown in Fig. 18.3. At some later instant, the N, = 2 molecules may be distributed differently over the g, = 3 states, say, leaving the top box vacant. This represents another, but a different microstate, labelled microstate II in Fig. 18.3. Arrangements of molecules in other vertical arrays of boxes between microstate I and II are shown in Fig. 18.3. Note that in both cases No equals 3, N, equals 2, etc. That is, the macrostate is the same. Thus, any one macrostate can have a number of different microstates depending on which of the degenerate states (the boxes in Fig. 18.3) are occupied by the molecules. In any given system of molecules, the microstates are constantly changing due to molecular collisions. Indeed, it is a central assumption of statistical thermodynamics that each microstate of a system occurs with equal probability. Therefore, we may say that the most probable macrostate is one which has the maximum number of microstates. If each microstate appears in the system with equal probability, and there is one particular macrostate that has

500 Fundamentals of Engineering Thermodynamics e2 N2 = 4

N=2 —

92= 5

g1 =3 ...

0

0 0

0

CI

0

0

0

0

eo M=2

No = 3

=3

9=5

Microstate

0 CI

❑ ❑

CI

❑

0 Microstate

0

❑

0 0 Figure 18.3 Illustration of microstates.

No. of microstate, f2

considerably more microstates than any other, then that is the macrostate which is seen in the system most of the time. This is indeed the situation is most real thermodynamic systems. The values of the number of microstates in different macrostates are shown schematically plotted in Fig. 18.4.

I

A

I

I

C

I

I

I

I

I

D E F G H

Macrostate The most probabe macrostate Figure 18.4 Illustration of the most probable macrostate.

Elements of Statistical Thermodynamics 501

It is shown in Fig. 18.4 that the macrostate D, the macrostate having the largest number of microstates is marked as the most probable macrostate. In fact, this is the macrostate that is usually seen, and constitutes the situation of thermodynamics equilibrium in the system. 'Therefore, to identify the most probable macrostate it is necessary to count the number of microstates of the macrostate under consideration. Once the most probable macrostate is identified, the equilibrium thermodynamics properties of the system can be computed. In the following section, we will show how we can count the number of microstates for a given macrostate.

18.2 COUNTING THE NUMBER OF MICROSTATES FOR A GIVEN MACROSTATE Molecules and atoms are made up of elementary particles, namely the electrons, protons, and neutrons. Quantum mechanics classifies molecules and atoms into two categories, depending on their number of elementary particles, as follows: 1. Molecules and atoms with an even number of elementary particles obey a certain statistical distribution called Bose-Einstein statistics. These particles are usually referred to as bosons. 2. Molecules and atoms with an odd number of elementary particles obey a certain other statistical distribution termed Fermi-Dirac statistics. These particles are called fermions. An important distinction between the bosons and fermions is the following: • For bosons, the number of molecules that can be in any one degenerate state (in any one of the boxes in Fig. 18.3) is unlimited (except, that it must be less than or equal to N1). • For fermions, only one molecule may be in any given degenerate state at any instant. This distinction has a major effect on the counting of microstates in a gas. For our counting of microstates, let us consider Bose—Einstein statistics. Also, let us consider only one energy level, say §, for the time being. This energy level has a gi degenerate states and Ni molecules. Consider the gj states as the gj containers, as shown in Fig. 18.5. Distribute the Ali molecules among the containers, such as 2 molecules in the fast container, 3 molecules in the second, etc. where the molecules are denoted by x in Fig. 18.5. The vertical bars in the figure are the partitions which separate one container from another. 1

2

3

g•

4 I ••• I

xx

Figure 18.5 Illustration of degenerate states. The distribution of molecules among these containers represents a distinct microstate. If a molecule is moved from container 1 to container 2, a different microstate is formed. To count the total number of different microstates possible, first note that the number of permutations between the symbols x and I is [Ni + (Si —

502 Fundamentals of Engineering Thermodynamics This is the number of distinct ways that the Ni molecules and the (g1 — 1) partitions can be arranged. However, the partitions are indistinguishable, therefore, the (gri — 1) partitions can be permuted in (g1 — 1)! different ways. The molecules are also indistinguishable, thus, they can also be permuted in Ni! different ways without changing the picture drawn in Fig. 18.5. Therefore, there are (gi — 1)!Ni! different permutations which yield the identical picture drawn as in Fig. 18.5, i.e. the same microstate. Thus, the number of different ways that NJ indistinguishable molecules can be distributed over gi states is (Ni + gi (gj — 1)!1Vj !

and gives the number of This expression applies to one energy level ej with population e i. Now let us consider the whole different microstates due to the different arrangements within set of Ni's distributed over the complete set of energy levels (note that the given set of Ni's defines a particular microstate). If the total number of microstates for a given macrostate is denoted by S2, then for our system consisting of N1 molecules, we have

S2 _

I)! (g1 - i)!N; !

(N. + g. —

(18.6)

where SI is a function of all the NJ values, L2 = 0(4 N2, ..., Ni, ...,). The quantity S2 is called the thermodynamics probability. The thermodynamic probability is obviously a measure of the "disorder" of the system. Thus, Eq. (18.6) is the way to count the number of microstates in a given macrostate as long as the molecules are bosons. Consider now the Fermi—Dirac statistics. We have seen that, for fermions only one molecule may be in any given degenerate state at any instant. That is, there cannot be more than one molecule in any container. This implies that gi Let us consider N1 molecules and gi containers. Let us put one of the molecules in one of the containers. There are g1 ways of doing this. Now let us put the second molecule in one of the remaining containers. But, now there are only gj — 1 choices, since one of the containers is already occupied. Continuing like this, we find that the number of ways Ni molecules can be distributed over gj boxes, with only one molecule (or no molecule) per container, is gj (gj — 1)(gj — 2) ••• [gj — (Ni — 1)] =

gi

(gi — N1 ) !

Note that the N1 molecules are indistinguishable, hence they can be permuted in Ni! different ways without changing the above picture. Therefore, the total number of microstates because of the different arrangements with energy level ei becomes

(g, — Nj )!Ni !

Thus, for all energy levels, the total number of microstates for a given macrostate for fennions is

- fl(g, - gi!

(18.7)

Elements of Statistical Thermodynamics 503 From the above discussion it is evident that for a given macrostate, Eq. (18.6) or (18.7) can be used to calculate the number of microstates possible for bosons or fermions, respectively. Further, it is seen that cl is a function of the molecular population Ni and hence different for different macrostates. Also, as sketched in Fig. 18.4, there will be a certain macrostate for which cl will be significantly larger than for any other macrostate. This state with the largest number of microstates is called the most probable macrostate.

18.3 THE MOST PROBABLE MACROSTATE The most probable macrostate is that macrostate which contains the maximum number of microstates, i.e. which has S2„,,„,. Let us examine systems consisting of a specific set of Ni's which allows the maximum S2, in order to identify the most probable macrostate. Let us fast take the case for bosons. Equation (18.6) can be rewritten as In S2 =

E [ln (Ni + g1 — 1)! — ln(gi — 1)! — In

(18.8)

At this stage, it is extremely important to note that the total energy of a molecule consists of translational, rotational, vibrational, and electronic energies. That is, for molecules e= emus + erot evil:. + However, for a simple atom, only the translational and electronic energies exist, that is, for atoms e = Evans + eei

To have a better understanding of the above energy components, let us have a close look at the microscopic picture of a gas. Let us assume that the gas consists of a large number of individual molecules. We know that a molecule is a collection of atoms bound together by a rather complex intramolecular force. A simple concept of a diatomic molecule (molecule with two atoms) is the "dumb-bell" model shown in Fig. 18.6(a). The molecules have many forms (modes) of energy. They are: • The translational energy et,„s. The translational kinetic energy of the centre of mass of the molecule is the source of this energy. As shown in Fig. 18.6(b), a molecule has three geometric degrees of freedom in translation. Since motion along x-, y-, and z—coordinate directions constitutes the total kinetic energy, the molecule is also said to have three thermal degrees of freedom. • The rotational energy ern,. The rotational energy is due to rotation of the molecule about the three orthogonal axes in space, as shown in Fig. 18.6(c). The source of ern, is the rotational kinetic energy associated with the molecule's rotational velocity and its moment of inertia. But for the diatomic molecule sketched in Fig. 18.6, the moment of inertia about the inter-nuclear axis (the z-axis) is very small, and therefore, the rotational kinetic energy about the z-axis is negligible compared to that about the x- and y-axis. Thus, the diatomic molecule has only two geometric and only two thermal degrees of freedom. The same is true for a linear polyatomic molecule such as CO2 shown in Fig. 18.6(d(i)). However, for a nonlinear molecule, such as H2O in Fig. 18.6(d(ii)), the number of geometric as well as thermal degrees of freedom in rotation is three.

504 Fundamentals of Engineering Thermodynamics

vr

Translational kinetic energy of the centre of mass (thermal degrees of freedom-3)

o. Vx •

• Source (a) Diatomic molecule

(b) Translational energy eta,

Y O

C

0

0 z (c) Rotational energy co

I-110.

.111— —I

•---M‘AAMA,AP---• (e) Vibrational energy Eva

•• H H (d) ---•-„ Electrons / ....... •,../ II 1 k 1 Kinetic energy of electrons in orbit 2 Potential energy of electrons in orbit Nucleus

‘ I.

7 ,' ;

(f) Electronic energy ea

Figure 18.6 Different energy modes of molecules. • The Vibrational energy ea,. The molecules and atoms vibrate with respect to an equilibrium location within the molecule. For a diatomic molecule, this vibration is modelled by a spring connecting the two atoms, as shown in Fig. 18.6(e). Thus, the molecule has vibrational energy ear The sources of this vibrational energy are (a) the kinetic energy of the linear motion of the atoms as they vibrate back and forth, and (b) the potential energy associated with the intramolecular force. Therefore, although a diatomic molecule vibrates along one direction, namely the inter-nuclear axis only and has only one geometric degree of freedom, it has two thermal degrees of freedom because of the contribution of both kinetic and potential energies. For polyatomic molecules, the vibrational motion is more complex, and numerous fundamental vibrational modes can occur, with a consequent large number of degrees of freedom. • The electronic energy ed. The electronic energy is because of the motion of electrons about the nucleus of each atom constituting the molecule, as shown in Fig. 18.6(1). The

1

Elements of Statistical Thermodynamics 505

electronic energy sources are (a) the kinetic energy owing to its translational motion throughout its orbit about the nucleus, and (b) the potential energy owing to its location in the electromagnetic force field established principally by the nucleus. The concepts of geometric and thermal degrees of freedom are usually not useful for describing electronic energy, since the overall electron motion is complex. The quantum mechanics results show that each of the above energies is quantized, i.e. it can exist only at certain discrete values, as illustrated in Fig. 18.7. Modes of molecular energy

Translation

Vibration

Rotation

Electronic E/, el

E2el El el et. rot

%rot

6/, trans

e2rot EOvib

e0rot

E0trans

E0e1

Figure 18.7 Quantized levels of different energy modes of molecules. Now it is clear to us that we are dealing with the combined translational, rotational, vibrational and electronic energies of a molecule, and that the closely spaced translational energy levels can be grouped into a number of degenerate states with essentially the same energy. Therefore, in Eq. (18.8), we can assume that Ni >> 1 and g; >> 1, and hence Afi + 1= + j and g1 — 1 = g1. Further, we can use Sterling's formula In x! = x In (x) — x

(18.9)

for the factorial terms in Eq. (18.8). Using Sterling's formula, we can write Eq. (18.8) as In S1=

[(N

+gi)In(V.i +g;)—(N

gi

Ni InNi

506 Fundamentals of Engineering Thermodynamics This can be rewritten as + gJ lni5:+11 gi

Ini2=E [N.J 141+.'51 NJ 1

(18.10)

Note that ln S2 =ftNi's) =fiNo, NI, N22

-9 NJ,

--)

Now for maximum value of Q, d(ln 12) = 0

(18.11)

We can write d(ln LI), using the chain rule, as

a(ln dN ' aN,

d(In LI) = 8(112)(iNo +

a(ln

+

(18.12)

From Eqs. (18.11) and (18.12), we get d(In

=

E 67(1.1) dNi = 0 c7Ni

(18.13)

From Eq. (18.10),

a(1n (1) _ aN;

N )

(18.14)

From Eqs. (18.13) and (18.14), we get d (ln f2) =

E [In (1 +

dN =0 NJ

J

(18.15)

It is important to note that in Eq. (18.15), the variation of Ni is not totally independent; dNi is governed by the following two physical constraints:

1.N =EN., = constant, and hence EdNj =0 I 2. E =

(18.16)

E FJ1VJ = constant, and hence J

(18.17) Assuming a and fi to be two Lagrange multipliers (two constants to be determined), Eqs. (18.16) and (18.17) can be written as —EadNi =0

(18.18)

and

- I /3ejdNi = 0

(18.19)

Elements of Statistical Thermodynamics 507 Addition of Eqs. (18.15), (18.18), and (18.19) yields

g-L)- a - fiej ldNi =0 ln( 1+-Nj It is a usual practice to define the Lagrange multipliers, term in brackets in Eq. (18.20) becomes zero, i.e.

14

a and

(18.20) in such a way that each

1+-1 )-a- fiej =0 N•

OT

1+

gi =eaefie N.

OT

* N

gi (18.21) eae - 1 The Nj in Eq. (18.21) is written as Nj* to denote that Ni* corresponds to maximum value of O. That is, Ni* corresponds to the most probable distribution of particles over the energy levels e. The set of values from Eq. (18.21) for all energy levels No*,

N1*, N2*, ..., Nj*,

is the most probable macrostate. That is, Eq. (18.21) gives the

most probable macrostate for

bosons. In a similar manner, the most probable distribution for

N• -1 - ea fie, e+1

fermions can be obtained as (18.22)

Note that the only difference between Eqs. (18.21) and (18.22) is the sign in the denominator.

18.3.1 The Boltzmann Distribution The Boltzmann distribution is the limiting case of the most probable distribution at high temperatures where the molecules are distributed over many energy levels. At very low temperatures, for instance less than 5 K, the molecules of a system are jammed together at or nearly the ground energy levels, and therefore, the degenerate states of these low-lying levels are highly populated (i.e. gi is large). As a result, the difference between the Bose-Einstein [Eq. (18.21)] and the Fermi-Dirac [Eq. (18.22)] statistics are important. However, at higher temperatures, the molecules are distributed over many energy levels, and therefore, the states are generally sparsely populated (i.e. Ni 10, and both are significant when 0.1 < Gr/(Re)2 < 10. Natural convection may augment or retard forced convection heat transfer, depending on the relative direction of buoyancy-induced and the forced convection motions. Based on the relative direction, as illustrated in Fig. 19.32, the flows are broadly classified as assisting flow, opposing flow, and transverse flow. Buoyant flow Cold plate

Hot plate Buoyant flow Forced flow

Buoyant flow

Forced flow t t

(a) Assisting flow

R..

Forced

(b) Opposing flow

(c) Transverse flow

Figure 19.32 Illustration of assisting, opposing, and transverse flow.

Elements of Heat Transfer 607 • In assisting flow, the buoyant motion is in the same direction to the forced motion. Therefore, the natural convection enhances the heat transfer by assisting forced convection. Upward forced flow over a hot surface shown in Fig. 19.32(a) is an example for this. • In opposing flow, the buoyant motion is in the opposite direction to the fluid motion. Therefore, the natural convection resists forced convection and decreases the heat transfer. Upward flow over a cold surface shown in Fig. 19.32(b) is an opposing flow. • In transverse flow, the buoyant motion is perpendicular to the forced motion, as shown in Fig. 19.32(c). This flow enhances fluid mixing and thus promotes heat transfer. To determine the heat transfer due to combined natural and forced convection conditions, it will appear that we should add the contributions of natural and forced convection in assisting flow and subtract them in opposing flow. But, the experience indicates differently. Experimental studies suggest a correlation of the form Nucombm • ed = (Nu—natural ± Nufo d

)l/n

where Nufoited and Nunatorm are determined from the correlations for pure forced and pure natural convection, respectively. The + is for assisting and transverse flows and — is for opposing flows. The exponent n varies between 3 and 4, depending on the geometry involved. It is proved that n = 3 correlates experimental data for vertical surface wall. Larger values of n are better suited for horizontal surfaces. EXAMPLE 19.27 The glass cover of a solar collector receives solar radiation at the rate of 700 W/m2. The glass transmits 88 per cent of the incident radiation and has an emissivity of 0.9. A collector of size 1.2 m high and 2 m wide is required to heat water which enters the tube attached to the collector plate at a rate of 1 kg/min. The temperature of the glass cover is measured to be 35°C on a day when the surrounding air temperature is 23°C and wind blows at 30 kmph. The effective sky temperature for radiation exchange between the glass cover and the open sky is — 40°C. Assuming the black surface collector plate to be perfectly insulated and the only heat loss is through the glass cover, determine (a) the total rate of heat loss from the collector, (b) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and (c) the temperature rise of water as it flows through the collector. Assume Prandtl number to be 0.707. Solution Total radiation energy incident on the glass is Grad = grad x A

= 700 x (2 x 1.2) =1680 W The glass wall is at 35°C and the sky is at — 40°C. Also, the emissivity of the glass is 0.90. Thus, the heat radiated by the glass to the sky is grA =0- x0.1 x (rs4 - rsty =5.67 x 10-8 x0.1(3084 — 2334 ) = 34.3 W/m2

608 Fundamentals of Engineering Thermodynamics Therefore, the total radiation loss from the glass surface of 2.4 m2 is ad*. = 4 x A = 34.3 x 2.4 = 82.4 W Thus, the net radiation received by the glass plate at steady state is

Orad, net = Orad — OradJoss =1680-82.3 =1597.7 W Out of this, 88 per cent is transmitted by the glass plate to the black plate below the glass. Thus, the heat energy received by the plate, is Opb, =1597.7 x 0.88 =1406 W Air blows at 30 km/h over the glass plate. The corresponding Reynolds number is ReL =

pVL

P = 1.177 x(30/3.6)x1.2 1.85x10-5 =6.36 x105 Here the density and viscosity are calculated for the fluid temperature of Tf =

'

Ts + To 35+ 23 = 29.0 , .... 2 2

The Reynolds number is greater than the critical value of 5 x 105. Therefore, the flow is turbulent. For turbulent flow over a flat plate, we have hL Nu =— = 0.037(ReL )415 (Pr)L'3 K

= 0.037 (16 x105 )415 (0.707)1/3 = 3029 3029 x 0.0263 h= = 40 W/(m2 °C) 2 Note that the Prandtl number is ,

Prr=

pCp 1.85 x 10-5 x 1005 =

K

K

= 0.707

This gives K = 0.0263. (a) The heat loss from the glass plate, by forced convection is

0.,=hA(Ts —T.,) = 40 x 2.4 x (35 — 23) = 1152W

Elements of Heat Transfer 609 (b) The collector efficiency is heat transferred to the water solar energy incident on the glass plate _ 1406 —1152 = 254 1597.7 1597.7 = 0.159

q=

(c) The heat transferred to the water is

Q

. = thCAT =

1 x 4178 x AT 60

Therefore, the temperature rise of water becomes AT —

254 x 60 = 3.6°C 4178

EXAMPLE 19.28 A train coach of length 8 m and width 2.8 m travels at 70 kmph. The top surface of the coach is absorbing solar radiation at the rate of 200 W/m2, and the temperature of the ambient air is 30°C. Assuming the roof of the coach to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to the convective heat transfer, determine the equilibrium temperature of the top surface. Take K for the top surface to be 0.026 W/(m °C). Solution The top surface of the coach can be assumed to be a flat plate. Therefore, the problem is essentially a flat plate exposed to a flow and solar heating. The density and viscosity of air at 30°C are

p =2—

RT 101325 — =1.165 kg/m3 287 x 303.15

pxec =1.46 x10

_6 T3/2 T +111

=1.46 x le x

(303.15)3/2 303.15 +111

=1.86x10 5 kg /(m s) The Reynolds and the Prandtl numbers of the flow are ReL —

pV L 1.165 x (70/3.6)x 8 — P 1.86 x10-5

=97.4 x105 PC 1.86 x 10-5 x 1005 P= K 0.026 = 0.719

Pr =

610 Fundamentals of Engineering Thermodynamics The Reynolds number is greater than the flat plate critical Reynolds number of 5 x 105 and thus the flow the turbulent. For turbulent flow over a flat plate, we have hL Nu = — = 0.037 (ReL )4/5 03015 = 0.037(97.4 x 105 )45 (0.719)15 =12920.9 Therefore, the film transfer coefficient becomes 12920.9 x 0.0261 = 42.15 W/(m2 °C) 8 The surface of the plate receives 200 W/m2 by radiation. Therefore, at steady state, the surface temperature 7; will assume a constant value and the heat received by the plate by radiation and the heat removed from the plate by the forced convection will balance each other, Thus,

h—

4rad = 4., 200 = h(7; — T„,,) Ts = 2+T

=

200 + 30 42.15 34.75°C

The top surface will be at an equilibrium temperature of 34.75°C.

19.20 RADIATION HEAT TRANSFER We have seen that in contrast to the mechanism of conduction and convection, where energy transfer through a material medium is involved, heat may also be transferred into regions where perfect vacuum exist. The mechanism in this case is electromagnetic radiation, which is propagated as a result of a temperature difference, and is called thermal radiation. The radiant energy leaving a surface varies with the direction. Therefore measures of radiation from a surface are specified, with the direction as an important parameter. One of the prime parameters in radiation heat transfer analysis is the intensity of radiation. Consider the elemental area radiating thermal energy as shown in Fig. 19.33. The Intensity of Radiation I is defined as the rate of energy leaving a surface in a given direction per unit projected area (area normal to the direction) per unit solid angle. That is, 1(fl, 9) — d2C568' dApdC

0) (19.144)

Elements of Heat Transfer 611

n2

Figure 19.33 Radiation from a surface. where

6113,

= rate of energy leaving the surface dAs in the solid angle dSI in the direction (AO

dAs = elemental surface area dA p = dAs cos fi = projected elemental area df2 = elemental solid angle = dA/R2 = angle measured from normal to the surface = azimuthal angle The brightness of a surface from a given direction of view depends on the radiation emitted per unit projected area in that direction. A surface which is equally bright in all the directions is called a diffuse surface. That is, the intensity of radiation is independent of the direction for such surfaces. The emissive power e of a surface is defined as the energy emission (in all directions) from the surface per unit surface area, i.e. e(TO=

dQe(Ts)

dAs

(19.145)

where OJT's) is the rate of energy emitted by the surface, which depends on its temperature Ts. The total energy Q leaving a surface, in Eq. (19.144), includes the energy emitted, Qe, the energy reflected, 0, and the energy transmitted, Q. Using Eqs. (19.144) and (19.145), we can write e(T,)= L Ie cos fi dfl where I, is the intensity of emission from the surface.

(19.146)

612 Fundamentals of Engineering Thermodynamics A blackbody is defined as a surface which absorbs all radiation incident on it. With this definition, it is possible to show that of all surfaces at a given temperature, the blackbody emits the maximum possible radiation and this radiation is called die radiation. By Stefan-Boltzmann law, the emissive power of a blackbody is given by eb(TJ = n2aTs4

(19.147)

where et, = emissive power of the blackbody at temperature 7', (W/m2) a = Stefan-Boltzmann constant, 5.669 x 104 W/(m2 K4) n = refractive index of the medium surrounding the surface; for gaseous mediums like air n 1.0.

19.20.1 Properties of Surfaces When radiant energy strikes a material surface, part of the radiation is reflected, part is absorbed, and part is transmitted as shown in Fig. 19.34. We define the reflectivity p as the fraction reflected, the absorptivity a as the fraction absorbed, and the transmissivity r as the fraction transmitted. Thus, (19.148) p+a+r=1 Incident radiation

•t.7.••

Reflection

•••••; • ••••1':.•

Transmitted

Figure 19.34 Radiant energy striking a material surface. Most solid bodies do not transmit radiation, so that for many applied problems the transmissivity may be taken as zero. Then, p+ a= 1 Two types of reflection phenomenon may be observed when radiation strikes a surface. If the angle of incidence is equal to the angle of reflection, the reflection is called specular reflection. When the incident beam gets distributed uniformly in all directions after reflection, the reflection is called diffuse reflection. Both types of reflections are illustrated in Fig. 19.35. The influence of surface roughness on thermal-radiation properties of materials is a matter of serious concern and remains a subject of active research. From the above discussion on properties of surfaces, it can be inferred that everything around us constantly emits radiation, and the emissivity represents the emission characteristics of these bodies. That is, every body is constantly bombarded by radiation coming from all directions over a range of wavelengths. The radiation energy incident on a surface per unit surface area per unit time is called irradiation and is denoted by G. When a radiation strikes a surface, a part of it is absorbed, part of it is reflected, and the

Elements of Heat Transfer 613 Source Reflected rays

(a) Specular

(b) Diffuse

Figure 1935 Specular and diffuse reflection. remaining part, if any, is transmitted, as illustrated is Fig. 19.34. The fraction of irradiation absorbed by the surface is called absorptivity a, the fraction reflected is termed reflectivity p, and the fraction transmitted is called transmissivity r. That is, a=

absorbed radiation _ Gabsorbed incident radiation reflected radiation Greflected

P _ incident radiation r=

transmitted radiation _ Gtransmitted incident radiation

But, in accordance with the first law of thermodynamics, the sum of absorbed, reflected, and transmitted radiation energy must be equal to the incident radiation, i.e. Gabsorbed

Gtransmitted = G

Greflected

Dividing each term by G, we get Eq. (19.148). The above definitions are for total hemispherical properties, since G represents the radiation energy incident on the surface from all directions and all wavelengths. Thus, a, p, and r are the average properties of the medium for all directions and all wavelengths. However, these properties can be defined for a specific wavelength and direction. For example, the spectral absorptivity, reflectivity, and transmissivity of a surface can be defined as a— PA

rA =

G2(absorbed)

G Gafteflected)

G

G A(transmitted) G

where GA is the radiation energy incident at the wavelength A and GA(absorbed), GA(reflected), and GA(transmitted) are the absorbed, reflected, and transmitted portions of it, respectively. Similar definitions can be given for directional properties in direction 19 by replacing all occurrences of the subscript A in the above equations of at, pA, and r2.

614 Fundamentals of Engineering Thermodynamics The average a, p, and r of a surface can also be determined in terms of their spectral counterparts as :aAGA d.1.

a- j

I:GA dA. P=

=

jo pAGAdi1 f:GA dA, rAGA cbt j: GA 61

It is essential to note that, among a, p, and r, the reflectivity alone is bidirectional in nature. That is, p of a surface depends not only on the direction of the incident radiation but also on the direction of reflection. In practice, for simplicity, surfaces are assumed to reflect in a perfectly specular or diffuse manner. Reflections from smooth and polished surfaces approximate specular reflection. In radiation analysis, smoothness is defined relative to wavelength. A surface is regarded smooth if the height of the surface roughness is much smaller than the wavelength of the incident radiation. Example 19.29 The coating on a plate is cured by exposing to an infrared lamp supplying 2 kW/m2 by irradiation. The plate absorbs 80 per cent of the incident energy and its emissivity is 0.5. The plate is exposed to a large surroundings at 30°C and air at 20°C, with convection coefficient of h = 15 W/(m2 K), flows over the plate surface. At a steady state, what will be the temperature at the surface of the coating? Solution At steady state whatever be the energy received by the coating should be transferred out to the air by convection and to the surrounding space by radiation. Thus, 4absorbed

= 4conv 4rad

aG = h (Ts — TO+ 06(7;4 —T„) 0.8 x 2000 =15(7; — 293) + 5.67 x 10-8 x 0.5 (7;4 —2934 ) 1600 =1575 — 4395 + 2.835 x 10-8 154 — 209 6204 =157; + 2.835 x10-8 7;4 Solving this by trial and error, we get Ts = 377 K = 104°C The surface of the coating will be at 104°C at steady state.

Elements of Heat Transfer 615

The greenhouse effect It can be noticed that when a car is left under direct sunlight on a sunny day, the interior of the car gets much warmer than the outside environment. The reason for this is the spectral transmissivity curve of the window glasses and wind-screen, which resembles an arc, as shown in Fig. 19.36.

Infrared radiation

Figure 19.36 A greenhouse traps energy by allowing the solar radiation to come in but not allowing the infrared radiation to go out. The glass at thickness encountered in practice transmits over 90 per cent of radiation in the visible range and is practically opaque to radiation in the larger-wavelength infrared region of the electromagnetic spectrum (A, > 3 pm). Therefore, glass has a transparent window in the wavelength range 0.3 pm < A < 3 pm in which over 90 per cent of solar radiation is emitted. On the other hand, the entire radiation emitted by surfaces at room temperature falls in the infrared region. Consequently, glass allows the solar radiation to enter but does not allow infrared radiation from the interior surfaces to leave. This causes a rise in the interior temperature as a result of energy build-up in the car. This heating effect, due to the non-gray characteristic of the glass (or clear plastic) is known as the greenhouse effect. On earth, the greenhouse effect is experienced on a larger scale. The earth surface, which warms up during the day as a result of absorption of solar energy, cools down at night by radiating its energy into space as infrared radiation. The combustion gases such as carbondioxide and water vapour in the atmosphere transmit the bulk of the solar radiation but absorb the infrared radiation emitted by the surface of the earth. Thus, there is concern that energy trapped on earth ultimately will cause global warming leading to drastic changes in weather patterns. In coastal areas with high humidity, there is no drastic change between day and night temperatures. This is because humidity acts as a barrier on the path of the infrared radiation coming from the earth, causing retardation of the cooling process duffing night. In a desert where the sky is clear, there is a large difference between day and night temperatures because of the absence of any barrier to infrared radiation.

616 Fundamentals of Engineering Thermodynamics

Blackbody radiation We saw that a body at a temperature above absolute zero emits radiation in all directions over a wide range of wavelengths. The amount of radiation energy emitted from a surface at a given wavelength depends on the material of the body and the condition of its surface as well as the surface temperature. Therefore, different bodies may emit different amounts of radiation per unit surface area, even when they are at the same temperature. Therefore, to identify the maximum amount of radiation that can be emitted by a surface at a given temperature, it is necessary to define an idealized body, termed blackbody, to serve as a standard against which the radiative properties of real surfaces can be compared. A blackbody is defined as a perfect emitter and absorber of radiation. At a specified temperature and wavelength, no surface can emit more energy than a blackbody. A blackbody absorbs all incident radiation, regardless of wavelength and direction. Also, a blackbody emits radiation energy uniformly in all directions. That is, a blackbody is a diffuse emitter. The radiation energy emitted by a blackbody per unit surface area per unit time is expressed as Eb = ars

where a = 5.67 x 104 W/(m2 K4) is the Stefan-Boltzmann constant. This relation is known as Stefan-Boltzmann law and Eb is called the blackbody emissive power. The Stefan-Boltzmann law gives the total blackbody emissive power Eb, which is the sum of the radiation emitted over all wavelengths. But the spectral blackbody emissive power, defined as the amount of radiation energy emitted by a blackbody at an absolute temperature T per unit time, per unit surface area, and per unit wavelength about the wavelength A. will be of interest. For example, the amount of radiation that an incandescent light bulb emits in the visible wavelength spectrum will be of interest rather than the total amount of radiation that the bulb emits. The relation for the spectral blackbody emissive power EbA was developed by Max Planck in 1901 in conjunction with his quantum theory. This relation is known as Planck's distribution law and is expressed as EbA(T) =

25 [exp (c,/AT) —1]

where ci = 2nlic2c, = 3.742 x 108 (Wµm4)/m2 and c, = hcolK = 1.439 x 104 (p.m x K). The temperature T is the absolute temperature of the surface, A is the wavelength of the radiation emitted, K = 1.3805 x 10-23 J/K is the Boltzmann constant, and co = 2.998 x 108 m/s is the speed of light in vacuum. This relation is valid for a surface in a vacuum or a gas. For other media, it needs to be modified by replacing c1 by c1ln2, where n is the index of refraction of the medium.

Emissive power The emissive power e of a body is defined as the energy emitted by the body per unit area and per unit time. The ratio of the emissive power of a body to that of a blackbody at the same temperature is defined as the emissivity t: (TS). That is, e(Ts)

eqs) eb(T,)

(19.149)

Since blackbody is a perfect emitter, the emissivity of a surface can vary only from 0 to 1.0.

Elements of Heat Transfer 617 It is important to note that the concept of blackbody is an idealization, i.e. a perfect blackbody does not exist—all surfaces reflect radiation to some extent. The properties p, a and r are strictly total properties since they represent ratios of total energies, i.e. energy integrated over all wavelengths. These properties when defined for every wavelength are called spectral properties. If all the spectral properties of a surface are independent of wavelength, then such a surface is called a gray surface. In other words, for a gray surface the spectral and total properties are identical. For the special case of a surface which is gray and diffuse it can be shown that a= e(7;)

(19.150)

That is, the total hemispherical emissivity of a surface at temperature 7; is equal to its total hemispherical absorptivity for radiation coming from a blackbody at the same temperature. This relation was first developed by Gustav Kirchhoff in 1860 and is called Kirchhoffs identity or Kirchhoffs law. It is important to realize that Kirchhoff's identity is derived under the condition that the source temperature is equal to the temperature of the source of radiation, and hence it should not be used when considerable difference, of the order of more than few hundred degrees, exits between the surface temperature and the temperature of the source of irradiation. For non-metallic surfaces the assumptions gray and diffuse are reasonably valid, while for metallic surfaces such assumptions lead to erroneous results. For an opaque gray-diffuse surface, Eqs. (19.148) and (19.150) can be combined to give p=I—e since

(19.151)

r = 0.

EXAMPLE 1930 A spherical ball of 10 cm diameter maintained at a constant temperature of 1100 K is suspended in air. Assuming the ball to closely approximate a blackbody, determine (a) the total blackbody emissive power, (b) the total amount of radiation emitted by the ball in 10 minutes, and (c) the spectral blackbody emissive power at a wavelength of 3 gm. Solution (a) The total blackbody emissive power is given by the Stefan-Boltzmann law as Eb =074 = (5.67 x 10-8 ) (11004 ) 83,014.5 W/m2 (b) The total amount of radiation emitted from the ball in 10 minutes is given by the product of the blackbody emissive power obtained as above and the surface area of the ball and the given time duration. Qrad = EbilAt = Eb x(ird 2 )x At =83,014.5 x (ff x 0.12)x (10 x 60) =1,564,786.5 Ws 1564.8 kJ

618 Fundamentals of Engineering Thermodynamics (c) The spectral blackbody emissive power at a wavelength of 3 gm can be determined from Planck's distribution law as Cl

EbA = it

s[

exp

( C ) 1] ;. A. 3.743 x 108

35

[ (1.4387 x104 ) exp 3 x 1100

19,944 W/(m2 gm)

19.20.2 The View or Configuration Factor In radiation heat transfer problems we wish to obtain a general expression for the energy exchange between surfaces when they are maintained at different temperatures. The problem becomes essentially one of determining the amount of energy which leaves one surface and reaches the other, as shown in Fig. 19.37.

Figure 19.37 Radiation exchange between elemental surfaces of area dil l and dA2. Let the surfaces dA 1 and dA2 be diffuse, gray and isothermal. The net energy leaving dil l and reaching dA2, given by Eq. (19.144) is dOt-42 =

cos fildflidAl

(19.152)

where dal is the solid angle subtended by the surfaces dA1 at dA2. The total energy leaving dA1 on one side is given by

=

2x

1, cos AdA,da,

(19.153)

Elements of Heat Transfer 619 This total energy leaving MI can be calculated by constructing an imaginary hemisphere of radius r centred at dA1 , as shown in Fig. 19.37. Then, noting that an identified area on a sphere is always normal to the radius of the sphere at that area, we can write dA dill = — sin A dA dOi r2 = Since the surface dA1 is diffuse, 4 in Eq. (19.153) can be taken out of the integration over the solid angle, to result in

A = x l'idAi

(19.154)

From Eqs. (19.152) and (19.154), we can express the fraction of energy leaving surface dAi and reaching surface dA2 as Fl-,2

= °4-.2 — cos A dill

(19.155)

but all — dA2 cos )62 R2 Therefore, we have cos /31 cos A dA2 R2

FI-42 -

(19.156)

F1.42 is called the configuration or view factor between the infinitesimal areas considered. Multiplying Eq. (19.156) by the area dAi , we get cos A cos A dA, dA2

dr4IFI._,,2 -

R2

(19.157)

From the symmetry of the right-hand side of Eq. (19.157), we can write dAlF1-2 = dA2F2-+I

(19.158)

This is called the reciprocity relation. The definition of F1.42, which is for energy exchange between two elemental areas, can be extended to define the energy exchange between two finite areas Al and A2, to give

L I °I-42 dAdA2 F1-)2 -

2 Air

J G dill A,

(19.159)

Using Eqs. (19.152) and (19.153), Eq. (19.159) can be expressed as /, cos A cos /32 dAldA2 F1-+2 -

fA f

24

R2 licliti

(19.160)

lc fAi

Now, assuming that the intensity of radiation leaving surface 1 is uniform over the entire surface, Eq. (19.160) can be simplified to

620

Fundamentals of Engineering Thermodynamics 1

FI-42 =— .1 A2 •,A1

Cos fl, Cm /32 dAICIA2 ItR2

(19.161)

From Eq. (19.161), we may define the view factor as the fraction of the radiation leaving surface A1, which is intercepted by surface A2. Similarly, the view factor F2_,1 is defined as the fraction of the radiation which leaves A2 and is intercepted by /11 . The same development then yields 1 F21 -*

f cos 1cos

f

dA1dA2

rR2

= A2 A2 'Ai

(19.162)

Either Eq. (19.161) or Eq. (19.162) may be used to determine the view factor associated with any two surfaces that are diffuse emitters and reflectors and have uniform radiosity.

View factor relation An important view factor relation is suggested by Eqs. (19.161) and (19.162). That is,

Ai

= A2F2,1

(19.163)

This is called the reciprocity relation. Another important view factor relation pertains to the surfaces of an enclosure like the one shown in Fig. 19.38.

T2

Figure 19.38 Radiation exchange in an enclosure. From the definition of the view factor, the summation rule N

:

j=1

=1

(19.164)

may be applied to each of the N surfaces in the enclosure. This rule follows from the conservation requirement that all the radiation leaving the surface Ai be intercepted by the enclosure surfaces. The term appearing in this summation represents the fraction of the radiation that leaves the surface Ai and is directly intercepted by A. If the surface is concave, it sees itself and F1....; is then nonzero. But for a plane or convex surface, Fi_,1 = 0.

Elements of Heat Transfer 621

To compute radiation exchange in an enclosure of N surfaces, a total number of N2 view factors is needed. However, all the view factors need not be calculated directly. A total of N view factors may be obtained from N equations associated with application of the summation rule, Eq. (19.164), to each of the surfaces in the enclosure. In addition, N(N — 1)/2 view factors may be obtained from the N(N — 1)/2 applications of the reciprocity relation, Eq. (19.163), which are possible for the enclosure. Thus, only [N2 — N — N(N — 1)/2] or N(N — 1)/2 view factors need to be determined directly. For example, in a three-surface enclosure this requirement corresponds to only 3(3 — 1)/2 = 3 view factors. The remaining six view factors may be obtained by solving the six equations that result from the use of Eqs. (19.163) and (19.164). To illustrate the above procedure, consider the two-surface enclosure involving the spherical surface shown in Fig. 19.39. Although the enclosure is characterized by N2 = 4 view

Figure 19.39 View factor for the enclosure formed by two spheres. factors, only N(N — 1)/2 = 1 view factor need to be determined directly. In this case, such a determination may by made by inspection. Note that all radiation leaving the inner surface must reach the outer surface; it therefore follows that F1 ,2 = 1. The same may be said of radiation leaving the outer surface, since this surface sees itself. However, from the reciprocity relation, Eq. (19.163), we get A, ) F2 1

) 2

A2

A2

Also, from the summation rule, Eq. (19.164), we get F13

F1-32

=1

But for the present case, F1 ,1 = 0, and F2-31 + F2-42 =

Thus, F2-42 = I -

A2

)

For more complicated geometries, the view factor must be determined by solving the double integral of Eq. (19.161).

622 Fundamentals of Engineering Thermodynamics

19.21 BLACKBODY RADIATION EXCHANGE We have seen that, in general, radiation may leave a surface because of both reflection and emission, and on reaching a second surface, experience reflection as well as absorption. However, the analysis of radiation exchange can be simplified for surfaces that may be approximated as blackbodies, since there is no reflection. Hence energy leaves only as a result of emission, and all incident radiation is absorbed. Examine the radiation exchange between two black surfaces of arbitrary shapes, shown in Fig. 19.40.

jr..11= Ebi A1, Ti

Ai, Ti Figure 19.40 Radiation exchange between two black surfaces of arbitrary shapes. Let qi_ti be the rate at which radiation leaves the surface A. and is intercepted by the surface A1. It follows that qt-y =

(19.165)

Since radiosity equals the emissive power for a black surface (Ji = Ebi), we have Eb,

(19.166)

= Ali ., Ebi

(19.167)

Similarly, qi

The net radiative change between two surfaces may then be defined as (19.168)

qy =

Substituting Eqs. (19.166) and (19.167) into Eq. (19.168), we get qy =

Ebi AjFj--oi Ebj

or from Eqs. (19.147) and (19.163), we get 47(7;4 71) q4

(19.169)

This gives the net rate at which radiation leaves the surface i as a result of its interaction with j, which is equal to the net rate at which j gains radiation owing to its interaction with i. Equation (19.169) may also be used to determine the net radiation transfer from any surface in an enclosure of black surfaces with N surfaces maintained at different temperatures. The net

Elements of Heat Transfer 623 transfer of radiation from the surface A. is due to exchange with the remaining surfaces and may be expressed as

N qi =E J.1

AiFi,, cr(Tis —71)

(19.170)

19.22 RADIATION EXCHANGE BETWEEN DIFFUSE, GRAY SURFACES IN AN ENCLOSURE Even though useful, the foregoing results have only a limited application, since they are only for blackbodies. The blackbody idealization, although closely approximated by some surfaces, is never precisely achieved. A major difficulty associated with radiation exchange between nonblack surfaces is due to surface reflection. Consider the enclosure shown in Fig. 19.41. In the GI Ai Ei Ai

\ Ai, El

T1 , A1 , et

PiGi A1

r

En;

--

il

C

t

11

C

Ti, Ai, ei (a)

1 — ei al Ai

cti Gi Ai

(b)

(c)

(d)

Figure 19.41 Radiation exchange in an enclosure of diffuse, gray surfaces with a nonparticipating medium: (a) schematic of the enclosure, (b) radiative balance according to Eq. (19.171), (c) radiative balance according to Eq. (19.172), and (d) network element representing the net radiation transfer from a surface. enclosure considered, radiation must experience multiple reflections of all surfaces, with partial absorption occurring at each surface. Analysis of radiation exchange in an enclosure may be simplified by making the following assumptions: • Each surface of the enclosure is assumed to be isothermal and characterized by uniform radiosity and irradiation. • All the surfaces are opaque, diffuse and gray. • The medium within the enclosure is non-participating. Usually, the problem is one in which the temperature Ti associated with each of the surfaces is known, and the objective is to determine the net radiative heat flux gi from each surface.

624 Fundamentals of Engineering Thermodynamics

19.22.1 Net Radiation Exchange at a Surface The net rate at which radiation leaves the surface i represents the net effect of radiative interaction occurring at the surface (Fig. 19.41(b)). It is the rate at which energy would have to be transferred to the surface by other means to maintain it at a constant temperature. It is equal to the difference between the surface radiosity and irradiation and may be expressed as (19.171)

q; = A; (J; — Gi) From Fig. 19.41(c) and the definition of the radiosity Ji, we have +

✓;

It is seen that the net radiative transfer from the surface may also be expressed in terms of the surface emissive power and the absorbed irradiation,• as q; = AXE; — c4Gi) Substituting from Eq. (19.149) and noting that pi = 1 — surface, the radiosity may also be expressed as

(19.172) = 1 — e; for an opaque, diffuse, gray

= EiEbi + (1 — OGi

(19.173)

Combining Eqs. (19.171) and (19.173), we get = Ai (J,

eiEbi )

1— ei

Or

qi

E • — J.1 1- e

(19.174)

This equation gives the net radiative heat transfer rate from a surface. This transfer may be represented by the network element of Fig. 19.41(d), with (Eb; — Ji) as the driving potential and (1 — as the surface radiative resistance. Hence, if the emissive power that the surface would have if it were black exceeds its radiosity, there is net radiation heat transfer from the surface; if the inverse is true, the net transfer is to the surface.

19.22.2 Radiation Exchange between Surfaces For using Eq. (19.174), the surface radiosityJi must be known. To determine Ji, it is necessary to consider radiation exchange between the surfaces of the enclosure. The irradiation of the surface i can be evaluated from the radiosities of all the surfaces in the enclosure. In particular, from the definition of the view factor, it follows that the total rate at which radiation reaches the surface i from all the surfaces, including i, is N

itiG; =

E

Elements of Heat Transfer 625 Or from the reciprocity relation, Eq. (19.158), we get N

AiGi =

E AiFfr.„.7;

Substituting this into Eq. (19.171), we get = Ai(Ji -

Using the summation rule, Eq. (19.164), this may be written as =

J-1

- J-1 E

if)

or N

N

j=1

1='

qi

(19.175)

That is, the net rate of radiation transfer from the surface , qi, is equal to the sum of components q, related to radiative exchange with the other surfaces. Each component may be represented by a network element for which (Ji - Ji) is the driving potential and (AiFi,j)-I is a space or geometrical resistance. Combining Eq. (19.174) and (19 175), we can write Ebi-Ji

(1 -

(19.176)

i =1 (A.F.,-,)

This radiative exchange may be represented as a network as shown in Fig. 19.42. As shown in Fig. 19.42, Eq. (19.175) represents the radiation balance for the mode associated with the surface Ai. O:ts O.&

`Pt

Jz

Ea

qi

Node corresponding to the surface i Figure 19.42

Network or radiative exchange between the surface i and the remaining surfaces in an enclosure.

626 Fundamentals of Engineering Thermodynamics It is essential to note that Eq. (19.176) is useful especially when the surface temperature Ti (hence EN) is known. Even though this situation is typical, it is not the case always. For example, a situation may arise for which the net radiation transfer rate at the surface q,, rather than the temperature Ti, is known. For such cases, Eq. (19.175) may be rearranged as J. —J.

N

=

E

(19.177)

J.1

19.22.3 The Two-Surface Enclosure Two surfaces that exchange radiation only with each other, as shown in Fig. 19.43(a), is the simplest example of an enclosure. A2,1-2,E2

APTI ,E1 (a) 1—c1 Eel

qi=

1

1-62

AiFi 2

e2A2

Ebi —Ji ( 1 — el )/elAl

Eb2

-92 = (1- e2)/e2A2

(b) Figure 19.43 Schematic of a two-surface enclosure and its network representation. Since there are only two surfaces, the net rate of radiation transfer from surface 1, q1, must be equal to the net rate of radiation transfer to surface 2, — q2, and both quantities must equal the net rate at which radiation is exchanged between the surfaces 1 and 2. Thus, qi _ q2 q12 From Fig. 19.43(b) it is seen that the total resistance to the radiation exchange between the surfaces 1 and 2 is made up of two surface resistances and the geometrical resistance. Therefore, the net radiation exchange between the surfaces may be expressed, with Eq. (19.147), as ql 2 = ql = q2 - 1 E1 Al

CrUi4 — 1 Al Fi_+2

1-62 e2 A2

This result is applicable to any two diffuse, gray surfaces which form an enclosure.

(19.178)

Elements of Heat Transfer 627 EXAMPLE 1931 Two long cylinders having diameters of 10 cm and 20 cm are kept concentric (one inside the other). The inner cylinder has T1 = 1000 K and el = 0.7 while the outer cylinder has T2 = 500 K and e2 = 0.3. Calculate the net heat transfer between the two cylinders per unit length. Solution This is a two-body problem. Since the cylinders are very long, very little radiation leaks out at the ends. All of the radiation leaving the inner cylinder arrives at the outer cylinder, thus, F12 = 1.0. Considering unit length, we have the surface area of the cylinders as A i = rd, = It x 0.1 = 0.314 m2 A2 = 7d2 = if x 0.2 = 0.628 m2 The resistance are 1 — el _

el Al 1 — e2 _ e2 A2 1 A1F12

—

1 — 0.7 0.7 x 0.314

—

1. 365

1 — 0.3 — 3.715 0.3 x 0.628 1 — 3.185 0.314 x 1.0

Ebi = cr4 = (5.67 x

10x moors = 56700 W/m2

Eb2 = o7 = (5.67 x 10)(500)4 = 3543.75 W/m2 The net heat transfer between the two cylinders per unit length is 9=

Ebl — Eb2 ..._

ER

56700 — 3543.75 _ 6.43 kW/m 1.365 + 3.715 + 3.185

EXAMPLE 1932 Two parallel plates of size 1 m x 2 m are spaced 1 m apart. One plate is maintained at 1500 K and the other at 1000 K. The emissivities of the plates are 0.3 and 0.7, respectively. The plates are placed in a large room whose walls are at 300 K. The plates exchange heat with each other and with the room, but only the plate surfaces facing each other are to be considered in the analysis. Calculate the net transfer of heat from each plate and to the room. The shape factors are F12 = F-,1 = 0.285. Solution This is basically a three-body problem. The radiation network is shown in the figure below.

628

Fundamentals of Engineering Thermodynamics

Eby = sTi4 •

1.167

1.754

0.214

1-E1

1

1 — E2

Al F12

£2 A2

w

• Eb2 = ST24

0.699

0.699 1 Ai F13

1 A2 F23

J3 = Eb3 = ST34

Given that, = 1500 K, A i = A2 = 2 m2, El = 0.3 T2 = 1000 K, £2 = 0.7 T3 = 300 K 3 Since the area of the room A3 is very large, the resistance 1 — E may be taken to be zero. Thus, E3A3 Eb3 = J3 F12 = F21 = 0.285 (given).

Therefore, F13 = 1 — F12 = 0.715 F23 = 1 — F21 = 0.715 The resistances in the network are calculated as 1— E1 A1

1 — 0.3 — 1.167 0.3 x 2

1 — £2 E2 A2

1 — 0.7 — 0.214 0.7 x 2

1 A1 F2

1 — 1.754 2 x 0.285

1 A I FI3 1 A,F23

1 — 0.699 2 x 0.715 1 — 0.699 2 x 0.715

To calculate the heat flow at each surface, we should determine the radiosities J1 and J2. The network is solved by setting the sum of the heat currents entering nodes J1 and J, to zero. Node Ji : Node J2:

Ebl

1.167 J i— -12

1.754

+

— 1.754 Eb3

0.699

_ 0

Eb3

0.699 Eb2 —

J, _ 0

0.214

Elements of Heat Transfer 629 Also, Ebi = 0'7;4 = (5.67 x 104)(1500)4 = 287.04 kW/m2 Eb2 = OT; = (5.67 x 10-8)(1000)4 = 56.700 kW/m2 Eb3 = crT: = J:, = (5.67 x 10-8)(300)4 = 0.4593 kW mWith these values, Eqs. (i) and (ii) become

-12 1.754

J1

287.04 0.4593 ( 1 + 1 4_ 1 + + =0 1.167 1.754 0.699) 1.167 0.699 1

1.754

J2(

+

1

+ 1

1.754 1.699 0.214)

+

0.4593 +

56.7 — 0.699 0.214

A

Solving, we get Ji = 96 kW/m2 = 47.4 kW/m2 The total heat lost by the plate 1 is Ehi - J 1

(1—

) 'EMI

287.04 — 96 1.167

163.7 kW

56.7 — 47.4 0.214

43.5 kW

Similarly, Q -

Eb2 — J 2 (1— e2 )1e2 A2

The total heat acquired by the room is

6 = 1 — J3

+ J2 — J 3 1 / AIFI3 1 / A2F23 47.4 — 0.4593 96 — 0.4593 0.699 0.699

203.84 kW

EXAMPLE 1933 The surfaces of the cylindrical furnace of radius 1 m and height 1 m, shown in the figure below, are maintained at constant temperatures. The side surface of the furnace closely approximates a blackbody. Determine the net rate of radiation heat transfer at each surface at steady-state, if the view factor F1, = 0.38. Solution The areas of the surfaces of the cylinder are A1 = A2 = irr2 =

x 1 = 3.14 m2, A3 = Drrh = 2ff x 1 x 1 = 6.28 m2

630 Fundamentals of Engineering Thermodynamics

h

The view factors are F11 +F 2 +F13 =1 F3 = 1 - F11 -F2 =1 - 0 - 0.38 = 0.62 Since the base surface is flat, F11 = 0. Also, the top and bottom surfaces are symmetric about the side surface, therefore, F21 = F12 = 0.38 and F23 = F32 = 0.62 The view factor F31 can be determined from the reciprocity rule. AI F13 = A3 F3 1

-4F F31

/13 13 =

3. 14 x 0.62 =0.31 6.28

Also, F13 = F31 = 0.31 due to symmetry. The radiation network associated with the problem is as shown in the figure below. I-E R, =—' v 4E1

E

a

Q3

Elements of Heat Transfer 631 The thermal resistances are 121 = R =

_ 1- 0.8 Ale, 1 -

3.14 x 0.8

= 0.08

0.4 = 0.48 3.14 x 0.4 1 -

A2e2

e3 R3 = 1- = 0 (since the surface 3 is a blackbody) A3€3 R12 -

1 1 = = 0.84 AI FI2 3.14 x 0.38

R13 =

1 1 = 0.51 AI F13 3.14 x 0.62

R23 =

1 1 = = 0.51 A2F23 3.14 x 0.62

The emissive powers of the surfaces are Eb1 =01;4 = 5.67 x10-8 x 7004 =13,614W/m2 Eb2 =a724 = 5.67 x 10118

x 5004 =3544W/m2

Eb3 =aT34 = 5.67 x 10-8 x 4004 =1452 W/m2

To calculate the heat flow at each surface we should determine the radiocities J1 , J2, and J3. The network is solved by setting the sum of the heat currents entering the nodes J1 and J2 to zero. Node J1

Ebl `11 + J2 - J

0.08

0.84 Eb2

Node J2 : .11 j2 0.84 0.48 Substituting the values of Ebi, Eb2 and Eb3, we have

Eb3

0.51 Eb3 - JZ

0.51

-0

13614-J12-J, 1452- J1 + =0 0.08 0.84 0.51 Ji 3544 - i2 1452 =0 0.08 0.48 0.51 Solving, we get J1 = 11,418 W/m2 and J2 = 4562 W/m2. Also, J3 = Eb3 = 1452 W/m2 The net rates of radiation heat transfer at the three surfaces are -

_13,614 -11,418 (1- 1)/(el A1 ) 0.08 Ebl

27, 450 W =

42 - J 2

3544 - 4562

(1- 52)1(52 A2)

0.48

-2121W

632

Fundamentals of Engineering Thermodynamics

= 1/(4F13 ) 1./(A2 F23 ) 1452 —11,418 1452 — 4562 0.51

0.51

— 25, 639 W Note that the direction of net radiation heat transfer is from the top surface to the base and side surfaces. The algebraic sum of these quantities must be zero, i.e. + 02 -I-

27,450 — 2121-25,639 =40 W

We get the sum as 40 W instead of zero. This is due to the accumulation of truncation error in the calculations at different steps.

19.22.4 Radiation Shields Radiation shields are surfaces meant for reducing the net radiation transfer between two surfaces. Radiation shields are constructed from low emissivity (high reflectivity) materials. Consider the radiation shield, surface 3, placed between the two large, parallel planes, as shown in Fig. 19.44(a). The net rate of radiation transfer between the surfaces 1 and 2, without the radiation shield, is given by 0171

1112

A

—

T24 )

1 , 1

, -

E1

62

412 --BP.

Radiation shield

413

C3,1 A1 ,

T1 , e1

A2 , T2, E2 A3 T3

(a) Ebl

J3,

Ch 1— £1

1

El Al

Ai F13

J3,2

1 — C3. 1

1 — E3. 2

£3.1A3 E3. 2 A3

1

1—E2

A 3F32

£2 A2

(b) Figure 19.44 Radiation exchange between two large parallel planes with a radiation shield placed in between them.

Elements of Heat Transfer 633

But, with the radiation shield, additional resistances are present, as shown in Fig. 19.44(b), and the heat transfer rate is reduced. It is important to note that the emissivity associated with one side of the shield (e3 ,1 ) may differ from that associated with the opposite side (€3.2) and the radiosities will always differ. Considering all the resistances and recognizing that Fl _.> 3 = F3-.>2 = 1, we can write

(7 (Ti — T2) 1 — e3.1 1— 1 +1 — + +

q12 .....

A

el

e2

63,1

(19.179)

E3,2

e3,2

The above procedure may be extended to deal with multiple radiation shields. For the special case of multiple radiation shields with all shields having equal emissivity, it can be shown that (for N shields) 1 0/12/N -

1 N+1

i kg12/10

(19.180)

where (q12)0 is the radiation transfer rate with no shield. EXAMPLE 19.34

Two large parallel planes with emissivities 0.4 and 0.7 exchange heat. Find the percentage reduction in heat transfer when a polished aluminium radiation sheld (e = 0.04) is placed between them. Solution The heat transfer without the shield is given by 912 = 6(14 — T24) — 0.341017;4 — V) 1 + _1 —1 A 61

62

The radiation network for the problem with the shield in place is shown in the figure below:

q/A —I.

(1 — Oki

Eb3

J3

J1

Eb

1/F13

J3

(1 — e3yE3 (1 — e3 )1€3

1 — e, _ 1 — 0.4 — 1.5 0.4 el 1 — e3 _ 1 — 0.04 _24 e3 0.04 1— e2

62 .... 1 -

0.7 _0.43 0.7

J2 1 / F32

E b2 (1 — E2)/ 62

634

Fundamentals of Engineering Thermodynamics

The total resistance is 1.5 +2 x 24+2 x 1+0.43 =51.93 and the heat transfer with the shield is q12 — a(74

A

T24) = 0.019266 (7'14 — T24 )

51.93

Thus, the reduction in heat transfer is 94.35 per cent

19.22.5 The Radiating Surface A radiating surface is an idealized surface characterized by zero net radiation transfer (q1 = 0). This assumption is common to many industrial applications. A radiating surface may only be approximated to such real surfaces that are well insulated on one side and for which convection effects may be neglected on the opposite (radiation) side. For the case with qi = 0, from Eqs. (19.171) and (19.174), we get Gi =Ji = Ebi

Hence, if the radiosity of a radiating surface is known, its temperature is readily determined. In an enclosure, the equilibrium temperature of a reradiating surface is detennined by its interaction with the other surfaces, and it is independent of the emissivity of the reradiating surface. Consider the three-surface enclosure with surface R as the reradiating surface, and the corresponding network as shown in Fig. 19.45. The surface R is assumed to be well insulated, and the convection effects are assumed to AR

TR ER

A2 T2 E.2

Figure 19.45 A three-surface enclosure with one surface radiating.

Elements of Heat Transfer 635 be negligible. Hence, with qR = 0, the net radiation transfer from surface 1 must be equal to the net radiation transfer to surface 2. The network is a simple series-parallel arrangement, and from its analysis, we get En; — Eb2 q1 =-q2 = 1 — 1 (19.181) 4_ el AI A, F, 2 + [(1/4 FIR ) (1 / A2 F2 R )1-1 + — e2 e2 A2

Knowing q1 = —q2, Eq. (19.174) may be applied to surfaces 1 and 2 to determine their radiosities J1 and A. Once J1 and J2 and the geometrical resistances are known, the radiosity of the reradiating surface JR may be determined from the requirement that — JR R)

(1/Ai

— J2 = 0 (1/ A2 F2 R )

(19.182)

The temperature of the reradiating surface may be determined from the requirement that arR4 = JR-

19.23 LIMITATIONS So far we have discussed the means for predicting radiation exchange between surfaces. Although, in our discussion, we developed methods to predict radiation exchange, it is important to realize the inherent limitations of these means. To start with, we considered isothermal, opaque and gray surfaces that emit and reflect diffusely and are characterized by uniform surface radiosity and irradiation. For enclosures, we considered the medium separating the surfaces to be nonparticipating, i.e. neither absorbing nor scattering the surface radiation (and emitting no radiation). Even though we made so many assumptions in their development, the foregoing conditions and the related equations may often be used to obtain reliable first estimates and, in some cases, highly accurate results for radiation transfer in an enclosure. However, sometimes, the assumptions are grossly inappropriate and more refined prediction methods are needed. For such methods, one may refer to books devoted to advanced treatment of radiation transfer, like, Thermal Radiation Heat Transfer, by Siegel R. and Howell J.R., McGraw-Hill, 1981.

SUMMARY In this chapter, we saw that the science of thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. But the science of heat transfer deals with the rate of heat transfer. Also, we know from thermodynamics that the total energy of a system consists of the internal, kinetic, and potential energies. The internal energy represents the molecular energy of a system, and it consists of sensible, latent, chemical, and nuclear forms. The sensible and latent forms of internal energy can be transferred from one medium to another as a result of temperature difference, and are referred to as heat or thermal energy. Thus, heat transfer is the exchange of the sensible and latent forms of internal energy between two mediums as a result of temperature difference. The amount of heat transferred per unit time is called heat transfer rate, Q . The rate of heat transfer per unit surface area is called heat flux, 9 .

636

Fundamentals of Engineering Thermodynamics

Heat can be transferred in three ways: conduction, convection, and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones through particle interactions. Fourier's law of heat conduction expresses the amount of rate of heat transfer by conduction as Q=

KA dT dx

where K is the thermal conductivity of the material, A is the area normal to the direction of heat transfer, and (IV& is the temperature gradient in that direction. The rate of heat conduction across a plane of thickness L is given by „ A AT Oconduction = 11-11

where AT = (T1 — T2) is the temperature difference across the plane. Convection is the mode of heat transfer between a solid and the adjacent fluid that is in motion. Thus, convection involves the combined effects of conduction and fluid motion. Newton's law of cooling expresses the convection heat transfer as

Oconvection = hA (Ts — T.) where h is the convection heat transfer coefficient, A is the surface area through which convecticn heat transfer is taking place, Ts is the surface temperature of the solid, and T., is the temperature of the fluid far from the surface. Radiation is the energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configuration of the atoms and molecules. The maximum rate of radiation that can be emitted from a surface at an absolute temperature Ts is given by the Stefan-Boltzmann law as Omit, max =

T54

where cr= 5.67 x 10-8 W/(m2 K4) is the Stefan-Boltzmann constant. When a body of surface area A and surface emissivity e at temperature Ts is enclosed by a dark black enclosure at a temperature Tsurr separated by a gas that does not intervene with radiation, the net rate of radiation heat transfer between these two surfaces is given by Qradiation = eaA Ts4 Tsurr Basically, heat transfer is the process of energy transfer owing to temperature difference. The general form of heat conduction equation is given by

q=—KVT=—Kl

. ar . ar , ar ) j— + 1C— ar ay aZ

1— +

For steady-state conduction through a plane wall, the thermal resistance is given by L KA If the wall is in contact with a fluid film at its surface at x = L, the resistance becomes p _ L 1 — th — KA + hA Rth =

Elements of Heat Transfer 637

The overall heat transfer coefficient for a composite wall, U, may be defined as 1

U

+ 1 4,_on hn

LI 2 + L23 + h1 K12 K23

For heat conduction across a cylindrical shell, the thermal resistance is given by

Rth —

ln r2 2ir LK

The critical thickness of insulation for heat transfer through cylindrical shells is given by K

rC •1 =

h

For steady-state conduction through a spherical shell, the thermal resistance is given by Rth =

1

(1

4x K r1

1) r2

The thermal resistance of a fm, is given by Rth —

KA m tank (mL)

In the absence of a heat source, the governing equation for the unsteady heat conduction becomes

Based on the nature of the flow, the convection heat transfer may be classified as • forced convection when the flow is caused by some external means, such as a fan, a

pump, or atmospheric winds. • free (or natural) convection when the flow is induced by buoyancy forces in the fluid.

Irrespective of the nature of the convection heat transfer mode, the appropriate rate equation is of the form q = h (T, — 71„,)

In functional form, the dimensionless temperature B can be expressed as 0= B [x*, y*, Re, Pr, Gr, Ec, 137',„„ g(Ts — Reynolds number, Re = pUL11.1, may be interpreted as the ratio of inertia forces to viscous forces. Prandtl number, may be interpreted as a ratio of momentum diffusivity v to thermal diffusivity a. The Prandtl number may be viewed as a measure of the relative effectiveness of momentum and energy transports by diffusion in the velocity and thermal boundary layer,

respectively.

638 Fundamentals of Engineering Thermodynamics Grashof number provides a measure of the ratio of buoyancy forces to viscous forces in the boundary layer. Its role in free convection is much the same as that of the Reynolds number in forced convection. Eckert number provides a measure of the kinetic energy of the flow relative to the enthalpy difference across the thermal boundary layer. It plays an important role in high-speed flows for which viscous dissipation is significant. It is important to note that, although similar in form, the Nusselt and Biot numbers differ in both definition and interpretation. The Nusselt number is defined in terms of the thermal conductivity of the fluid, whereas the Biot number is based on the thermal conductivity of the solid. In contrast to the mechanism of conduction and convection, where energy transfer through a material medium is involved, heat may also be transferred into regions where perfect vacuum exists. The maximum flux (W/m2) at which radiation may be emitted from a surface is given by the Stefan-Boltzmann law. = q The intensity of radiation I is defined as the rate of energy leaving a surface in a given direction per unit projected area (area normal to the direction) per unit solid angle. That is, 1(13, 0) —

d2 )(fi, 0) dAPdcI

A blackbody is defined as a surface which absorbs all radiation incident on it. With this definition, it is possible to show that of all surfaces at a given temperature, the blackbody emits the maximum possible radiation and this radiation is called diffuse radiation. By StefanBoltzmann law, the emissive power of a blackbody is given by eb(Ts) = n2cir54

PROBLEMS 19.1 The wall of an industrial furnace is constructed from 20 cm thick fire-clay bricks having thermal conductivity of 1.75 W/(m K). Measurements made during steady-state operation reveal temperatures of 1200 and 900 K at the inner and outer surfaces, respectively. If the wall is 2.5 m by 0.75 m on one side, determine the rate of heat loss through the wall. [Ans. 4921.88 W] 19.2 An insulated steam pipe passes through a room in which the air and walls are both at 20°C. The outer diameter of the pipe is 100 mm,' and its surface temperature and emissivity are 250°C and 0.8, respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 15 W/(m2 K), compute the rate of heat loss from the surface per unit length of the pipe. [Ans. 2046 W/m] 19.3 The hot combustion gases of a furnace are separated from the ambient air and its surroundings, which is at 22°C, by a 20 cm thick brick wall. The brick has thermal conductivity of 1.22 W/(m K) and surface emissivity of 0.8. Under steady-state conduction, the temperature on the outer surface of the brick is measured to be of 100°C. Free convection heat transfer to the air adjoining this surface is characterized by a

Elements of Heat Transfer 639 convection coefficient of h = 20 W/(m2 K). Determine the temperature of the inner surface of the brick. [Ans. 443.48°C] 19.4 A 100 mm diameter steam pipe is covered with two layers of insulation. The inner layer is 50 mm thick and has K1 = 0.08 W/(m K). The outer layer is 30 mm thick and has K2 = 1.1 W/(m K). The pipe conveys steam at a pressure of 2 MPa with 30°C superheat. The outside temperature of the insulation is 30°C. If the steam pipe is 25 m long, determine (a) the heat lost per hour, and (b) the interface temperature of the insulation, neglecting the temperature drop across the steam pipe. [Ans. (a) 13.49 MJ, (b) 35.69°C] 19.5 A wall of a room measuring 3 m by 6 m is made up of concrete material. The thickness of the wall is 300 mm. The inside temperature of the wall is 20°C and the outside temperature is 3°C. Determine the heat loss through the wall, taking the thermal conductivity K for the concrete to be 0.8 W/(m °C). [Ans. 816 W] 19.6 An aluminium fin 2.5 mm thick, 100 mm wide and 9 cm long protrudes from a wall. The base is maintained at 270°C, and the ambient temperature is 30°C with h = 10 W/(m2 K). If the thermal conductivity of aluminium is 200 W/(m K), calculate the heat loss from the fin. [Ans. 39.95 W] 19.7 A 250 mm thick brick wall has temperatures of 40°C and 20°C on either side of it. If the thermal conductivity of the brick is 0.52 W/(m K), calculate the rate of heat transfer per unit area of the wall surface. [Ans. 41.6 W/m2] 19.8 A mild steel tank of wall thickness 12 mm contains water at 70°C and the atmospheric temperature is 20°C. The thermal conductivity of mild steel is 50 W/(m K). The heat transfer coefficients for the inside and outside surfaces of the tank are 2800 and 11 W/(m2 K), respectively. Calculate (a) the rate of heat loss per unit area of the tank surface, (b) the temperature of the outside surface of the tank. [Ans. (a) 546.45 W/m2, (b) 69.7°C] 19.9 A furnace wall consists of 125 mm wide refractory bricks and 125 mm wide insulating bricks, separated by an air gap. The outside wall is covered with 12 mm thick plaster. The furnace temperature is 1100°C and the room temperature is 25°C. The heat transfer coefficient from the outside wall surface to the air in the room is 17 W/(m2 K), and the resistance to heat flow of the air gap is 0.16 K/W. The thermal conductivities of the refractory brick, the insulating brick, and plaster are 1.6, 0.3, and 0.14 W/(m K), respectively. Calculate (a) the rate of heat loss per unit area of the furnace wall surface, (b) the temperature at each interface throughout the wall, (c) the temperature at the outside surface of the wall. [Ans. (a) 1344.93 W, (b) 7'1 = 994.96°C, 7', = 779.77°C, 7'3 = 219.34°C, T4 = 104.08°C, (c) T4 = 104.08°C] 19.10 A steel pipe of 100 mm inside diameter and 114 mm outside diameter carries steam at 260°C. The pipe has two layers of insulation. The inner layer is 40 mm thick having K = 0.09 W/(m K), and the outer layer is 60 mm thick having K = 0.07 W/(m K). The heat

640 Fundamentals of Engineering Thermodynamics transfer coefficients for the inside of the pipe and the outer layer of insulation are 550 and 15 W/(m2 K), respectively. The thermal conductivity of steel is 50 W/(m K). If the atmospheric temperature is 15°C, calculate (a) the rate of heat loss by the steam per unit length of pipe, and (b) the temperature at the outside surface of the insulation. [Ans. (a) 116.18 W, (b) To, = 22.85°C] 19.11 A 55 mm thick thermoplastic sheet with K = 5 W/(m K) and a = 4 x 10-7 m2/s is kept in an oven maintained at uniform temperature of 100°C. What should be the least time of exposure for the sheet to attain a minimum temperature of 75°C everywhere, if its initial temperature is 25°C. For the oven, take h = 25 W/(m2 K). [Ans. 117.2 min] 19.12 A large aluminum plate 50 mm thick and initially at 200°C is suddenly exposed to a convection-surface environment at 70°C with a heat-transfer coefficient of 525 W/(m2 K). Calculate the temperature at a depth of 12.5 mm from one of the faces of the plate after a lapse of 1 minute. Take a = 8.4 x 10-5 m2/s, K = 215 W/(m K). [Ans. 147.71°C] 19.13 An aluminium fin 2 mm thick and 100 mm long and 70 mm wide is attached to a body to be cooled. The base temperature of the fin Tb = 270°C. If the thermal conductivity of aluminium is 190 W/(m K), fmd the cooling capacity of the fin if the conductivity of the atmospheric air is 10 W/(m2 K) and the atmospheric temperature is 20°C. [Ans. 0.07 W] 19.14 Air at 27°C and 1 atm flows over a flat plate at a speed of 5 m/s. Calculate the boundary layer thickness at distances of 25 cm and 50 cm from the leading edge of the plate. The viscosity of air at 27°C is 1.98 x 10-5 kg/(m s). [Ans. 4.59 mm, 6.49 mm] 19.15 The top and side surfaces of a3 m x 3 m x 3 m cubical furnace closely approximate black surfaces. The temperature of these surfaces is 1200 K. If the bottom surface maintained at 800 K transfers 594.4 kW of heat by radiation, determine the emissivity of the bottom surface. [Ans. 0.7] 19.16 A passive solar house is heated by 50 glass containers each containing 20 litres of water heated to 80°C during the day by absorbing solar energy. The house has to be maintained at 22°C all times during a winter night for 10 hours. During this period heat is lost to the outdoors at an average of 50,000 kJ/h. A thermostat-controlled 15 kW back-up electrical resistance heater turns on whenever necessary to keep the house at 22°C. (a) How long did the electrical heating system run that night? (b) How long would the electric heater have to run that night if the house had no solar heating? Assume 1 litre of water is 1 kg mass and the average specific heat of water is 4.2 kJ/(kg °C). [Ans. (a) 4.748 h, (b) 9.26 h] 19.17 An electric iron box is left on board with its base exposed to the ambient atmosphere air at 20°C. The convection heat transfer coefficient between the base and surrounding air is 35 W/(m2 °C). If the base has an emissivity of 0.6 and a surface area of 0.02 m2, determine the heat supplied to the base if it is at a steady temperature of 674°C. [Ans. 1000 W]

Elements of Heat Transfer 641 19.18 An electric wire of diameter 4 mm and length 1.5 m extends across a room that is at a uniform temperature of 25°C. Heat is generated in the wire as a result of resistance heating, and the surface temperature of the wire is measured to be 230°C in steady operation. Also, the voltage drop and electric current through the wire are measured to be 180 V and 2 A, respectively. Neglecting heat transfer by radiation, determine the convection heat transfer coefficient of the heat transfer between the outer surface of the wire and the air in the room. [Ans. 9.32 W/(m2 °C)] 19.19 Consider a man of exposed surface area of 1.6 m2, emissivity 0.75, and surface temperature 37°C standing in a large room having walls at a surface temperature of (a) 35°C and (b) 25°C. Determine the rate of heat loss from that person by radiation. [Ans. (a) 16.06 W, (b) 91.79 W] 19.20 12 kg of liquid water initially at 10°C is to be heated to 90°C in a copper vessel, by burning wood of specific heat 0.65 kJ/(kg °C) whose combustion liberates heat steadily at 200°C. The vessel weighs 1.2 kg and has an average specific heat of 385 J/(kg °C). Neglecting the heat loss from the vessel, determine how long it will take to heat the water. Take specific heat of water as 4.18 kJ/(kg °C) and wood burnt to be 50 grams per second. [Ans. 11.23 minutes] 19.21 The rate of heat transfer through a 3 m x 4 m brick wall of thickness 40 cm is 800 W. If the inner and outer surfaces of the wall are maintained at 25°C and 10°C, respectively, determine the thermal conductivity of the brick wall. [Ans. 0.988 W/(m °C)] 19.22 The inner and outer surfaces of a 2 m x 2 m window with 6 mm thick glass are at 10°C and 2°C, respectively, in winter. If the thermal conductivity of the glass is 0.78 W/(m °C), determine the amount of heat loss, in kJ, through the window over a period of 8 hours. How much would be the reduction in heat loss if the window thickness is doubled? [Ans. 119808 kJ, 59904 kJ] 19.23 In a nuclear reactor, cylindrical uranium fuel rods of diameter 30 mm generate heat at a rate of 5 x 107 W/m3. If the thermal conductivity of uranium is 27.6 W/(m K), determine the temperature difference between the centre and surface of the fuel rod. [Ans. 101.9°C] 19.24 The two sides of a large plane wall of thickness 0.3 m, surface area 10 m2 and thermal conductivity 1.2 W/(m °C) are maintained at constant temperatures of 100°C and 60°C. Determine the temperature at the middle of the wall and the rate of heat conduction through the wall under steady conditions. [Ans. 80°C, 1600 W] 19.25 Hot air at 100°C blows over a flat plate surface area of 12 m2 maintained at a temperature of 25°C. If the rate of heat transfer from air to the plate is 49.5 kW, determine the average convective heat transfer coefficient. [Ans. 55 W/(m2 °C)] 19.26 A heat flux meter attached to the inner surface of a refrigerator door indicates a heat flux of 30 W/m2 through the door. The average thermal conductivity of the refrigerator door is

642 Fundamentals of Engineering Thermodynamics 0.09 W/(m °C). If the temperatures of the inner and outer surfaces of the door are 5°C and 20°C, respectively, find the door thickness. [Ans. 4.5 cm] 19.27 Consider a large plane wall of thickness 0.3 m and thermal conductivity 2.5 W/(m °C). The left side of the wall is subjected to a net heat flux of 9.25 W/m2, while the temperature at that surface is 92°C. Assuming constant thermal conductivity and no heat generation in the wall, determine the temperature on the right surface of the wall. [Ans.-19°C] 19.28 In a nuclear ractor, heat is generated unifomily at a rate of 6.5 x 107 W/m3 by cylindrical uranium rods of 6 cm diameter and 1.2 m length. Detemine the rate of heat generation in each rod. [Ans. 220.5 kW] 19.29 Heat at a uniform rate of 5 x 106 W/m2 is generated in a large stainless steel plate of thickness 0.3 m. Assuming that the plate is loosing heat from both sides, determine the heat flux on the surface of the plate during a steady heat transfer process. [Ans. 75000 W/m2] 19.30 70 kg of ice at —4°C is added to 900 kg of water at 21°C in order to cool the water. If the melting temperature and heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, determine the final equlibrium temperature of the mixture. Take specific heat of water as 4.2 kJ/(kg °C). [Ans. 13.46°C] 19.31 The temperature of a gas stream at 100°C is to be measured by a thermocouple whose junction can be approximated as a sphere of diameter 1 mm. The properties of the junction are p= 8500 kg/m3, K = 35 W/(m °C), and C = 320 J/(kg °C), and the convection heat transfer coefficient between the junction and the gas is h = 210 W/(m2 °C). Neglecting the radiation losses from the junction, determine how long it will take for the thermocouple to read 99 per cent of the initial temperature difference if the thermocouple is initially at 20°C. [Ans. 2.9 s] 19.32 A house is maintained at 22°C at all times. The walls of the house are insulated with a material having an L/K value or a thermal resistance of 3.38 (m2 °C)/W. During a winter night, the outside air temperature is 4°C and wind at 50 km/h is blowing parallel to a 3 m high and 8 m long wall of the house. If the heat transfer coefficient on the interior surface of the wall is 8 W/(m2 °C), determine the rate of heat loss from that wall of the house. Assume the radiaton heat transfer to be negligible. [Ans. 123.12 W] 19.33 A circuit board of 18 cm length, 12 cm height, and 3 mm thickness houses 80 closely spaced logic chips, each dissipating 0.04 W, on one side. The board is impregnated with copper fillings and has an effective thermal conductivity across the circuit board and heat is dissipated from the back side of the board to the ambient air at 40°C, which is forced to flow over the surface by a fan at a free stream velocity of 6.7 m/s. Determine the temperatures on the two sides of the circuit board. Take Prandtl number at 40°C to be 0.711. [Ans. 46.25°C on the back surface, 46.53°C on the front surface]

Elements of Heat Transfer

643

19.34 A thermocouple junction of spherical shape is used to measure temperature in a gas stream. The convection heat transfer coefficient between the junction surface and the gas is 400 W/(m2 K). The thermophysical properties of the junction material are p = 8500 kg/m2, C = 400 J/(kg K) and K = 20 W/(m K). (a) Determine the junction diameter required for the thermocouple to have a response time of 1 second. (b) If the junction is at 25°C and is placed in a gas stream of 200°C, how long will it take for the junction to reach 199°C? [Ans. (a) 0.706 mm, (b) 5.17 s] 19.35 Air at 6 kPa and 300°C flows over a 0.5 m long plate with a velocity of 10 m/s. Determine the cooling rate per unit width of the plate required to maintain the plate surface at 27°C. Assume K = 0.0364 W/(m K) and Pr = 0.687 for air. [Ans. 455.91 W/m] 19.36 Air at 120°C flows with a velocity of 3 m/s over a 3 m long flat plate whose temperature is 20°C. Neglecting the radiation effects, determine the total drag per unit width of the plate. [Ans. 0.054 N] 19.37 Determine the minimum depth below which water pipe lines should be laid to avoid freezing of water in a place where the surface initially at 22°C will experience a constant temperature of —22°C for 45 days in winter. The soil properties are p = 2050 kg/m3, C = 1840 J/(kg K) and K = 0.52 W/(m K). [Ans. 0.7 m] 19.38 Air at 90 kPa and 22°C flows at 10 m/s over the top surface of a flat plate of 1 m width and 5 m length. The plate is maintained at a constant temperature of 130°C. Taking K = 0.03 W/(m °C) and Pr = 0.7 for air, determine the rate of heat transfer from the plate if the air flows parallel to either (a) 5 m length side or (b) 1 m width side. [Ans. (a) 9973.8 W, (b) 6291 W] 19.39 A thin square silicon chip of side 10 mm is fixed on an aluminium plate of same crosssection and 8 mm thickness, with epoxy layer of thickness 0.02 mm. The silicon chip top surface and aluminium plate bottom surface are cooled by air at 25°C and convection heat transfer coefficient h = 100 W/(m2 K). If the chip dissipates 104 W/m2 under steadystate condition, determine the temperature at the chip surface. Assume K = 238 W/(m K) for aluminium and the conduction to be one-dimensional and the resistance to heat transfer through the expoxy layer to be 0.9 x le (m2 K)/W. [Ans. 124°C] 19.40 A furnace cavity, in the form of a cylinder of 100 mm diameter and 200 mm length, is open at one end to a large surrounding at 30°C. The sides and bottom are maintained at constant temperatures of 1200°C and 1500°C, respectively, by electrical heating. All the surfaces are perfectly insulated. Assuming the sides and bottom as blackbodies, determine the power required to maintain the sides at constant temperatures. Assume the view factor between the sides and the open top to be 0.06. [Ans. 2202.4 W] 19.41 Two large parallel plates maintained at uniform temperatures of Ti = 1000 K and T2 = 700 K have emissivities = 0.3 and 62 = 0.8, respectively. Determine the net rate of radiation heat transfer between the two surfaces per unit surface area of the plate. [Ans. 12025 W/m2]

644

Fundamentals of Engineering Thermodynamics

19.42 The surfaces of a cubical furnace of side 5 m are maintained at uniform temperature shown in the figure below. If the surfaces closely approximate the black surface, determine (a) the net rate of radiation heat transfer between the base and the side surfaces, (b) the net rate of radiation heat transfer between the base and top surfaces, and (c) the net rate of radiation heat transfer from the base surface. Take the view factor F12 = 0 2 • [Ans. (a) 987033.6 W, (b) —304365.6 W, (c) 682668 W] T2 =

1200 K

= 600 K

19.43 A flat-plate solar collector with no cover plate has a selective absorber surface of emissivity 0.1 and solar absorptivity 0.95. At a given time of a day the absorber surface temperature 7; = 120°C when the solar irradiation is 750 W/m2, the effective sky temperature is —10°C, and the ambient air temperature T., = 30°C. The convection heat transfer coefficient closely approximates = 0.22 (Ts — Ta.3413 W/(m2 K) Calculate the useful heat that can be removed from the collector and also fmd the efficiency of the collector. Assume the absorptivity of the collector for irradiation from surrounding to be 0.1. [Ans. 515.6 W/m2, 0.687] 19.44 An enclosure is made up of 15 surfaces. How many view factors does this geometry involve? How many of these view factors can be detemined using the reciprocity and summation rules? [Ans. 225, 120] 19.45 A hemispherical furnace has a flat circular base of diameter D. Determine the view factors from the dome of this furnace to its base. [Ans. 0.5] 19.46 Find the view factors from the base of a cube to each of the other 5 sides. [Ans. 0.2]

Elements of Heat Transfer

645

19.47 A hemispherical furnace of diameter 5 m has a flat base. The dome of the furnace closely approximates a black surface and the base has an emissivity of 0.7. The dome and base are maintained at uniform temperatures of 1000 K and 400 K, respectively. Determine the net rate of radiation heat transfer from the dome to the base surface during steady operation. [Ans. 759.4 kW] 19.48 Two parallel disks of diameter 0.6 m separated by a distance 0.6 m are located directly on top of each other. Both disks are black and maintianed at 1000 K. The back side of the disks are insulated, and the disks are in an environment at 300 K. Treating the environment to be a blackbody, determine the net rate of radiation heat transfer from the disks to the environment. [Ans. 15901.6 W] 19.49 A thermocouple measures the temperature of hot air flowing in a duct as 600 K. The walls of the duct are maintianed at 450 K. Assuming the convection heat transfer coefficient to be 80 W/(m2 °C), determine the actual temperature of the air if the emissivity of the thermocouple junction is 0.7. [Ans. 644 K] 19.50 A thermocouple meaures the temperature of hot air flowing in a duct whose walls are maintianed at 500 K and 850 K. If the actual- temperature of the hot air is 1000 K, determine the convection heat transfer coefficient of the air, assuming the emissivity of the thermocouple junction to be 0.6. [Ans. 104.21 W/(m2 °C)] 19.51 A thin aluminium sheet with an emissivity of 0.1 on both sides is palced between two large parallel plates maintained at uniform temperatures of 600 K and 300 K, as shown in the figure below. The emmissivity values of the plates are 0.3 and 0.6. Determine the net rate of radiation heat transfer between the two plates per unit surface area. What will be the percentage incerase in heat transfer if the shield is removed. T1 = 600 K

T2 = 300 K

e2 = 0.6

el = 0.3

912 Po-

1

P E3 = 0.1 2

Shield

[Ans. 299.5 W/m2, 491.8%]

Appendix Table 1

Substance

Molar mass, gas constant, and critical-point properties

Formula

Ammonia NH3 Argon Ar Bromine Br2 Carbon dioxide CO2 Carbon monoxide CO Chlorine Cl2 Deuterium (normal) D2 He Helium Hydrogen (normal) H2 Krypton Kr Neon Ne Nitrogen N2 N20 Nitrous oxide Oxygen 02 Sulfur dioxide SO2 Water H2O Xenon Xe Benzene C6H6 n-Butane C4H I o Carbon tetrachloride CCI4 Chloroform CHC13 Dichlorodifluoromethane (R-12) CC12F2 Dichlorofluoromethane CHCl2F Ethane C2H6 Ethyl alcohol C2H5OH Ethylene C2H4 n-Hexane C6HI4 Methane CH4 Methyl alcohol CH3OH Methyl chloride CH3CI C3H13 Propane Propene C3H6 C3H4 Propyne Trichlorofluoromethane CC13F Air

Molar mass lcg/kmol

R Temperature K k1/(kg • K)*

Pressure Volume MPa m3/kmol

17.03 39.948 159.808 44.01 28.011 70.906 4.00 4.003 2.016 83.80 20.183 28.013 44.013 31.999 64.063 18.015 131.30 78.115 58.124 153.82 119.38

0.4882 0.2081 0.0520 0.1889 0.2968 0.1173 2.0785 2.0769 4.1240 0.09921 0.4119 0.2968 0.1889 0.2598 0.1298 0.4615 0.06332 0.1064 0.1430 0.05405 0.06964

405.5 151 584 304.2 133 417 38.4 5.3 33.3 209.4 44.5 126.2 309.7 154.8 430.7 647.3 289.8 562 425.2 556.4 536.6

11.28 4.86 10.34 7.39 3.50 7.71 1.66 0.23 1.30 5.50 2.73 3.39 7.27 5.08 7.88 22.09 5.88 4.92 3.80 4.56 5.47

0.0724 0.0749 0.1355 0.0943 0.0930 0.1242

120.91 102.92 30.070 46.07 28.054 86.178 16.043 32.042 50.488 44.097 42.081 40.065 137.37 28.97

0.06876 0.08078 0.2765 0.1805 0.2964 0.09647 0.5182 0.2595 0.1647 0.1885 0.1976 0.2075 0.06052 0.2870

384.7 451.7 305.5 516 282.4 507.9 191.1 513.2 416.3 370 365 401 471.2

4.01 5.17 4.88 6.38 5.12 3.03 4.64 7.95 6.68 4.26 4.62 5.35 4.38

0.2179 0.1973 0.1480 0.1673 0.1242 0.3677 0.0993 0.1180 0.1430 0.1998 0.1810

0.0578 0.0649 0.0924 0.0417 0.0899 0.0961 0.0780 0.1217 0.0568 0.1186 0.2603 0.2547 0.2759 0.2403

0.2478

*The unit ki/kg • K) is equivalent to kPa • m3/(kg • K). The gas constant is calculated from R= R,/M, where 8.314 kJ/(kmol • K) and M is the molar mass. Reproduced with permission from Thermodynamics. • An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994. 647

Ru =

648 Appendix Table 2 Ideal-gas specific heats of various common gases At 300 K Gas Air Argon Butane Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Methane Neon Nitrogen Octane Oxygen Propane Steam

Formula Ar C4H i o CO2 CO C2H6 C2H4 He H2

CH4 Ne N2

C8H18 02

C31-18 H2O

Gas constant R kJ/(kg • K)

Co kJ/(kg • K)

Go kJ/(kg • K)

y

0.2870 0.2081 0.1433 0.1889 0.2968 0.2765 0.2964 2.0769 4.1240 0.5182 0.4119 0.2968 0.0729 0.2598 0.1885 0.4615

1.005 0.5203 1.7164 0.846 1.040 1.7662 1.5482 5.1926 14.307 2.2537 1.0299 1.039 1.7113 0.918 1.6794 1.8723

0.718 0.3122 1.5734 0.657 0.744 1.4897 1.2518 3.1156 10.183 1.7354 0.6179 0.743 1.6385 0.658 1.4909 1.4108

1.400 1.667 1.091 1.289 1.400 1.186 1.237 1.667 1.405 1.299 1.667 1.400 1.044 1.395 1.126 1.327

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, V.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

Table 3 Saturated water-Temperature table

Temp. °C

Sat. press. kPa

T

P.,

Specific volume m3/kg Sat. Sat. liquid vapour

of

vg

0.01

0.6113

0.001000

206.14

Enthalpy kJ/kg

Internal energy kJ/kg

Entropy kJ/(kg • K)

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

of

uA,

u8

hf

hA

hg

sf

sfg

s8

0.0

2375.3

2375.3

0.01

2501.4

2501.4

0.000

9.1562

9.1562

5

0.8721

0.001000

147.12

20.97

2361.3

2382.3

20.98

2489.6

2510.6

0.0761

8.9496

9.0257

10

1.2276

0.001000

106.38

42.00

2347.2

2389.2

42.01

2477.7

2519.8

0.1510

8.7498

8.9008

15

1.7051

0.001001

77.93

62.99

2333.1

2396.1

62.99

2465.9

2528.9

0.2245

8.5569

8.7814

20

2.339

0.001002

57.79

83.95

2319.0

2402.9

83.96

2454.1

2538.1

0.2966

8.3706

8.6672

25

3.169

0.001003

43.36

104.88

2409.8

2409.8

104.89

2442.3

2547.2

0.3674

8.1905

8.5580

30

4.246

0.001004

32.89

125.78

2290.8

2416.6

125.79

2430.5

2556.3

0.4369

8.0164

8.4533

35

5.628

0.001006

25.22

146.67

2276.7

2423.4

146.68

2418.6

2565.3

0.5053

7.8478

8.3531

40

7.384

0.001008

19.52

167.56

2262.6

2430.1

167.57

2406.7

2574.3

0.5725

7.6845

8.2570

45

9.593

0.001010

15.26

188.44

2248.4

2436.8

188.45

2394.8

2583.2

0.6387

7.5261

8.1648 8.0763

209.32

2234.2

2443.5

209.33

2382.7

2592.1

0.7038

7.3725

9.568

230.21

2219.9

2450.1

230.23

2370.7

2600.9

0.7679

7.2234

7.9913

0.001017

7.671

251.11

2205.5

2456.6

251.13

2358.5

2609.6

0.8312

7.0784

7.9096

25.03

0.001020

6.197

272.02

2191.1

2463.1

272.06

2346.2

2618.3

0.8935

6.9375

7.8310

31.19

0.001023

5.042

292.95

2176.6

2469.6

292.98

2333.8

2626.8

0.9549

6.8004

7.7553

2162.0

2475.9

313.93

2321.4

2635.3

1.0155

6.6669

7.6824

50

12.349

0.001012

55

15.758

0.001015

60

19.940

65 70

12.03

75

38.58

0.001026

4.131

313.90

80

47.39

0.001029

3.407

334.86

2147.4

2482.2

334.91

2308.8

2643.7

1.0753

6.5369

7.6122

85

57.83

0.001033

2.828

355.84

2132.6

2488.4

355.90

2296.0

2651.9

1.1343

6.4102

7.5445

90

70.14

0.001036

2.361

376.85

2117.7

2494.5

376.92

2283.2

2660.1

1.1925

6.2866

7.4791

95

84.55

0.001040

1.982

397.88

2102.7

2500.6

397.96

2270.2

2668.1

1.2500

6.1659

7.4159 Cont.

Table 3 Saturated water-Temperature table (Cont.)

Temp. °C T

Sat. press. MPa

Specific volume m3/kg Sat. Sat. liquid vapour of vs

Internal energy kJ/kg Sat. liquid :if

Evap. ufg

Sat. vapour u8

Enthalpy kJ/kg

Entropy kJ/(kg • K)

Sat. liquid hf

Evap. hfg

Sat. vapour hg

Sat. liquid sf

Evap. sfg

Sat. vapour s8

100

0.10135

0.001044

1.6729

418.94

2087.6

2506.5

419.04

2257.0

2676.1

1.3069

6.0480

7.3549

105

0.12082

0.001048

1.4194

440.02

2072.3

2512.4

440.15

2243.7

2683.8

1.3630

5.9328

7.2958

110

0.14327

0.001052

1.2102

461.14

2057.0

2518.1

461.30

2230.2

2691.5

1.4185

5.8202

7.2387

115

0.16906

0.001056

1.0366

482.30

2041.4

2523.7

482.48

2216.5

2699.0

1.4734

5.7100

7.1833

120

0.19853

0.001060

0.8919

503.50

2025.8

2529.3

503.71

2202.6

2706.3

1.5276

5.6020

7.1296

125

0.2321

0.001065

0.7706

524.74

2009.9

2534.6

524.99

2188.5

2713.5

1.5813

5.4962

7.0775

130

0.2701

0.001070

0.6685

546.02

1993.9

2539.9

546.31

2174.2

2720.5

1.6344

5.3925

7.0269

135

0.3130

0.001075

0.5822

567.35

1977.7

2545.0

567.69

2159.6

2727.3

1.6870

5.2907

6.9777

140

0.3613

0.001080

0.5089

588.74

1961.3

2550.0

589.13

2144.7

2733.9

1.7391

5.1908

6.9299

145

0.4154

0.001085

0.4463

610.18

1944.7

2554.9

610.63

2129.6

2740.3

1.7907

5.0926

6.8833

150

0.4758

0.001091

0.3928

631.68

1927.9

2559.5

632.20

2114.3

2746.5

1.8418

4.9960

6.8379

155

0.5431

0.001096

0.3468

653.24

1910.8

2564.1

653.84

2098.6

2752.4

1.8925

4.9010

6.7935

160

0.6178

0.001102

0.3071

674.87

1893.5

2568.4

675.55

2082.6

2758.1

1.9427

4.8075

6.7502

165

0.7005

0.001108

0.2727

696.56

1876.0

2572.5

697.34

2066.2

2763.5

1.9925

4.7153

6.7078

170

0.7917

0.001114

0.2428

718.33

1858.1

2576.5

719.21

2049.5

2768.7

2.0419

4.6244

6.6663

175

0.8920

0.001121

0.2168

740.17

1840.0

2580.2

741.17

2032.4

2773.6

2.0909

4.5347

6.6256

180

1.0021

0.001127

0.19405 762.09

1821.6

2583.7

763.22

2015.0

2778.2

2.1396

4.4461

6.5857

185

1.1227

0.001134

0.17409 784.10

1802.9

2587.0

785.37

1997.1

2782.4

2.1879

4.3586

6.5465

190

1.2544

0.001141

0.15654 806.19

1783.8

2590.0

807.62

1978.8

2786.4

2.2359

4.2720

6.5079

195

1.3978

0.001149

0.14105 828.37

1764.4

2592.8

829.98

1960.0

2790.0

2.2835

4.1863

6.4698 Cont.

Table 3 Saturated water-Temperature table (Cont.) Internal energy kJ/kg

Enthalpy kJ/kg

Entropy kJ/(kg • K)

Temp. °C

Sat. liquid

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

T

Psa,

of

vg

of

ufg

u8

lif

hfg

hg

sf

sfg

s8

850.65 873.04 895.53 918.14 940.87 963.73 986.74 1009.89 1033.21 1056.71 1080.39 1104.28 1128.39 1152.74 1177.36 1202.25 1227.46 1253.00 1278.92 1305.2 1332.0 1359.3 1387.1 1415.5 1444.6 1505.3 1570.3 1641.9 1725.2 1844.0 2029.6

1744.7 1724.5 1703.9 1682.9 1661.5 1639.6 1617.2 1594.2 1570.8 1546.7 1522.0 1596.7 1470.6 1443.9 1416.3 1387.9 1358.7 1328.4 1297.1 1264.7 1231.0 1195.9 1159.4 1121.1 1080.9 993.7 894.3 776.6 626.3 384.5 0

2595.3 2597.5 2599.5 2601.1 2602.4 2603.3 2603.9 2604.1 2604.0 2603.4 2602.4 2600.9 2599.0 2596.6 2593.7 2590.2 2586.1 2581.4 2576.0 2569.9 2563.0 2555.2 2546.4 2536.6 2525.5 2498.9 2464.6 2418.4 2351.5 2228.5 2029.6

1940.7 1921.0 1900.7 1879.9 1858.5 1836.5 1813.8 1790.5 1766.5 1741.7 1716.2 1689.8 1662.5 1634.4 1605.2 1574.9 1543.6 1511.0 1477.1 1441.8 1404.9 1366.4 1326.0 1283.5 1238.6 1140.6 1027.9 893.4 720.3 441.6 0

2793.2 2796.0 2798.5 2800.5 2802.1 2803.3 2804.0 2804.2 2803.8 2803.0 2801.5 2799.5 2796.9 2793.6 2789.7 2785.0 2779.6 2773.3 2766.2 2758.1 2749.0 2738.7 2727.3 2714.5 2700.1 2665.9 2622.0 2563.9 2481.0 2332.1 2099.3

2.3309 2.3780 2.4248 2.4714 2.5178 2.5639 2.6099 2.6558 2.7015 2.7472 2.7927 2.8383 2.8838 2.9294 2.9751 3.0208 3.0668 3.1130 3.1594 3.2062 3.2534 3.3010 3.3493 3.3982 3.4480 3.5507 3.6594 3.7777 3.9147 4.1106 4.4298

4.1014 4.0172 3.9337 3.8507 3.7683 3.6863 3.6047 3.5233 3.4422 3.3612 3.2802 3.1992 3.1181 3.0368 2.9551 2.8730 2.7903 2.7070 2.6227 2.5375 2.4511 2.3633 2.2737 2.1821 2.0882 1.8909 1.6763 1.4335 1.1379 0.6865 0

6.4323 6.3952 6.3585 6.3221 6.2861 6.2503 6.2146 6.1791 6.1437 6.1083 6.0730 6.0375 6.0019 5.9662 5.9301 5.8938 5.8571 5.8199 5.7821 5.7437 5.7045 5.6643 5.6230 5.5804 5.5362 5.4417 5.3357 5.2112 5.0526 4.7971 4.4298

200 205 210 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 330 340 350 360 370 374.14

1.5538 1.7230 1.9062 2.104 2.318 2.548 2.795 3.060 3.344 3.648 3.973 4.319 4.688 5.081 5.499 5.942 6.412 6.909 7.436 7.993 8.581 9.202 9.856 10.547 11.274 12.845 14.586 16.513 18.651 21.03 22.09

0.001157 0.001164 0.001173 0.001181 0.001190 0.001199 0.001209 0.001219 0.001229 0.001240 0.001251 0.001263 0.001276 0.001289 0.001302 0.001317 0.001332 0.001348 0.001366 0.001384 0.001404 0.001425 0.001447 0.001472 0.001499 0.001561 0.001638 0.001740 0.001893 0.002213 0.003155

0.12736 0.11521 0.10441 0.09479 0.08619 0.07849 0.07158 0.06537 0.05976 0.05471 0.05013 0.04598 0.04221 0.03877 0.03564 0.03279 0.03017 0.02777 0.02557 0.02354 0.02167 0.019948 0.018350 0.016867 0.015488 0.012996 0.010797 0.008813 0.006945 0.004925 0.003155

852.45 875.04 897.76 920.62 943.62 966.78 990.12 1013.62 1037.32 1061.23 1085.36 1109.73 1134.37 1159.28 1184.51 1210.07 1235.99 1262.31 1289.07 1316.3 1344.0 1372.4 1401.3 1431.0 1461.5 1525.3 1594.2 1670.6 1760.5 1890.5 2099.3

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

TS9 xypitaddy

Specific volume m3/kg

Sat. press. MPa

Press. kPa P

Sat. temp. °C T.,

Specific volume m3/kg Sat. Sat. liquid vapour of vs

Internal energy kJ/kg Sat. liquid of

Evap. ujk

Sat. vapour us,

zs9

Enthalpy kJ/kg

Entropy kJ/(kg • K)

Sat. liquid hf

Evap. hfs

Sat. vapour Its

Sat. liquid sf

Evap. sr,

Sat. vapour ss

0.6113

0.01

0.001000

206.14

0.00

2375.3

2375.3

0.01

2501.3

2501.4

0.0000

9.1562

9.1562

1.0

6.98

0.001000

129.21

29.30

2355.7

2385.0

29.30

2484.9

2514.2

0.1059

8.8697

8.9756

1.5

13.03

0.001001

87.98

54.71

2338.6

2393.3

54.71

2470.6

2525.3

0.1957

8.6322

8.8279

2.0

17.50

0.001001

67.00

73.48

2326.0

2399.5

73.48

2460.0

2533.5

0.2607

8.4629

8.7237

2.5

21.08

0.001002

54.25

88.48

2315.9

2404.4

88.49

2451.6

2540.0

0.3120

8.3311

8.6432

3.0

24.08

0.001003

45.67

101.04

2307.5

2408.5

101.05

2444.5

2545.5

0.3545

8.2231

8.5776

4.0

28.96

0.001004

34.80

121.45

2293.7

2415.2

121.46

2432.9

2554.4

0.4226

8.0520

8.4746

5.0

32.88

0.001005

28.19

137.81

2282.7

2420.5

137.82

2423.7

2561.5

0.4764

7.9187

8.3951

7.5

40.29

0.001008

19.24

168.78

2261.7

2430.5

168.79

2406.0

2574.8

0.5764

7.6750

8.2515

10

45.81

0.001010

14.67

191.82

2246.1

2437.9

191.83

2392.8

2584.7

0.6493

7.5009

8.1502

15

53.97

0.001014

10.02

225.92

2222.8

2448.7

225.94

2373.1

2599.1

0.7549

7.2536

8.0085

20

60.06

0.001017

7.649

251.38

2205.4

2456.7

251.40

2358.3

2609.7

0.8320

7.0766

7.9085

25

64.97

0.001020

6.204

271.90

2191.2

2463.1

271.93

2346.3

2618.2

0.8931

6.9383

7.8314

30

69.10

0.001022

5.229

289.20

2179.2

2468.4

289.23

2336.1

2625.3

0.9439

6.8247

7.7686

40

75.87

0.001027

3.993

317.53

2159.5

2477.0

317.58

2319.2

2636.8

1.0259

6.6441

7.6700

50

81.33

0.001030

3.240

340.44

2143.4

2483.9

340.49

2305.4

2645.9

1.0910

6.5029

7.5939

75

91.78

0.001037

2.217

384.31

2112.4

2496.7

384.39

2278.6

2663.0

1.2130

6.2434

7.4564 Cont.

lipuaddv

Table 4 Saturated water-Pressure table

Table 4 Saturated water-Pressure table (Cont.)

Press. MPa P

Sat. temp. °C Tin

Sat. liquid of

0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300 0.325 0.350 0.375 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.10 1.20 1.30

99.63 105.99 111.37 116.06 120.23 124.00 127.44 130.60 133.55 136.30 138.88 141.32 143.63 147.93 151.86 155.48 158.85 162.01 164.97 167.78 170.43 172.96 175.38 177.69 179.91 184.09 187.99 191.64

0.001043 0.001048 0.001053 0.001057 0.001061 0.001064 0.001067 0.001070 0.001073 0.001076 0.001079 0.001081 0.001084 0.001088 0.001093 0.001097 0.001101 0.001104 0.001108 0.001112 0.001115 0.001118 0.001121 0.001124 0.001127 0.001133 0.001139 0.001144

Sat. vapour v8 1.6940 1.3749 1.1593 1.0036 0.8857 0.7933 0.7187 0.6573 0.6058 0.5620 0.5243 0.4914 0.4625 0.4140 0.3749 0.3427 0.3157 0.2927 0.2729 0.2556 0.2404 0.2270 0.2150 0.2042 0.19444 0.17753 0.16333 0.15125

Entropy kJ/(kg • K)

Enthalpy kJ/kg

Internal energy kJ/kg Sat. liquid of

Evap. ufg

Sat. vapour u8

Sat. liquid hf

Evap. hfs

Sat. vapour hg

Sat. liquid sf

Evap. S1

Sat. vapour s8

417.36 444.19 466.94 486.80 504.49 520.47 535.10 548.59 561.15 572.90 583.95 594.40 604.31 622.77 639.68 655.32 669.90 683.56 696.44 708.64 720.22 731.27 741.83 751.95 761.68 780.09 797.29 813.44

2088.7 2069.3 2052.7 2038.1 2025.0 2013.1 2002.1 1991.9 1982.4 1973.5 1965.0 1956.9 1949.3 1934.9 1921.6 1909.2 1897.5 1886.5 1876.1 1866.1 1856.6 1847.4 1838.6 1830.2 1822.0 1806.3 1791.5 1777.5

2506.1 2513.5 2519.7 2524.9 2529.5 2533.6 2537.2 2540.5 2543.6 2546.4 2548.9 2551.3 2553.6 2557.6 2561.2 2564.5 2567.4 2570.1 2572.5 2574.7 2576.8 2578.7 2580.5 2582.1 2583.6 2586.4 2588.8 2591.0

417.46 444.32 467.11 486.99 504.70 520.72 535.37 548.89 561.47 573.25 584.33 594.81 604.74 623.25 640.23 665.93 670.56 684.28 697.22 709.47 721.11 732.22 742.83 753.02 762.81 781.34 798.65 814.93

2258.0 2241.0 2226.5 2213.6 2201.9 2191.3 2181.5 2172.4 2163.8 2155.8 2148.1 2140.8 2133.8 2120.7 2108.5 2097.0 2086.3 2076.0 2066.3 2057.0 2048.0 2039.4 2031.1 2023.1 2015.3 2000.4 1986.2 1972.7

2675.5 2685.4 2693.6 2700.6 2706.7 2712.1 2716.9 2721.3 2725.3 2729.0 2732.4 2735.6 2738.6 2743.9 2748.7 2753.0 2756.8 2760.3 2763.5 2766.4 2769.1 2771.6 2773.9 2776.1 2778.1 2781.7 2784.8 2787.6

1.3026 1.3740 1.4336 1.4849 1.5301 1.5706 1.6072 1.6408 1.6718 1.7006 1.7275 1.7528 1.7766 1.8207 1.8607 1.8973 1.9312 1.9627 1.9922 2.0200 2.0462 2.0710 2.0946 2.1172 2.1387 2.1792 2.2166 2.2515

6.0568 5.9104 5.7897 5.6868 5.5970 5.5173 5.4455 5.3801 5.3201 5.2646 5.2130 5.1647 5.1193 5.0359 4.9606 4.8920 4.8288 4.7703 4.7158 4.6647 4.6166 4.5711 4.5280 4.4869 4.4478 4.3744 4.3067 4.2438

7.3594 7.2844 7.2233 7.1717 7.1271 7.0878 7.0527 7.0209 6.9919 6.9652 6.9405 6.9175 6.8959 6.8565 6.8213 6.7893 6.7600 6.7331 6.7080 6.6847 6.6628 6.6421 6.6226 6.6041 6.5865 6.5536 6.5233 6.4953 Cont.

£S9 weaddy

Specific volume m3/kg

Table 4 Saturated water-Pressure table (Cont.)

Press. MPa

Sat. temp. ec

P

Tsai

1.40 1.50 1.75 2.00 2.25 2.5 3.0 3.5 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 22.09

195.07 198.32 205.76 212.42 218.45 223.99 233.90 242.60 250.40 263.99 275.64 285.88 295.06 303.40 311.06 318.15 324.75 330.93 336.75 342.24 347.44 352.37 357.06 361.54 365.81 369.89 373.80 374.12

Specific volume m3/kg Sat. Sat. liquid vapour of

0.001149 0.001154 0.001166 0.001177 0.001187 0.001197 0.001217 0.001235 0.001252 0.001286 0.001319 0.001351 0.001384 0.001418 0.001452 0.001489 0.001527 0.001567 0.001611 0.001658 0.001711 0.001770 0.001840 0.001924 0.002036 0.002207 0.002742 0.003155

vs

0.14084 0.13177 0.11349 0.09963 0.08875 0.07998 0.06668 0.05707 0.04978 0.03944 0.03244 0.02737 0.02352 0.02048 0.018026 0.015987 0.014263 0.012780 0.011485 0.010337 0.009306 0.008364 0.007489 0.006657 0.005834 0.004952 0.003568 0.003155

Internal energy kJ/kg

Entropy kJ/(kg • K)

Enthalpy kJ/kg

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

Sat. liquid

Evap.

Sat. vapour

of

ufs

us

hf

hfs

hs

sf

sfs

s5

828.70 843.16 876.46 906.44 933.83 959.11 1004.78 1045.43 1082.31 1147.81 1205.44 1257.55 1305.57 1350.51 1393.04 1433.7 1473.0 1511.1 1548.6 1585.6 1622.7 1660.2 1698.9 1739.9 1785.6 1842.1 1961.9 2029.6

1764.1 1751.3 1721.4 1693.8 1668.2 1644.0 1599.3 1558.3 1520.0 1449.3 1384.3 1323.0 1264.2 1207.3 1151.4 1096.0 1040.7 985.0 928.2 869.8 809.0 744.8 675.4 598.1 507.5 388.5 125.2 0

2592.8 2594.5 2597.8 2600.3 2602.0 2603.1 2604.1 2603.7 2602.3 2597.1 2589.7 2580.5 2569.8 2557.8 2544.4 2529.8 2513.7 2496.1 2476.8 2455.5 2431.7 2405.0 2374.3 2338.1 2293.0 2230.6 2087.1 2029.6

1959.7 1947.3 1917.9 1890.7 1865.2 1841.0 1795.7 1753.7 1714.1 1640.1 1571.0 1505.1 1441.3 1378.9 1317.1 1255.5 1193.3 1130.7 1066.5 1000.0 930.6 856.9 777.1 688.0 583.4 446.2 143.4 0

2790.0 2792.2 2796.4 2799.5 2801.7 2803.1 2804.2 2803.4 2801.4 2794.3 2784.3 2772.1 2758.0 2742.1 2724.7 2705.6 2684.9 2662.2 2637.6 2610.5 2580.6 2547.2 2509.1 2464.5 2409.7 2334.6 2165.6 2099.3

2.2842 2.3150 2.3851 2.4474 2.5035 2.5547 2.6457 2.7253 2.7964 2.9202 3.0267 3.1211 3.2068 3.2858 3.3596 3.4295 3.4962 3.5606 3.6232 3.6848 3.7461 3.8079 3.8715 3.9388 4.0139 4.1075 4.3110 4.4298

4.1850 4.1298 4.0044 3.8935 3.7937 3.7028 3.5412 3.4000 3.2737 3.0532 2.8625 2.6922 2.5364 2.3915 2.2544 2.1233 1.9962 1.8718 1.7485 1.6249 1.4994 1.3698 1.2329 1.0839 0.9130 0.6938 0.2216 0

6.4693 6.4448 6.3896 6.3409 6.2972 6.2575 6.1869 6.1253 6.0701 5.9734 5.8892 5.8133 5.7432 5.6722 5.6141 5.5527 5.4924 5.4323 5.3717 5.3098 5.2455 5.1777 5.1044 5.0228 4.9269 4.8013 4.5327 4.4298

830.30 844.89 878.50 908.79 936.49 962.11 1008.42 1049.75 1087.31 1154.23 1213.35 1267.00 1316.64 1363.26 1407.56 1450.1 1491.3 1531.5 1571.1 1610.5 1650.1 1690.3 1732.0 1776.5 1826.3 1888.4 2022.2 2099.3

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 0.01 MPa (45.81°C)* Sat.t 50 100 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

14.674 14.869 17.196 19.512 21.825 24.136 26.445 31.063 35.679 40.295 44.911 49.526 54.141 58.757 63.372 67.987 72.602

2437.9 2443.9 2515.5 2587.9 2661.3 2736.0 2812.1 2968.9 3132.3 3302.5 3479.6 3663.8 3855.0 4053.0 4257.5 4467.9 4683.7

2584.7 2592.6 2687.5 2783.0 2879.5 2977.3 3076.5 3279.6 3489.1 3705.4 3928.7 4159.0 4396.4 4640.6 4891.2 5147.8 5409.7

8.1502 8.1749 8.4479 8.6882 8.9038 9.1002 9.2813 9.6077 9.8978 0.1608 0.4028 0.6281 0.8396 1.0393 1.2287 1.4091 1.5811

0.8857 0.9596 1.0803 1.1988 1.3162 1.5493 1.7814 2.013 2.244 2.475 2.705 2.937 3.168 3.399 3.630

2529.5 2576.9 2654.4 2731.2 2808.6 2966.7 3130.8 3301.4 3478.8 3663.1 3854.5 4052.5 4257.0 4467.5 4683.2

2706.7 2768.8 2870.5 2971.0 3071.8 3276.6 3487.1 3704.0 3927.6 4158.2 4395.8 4640.0 4890.7 5147.5 5409.3

7.1272 7.2795 7.5066 7.7086 7.8926 8.2218 8.5133 8.7770 9.0194 9.2449 9.4566 9.5663 9.8458 10.0262 10.1982

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 0.05 MPa (81.33°C)

u kJ/kg

h $ kJ/kg kJ/(kg • K)

P = 0.10 MPa (99.63°C)

3.240

2483.9

2645.9

7.5939

1.6940

2506.1

2675.5

7.3594

3.418 3.889 4.356 4.820 5.284 6.209 7.134 8.057 8.981 9.904 10.828 11.751 12.674 13.597 14.521

2511.6 2585.6 2659.9 2735.0 2811.3 2968.5 3132.0 3302.2 3479.4 3663.6 3854.9 4052.9 4257.4 4467.8 4683.6

2682.5 2780.1 2877.7 2976.0 3075.5 3278.9 3488.7 3705.1 3928.5 4158.0 4396.3 4640.5 4891.1 5147.7 5409.6

7.6947 7.9401 8.1580 8.3556 8.5373 8.8642 9.1546 9.4178 9.659§ 9.8852 10.0967 10.2964 10.4859 10.6662 10.8382

1.6958 1.9364 2.172 2.406 2.639 3.103 3.565 4.028 4.490 4.952 5.414 5.875 6.337 6.799 7.260

2506.7 2582.8 2658.1 2733.7 2810.4 2967.9 3131.6 3301.9 3479.2 3663.5 3854.8 4052.8 4257.3 4467.7 4683.5

2676.2 2776.4 2875.3 2974.3 3074.3 3278.2 3488.1 3704.4 3928.2 4158.6 4396.1 4640.3 4891.0 5147.6 5409.5

7.3614 7.6134 7.8343 8.0333 8.2158 8.5435 8.8342 9.0976 9.3398 9.5652 9.7767 9.9764 10.1659 10.3463 10.5183

P = 0.20 MPa (120.23°C) Sat. 150 200 250 300 400 500 600 700 800 900 1000 1100 1200 1300

u kJ/kg

P = 0.30 MPa (133.55°C) 0.6058 0.6339 0.7163 0.7964 0.8753 1.0315 1.1867 1.3414 1.4957 1.6499 1.8041 1.9581 2.1121 2.2661 2.4201

2543.6 2570.8 2650.7 2728.7 2806.7 2965.6 3130.0 3300.8 3478.4 3662.9 3854.2 4052.3 4256.8 4467.2 4683.0

2725.3 2761.0 2865.6 2967.6 3069.3 3275.0 3486.0 3703.2 3927.1 4157.8 4395.4 4639.7 4890.4 5147.1 5409.0

The temperature in parentheses is the saturation temperature at the specified pressure.

6.9919 7.0778 7.3115 7.5166 7.7022 8.0330 8.3251 8.5892 8.8319 9.0576 9.2692 9.4690 9.6585 9.8389 10.0110

P = 0.40 MPa (143.63°C) 0.4625 0.4708 0.5342 0.5951 0.6548 0.7726 0.8893 1.0055 1.1215 1.2372 1.3529 1.4685 1.5840 1.6996 1.8151

2553.6 2564.5 2646.8 2726.1 2804.8 2964.4 3129.2 3300.2 3477.9 3662.4 3853.9 4052.0 4256.5 4467.0 4682.8

2738.6 2752.8 2860.5 2964.2 3066.8 3273.4 3484.9 3702.4 3926.5 4157.3 4395.1 4639.4 4890.2 5146.8 5408.8

6.8959 6.9299 7.1706, 7.3789 7.5662 7.8985 8.1913 8.4558 8.6987 8.9244 9.1362 9.3360 9.5256 9.7060 9.8780 Cont.

x?PuaddV

T °C

Superheated water

SS9

Table 5

Table 5 Superheated water (Cont.) T °C

v m3/kg

h s u kJ/kg kJ/kg kJ/(kg • K)

P = 0.50 MPa (151.86°C) Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.3749 2561.2 2748.7 6.8213 0.4249 2642.9 2855.4 7.0592 0.4744 2723.5 2960.7 7.2709 0.5226 2802.9 3064.2 7.4599 0.5701 2882.6 3167.7 7.6329 0.6173 2963.2 3271.9 7.7938 0.7109 3128.4 3483.9 8.0873 0.8041 3299.6 3701.7 7.3522 0.8969 3477.5 3925.9 8.5952 0.9896 3662.1 4156.9 8.8211 1.0822 3853.6 4394.7 9.0329 1.1747 4051.8 4639.1 9.2328 1.2672 4256.3 4889.9 9.4224 1.3596 4466.8 5146.6 9.6029 1.4521 4682.5 5408.6 9.7749

v m3/kg

u h s v kJ/kg kJ/kg kJ/(kg • K) m3/kg

P = 0.60 MPa (158.85°C) 0.3157 2567.4 2756.8 6.7600 0.3520 2638.9 2850.1 6.9665 0.3938 2720.9 2957.2 7.1816 0.4344 2801.0 3061.6 7.3724 0.4742 2881.2 3165.7 7.5464 0.5137 2962.1 32703 7.7079 0.5920 3127.6 3482.8 8.0021 0.6697 3299.1 3700.9 8.2674 0.7472 3477.0 3925.3 8.5107 0.8215 3661.8 4156.5 8.7367 0.9017 3853.4 4394.4 8.9486 0.9788 4051.5 4638.8 9.1485 1.0559 4256.1 4889.6 9.3381 1.1330 4466.5 5146.3 9.5185 1.2101 4682.3 5408.3 9.6906

P = 1.00 MPa (179.91°C) Sat. 200 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.19444 2583.6 2778.1 6.5865 0.2060 2621.9 2827.9 6.6940 0.2327 2709.9 2942.6 6.9247 0.2579 2793.2 3051.2 7.1229 0.2825 2875.2 3157.7 7.3011 0.3066 2957.3 3263.9 7.4651 0.3541 3124.4 3478.5 7.7622 0.4011 3296.8 3697.9 8.0290 0.4478 3475.3 3923.1 8.2731 0.4943 3660.4 4154.7 8.4996 0.5407 3852.2 4392.9 8.7118 0.5871 4050.5 4637.6 8.9119 0.6335 4255.1 4888.6 9.1017 0.6798 4465.6 5145.4 9.2822 0.7261 4681.3 5407.4 9.4543

P = 1.20 MPa (187.99°C) 0.16333 0.16930 0.19234 0.2138 0.2345 0.2548 0.2946 0.3339 0.3729 0.4118 0.4505 0.4892 0.5278 0.5665 0.6051

2588.8 2612.8 2704.2 2789.2 2872.2 2954.9 3122.8 3295.6 3474.4 3659.7 3851.6 4050.0 4254.6 4465.1 4680.9

2784.8 2815.9 2935.0 3045.8 3153.6 3260.7 3476.3 3696.3 3922.0 4153.8 4392.2 4637.0 4888.0 5144.9 5407.0

6.5233 6.5898 6.8294 7.0317 7.2121 7.3774 7.6759 7.9435 8.1881 8.4148 8.6272 8.8274 9.0172 9.1977 9.3698

u h s kJ/kg kJ/kg kJ/(kg • K)

P = 0.80 MPa (170.43°C) 0.2404 0.2608 0.2931 0.3241 0.3544 0.3843 0.4433 0.5018 0.5601 0.6181 0.6761 0.7340 0.7919 0.8497 0.9076

2576.8 2630.6 2715.5 2797.2 2878.2 2959.7 3126.0 3297.9 3476.2 3661.1 3852.8 4051.0 4255.6 4466.1 4681.8

2769.1 2839.3 2950.0 3056.5 3161.7 3267.1 3480.6 3699.4 3924.2 4155.6 4393.7 4638.2 4889.1 5145.9 5407.9

6.6628 6.8158 7.0384 7.2328 7.4089 7.5716 7.8673 8.1333 8.3770 8.6033 8.8153 9.0153 9.2050 9.3855 9.5575

P = 1.40 MPa (195.70°C) 0.14084 0.14302 0.16350 0.18228 0.2003 0.2178 0.2521 0.2860 0.3195 0.3528 0.3861 0.4192 0.4524 0.4855 0.5186

2592.8 2603.1 2698.3 2785.2 2869.2 2952.5 3121.1 3294.4 3473.6 3659.0 3851.1 4049.5 4254.1 4464.7 4680.4

2790.0 2803.3 2927.2 3040.4 3149.5 3257.5 3474.1 3694.8 3920.8 4153.0 4391.5 4636.4 4887.5 5144.4 5406.5

6.4693 6.4975 6.7467 6.9534 7.1360 7.3026 7.6027 7.8710 8.1160 8.3431 8.5556 8.7559 8.9457 9.1262 9.2984 Cont.

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 1.60 MPa (201.41°C) Sat 225 250 300 350 400 500 600 700 800 900 1000 1100 1200 1300

0.12380 0.13287 0.14184 0.15862 0.17456 0.19005 0.2203 0.2500 0.2794 0.3086 0.3377 0.3668 0.3958 0.4248 0.4538

2596.0 2644.7 2692.3 2781.1 2866.1 2950.1 3119.5 3293.3 3472.7 3658.3 3850.5 4049.0 4253.7 4464.2 4679.9

2794.0 2857.3 2919.2 3034.8 3145.4 3254.2 3472.0 3693.2 3919.7 4152.1 4390.8 4635.8 4887.0 5143.9 5406.0

6.4218 6.5518 6.6732 6.8844 7.0694 7.2374 7.5390 7.8080 8.0535 8.2808 8.4935 8.6938 8.8837 9.0643 9.2364

0.07998 0.08027 0.08700 0.09890 0.10976 0.12010 0.13014 0.13993 0.15930 0.17832 0.19716 0.21590 0.2346 0.2532 0.2718 0.2905

2603.1 2605.6 2662.6 2761.6 2851.9 2939.1 3025.5 3112.1 3288.0 3468.7 3655.3 3847.9 4046.7 4251.5 4462.1 4677.8

2803.1 2806.3 2880.1 3008.8 3126.3 3239.3 3350.8 3462.1 3686.3 3914.5 4148.2 4387.6 4633.1 4884.6 5141.7 5404.0

6.2575 6.2639 6.4085 6.6438 6.8403 7.0148 7.1746 7.3234 7.5960 7.8435 8.0720 8.2853 8.4861 8.6762 8.8569 9.0291

h kJ/kg

s kJ/(kg • K)

P = 1.80 MPa (207.15°C) 0.11042 0.11673 0.12497 0.14021 0.15457 0.16847 0.19550 0.2220 0.2482 0.2742 0.3001 0.3260 0.3518 0.3776 0.4034

2598.4 2636.6 2686.0 2776.9 2863.0 2947.7 3117.9 3292.1 3471.8 3657.6 3849.9 4048.5 4253.2 4463.7 4679.5

2797.1 2846.7 2911.0 3029.2 3141.2 3250.9 3469.8 3691.7 3971.8 4151.2 4390.1 4635.2 4886.4 5143.4 5405.6

6.3794 6.4808 6.6066 6.8226 7.0100 7.1794 7.4825 7.7523 7.9983 8.2258 8.4386 8.6391 8.8290 9.0096 9.1818

P = 3.00 MPa (233.90°C)

P = 2.50 MPa (223.99°C) Sat. 225 250 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

u kJ/kg

v m3/kg

u kJ/kg

h s kJ/kg kJ/(kg • K)

P = 2.00 MPa (212.42°C) 0.09963 0.10377 0.11144 0.12547 0.13857 0.15120 0.17568 0.19960 0.2232 0.2467 0.2700 0.2933 0.3166 0.3398 0.3631

P

2600.3 2628.3 2679.6 2772.6 2859.8 2945.2 3116.2 3290.9 3470.9 3657.0 3849.3 4048.0 4252.7 4463.3 4679.0

2799.5 2835.8 2902.5 3023.5 3137.0 3247.6 3467.6 3690.1 3917.4 4150.3 4389.4 4634.6 4885.9 5142.9 5405.1

6.3409 6.4147 6.5453 6.7664 6.9563 7.1271 7.4317 7.7024 7.9487 8.1765 8.3895 8.5901 8.7800 8.9607 9.1329

3.50 MPa (242.60°C)

0.06668

2604.1

2804.2

6.1869

0.05707

2603.7

2803.4

6.1253

0.07058 0.08114 0.09053 0.09936 0.10787 0.11619 0.13243 0.14838 0.16414 0.17980 0.19541 0.21098 0.22652 0.24206

2644.0 2750.1 2843.7 2932.8 3020.4 3108.0 3285.0 3466.5 3653.5 3846.5 4045.4 4250.3 4460.9 4676.6

2855.8 2993.5 3115.3 3230.9 3344.0 3456.5 3682.3 3911.7 4145.9 4385.9 4631.6 4883.3 5140.5 5402.8

6.2872 6.5390 6.7428 6.9212 7.0834 7.2338 7.5085 7.7571 7.9862 8.1999 8.4009 8.5912 8.7720 8.9442

0.05872 0.06842 0.07678 0.08453 0.09196 0.09918 0.11324 0.12699 0.14056 0.15402 0.16743 0.18080 0.19415 0.20749

2623.7 2738.0 2835.3 2926.4 3015.3 3103.0 3282.1 3464.3 3651.8 3845.0 4044.1 4249.2 4459.8 4675.5

2829.2 2977.5 3104.0 3222.3 3337.2 3450.9 3678.4 3908.8 4143.7 4384.1 4630.1 4881.9 5139.3 5401.7

6.1749 6.4461 6.6579 6.8405 7.0052 7.1572 7.4339 7.6837 7.9134 8.1276 8.3288 8.5192 8.7000 8.8723 Cont.

mpuaddy

T °C

Superheated water (Cont.)

is9

Table 5

Table 5 Superheated water (Cont.) T °C

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 4.0 MPa (250.40°C) Sat. 275 300 350 400 450 500 600 700 800 900 1000 1100 1200 1300

0.04978 0.05457 0.05884 0.06645 0.07341 0.08002 0.08643 0.09885 0.11095 0.12287 0.13469 0.14645 0.15817 0.16987 0.18156

2602.3 2667.9 2725.3 2826.7 2919.9 3010.2 3099.5 3279.1 3462.1 3650.0 3843.6 4042.9 4248.0 4458.6 4674.3

2801.4 2886.2 2960.7 3092.5 3213.6 3330.3 3445.3 3674.4 3905.9 4141.5 4382.3 4628.7 4880.6 5138.1 5400.5

6.0701 6.2285 6.3615 6.5821 6.7690 6.9363 7.0901 7.3688 7.6198 7.8502 8.0647 8.2662 8.4567 8.6376 8.8100

0.03244 0.03616 0.04223 0.04739 0.05214 0.05665 0.06101 0.06525 0.07352 0.08160 0.08958 0.09749 0.10536 0.11321 0.12106

2589.7 2667.2 2789.6 2892.9 2988.9 3082.2 3174.6 3266.9 3453.1 3643.1 3837.8 4037.8 4243.3 4454.0 4669.6

2784.3 2884.2 3043.0 3177.2 3301.8 3422.2 3540.6 3658.4 3894.2 4132.7 4375.3 4622.7 4875.4 5133.3 5396.0

5.8892 6.0674 6.3335 6.5408 6.7193 6.8803 7.0288 7.1677 7.4234 7.6566 7.8727 8.0751 8.2661 8.4474 8.6199

h kJ/kg

s kJ/(kg • K)

P = 4.5 MPa (257.49°C) 0.04406 0.04730 0.05135 0.05840 0.06475 0.07074 0.07651 0.08765 0.09847 0.10911 0.11965 0.13013 0.14056 0.15098 0.16139

P = 6.0 MPa (275.64°C) Sat. 300 350 400 450 500 550 600 700 800 900 1000 1100 1200 1300

u kJ/kg

2600.1 2650.3 2712.0 2817.8 2913.3 3005.0 3095.3 3276.0 3459.9 3648.3 3842.2 4041.6 4246.8 4457.5 4673.1

2798.3 2863.2 2943.1 3080.6 3204.7 3323.3 3439.6 3670.5 3903.0 4139.3 4380.6 4627.2 4879.3 5136.9 5399.4

6.0198 6.1401 6.2828 6.5131 6.7047 6.8746 7.0301 7.3110 7.5631 7.7942 8.0091 8.2108 8.4015 8.5825 8.7549

P = 7.0 MPa (285.88°C) 0.02737 0.02947 0.03524 0.03993 0.04416 0.04814 0.05195 0.05565 0.06283 0.06981 0.07669 0.08350 0.09027 0.09703 0.10377

2580.5 2632.2 2769.4 2878.6 2978.0 3073.4 3167.2 3260.7 3448.5 3639.5 3835.0 4035.3 4240.9 4451.7 4667.3

2772.1 2838.4 3016.0 3458.1 3287.1 3410.3 3530.9 3650.3 3888.3 4128.2 4371.8 4619.8 4872.8 5130.9 5393.7

5.8133 5.9305 6.2283 6.4478 6.6327 6.7975 6.9486 7.0894 7.3476 7.5822 7.7991 8.0020 8.1933 8.3747 8.5475

v m3/kg

u kJ/kg

h s kJ/kg kJ/(kg • K)

P = 5.0 MPa (263.99°C) 0.03944 0.04141 0.04532 0.05194 0.05781 0.06330 0.06857 0.07869 0.08849 0.09811 0.10762 0.11707 0.12648 0.13587 0.14526

2597.1 2631.3 2698.0 2808.7 2906.6 2999.7 3091.0 3273.0 3457.6 3646.6 3840.7 4040.4 4245.6 4456.3 4672.0

2794.3 2838.3 2924.5 3068.4 3195.7 3316.2 3433.8 3666.5 3900.1 4137.1 4378.8 4625.7 4878.0 5135.7 5398.2

5.9734 6.0544 6.2084 6.4493 6.6459 6.8186 6.9759 7.2589 7.5122 7.7440 7.9593 8.1612 8.3520 8.5331 8.7055

P = 8.0 MPa (295.06°C) 0.02352 0.02426 0.02995 0.03432 0.03817 0.04175 0.04516 0.04845 0.05481 0.06097 0.06702 0.07301 0.07896 0.08489 0.09080

2569.0 2590.9 2747.7 2863.8 2966.7 3064.3 3159.8 3254.4 3443.9 3636.0 3832.1 4032.8 4238.6 4449.5 4665.0

2758.0 2785.0 2987.3 3138.3 3272.0 3398.3 3521.0 3642.0 3882.4 4123.8 4368.3 4616.9 4870.3 5128.5 5391.5

5.7432 5.7906 6.1301 6.3634 6.5551 6.7240 6.8778 7.0206 7.2812 7.5173 7.7351 7.9384 8.1300 8.3115 8.4842 Cont.

Table 5 Superheated water (Cont.) T °C

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

P = 9.0 MPa (303.40°C) Sat. 325 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.02048 0.02327 0.02580 0.02993 0.03350 0.03677 0.03987 0.04285 0.04574 0.04857 0.05409 0.05950 0.06485 0.07016 0.07544 0.08072

2557.8 2646.6 2724.4 2848.4 2955.2 3055.2 3152.2 3248.1 3343.6 3439.3 3632.5 3829.2 4030.3 4236.3 4447.2 4662.7

2742.1 2856.0 2956.6 3117.8 3256.6 3386.1 3511.0 3633.7 3755.3 3876.5 4119.3 4364.8 4614.0 4867.7 5126.2 5389.2

5.6772 5.8712 6.0361 6.2854 6.4844 6.6576 6.8142 6.9589 7.0943 7.2221 7.4596 7.6783 7.8821 8.0740 8.2556 8.4284

P = 15.0 MPa (342.24°C) Sat. 350 400 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.010337 0.011470 0.015649 0.018445 0.02080 0.02293 0.02491 0.02680 0.02861 0.03210 0.03546 0.03875 0.04200 0.04523 0.04845

2455.5 2520.4 2740.7 2879.5 2996.6 3104.7 3208.6 3310.3 3410.9 3610.9 3811.9 4015.4 4222.6 4433.8 4649.1

2610.5 2692.4 2975.5 3156.2 3308.6 3448.6 3582.3 3712.3 3840.1 4092.4 4343.8 4596.6 4852.6 5112.3 5376.0

5.3098 5.4421 5.8811 6.1404 6.3443 6.5199 6.6776 6.8224 6.9572 7.2040 7.4279 7.6348 7.8283 8.0108 8.1840

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

P = 10.0 MPa (311.06°C) 0.018026 0.019861 0.02242 0.02641 0.02975 0.03279 0.03564 0.03837 0.04101 0.04358 0.04859 0.05349 0.05832 0.06312 0.06789 0.07265

2544.4 2610.4 2699.2 2832.4 2943.4 3045.8 3144.6 3241.7 3338.2 3434.7 3628.9 3826.3 4027.8 4234.0 4444.9 4460.5

2724.7 2809.1 2923.4 3096.5 3240.9 3373.7 3500.9 3625.3 3748.2 3870.5 4114.8 4361.2 4611.0 4865.1 5123.8 5387.0

5.6141 5.7568 5.9443 6.2120 6.4190 6.5966 6.7561 6.9029 7.0398 7.1687 7.4077 7.6272 7.8315 8.0237 8.2055 8.3783

P = 17.5 MPa (354.75°C)

v m3/kg

u kJ/kg

h s kJ/kg kJ/(kg • K)

P = 12.5 MPa (327.89°C) 0.013495

2505.1

2673.8

.5.4624

0.016126 0.02000 0.02299 0.02560 0.02801 0.03029 0.03248 0.03460 0.03869 0.04267 0.04658 0.05045 0.05430 0.05813

2624.6 2789.3 2912.5 3021.7 3125.0 3225.4 3324.4 3422.9 3620.0 3819.1 4021.6 4228.2 4439.3 4654.8

2826.2 3039.3 3199.8 3341.8 3475.2 3604.0 3730.4 3855.3 4103.6 4352.5 4603.8 4858.8 5118.0 5381.4

5.7118 6.0417 6.2719 6.4618 6.6290 6.7810 6.9218 7.0536 7.2965 7.5182 7.7237 7.9165 8.0937 8.2717

P = 20.0 MPa (365.81°C)

0.007920

2390.2

2528.8

5.1419

0.005834

2293.0

2409.7

4.9269

0.012447 0.015174 0.017358 0.019288 0.02106 0.02274 0.02434 0.02738 0.03031 0.03316 0.03597 0.03876 0.04154

2685.0 2844.2 2970.3 3083.9 3191.5 3296.0 3398.7 3601.8 3804.7 4009.3 4216.9 4428.3 4643.5

2902.9 3109.7 3274.1 3421.4 3560.1 3693.9 3824.6 4081.1 4335.1 4589.5 4846.4 5106.6 5370.5

5.7213 6.0184 6.2383 6.4230 6.5866 6.7357 6.8736 7.1244 7.3507 7.5589 7.7531 7.9360 8.1093

0.009942 0.012695 0.014768 0.016555 0.018178 0.019693 0.02113 0.02385 0.02645 0.02897 0.03145 0.03391 0.03636

2619.3 2806.2 2942.9 3062.4 3174.0 3281.4 3386.4 3592.7 3797.5 4003.1 4211.3 4422.8 4638.0

2818.1 3060.1 3238.2 3393.5 3537.6 3675.3 3809.0 4069.7 4326.4 4582.5 4840.2 5101.0 5365.1

5.5540 5.9017 6.1401 6.3348 6.5048 6.6582 6.7993 7.0544 7.2830 7.4925 7.6874 7.8707 8.0442 Cont.

Table S Superheated water (Cont.)

T °C

v m3/itg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 25.0 MPa 375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

0.0019731 0.006004 0.007881 0.009162 0.011123 0.012724 0.014137 0.015433 0.016646 0.018912 0.021045 0.02310 0.02512 0.02711 0.02910

1798.7 2430.1 2609.2 2720.7 2884.3 3017.5 3137.9 3251.6 3361.3 3574.3 3783.0 3990.9 4200.2 4412.0 4626.9

1848.0 2580.2 2806.3 2949.7 3162.4 3335.6 3491.4 3637.4 3777.5 4047.1 4309.1 4568.5 4828.2 5089.9 5354.4

0.0016407 0.0019077 0.002532 0.003693 0.005622 0.006984 0.008094 0.009063 0.009941 0.011523 0.012962 0.014324 0.015642 0.016940 0.018229

1677.1 1854.6 2096.9 2365.1 2678.4 2869.7 3022.6 3158.0 3283.6 3517.8 3739.4 3954.6 4167.4 4380.1 4594.3

1742.8 1930.9 2198.1 2512.8 2903.3 3149.1 3346.4 3520.6 3681.2 3978.7 4257.9 4527.6 4793.1 5057.7 5323.5

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 30.0 MPa 4.0320 5.1418 5.4723 5.6744 5.9592 6.1765 6.3602 6.5229 6.6707 6.9345 7.1680 7.3802 7.5765 7.7605 7.9342

0.0017892 0.002790 0.005303 0.006735 0.008678 0.010168 0.011446 0.012596 0.013661 0.015623 0.017448 0.019196 0.020903 0.022589 0.024266

P = 40.0 MPa 375 400 425 450 500 550 600 650 700 800 900 1000 1100 1200 1300

u kJ/kg

1737.8 2067.4 2455.1 2619.3 2820.7 2970.3 3100.5 3221.0 3335.8 3555.5 3768.5 3978.8 4189.2 4401.3 4616.0

1791.5 2151.1 2614.2 2821.4 3081.1 3275.4 3443.9 3598.9 3745.6 4024.2 4291.9 4554.7 4816.3 5079.0 5344.0

0.0015594 0.0017309 0.002007 0.002486 0.003892 0.005118 0.006112 0.006966 0.007727 0.009076 0.010283 0.011411 0.012496 0.013561 0.014616

1638.6 1788.1 1959.7 2159.6 2525.5 2763.6 2942.0 3093.5 3230.5 3479.8 3710.3 3930.5 4145.7 4359.1 4572.8

1716.6 1874.6 2060.0 2284.0 2720.1 3019.5 3247.6 3441.8 3616.8 3933.6 4224.4 4501.1 4770.5 5037.2 5303.6

h kJ/kg

s kJ/(kg • K)

P = 35.0 MPa 3.9305 4.4728 5.1504 5.4424 5.7905 6.0342 6.2331 6.4058 6.5606 6.8332 7.0718 7.2867 7.4845 7.6692 7.8432

0.0017003 0.002100 0.003428 0.004961 0.006927 0.008345 0.009527 0.010575 0.011533 0.013278 0.014883 0.016410 0.017895 0.019360 0.020815

P = 50.0 MPa 3.8290 4.1135 4.5029 4.9459 5.4700 5.7785 6.0144 6.2054 6.3750 6.6662 6.9150 7.1356 7.3364 7.5224 7.6969

u kJ/kg

1702.9 1914.1 2253.4 2498.7 2751.9 2921.0 3062.0 3189.8 3309.8 3536.7 3754.0 3966.7 4178.3 4390.7 4605.1

1762.4 1987.6 2373.4 2672.4 2994.4 3213.0 3395.5 3559.9 3713.5 4001.5 4274.9 4541.1 4804.6 5068.3 5333.6

3.8722 4.2126 4.7747 5.1962 5.6282 5.9026 6.1179 6.3010 6.4631 6.7450 6.9386 7.2064 7.4037 7.5910 7.7653

P = 60.0 MPa 3.7639 4.0031 4.2734 4.5884 5.1726 5.5485 5.8178 6.0342 6.2189 6.5290 6.7882 7.0146 7.2184 7.4058 7.5808

0.0015028 0.0016335 0.0018165 0.002085 0.002956 0.003956 0.004834 0.005595 0.006272 0.007459 0.008508 0.009480 0.010409 0.011317 0.012215

1609.4 1745.4 1892.7 2053.9 2390.6 2658.8 2861.1 3028.8 3177.2 3441.5 3681.0 3906.4 4124.1 4338.2 4551.4

1699.5 1843.4 2001.7 2179.0 2567.9 2896.2 3151.2 3364.5 3553.5 3889.1 4191.5 4475.2 4748.6 5017.2 5284.3

•The temperature in parentheses is the saturation temperature at the specified pressure. 'Properties of saturated vapour at the specified pressure. Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, V.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

3.7141 3.9318 4.1626 4.4121 4.9321 5.3441 5.6452 5.8829 6.0824 6.4109 6.6805 6.9127 7.1195 7.3083 7.4837

Table 6 Compressed liquid water v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

P = 5 MPa (263.99°C) Sat. 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340

0.0012859 0.0009977 0.0009995 0.0010056 0.0010149 0.0010268 0.0010410 0.0010576 0.0010768 0.0010988 0.0011240 0.0011530 0.0011866 0.0012264 0.0012749

1147.8 1154.2 0.04 5.04 88.65 83.65 166.95 171.97 250.23 255.30 333.72 338.85 417.52 422.72 501.80 507.09 586.76 592.15 672.62 678.12 759.63 765.25 848.1 853.9 938.4 944.4 1031.4 1037.5 1127.9 1134.3

2.9202 0.0001 0.2956 0.5705 0.8285 1.0720 1.3030 1.5233 1.7343 1.9375 2.1341 2.3255 2.5128 2.6979 2.8830

u kJ/kg

v m3/kg

h kJ/kg

s kJ/(kg • K)

P = 10 MPa (311.06°C) 0.0014524 0.0009952 0.0009972 0.0010034 0.0010127 0.0010245 0.0010385 0.0010549 0.0010737 0.0010953 0.0011199 0.0011480 0.0011805 0.0012187 0.0012645 0.0013216 0.0013972

1393.0 0.09 83.36 166.35 249.36 332.59 416.12 500.08 584.68 670.13 756.65 844.5 934.1 1026.0 1121.1 1220.9 1328.4

1407.6 10.04 93.33 176.38 259.49 342.83 426.50 510.64 595.42 681.08 767.84 856.0 945.9 1038.1 1133.7 1234.1 1342.3

v m3/kg

u kJ/kg

h s kJ/kg kJ/(kg • K)

P = 15 MPa (342.24°C) 3.3596 0.0002 0.2945 0.5686 0.8258 1.0688 1.2992 1.5189 1.7292 1.9317 2.1275 2.3178 2.5039 2.6872 2.8699 3.0548 3.2469

0.0016581 0.0009928 0.0009950 0.0010013 0.0010105 0.0010222 0.0010361 0.0010522 0.0010707 0.0010918 0.0011159 0.0011433 0.0011748 0.0012114 0.0012550 0.0013084 0.0013770 0.0014724 0.0016311

1585.6 0.15 83.06 165.76 248.51 331.48 414.74 498.40 582.66 667.71 753.76 841.0 929.9 1020.8 1114.6 1212.5 1316.6 1431.1 1567.5

1610.5 15.05 97.99 180.78 263.67 346.81 430.28 514.19 598.72 684.09 770.50 858.2 947.5 1039.0 1133.4 1232.1 1337.3 1453.2 1591.9

3.6848 0.0004 0.2934 0.5666 0.8232 1.0656 1.2955 1.5145 1.7242 1.9260 2.1210 2.3104 2.4953 2.6771 2.8576 3.0393 3.2260 3.4247 3.6546 Cont.

T99 mpsiaddv

T °C

Table 6 T °C

v m3/kg

u kJ/kg

h kJ/kg

s kJ/(kg • K)

Compressed liquid water (Cont.) v m3/kg

P = 20.0 MPa (365.81°C) Sat. 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300 320 340 360 380

0.002036 0.0009904 0.0009928 0.0009992 0.0010084 0.0010199 0.0010337 0.0010496 0.0010678 0.0010885 0.0011120 0.0011388 0.0011695 0.0012046 0.0012462 0.0012965 0.0013596 0.0014437 0.0015684 0.0018226

1785.6 0.19 82.77 165.17 247.68 330.40 413.39 496.76 580.69 665.35 750.95 837.7 925.9 1016.0 1108.6 1204.7 1306.1 1415.7 1539.7 1702.8

1826.3 20.01 102.62 185.16 267.85 350.80 434.06 517.76 602.04 687.12 773.20 860.5 949.3 1040.0 1133.5 1230.6 1333.3 1444.6 1571.0 1739.3

4.0139 0.0004 0.2923 0.5646 0.8206 1.0624 1.2917 1.5102 1.7193 1.9204 2.1147 2.3031 2.4870 2.6674 2.8459 3.0248 3.2071 3.3979 3.6075 3.8772

u kJ/kg

h kJ/kg

s kJ/(kg • K)

v m3/kg

P = 30 MPa 0.0009856 0.0009886 0.0009951 0.0010042 0.0010156 0.0010290 0.0010445 0.0010621 0.0010821 0.0011047 0.0011302 0.0011590 0.0011920 0.0012303 0.0012755 0.0013304 0.0013997 0.0014920 0.0016265 0.0018691

0.25 82.17 164.04 246.06 328.30 410.78 493.59 576.88 660.82 745.59 831.4 918.3 1006.9 1097.4 1190.7 1287.9 1390.7 1501.7 1626.6 1781.4

29.82 111.84 193.89 276.19 358.77 441.66 524.93 608.75 693.28 778.73 865.3 953.1 1042.6 1134.3 1229.0 1327.8 1432.7 1546.5 1675.4 1837.5

u kJ/kg

h kJ/kg

s kJ/(kg • K)

P = 50 MPa 0.0001 0.2899 0.5607 0.8154 1.0561 1.2844 1.5018 1.7098 1.9096 2.1024 2.2893 2.4711 2.6490 2.8243 2.9986 3.1741 3.3539 3.5426 3.7494 4.0012

0.0009766 0.0009804 0.0009872 0.0009962 0.0010073 0.0010201 0.0010348 0.0010515 0.0010703 0.0010912 0.0011146 0.0011408 0.0011702 0.0012034 0.0012415 0.0012860 0.0013388 0.0014032 0.0014838 0.0015884

0.20 81.00 161.86 242.98 324.34 405.88 487.65 569.77 652.41 735.69 819.7 904.7 990.7 1078.1 1167.2 1258.7 1353.3 1452.0 1556.0 1667.2

49.03 130.02 211.21 292.79 374.70 456.89 539.39 622.35 705.92 790.25 875.5 961.7 1049.2 1138.2 1229.3 1323.0 1420.2 1522.1 1630.2 1746.6

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

0.0014 0.2848 0.5527 0.8052 1.0440 1.2703 1.4857 1.6915 1.8891 2.0794 2.2634 2.4419 2.6158 2.7860 2.9537 3.1200 3.2868 3.4557 3.6291 3.8101

Table 7 Saturated ice-water vapour

Temp. T °C 0.01 0 -2 -4 -6 -8 -10 -12 -14 -16 -18 -20 -22 -24 -26 -28 -30 -32 -34 -36 -38 -40

Sat. press. P., kPa

Sat. ice vi x 103

0.6113 0.6108 0.5176 0.4375 0.3689 0.3102 0.2602 0.2176 0.1815 0.1510 0.1252 0.1035 0.0853 0.0701 0.0574 0.0469 0.0381 0.0309 0.0250 0.0201 0.0161 0.0129

1.0908 1.0908 1.0904 1.0901 1.0898 1.0894 1.0891 1.0888 1.0884 1.0881 1.0878 1.0874 1.0871 1.0868 1.0864 1.0861 1.0858 1.0854 1.0851 1.0848 1.0844 1.0841

Sat. vapour vg 206.1 206.3 241.7 283.8 334.2 394.4 466.7 553.7 658.8 786.0 940.5 1128.6 1358.4 1640.1 1986.4 2413.7 2943 3600 4419 5444 6731 8354

Sat. ice ui

Subl. uis,

-333.40 -333.43 -337.62 -341.78 -345.91 -350.02 -354.09 -358.14 -362.15 -366.14 -370.10 -374.03 -377.93 -381.80 -385.64 -389.45 -393.23 -396.98 -400.71 -404.40 -408.06 -411.70

2708.7 2708.8 2710.2 2711.6 2712.9 2714.2 2715.5 2716.8 2718.0 2719.2 2720.4 2721.6 2722.7 2723.7 2724.8 2725.8 2726.8 2727.8 2728.7 2729.6 2730.5 2731.3

Sat. vapour ug 2375.3 2375.3 2372.6 2369.8 2367.0 2364.2 2361.4 2358.7 2355.9 2353.1 2350.3 2347.5 2344.7 2342.0 2339.2 2336.4 2333.6 2330.8 2328.0 2325.2 2322.4 2319.6

Entropy kJ/(kg • K)

Enthalpy kJ/kg

Internal energy kJ/kg Sat. ice hi -333.40 -333.43 -337.62 -341.78 -345.91 -350.02 -354.09 -358.14 -362.15 -366.14 -370.10 -374.03 -377.93 -381.80 -385.64 -389.45 -393.23 -396.98 -400.71 -404.40 -408.06 -411.70

Subi. hig

Sat. vapour hg

Sat. ice Si

Subl. gig

2834.8 2834.8 2835.3 2835.7 2836.2 2836.6 2837.0 2837.3 2837.6 2837.9 2838.2 2838.4 2838.6 2838.7 2838.9 2839.0 2839.0 2839.1 2839.1 2839.1 2839.0 2839.9

2501.4 2501.3 2497.7 2494.0 2490.3 2486.6 2482.9 2479.2 2475.5 2471.8 2468.1 2464.3 2460.6 2456.9 2453.2 2449.5 2445.8 2442.1 2438.4 2434.7 2430.9 2427.2

-1.221 -1.221 -1.237 -1.253 -1.268 -1.284 -1.299 -1.315 -1.331 -1.346 -1.362 -1.377 -1.393 -1.408 -1.424 -1.439 -1.455 -1.471 -1.486 -1.501 -1.517 -1.532

10.378 10.378 10.456 10.536 10.616 10.698 10.781 10.865 10.950 11.036 11.123 11.212 11.302 11.394 11.486 11.580 11.676 11.773 11.872 11.972 12.073 12.176

Sat. vapour sg 9.156 9.157 9.219 9.283 9.348 9.414 9.481 9.550 9.619 9.690 9.762 9.835 9.909 9.985 10.062 10.141 10.221 10.303 10.386 10.470 10.556 10.644

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, ,Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

ddv £99 AV ua

Specific volume m3/kg

664 Appendix Table 8 Ideal-gas properties of air T

h

K

kJ/kg

Pr

14'

vr

kJ/kg

-o

T

h

kJ/(kg • K)

K

kJ/kg

Pr

u

yr

kJ/kg

-o kJ/(kg • K)

200

199.97 0.3363 142.56 1707.0

1.29559

450

451.80

5.775 322.62 223.6

2.11161

210

209.97 0.3987 149.69 1512.0

1.34444

460

462.02

6.245 329.97 211.4

2.13407

220

219.97 0.4690 156.82 1346.0

1.39105

470

472.24

6.742 337.32 200.1

2.15604

230

230.02 0.5477 164.00 1205.0

1.43557

480

482.49

7.268 344.70 189.5

2.17760

240

240.02 0.6355 171.13 1084.0

1.47824

490

492.74

7.824 352.08 179.7

2.19876

250

250.05 0.7329 178.28

979.0

1.51917

500

503.02

8.411 359.49 170.6

2.21952

260

260.09 0.8405 185.45

887.8

1.55848

510

513.32

9.031 366.92 162.1

2.23993

270

270.11 0.9590 192.60

808.0

1.59634

520

523.63

9.684 374.36 154.1

2.25997

280

280.13 1.0889 199.75

738.0

1.63279

530

533.98 10.37

381.84 146.7

2.27967

285

285.14 1.1584 203.33

706.1

1.65055

540

544.35 11.10

389.34 139.7

2.29906

290

290.16 1.231 I

206.91

676.1

1.66802

550

554.74 11.86

396.86 133.1

2.31809

295

295.17 1.3068 210.49

647.9

1.68515

560

565.17 12.66

404.42 127.0

2.33685

300

300.19 1.3860 214.07

621.2

1.70203

570

575.59 13.50

411.97 121.2

2.35531

305

305.22 1.4686 217.67

596.0

1.71865

580

586.04 14.38

419.55 115.7

2.37348

310

310.24 1.5546 221.25

572.3

1.73498

590

596.52 15.31

427.15 110.6

2.39140

315

315.27 1.6442 224.85

549.8

1.75106

600

607.02 16.28

434.78 105.8

2.40902

320

320.29 1.7375 228.42

528.6

1.76690

610

617.53 17.30

442.42 101.2

2.42644

325

325.31 1.8345 232.02

508.4

1.78249

620

628.07 18.36

450.09

96.92 2.44356

330

330.34 1.9352 235.61

489.4

1.79783

630

638.63 19.84

457.78

92.84 2.46048

340

340.42

2.149

242.82

454.1

1.82790

640

649.22 20.64

465.50

88.99 2.47716

350

350.49

2.379

250.02

422.2

1.85708

650

659.84 21.86

473.25

85.34 2.49364

360

360.58

2.626

257.24

393.4

1.88543

660

670.47 23.13

481.01

81.89 2.50985

370

370.67

2.892

264.46

367.2

1.91313

670

681.14 24.46

488.81

78.61 2.52589

380 ' 380.77

3.176

271.69

343.4

1.94001

680

691.82 25.85

496.62

75.50 2.54175

390

390.88

3.481

278.93

321.5

1.96633

690

702.52 27.29

504.45

72.56 2.55731

400

400.98

3.806

286.16

301.6

1.99194

700

713.27 28.80

512.33

69.76 2.57277

410

411.12

4.153

293.43

283.3

2.01699

710

724.04 30.38

520.23

67.07 2.58810

420

421.26

4.522

300.69

266.6

2.04142

720

734.82 32.02

528.14

64.53 2.60319

430

431.43

4.915

307.99

251.1

2.06533

730

745.62 33.72

536.07

62.13 2.61803

440

441.61

5.332

315.30

236.8

2.08870

740

756.44 35.50

544.02

59.82 2.63280 Cont.

Appendix 665 Table 8 v,

Ideal-gas properties of air (Cont.) -o

s

T

h

kJ/(kg • K)

K

kJ/kg

-o

T K

h kJ/kg

750

767.29

37.35 551.99

57.63

2.64737

1340 1443.60 375.3 1058.94 10.247 3.30959

760

778.18

39.27 560.01

55.54

2.66176

1360 1467.49 399.1 1077.10 9.780 3.32724

780

800.03

43.35 576.12

51.64

2.69013

1380 1491.44 424.2 1095.26 9.337 3.34474

800

821.95

47.75 592.30

48.08

2.71787

1400 1515.42 450.5 1113.52 8.919 3.36200

820

843.98

52.59 608.59

44.84

2.74504

1420 1539.44 478.0 1131.77 8.526 3.37901

840

866.08

57.60 624.95

41.85

2.77170

1440 1563.51 506.9 1150.13 8.153 3.39586

860

888.27

63.09 641.40

39.12

2.79783

1460 1587.63 537.1 1168.49 7.801 3.41247

880

910.56

68.98 657.95

36.61

2.82344

1480 1611.79 568.8 1186.95 7.468 3.42892

900

932.93

75.29 674.58

34.31

2.84856

1500 1635.97 601.9 1205.41 7.152 3.44516

920

955.38

82.05 691.28

32.18

2.87324

1520 1660.23 636.5 1223.87 6.854 3.46120

940

977.92

89.28 708.08

30.22

2.89748

1540 1684.51 672.8 1242.43 6.569 3.47712

960 1000.55

97.00 725.02

28.40

2.92128

1560 1708.82 710.5 1260.99 6.301 3.49276

Pr kJ/kg

Pr

Vr kJ/kg

kJ/(kg • K)

980 1023.25 105.2

741.98

26.73

2.94468

1580 1733.17 750.0 1279.65 6.046 3.50829

1000 1046.04 114.0

758.94

25.17

2.96770

1600 1757.57 791.2 1298.30 5.804 3.52364

1020 1068.89 123.4

776.10

23.72

2.99034

1620 1782.00 834.1 1316.96 5.574 3.53879

1040 1091.85 133.3

793.36

22.39

3.01260

1640 1806.46 878.9 1335.72 5.355 3.55381

1060 1114.86 143.9

810.62

21.14

3.03449

1660 1830.96 925.6 1354.48 5.147 3.56867

1080 1137.89 155.2

827.88

19.98

3.05608

1680 1855.50 974.2 1373.24 4.949 3.58335

1100 1161.07 167.1

845.33 18.896

3.07732

1700 1880.1 1025 1392.7

4.761 3.5979

1120 1184.28 179.7

862.79 17.886

3.09825

1750 1941.6 1161

1439.8

4.328 3.6336

1140 1207.57 193.1

880.35 16.946

3.11883

1800 2003.3 1310 1487.2

3.944 3.6684

1160 1230.92 207.2

897.91 16.064

3.13916

1850 2065.3 1475 1534.9

3.601 3.7023

1180 1254.34 222.2

915.57 15.241

3.15916

1900 2127.4 1655 1582.6

3.295 3.7354

1200 1277.79 238.0

933.33 14.470

3.17888

1950 2189.7 1852 1630.6

3.022 3.7677

1220 1301.31 254.7

951.09 13.747

3.19834

2000 2252.1 2068 1678.7

2.776 3.7994

1240 1324.93 272.3

968.95 13.069

3.21751

2050 2314.6 2303 1726.8

2.555 3.8303

1260 1348.55 290.8

986.90 12.435

3.23638

2100 2377.4 2559 1775.3

2.356 3.8605

1280 1372.24 310.4 1004.76 11.835

3.25510

2150 2440.3 2837 1823.8

2.175 3.8901

1300 1395.97 330.9 1022.82 11.275

3.27345

2200 2503.2 3138 1872.4

2.012 3.9191

1320 1419.76 352.5 1040.88 10.747

3.29160

2250 2566.4 3464 1921.3

1.864 3.9474

Reproduced with permission from Thermodynamics : An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

666 Appendix Table 9 Ideal-gas properties of nitrogen, N2 _o K

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(unol • K)

kJ/kmol

kJ/kmol

0 220 230 240 250

0 6,391 6,683 6,975 7,266

0 4,562 4,770 4,979 5,188

0 182.639 183.938 185.180 186.370

600 610 620 630 640

17,563 17,864 18,166 18,468 18,772

12,574 12,792 13,011 13,230 13,450

212.066 212.564 213.055 213.541 214.018

260 270 280 290 298

7,558 7,849 8,141 8,432 8,669

5,396 5,604 5,813 6,021 6,190

187.514 188.614 189.673 190.695 191.502

650 660 670 680 690

19,075 19,380 19,685 19,991 20,297

13,671 13,892 14,114 14,337 14,560

214.489 214.954 215.413 215.866 216.314

300 310 320 330 340

8,723 9,014 9,306 9,597 9,888

6,229 6,437 6,645 6,853 7,061

191.682 192.638 193.562 194.459 195.328

700 710 720 730 740

20,604 20,912 21,220 21,529 21,839

14,784 15,008 15,234 15,460 15,686

216.756 217.192 217.624 218.059 218.472

350 360 370 380 390

10,180 10,471 10,763 11,055 11,347

7,270 7,478 7,687 7,895 8,104

196.173 196.995 197.794 198.572 199.331

750 760 770 780 790

22,149 22,460 22,772 23,085 23,398

15,913 16,141 16,370 16,599 16,830

218.889 219.301 219.709 220.113 220.512

400 410 420 430 440

11,640 11,932 12,225 12,518 12,811

8,314 8,523 8,733 8,943 9,153

200.071 200.794 201.499 202.189 202.863

800 810 820 830 840

23,714 24,027 24,342 24,658 24,974

17,061 17,292 17,524 17,757 17,990

220.907 221.298 221.684 222.067 222.447

450 460 470 480 490

13,105 13,399 13,693 13,988 14,285

9,363 9,574 9,786 9,997 10,210

203.523 204.170 204.803 205.424 206.033

850 860 870 880 890

25,292 25,610 25,928 26,248 26,568

18,224 18,459 18,695 18,931 19,168

222.822 223.194 223.562 223.927 224.288

500 510 520 530 540

14,581 14,876 15,172 15,469 15,766

10,423 10,635 10,848 11,062 11,277

206.630 207.216 207.792 208.358 208.914

900 910 920 930 940

26,890 27,210 27,532 27,854 28,178

19,407 19,644 19,883 20,122 20,362

224.647 225.002 225.353 225.701 226.047

550 560 570 580 590

16,064 16,363 16,662 16,962 17,262

11,492 11,707 11,923 12,139 12,356

209.461 209.999 210.528 211.049 211.562

950 960 970 980 990

28,501 28,826 29,151 29,476 29,803

20,603 20,844 21,086 21,328 21,571

226.389 226.728 227.064 227.398 227.728 Cont.

Appendix

667

Table 9 Ideal-gas properties of nitrogen, N2 (Cont.)

So

7' K

ii kJ/kmol

ii kJ/kmol

So

T

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

1000 1020 1040 1060 1080

30,129 30,784 31,442 32,101 32,762

21,815 22,304 22,795 23,288 23,782

228.057 228.706 229.344 229.973 230.591

1760 1780 1800 1820 1840

56,227 56.938 57,651 58,363 59,075

41,594 42,139 42,685 43,231 43,777

247.396 247.798 248.195 248.589 248.979

1100 1120 1140 1160 1180

33,426 34,092 34,760 35,430 36,104

24,280 24,780 25,282 25,786 26,291

231.199 231.799 232.391 232.973 233.549

1860 1880 1900 1920 1940

59,790 60,504 61,220 61,936 62,654

44,324 44,873 45423 45,973 46,524

249.365 249.748 250.128 250.502 250.874

1200 1220 1240 1260 1280

36,777 37,452 38,129 38,807 39,488

26,799 27,308 27,819 28,331 28,845

234.115 234.673 235.223 235.766 236.302

1960 1980 2000 2050 2100

63,381 64,090 64,810 66,612 68,417

47,075 47,627 48,181 49,567 50,957

251.242 251.607 251.969 252.858 253.726

1300 1320 1340 1360 1380

40,170 40,853 41,539 42,227 42,915

29,361 29,378 30,398 30,919 31,441

236.831 237.353 237.867 238.376 238.878

2150 2200 2250 2300 2350

70,226 72,040 73,856 75,676 77,496

52,351 53,749 55,149 56,553 57,958

254.578 255.412 256.227 257.027 257.810

1400 1420 1440 1460 1480

43,605 44,295 44,988 45,682 46,377

31,964 32,489 33,014 33,543 34,071

239.375 239.865 240.350 240.827 241.301

2400 2450 2500 2550 2600

79,320 81,149 82,981 84,814 86,650

59,366 60,779 62,195 63,613 65,033

258.580 259.332 260.073 260.799 261.512

1500 1520 1540 1560 1580

47,073 47,771 48,470 49,168 49,869

34,601 35,133 35,665 36,197 36,732

241.768 242.228 242.685 243.137 243.585

2650 2700 2750 2800 2850

88,488 90,328 92,171 94,014 95,859

66,455 67,880 69,306 70,734 72,163

262.213 262.902 263.577 264.241 264.895

1600 1620 1640 1660 1680

50,571 51,275 51,980 52,686 53,393

37,268 37,806 38,344 38,884 39,424

244.028 244.464 244.896 245.324 245.747

2900 2950 3000 3050 3100

97,705 99,556 101,407 103,260 105,115

73,593 75,028 76,464 77,902 79,341

265.538 266.170 266.793 267.404 268.007

1700 1720 1740

54,099 54,807 55,516

39,965 40,507 41,049

246.166 246.580 246.990

3150 3200 3250

106,972 108,830 110,690

80,782 82,224 83,668

268.601 269.186 269.763

ii

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

668 Appendix Table 10

Ideal-gas properties of oxygen, 02 _o

T

T

kJ/kmol

17 kJ/kmol

_o

K

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

0 220 230 240 250

0 6,404 6,694 6,984 7,275

0 4,575 4,782 4,989 5,197

0 196.171 197.461 198.696 199.885

600 610 620 630 640

17,929 18,250 18,572 18,895 19,219

12,940 13,178 13,417 13,657 13,898

226.346 226.877 227.400 227.918 228.429

260 270 280 290 298

7,566 7,858 8,150 8,443 8,682

5,405 5,613 5,822 6,032 6,203

201.027 202.128 203.191 204.218 205.033

650 660 670 680 690

19,544 19,870 20,197 20,524 20,854

14,140 14,383 14,626 14,871 15,116

228.932 229.430 229.920 230.405 230.885

300 310 320 330 340

8,736 9,030 9,325 9,620 9,916

6,242 6,453 6,664 6,877 7,090

205.213 206.177 207.112 208.020 208.904

700 710 720 730 740

21,184 21,514 21,845 22,177 22,510

15,364 15,611 15,859 16,107 16,357

231.358 231.827 232.291 232.748 233.201

350 360 370 380 390

10,213 10,511 10,809 11,109 11,409

7,303 7,518 7,733 7,949 8,166

209.765 210.604 211.423 212.222 213.002

750 760 770 780 790

22,844 23,178 23,513 23,850 24,186

16,607 16,859 17,111 17,364 17,618

233.649 234.091 234.528 234.960 235.387

400 410 420 430 440

11,711 12,012 12,314 12,618 12,923

8,384 8,603 8,822 9,043 9,264

213.765 214.510 215.241 215.955 216.656

800 810 820 830 840

24,523 24,861 25,199 25,537 25,877

17,872 18,126 18,382 18,637 18,893

235.810 236.230 236.644 237.055 237.462

450 460 470 480 490

13,228 13,525 13,842 14,151 14,460

9,487 9,710 9,935 10,160 10,386

217.342 218.016 218.676 219.326 219.963

850 860 870 880 890

26,218 26,559 26,899 27,242 27,584

19,150 19,408 19,666 19,925 20,185

237.864 238.264 238.660 239.051 239.439

500 510 520 530 540

14,770 15,082 15,395 15,708 16,022

10,614 10,842 11,071 11,301 11,533

220.589 221.206 221.812 222.409 222.997

900 910 920 930 940

27,928 28,272 28,616 28,960 29,306

20,445 20,706 20,967 21,228 21,491

239.823 240.203 240.580 240.953 241.323

550 560 570 580 590

16,338 16,654 16,971 17,290 17,609

11,765 11,998 12,232 12,467 12,703

223.576 224.146 224.708 225.262 225.808

950 960 970 980 990

29,652 29,999 30,345 30,692 31,041

21,754 22,017 22,280 22,544 22,809

241.689 242.052 242.411 242.768 243.120

s

Cont.

Appendix 669 Table 10

Ideal-gas properties of oxygen, 02 (Cont.)

so

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

1000 1020 1040 1060 1080

31,389 32,088 32,789 33,490 34,194

23,075 23,607 24,142 24,677 25,214

243.471 244.164 244.844 245.513 246.171

1760 1780 1800 1820 1840

58,880 59,624 60,371 61,118 61,866

44,247 44,825 45,405 45,986 46,568

263.861 264.283 264.701 265.113 265.521

1100 1120 1140 1160 1180

34,899 35,606 36,314 37,023 37,734

25,753 26,294 26,836 27,379 27,923

246.818 247.454 248.081 248.698 249.307

1860 1880 1900 1920 1940

62,616 63,365 64,116 64,868 65,620

47,151 47,734 48,319 48,904 49,490

265.925 266.326 266.722 267.115 267.505

1200 1220 1240 1260 1280

38,447 39,162 39,877 40,594 41,312

28,469 29,018 29,568 30,118 30,670

249.906 250.497 251.079 251.653 252.219

1960 1980 2000 2050 2100

66,374 67,127 67,881 69,772 71,668

50,078 50,665 51,253 52,727 54,208

267.891 268.275 268.655 269.588 270.504

1300 1320 1340 1360 1380

42,033 42,753 43,475 44,198 44,923

31,224 31,778 32,334 32,891 33,449

252.776 253.325 253.868 254.404 • 254.932

2150 2200 2250 2300 2350

73,573 75,484 77,397 79,316 81,243

55,697 57,192 58,690 60,193 61,704

271.399 272.278 273.136 273.891 274.809

1400 1420 1440 1460 1480

45,648 46,374 47,102 47,831 48,561

34,008 34,567 35,129 35,692 36,256

255.454 255.968 256.475 256.978 257.474

2400 2450 2500 2550 2600

83,174 85,112 87,057 89,004 90,956

63,219 64,742 66,271 67,802 69,339

275.625 276.424 277.207 277.979 278.738

1500 1520 1540 1560 1580

49,292 50,024 50,756 51,490 52,224

36,821 37,387 37,952 38,520 39,088

257.965 258.450 258.928 259.402 259.870

2650 2700 2750 2800 2850

92,916 94,881 96,852 98,826 100,808

70,883 72,433 73,987 75,546 77,112

279.485 280.219 280.942 281.654 282.357

1600 1620 1640 1660 1680

52,961 53,696 54,434 55,172 55,912

39,658 40,227 40,799 41,370 41,944

260.333 260.791 261.242 261.690 262.132

2900 2950 3000 3050 3100

102,793 104,785 106,780 108,778 110,784

78,682 80,258 81,837 83,419 85,009

283.048 283.728 284.399 285.060 285.713

1700 1720 1740

56,652 57,394 58,136

42,517 43,093 43,669

262.571 263.005 263.435

3150 3200 3250

112,795 114,809 116,827

86,601 88,203 89,804

286.355 286.989 287.614

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y. A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

1

670

Appendix

Table 11 T K

ti kJ/kmol

17 kJ/kmol

Ideal-gas properties of carbon dioxide, CO2

i0 kJ/(kmol - K)

T K

kJ/kmol

17 kJ/kmol

_ s kJ/(kmol • K)

ti

0

0

0

220 230 240 250

6,601 6,938 7,280 7,627

4,772 5,026 5.285 5,548

202.966 204.464 205.920 207.337

600 610 620 630 640

22,280 22,754 23,231 23,709 24,190

17,291 17,683 18,076 18,471 18,869

243.199 243.983 244.758 245.524 246.282

260 270 280 290 298

7,979 8,335 8,697 9,063 9,364

5,817 6,091 6,369 6,651 6,885

208.717 210.062 211.376 212.660 213.685

650 660 670 680 690

24,674 25,160 25,648 26,138 26,631

19,270 19,672 20,078 20,484 20,894

247.032 247.773 248.507 249.233 249.952

300 310 320 330 340

9,431 9,807 10,186 10,570 10,959

6,939 7,230 7,526 7,826 8,131

213.915 215.146 216.351 217.534 218.694

700 710 720 730 740

27,125 27,622 28,121 28,622 29,124

21,305 21,719 22,134 22,552 22,972

250.663 251.368 252.065 252.755 253.439

350 360 370 380 390

11,351 11,748 12,148 12,552 12,960

8,439 8,752 9,068 9,392 9,718

219.831 220.948 222.044 223.122 224.182

750 760 770 780 790

29,629 20,135 30,644 31,154 31,665

23,393 23,817 24,242 24,669 25,097

254.117 254.787 255.452 256.110 256.762

400 410 420 430 440

13,372 13,787 14,206 14,628 15,054

10,046 10,378 10,714 11,053 11,393

225.225 226.250 227.258 228.252 229.230

800 810 820 830 840

32,179 32,694 33,212 33,730 34,251

25,527 25,959 26,394 26,829 27,267

257.408 258.048 258.682 259.311 259.934

450 460 480 490

15,483 15,916 16,351 16,791 17,232

11,742 12,091 12,444 12,800 13,158

230.194 231.144 232.080 233.004 233.916

850 860 870 880 890

34,773 35,296 35,821 36,347 36,876

27,706 28,125 28,588 29,031 29,476

260.551 261.164 261.770 262.371 262.968

500 510 520 530 540

17,678 18,126 18,576 19,029 19,485

13,521 13,885 14,253 14,622 14,996

234.814 235.700 236.575 237.439 238.292

900 910 920 930 940

37,405 37,935 38,467 39,000 39,535

29,922 30,369 30,818 31,268 31,719

263.559 264.146 264.728 265.304 265.877

550 560 570 580 590

19,945 20,407 20,870 21,337 21,807

15,372 15,751 16,131 16,515 16,902

239.135 239.962 240.789 241.602 242.405

950 960 970 980 990

40,070 40,607 41,145 41,685 42,226

32,171 32,625 33,081 33,537 33,995

266.444 267.007 267.566 268.119 268.670

an

0

Cont.

Appendix

671

Table 11 Ideal-gas properties of carbon dioxide, CO2 (Conn.) _o

-o s

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

1000 1020 1040 1060 1080

42,769 43,859 44,953 46,051 47,153

34,455 35,378 36,306 37,238 38,174

269.215 270.293 271.354 272.400 273.430

1760 1780 1800 1820 1840

86,402 87,612 88,806 90,000

71,787

301.543

72,812 73,840 74,868

302.217 302.884 303.544

91,196

75,897

304.198

1100 1120 1140 1160 1180

48,258 49,369 50,484 51,602 52,724

39,112 40,057 41,006 41,957 42,913

274.445 275.444 276.430 277.403 278.361

1860 1880 1900 1920 1940

92,394 93,593 94,793 95,995 97,197

76,929 77,962 78,996 80,031 81,067

304.845 305.487 306.122 306.751 307.374

1200 1220 1240 1260 1280

53,848 54,977 56,108 57,244 58,381

43,871 44,834 45,799 46,768 47,739

279.307 280.238 281.158 282.066 282.962

1960 1980 2000 2050 2100

98,401 99,606 100,804 103,835 106,864

82,105 83,144 84,185 86,791 89,404

307.992 308.604 309.210 310.701 312.160

1300 1320 1340 1360 1380

59,522 60,666 61,813 61,963 64,116

48,713 49,691 50,672 51,656 52,643

283.847 284.722 285.586 286.439 287.283

2150 2200 2250 2300 2350

109,898 112,939 115,984 119,035 122,091

92,023 94,648 97,277 .99,912 102,552

313.589 314.988 316.356 317.695 319.011

1400 1420 1440 1460 1480

65,271 66,427 67,586 68,748 69,911

53,631 54,621 55,614 56,609 57,606

288.106 288.934 289.743 290.542 291.333

2400 2450 2500 2550 2600

125,152 128,219 131,290 134,368 137,449

105,197 107,849 110,504 113,166 115,832

320.302 321.566 322.808 324.026 325.222

1500 1520 1540 1560 1580

71,078 72,246 73,417 74,590 76,767

58,606 59,609 60,613 61,620 62,630

292.114 292.888 292.654 294.411 295.161 •

2650 2700 2750 2800 2850

140,533 143,620 146,713 149,808 152,908

118,500 121,172 123,849 126,528 129,212

326.396 327.549 328.684 329.800 330.896

1600 1620 1640 1660 1680

76,944 78,123 79,303 80,486 81,670

63,741 64,653 65,668 66,592 67,702

295.901 296.632 297.356 298.072 298.781

2900 2950 3000 3050 3100

156,009 159,117 162,226 165,341 168,456

131,898 134,589 137,283 139,982 142,681

331.975 333.037 334.084 335.114 336.126

1700 1720 1740

82,856 84,043 85,231

68,721 69,742 70,764

299.482 300.177 300.863

3150 3200 3250

171,576 174,695 177,822

145,385 148,089 150,801

337.124 338.109 339.069

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

672

Appendix Table 12 Ideal-gas properties of carbon monoxide, CO _o

K

kJ/kmol kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

0 220 230 240 250

0 6,391 6,683 6,975 7,266

0 4,562 4,771 4,979 5,188

0 188.683 189.980 191.221 192.411

600 610 620 630 640

17,611 17,915 18,221 18,527 18,833

12,622 12,843 13,066 13,289 13,512

218,204 218.708 219.205 219.695 220.179

260 270 280 290 298

7,558 7,849 8,140 8,432 8,669

5,396 5,604 5,812 6,020 6,190

193.554 194.654 195.713 196.735 197.543

650 660 670 680 690

19,141 19,449 19,758 20.068 20,378

13,736 13,962 14,187 14,414 14,641

220.656 221.127 221.592 222.052 222.505

300 310 320 330 340

8,723 9,014 9,306 9,597 9,889

6,229 6,437 6,645 6,854 7,062

197.723 198.678 199.603 200.500 201.371

700 710 720 730 740

20,690 21,002 21,315 21,628 21,943

14,870 15,099 15,328 15,558 15,789

222.953 223.396 223.833 224.265 224.692

350 360 370 380 390

10,181 10,473 10,765 11,058 11,351

7,271 7,480 7,689 7,899 8,108

202.217 203.040 203.842 204.622 205.383

750 760 770 780 790

22,258 22,573 22,890 23,208 23,526

16,022 16,255 16,488 16,723 16,957

225.115 225.533 225.947 226.357 226.762

400 410 420 430 440

11,644 11,938 12,232 12,526 12,821

8,319 8,529 8,740 8,951 9,163

206.125 206.850 207.549 208.252 208.929

800 810 820 830 840

23,844 24,164 24,483 24,803 25,124

17,193 17,429 17,665 17,902 18,140

227.162 227.559 227.952 228.339 228.724

450 460 470 480 490

13,116 13,412 13,708 14,005 14,302

9,375 9,587 9,800 10,014 10,228

209.593 210.243 210.880 211.504 212.117

850 860 870 880 890

25,446 25,768 26,091 26,415 26,740

18,379 18,617 18,858 19,099 19,341

229.106 229.482 229.856 230.227 230.593

500 510 520 530 540

14,600 14,898 15,197 15,497 15,797

10,443 10,658 10,874 11,090 11,307

212.719 213.310 213.890 214.460 215.020

900 910 920 930 940

27,066 27,392 27,719 28,046 28,375

19,583 19,826 20,070 20,314 20,559

230.957 231.317 231.674 232.028 232.379

550 560 570 580 590

16,097 16,399 16,701 17,003 17,307

11,524 11,743 11,961 12,181 12,401

215.572 216.115 216.649 217.175 217.693

950 960 970 980 990

28,703 29,033 29,362 29,693 30,024

20,805 21,051 21,298 21,545 21,793

232.727 233.072 233.413 233.752 234.088 Cont.

Appendix

Table 12

673

Ideal-gas properties of carbon monoxide, CO (Cont.) -o

ti

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

.1° kJ/(kmol • K)

1000 1020 1040 1060 1080

30,355 31,020 31,688 32,357 33,029

22,041 22,540 23,041 23,544 24,049

234.421 235.079 235.728 236.364 236.992

1760 1780 1800 1820 1840

56,756 57,473 58,191 58,910 59,629

42,123 42,673 43,225 43,778 44,331

253.991 254.398 254.797 255.194 255.587

1100 1120 1140 1160 1180

33,702 34,377 35,054 35,733 36,406

24,557 25,065 25,575 26,088 26,602

237.609 238.217 238.817 239.407 239.989

1860 1880 1900 1920 1940

60,351 61,072 61,794 62,516 63,238

44,886 45,441 45,997 46,552 47,108

255.976 256.361 256.743 257.122 257.497

1200 1220 1240 1260 1280

37,095 37,780 38,466 39,154 39,844

27,118 27,637 28,426 28,678 29,201

240.663 241.128 241.686 242.236 242.780

1960 1980 2000 2050 2100

63,961 64,684 65,408 67,224 69,044

47,665 48,221 48,780 50,179 51,584

257.868 258.236 258.600 259.494 260.370

1300 1320 1340 1360 1380

40,534 41,226 41,919 42,613 43,309

29,725 30,251 30,778 31,306 31,836

243.316 243.844 244.366 244.880 245.388

2150 2200 2250 2300 2350

70,864 72,688 74,516 76,345 78,178

52,988 54,396 55,809 57,222 58,640

261.226 262.065 262.887 263.692 264.480

1400 1420 1440 1460 1480

44,007 44,707 45,408 46,110 46,813

32,367 32,900 33,434 33,971 34,508

245.889 246.385 246.876 247.360 247.839

2400 2450 2500 2550 2600

80,015 81,852 83,692 85,537 87,383

60,060 61,482 62,906 64,335 65,766

265.253 266.012 266.755 267.485 268.202

1500 1520 1540 1560 1580

47,517 48,222 48,928 49,635 50,344

35,046 35,584 36,124 36,665 37,207

248.312 248.778 249.240 249.695 250.147

2650 2700 2750 2800 2850

89,230 91,077 92,930 94,784 96,639

67,197 68,628 70,066 71,504 72,945

268.905 269.596 270.285 270.943 271.602

1600 1620 1640 1660 1680

51,053 51,763 52,472 53,184 53,895

37,750 38,293 38,837 39,382 39,927

250.592 251.033 251.470 251.901 252.329

2900 2950 3000 3050 3100

98,495 100,352 102,210 104,073 105,939

74,383 75,825 77,267 78,715 80,164

272.249 272.884 273.508 274.123 274.730

1700 1720 1740

54,609 55,323 56.039

40,474 41,023 41,572

252.751 253.169 253.582

3150 3200 3250

107,802 109,667 111,534

81,612 83,061 84,513

275.326 275.914 276.494

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

iii:

674 Appendix Table 13 Ideal-gas properties of hydrogen, H2

s°

s° kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

0 5,209 5,412 5,617 5,822

0 126.636 127.719 128.765 129.775

1440 1480 1520 1560 1600

42,808 44,091 45,384 46,683 47,990

30,835 31,786 32,746 33,713 34,687

177.410 178.291 179.153 179.995 180.820

8,468 8,522 9,100 9,680 10,262

5,989 6,027 6,440 6,853 7,268

130.574 130.754 132.621 134.378 136.039

1640 1680 1720 1760 1800

49,303 50,622 51,947 53,279 54,618

35,668 36,654 37,646 38,645 39,652

181.632 182.428 183.208 183.973 184.724

380 400 420 440 460

10,843 11,426 12,010 12,594 13,179

7,684 8,100 8,518 8,936 9,355

137.612 139.106 140.529 141.888 143.187

1840 1880 1920 1960 2000

55,962 57,311 58,668 60,031 61,400

40,663 41,680 42,705 43,735 44,771

185.463 186.190 186.904 187.607 188.297

480 500 520 560 600

13,764 14,350 14,935 16,107 17,280

9,773 10,193 10,611 11,451 12,291

144.432 145.628 146.775 148.945 150.968

2050 2100 2150 2200 2250

63,119 64,847 66,584 68,328 70,080

46,074 47,386 48,708 50,037 51,373

189.148 189.979 190.796 191.598 192.385

640 680 720 760 800

18,453 19,630 20,807 21,988 23,171

13,133 13,976 14,821 15,669 16,520

152.863 154.645 156.328 157.923 159.440

2300 2350 2400 2450 2500

71,839 73,608 75,383 77,168 78,960

52,716 54,069 55,429 56,798 58,175

193.159 193.921 194.669 195.403 196.125

840 880 920 960 1000

24,359 25,551 26,747 27,948 29,154

17,375 18,235 19,098 19,966 20,839

160.891 162.277 163.607 164.884 166.114

2550 2600 2650 2700 2750

80,755 82,558 84,368 86,186 88,008

59,554 60,941 62,335 63,737 65,144

196.837 197.539 198.229 198.907 199.575

1040 1080 1120 1160 1200

30,364 31,580 32,802 34,028 35,262

21,717 22,601 23,490 24,384 25,284

167.300 168.449 169.560 170.636 171.682

2800 2850 2900 2950 3000

89,838 91,671 93,512 95,358 97,211

66,558 67,976 69,401 70,831 72,268

200.234 200.885 201.527 202.157 202.778

1240 1280 1320 1360 1400

36,502 37,749 39,002 40,263 41,530

26,192 27,106 28,027 28,955 29,889

172.698 173.687 174.652 175.593 176.510

3050 3100 3150 3200 3250

99,065 100,926 102,793 104,667 106,545

73,707 75,152 76,604 78,061 79,523

203.391 203.995 204.592 205.181 205.765

K

kJ/kmol

kJ/kmol

0 260 270 280 290

0 7,370 1,657 7,945 8,233

298 300 320 340 360

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

Appendix 675 Table 14 Ideal-gas properties of water vapour, H2O

K

kJ/kmol

-o

T

17

u

-o

kJ/kmol

kJ/(1cmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

0 220 230 240 250

0 7,295 7,628 7,961 8,294

0 5,466 5,715 5,965 6,215

0 178.576 180.054 181.471 182.831

600 610 620 630 640

20,402 20,765 21,130 21,495 21,862

15,413 15,693 15,975 16,257 16,541

212.920 213.529 214.122 214.707 215.285

260 270 280 290 298

8,627 8,961 9,296 9,631 9,904

6,466 6,716 6,968 7,219 7,425

184.139 185.399 186.616 187.791 188.720

650 660 670 680 690

22,230 22,600 22,970 23,342 23,714

16,826 17,112 17,399 17,688 17,978

215.856 216.419 216.976 217.527 218.071

300 310 320 330 340

9,966 10,302 10,639 10,976 11,314

7,472 7,725 7,978 8,232 8,487

188.928 190.030 191.098 192.136 193.144

700 710 720 730 740

24,088 24,464 24,840 25,218 25,597

18,268 18,561 18,854 19,148 19,444

218.610 219.142 219.668 220.189 220.707

350 360 370 380 390

11,652 11,992 12,331 12,672 13,014

8,742 8,998 9,255 9,513 9,771

194.125 195.081 196.012 196.920 197.807

750 760 770 780 790

25,977 26,358 26,741 27,125 27,510

19,741 20,039 20,339 20,639 20,941

221.215 221.720 222.221 222.717 223.207

400 410 420 430 440

13,356 13,699 14,043 14,388 14,734

10,030 10,290 10,551 10,813 11,075

198.673 199.521 200.350 201.160 201.955

800 810 820 830 840

27,896 28,284 28,672 29,062 29,454

21,245 21,549 21,855 22,162 22,470

223.693 224.174 224.651 225.123 225.592

450 460 470 480 490

15,080 15,428 15,777 16,126 16,477

11,339 11,603 11,869 12,135 12,403

202.734 203.497 204.247 204.982 205.705

850 860 870 880 890

29,846 30,240 30,635 31,032 31,429

22,779 23,090 23,402 23,715 24,029

226.057 226.517 226.973 227.426 227.875

500 510 520 530 540

16,828 17,181 17,534 17,889 18,245

12,671 12,940 13,211 13,482 13,755

206.413 207.112 207.799 208.475 209.139

900 910 920 930 940

31,828 32,228 32,629 33,032 33,436

24,345 24,662 24,980 25,300 25,621

228.321 228.763 229.202 229.637 230.070

550 560 570 580 590

18,601 18,959 19,318 19,678 20,039

14,028 14,303 14,579 14,856 15,134

209.795 210.440 211.075 211.702 212.320

950 960 970 980 990

33,841 34,247 34,653 35,061 35,472

25,943 26,265 26,588 26,913 27,240

230.499 230.924 231.347 231.767 232.184 Cont.

676 Appendix Table 14 Ideal-gas properties of water vapour, H2O (Cont.) -o s

-o

T K

h kJ/kmol

u kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

kJ/(kmol • K)

1000 1020 1040 1060 1080

35,882 36,709 37,542 38,380 39,223

27,568 28,228 28,895 29,567 30,243

232.597 233.415 234.223 235.020 235.806

1760 1780 1800 1820 1840

70,535 71,523 72,513 73,507 74,506

55,902 56,723 57,547 58,375 59,207

258.151 258.708 259.262 259.811 260.357

1100 1120 1140 1160 1180

40,071 40,923 41,780 42,642 43,509

30,925 31,611 32,301 32,997 33,698

236.584 237.352 238.110 238.859 239.600

1860 1880 1900 1920 1940

75,506 76,511 77,517 78,527 79,540

60,042 60,880 61,720 62,564 63,411

260.898 261.436 261.969 262.497 263.022

1200 1220 1240 1260 1280

44,380 45,256 46,137 47,022 47,912

34,403 35,112 35,827 36,546 37,270

240.333 241.057 241.773 242.482 243.183

1960 1980 2000 2050 2100

80,555 81,573 82,593 85,156 87,735

64,259 65,111 65,965 68,111 70,275

263.542 264.059 264.571 265.838 267.081

1300 1320 1340 1360 1380

48,807 49,707 50,612 51,521 52,434

38,000 38,732 39,470 40,213 40,960

243.877 244.564 245.243 245.915 246.582

2150 2200 2250 2300 2350

90,330 92,940 95,562 98,199 100,846

72,454 74,649 76,855 79,076 81,308

268.301 269.500 270.679 271.839 272.978

1400 1420 1440 1460 1480

53,351 54,273 55,198 56,128 57,062

41,711 42,466 43,226 43,989 44,756

247.241 247.895 248.543 249.185 249.820

2400 2450 2500 2550 2600

103,508 106,183 108,868 111,565 114,273

83,553 85,811 88,082 90,364 92,656

274.098 275.201 276.286 277.354 278.407

1500 1520 1540 1560 1580

57,999 58,942 59,888 60,838 61,792

45,528 46,304 47,084 47,868 48,655

250.450 251.074 251.693 252.305 252.912

2650 2700 2750 2800 2850

116,991 119,717 122,453 125,198 127,952

94,958 97,269 99,588 101,917 104,256

279.441 280.462 281.464 282.453 283.429

1600 1620 1640 1660 1680

62,748 63,709 64,675 65,643 66,614

49,445 50,240 51,039 51,841 52,646

253.513 254.111 254.703 255.290 255.873

2900 2950 3000 3050 3100

130,717 133,486 136,264 139,051 141,846

106,605 108,959 111,321 113,692 116,072

284.390 285.338 286.273 287.194 288.102

1700 1720 1740

67,589 68,567 69,550

53,455 54,267 55,083

256.450 257.022 257.589

3150 3200 3250

144,648 147,457 150,272

118,458 120,851 123,250

288.999 289.884 290.756

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, V.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

Appendix

Table 15

K 0 298 300 500 1000 1500 1600 1700 1800 1900 2000 2050 2100 2150 2200 2250 2300 2350

677

Ideal-gas properties of monatomic oxygen, 0

kJ/kmol

k.likmol

kJ/(kmol • K)

K

kJ/kmol

i7 kJ/kmol

kJ/(kmol • K)

0 6,852 6,892 11,197 21,713 32,150 34,234 36,317 38,400 40,482 42,564 43,605 44,646 45,687 46,728 47,769 48,811 49,852

0 4,373 4,398 7,040 13,398 19,679 20,931 22,183 23,434 24,685 25,935 26,560 27,186 27,811 28,436 29,062 29,688 30,314

0 160.944 161.079 172.088 186.678 195.143 196.488 197.751 198.941 200.067 201.135 201.649 202.151 202.641 203.119 203.588 204.045 204.493

2400 2450 2500 2550 2600 2650 2700 2750 2800 2850 2900 2950 3000 3100 3200 3300 3400 3500

50,894 51,936 52,979 54,021 55,064 56,108 57,152 58,196 59,241 60,286 61,332 62,378 63,425 65,520 67,619 69,720 71,824 73,932

30,940 31,566 32,193 32,820 33,447 34,075 34,703 35,332 35,961 36,590 37,220 37,851 38,482 39,746 41,013 42,283 43,556 44,832

204.932 205.362 205.783 206.196 206.601 206.999 207,389 207.772 208.148 208.518 208.882 209.240 209.592 210.279 210.945 211.592 212.220 212.831

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994. Table 16

Ideal-gas properties of hydroxyl, OH

kJ/kmol

kJ/kmol

kJ/(kmol • K)

K

kJ/kmol

kJ/kmol

s-0 kJ/(kmol • K)

0 9,188 9,244 15,181 30,123 46,046 49,358 52,706 56,089 59,505 62,952 64,687 66,428 68,177 69,932 71,694 73,462 75,236

0 6,709 6,749 11,024 21,809 33,575 36,055 38,571 41,123 43,708 46,323 47,642 48,968 50,301 51,641 52,987 54,339 55,697

0 183.594 183.779 198.955 219.624 232.506 234.642 236.672 238.606 240.453 242.221 243.077 243.917 244.740 245.547 246.338 247.116 247.879

2400 2450 2500 2550 2600 2650 2700 2750 2800 2850 2900 2950 3000 3100 3200 3300 3400 3500

77,015 78,801 80,592 82,388 84,189 85,995 87,806 89,622 91,442 93,266 95,095 96,927 98,763 102,447 106,145 109,855 113,578 117,312

57,061 58,431 59,806 61,186 62,572 63,962 65,358 66,757 68,162 69,570 70,983 72,400 73,820 76,673 79,539 82,418 85,309 88,212

248.628 249.364 250.088 250.799 251.499 252.187 252.864 253.530 254.186 254.832 255.468 256.094 256.712 257.919 259.093 260.235 261.347 262.429

-o

K 0 298 300 500 1000 1500 1600 1700 1800 1900 2000 2050 2100 2150 2200 2250 2300 2350

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

678

Appendix

Table 17

Enthalpy of formation, Gibbs function of formation, and absolute entropy at 25°C, 1 atm

Substance

ho

-o

Formula

kJ/kmol

kJ/kmol

C(s) H2(g) N2(g) 02(g) CO(g) CO2(g) H20(g) H20(/ )

0

5.74

0

130.68

0

191.61

-o

gf

kJ/(kmol • K)

Methyl alcohol

CH3OH(g)

0 0 0 0 -110,530 -393,520 -241,820 -285,830 -136,310 -46,190 -74,850 +226,730 +52,280 -84,680 +20,410 -103,850 -126,150 -208,450 -249,950 -291,010 +82,930 -200,670

Methyl alcohol

CH3OH(/)

-238,660

-166,360

126.80

Ethyl alcohol

C2H3OH(g)

-235,310

-168,570

282.59

Ethyl alcohol

C2H5OH(1)

-277,690

Oxygen

0(g)

+249,190

-174,890 +231,770

161.06

Hydrogen

H(g)

+218,000

+203,290

114.72

Nitrogen

N(g)

+472,650

+455,510

153.30

Hydroxyl

OH(g)

+39,460

+34,280

183.70

Carbon Hydrogen Nitrogen Oxygen Carbon monoxide Carbon dioxide Water vapour Water Hydrogen peroxide Ammonia

H202(g) NH3(g)

Methane

CH4(g)

Acetylene Ethane

C2H2(g) C2H4(g) C2H6(g)

Propylene

C3H6(g)

Propane

C3H8(g)

Ethylene

n-Butane

C4f110(g)

n-Octane

C8His(g)

n-Octane

C8H B(/)

n-Dodecane

C121126(g)

Benzene

C6H6(g)

Reproduced with permission from Thermodynamics: Boles, M.A., 2nd ed., McGraw-Hill, 1994.

0

205.04

-137,150

197.65

-394,360

213.80

-228,590

188.83

-237,180

69.92

-105,600

232.63

-16,590

192.33

-50,790

186.16

+209,170

200.85

+68,120

219.83

-32,890

229.49

+62,720

266.94

-23,490

269.91

-15,710

310.12

+16,530 +6,610 +50,150

466.73

+129,660

269.20

-162,000

239.70

An Engineering Approach by

360.79 622.83

160.70

Cengel, Y.A. and

Appendix

679

Table 18 Enthalpy of combustion and enthalpy of vaporization at 25°C, 1 atm (Water appears as a liquid in the products of combustion) Alic° = —HHV

Substance

Formula

kJ/kmol

Hydrogen

H2(g)

—285,840

Carbon

C(s)

—393,520

Carbon monoxide

CO(g)

—282,990

Methane

CH4(g)

Acetylene

C2H2(g)

—1,299,600

Ethylene

C21-14(g)

—1,410,970

Ethane

—1,559,900

Propylene

C2H6(g) C3H6(g)

Propane

C3H8(g)

—2,220,000

15,060

n-Butane

G$H10(8)

—2,877,100

21,060

n-Pentane

C5H12(g)

—3,536,100

26,410

n-Hexane

C6H14(g)

—4,194,800

31,530

n-Heptane

C7Hidg)

—4,853,500

36,520

kJ/kmol

—890,360

—2,058,500

n-Octane

C8H18(g)

—5,512,200

41,460

Benzene

C6H6(g)

—3,301,500

33,830

Toluene

C7H8(g)

—3,947,900

39,920

Methyl alcohol

C1-130H(g)

Ethyl alcohol

C2H5OH(g)

—764,540

37,900

—1,409,300

42,340

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

Table 19 Logarithms to base e of the equilibrium constant K,, The equilibrium constant Kp for the reaction vAA + v8B

Temp. K

H2 c=-1 2H

02 20

N2 k-=-1 2N

H2O = H2 + 1 02

vcC + v0D is defined as

H2O

1 H2 + OH CO2

K1,

E

PD" v v PA ID

CO + 102

1N2 + 102 c='-. NO

298

-164.005

-186.975

-367.480

-92.208

-106.208

-103.762

-35.052

500

-92.827

-105.630

-213.372

-52.691

-60.281

-57.616

-20.295

1000

-39.803

-45.150

-99.127

-23.163

-26.034

-23.529

-9.388

1200

-30.874

-35.005

-80.011

-18.182

-20.283

-17.871

-7.569

1400

-24.463

-27.742

-66.329

-14.609

-16.099

-13.842

-6.270

1600

-19.637

-22.285

-56.055

-11.921

-13.066

-10.830

-5.294

1800

-15.866

-18.030

-48.051

-9.826

-10.657

-8.497

-4.536

2000

-12.840

-14.622

-41.645

-8.145

-8.728

-6.635

-3.931

2200

-10.353

-11.827

-36.391

-6.768

-7.148

-5.120

-3.433

2400

-8.276

-9.497

-32.011

-5.619

-5.832

-3.860

-3.019

2600

-6.517

-7.521

-28.304

-4.648

-4.719

-2.801

-2.671

2800

-5.002

-5.826

-25.117

-3.812

-3.763

-1.894

-2.372

3000

-3.685

-4.357

-22.359

-3.086

-2.937

-1.111

-2.114

3200

-2.534

-3.072

-19.937

-2.451

-2.212

-0.429

-1.888

3400

-1.516

-1.935

-17.800

-1.891

-1.576

0.169

-1.690

3600

-0.609

-0.926

-15.898

-1.392

-1.088

0.701

-1.513

3800

0.202

-0.019

-14.199

-0.945

-0.501

1.176

-1.356

4000

0.934

0.796

-12.660

-0.542

-0.044

1.599

-1.216

4500

2.486

2.513

-9.414

0.312

0.920

2.490

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5000

3.725

3.895

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0.996

1.689

3.197

-0.686

5500

4.743

5.023

-4.666

1.560

2.318

3.771

-0.497

6000

5.590

5.963

-2.865

2.032

2.843

4.245

-0.341

Reproduced with permission from Thermodynamics: An Engineering Approach by Cengel, Y.A. and Boles, M.A., 2nd ed., McGraw-Hill, 1994.

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Appendix 683 mu, . imisp6C- ion J1.a MAO • III

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